tag:blogger.com,1999:blog-76298694103691329092020-08-26T02:58:21.815-07:00THE INDIAN MATHEMATICIAN, NBHM CSIR GATE EXAM PREVIOUS YEAR QUESTIONS SOLUTION THROUGH CONCEPTSTHE CONVERGING POINT FOR CLEARING COMPETITIVE EXAMS (M.Sc., Ph.D. MATHEMATICS ENTRANCE EXAMS QUESTION PAPER SOLUTIONS) (EVERYDAY NEW PROBLEMS)THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.comBlogger78125TheIndianMathematicianhttps://feedburner.google.comtag:blogger.com,1999:blog-7629869410369132909.post-41064237009667211542020-05-10T12:27:00.001-07:002020-05-10T12:30:45.359-07:00NBHM 2020 PART C Question 31 Solution (continuous images of $G = GL_2(\Bbb R)$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); padding: 10px;"><div class="p1" style="font-family: Helvetica; font-stretch: normal; font-variant-east-asian: normal; font-variant-numeric: normal; line-height: normal;"><span style="font-size: large;">Let $G = GL_2(\Bbb R)$ be the set of all $2 \times 2$ invertible real matrices. Elements of $G$<span class="s2" style="font-stretch: normal; font-variant-east-asian: normal; font-variant-numeric: normal; line-height: normal;">&nbsp;</span>can be identified with vectors in $\Bbb R^4$&nbsp;and this makes $G$ a metric space. Which of the following sets can be obtained as the image of a continuous onto function from $G$?&nbsp;</span></div><div class="p1" style="font-family: Helvetica; font-stretch: normal; font-variant-east-asian: normal; font-variant-numeric: normal; line-height: normal;"><span style="font-size: large;">1. The real line.&nbsp;</span></div><div class="p1" style="font-family: Helvetica; font-stretch: normal; font-variant-east-asian: normal; font-variant-numeric: normal; line-height: normal;"><span style="font-size: large;">2. $A = \{(x,\frac{1}{x}) : x \in \Bbb R \text{ and } x \ne 0\}$.</span></div><div class="p1" style="font-family: Helvetica; font-stretch: normal; font-variant-east-asian: normal; font-variant-numeric: normal; line-height: normal;"><span style="font-size: large;">3. $\Bbb R^2 \backslash A$.&nbsp;</span></div><div class="p1" style="font-family: Helvetica; font-stretch: normal; font-variant-east-asian: normal; font-variant-numeric: normal; line-height: normal;"><span style="font-size: large;">4. $\{(x,y) \in \Bbb R^2 : |x|^2 + |y^2| \le 1\}$</span></div><div style="color: black;"><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b></div><div style="color: black;"><b><span style="font-size: large;">Solution:</span></b></div><div style="color: black;"><span style="font-size: large;"><b>Option 1:(True) </b>Consider the determinant map $\text{det }: GL_2(\Bbb R) \to \Bbb R^*$ where $\Bbb R^{*} = \Bbb R \backslash \{0\}$ is a group under multiplication. Now $$\text{det}(AB) = \text{det}(A)\text{det}(B)$$ Therefore this defines a homomorphism. We have determinant is a continuous map, because it is a polynomial in the matrix entries.</span></div><div style="color: black;"><span style="font-size: large;">Let $\alpha \in \Bbb R^*$ then the matrix $\begin{bmatrix}\alpha &amp; 0 \\ 0 &amp; 1\end{bmatrix} \in GL_2(\Bbb R)$ has determinant $\alpha$. Therefore determinant is a surjective map. Now, the function $\phi: \Bbb R^* \to \Bbb R$ defined by $\phi(x) = |\text{log}(x)|$ is a continuous surjective map. Hence the composition $\phi \circ \text{det} : GL_2(\Bbb R) \to \Bbb R$ is the required continuous surjection. Note that this map is also a composition of group homomorphisms and hence is a group homomorphism. Therefor there exists a continuous surjective group homomorphism between $GL_2(\Bbb R)$ to $\Bbb R$.&nbsp;</span></div><div style="color: black;"><span style="font-size: large;"><b>Option 2:(True) </b>Consider the map $\phi : GL_2(\Bbb R) \to A = \{(x,\frac{1}{x}: x \in \Bbb R)\}$ defined by $\phi(A) = (\text{det}(A),\frac{1}{\text{det}(A)})$. We have seen that $\text{det}$ is a continuous map. Since $\text{det}(A) \ne 0$ for any $A \in GL_2(R)$, the function $A$ goes to $\frac{1}{\text{det}(A)}$ is also continuous. This shows that, the function $\phi$ is continuous since each of its coordinate map is continuous. Let $(\alpha,\frac{1}{\alpha}) \in A$ then</span><span style="font-size: large;">&nbsp;the matrix $M$ defined by $\begin{bmatrix}\alpha &amp; 0 \\ 0 &amp; 1\end{bmatrix} \in GL_2(\Bbb R)$ has determinant $\alpha$ and $\phi (M) = (\alpha,\frac{1}{\alpha})$. Therefore $\phi$ is surjective.</span><br /><span style="font-size: large;"><b>option(3):&nbsp;</b></span><br /><span style="font-size: large;"><b>Result: </b>Let $X$ and $Y$ be two metric spaces such that $X$ has $m$ connected components and $Y$ has $n$ connected components with $m &lt; n$. Then there cannot be a surjective continuous map from $X$ to $Y$.</span><br /><span style="font-size: large;">Proof: Let $f$ be a continuous function from $X$ to $Y$ and $X_1,X_2,\dots,X_m$ be the connected components of $X$. We can restrict $f$ to each of these connected component and the resulting map is still continuous. Also, the images $f(X_i)$ is connected for each $i$. Let $Y_i$ be the connected component of $Y$ such that $f(X_i) \subseteq Y_i$&nbsp; (Note that, different connected components $X_i$ could have mapped to the same connected component $Y_i$). This shows that $f(X) \subseteq \cup_{i=1}^m Y_i$. This means that $f(X)$ doesn't intersect remaining $n-m$ connected components of $Y$. Therefore $Y$ is not surjective.</span><br /><span style="font-size: large;"><b>&nbsp;</b>$GL_2(\Bbb R)$ is disconnected and has $2$ connected components. The set $\Bbb R^2 \backslash A$ has $3$ connected components. Now, the above result shows that there cannot be continuous surjection between these sets.</span><br /><span style="font-size: large;"><b>option 4</b>:(<b>True</b>)</span><br /><span style="font-size: large;">Consider the function $f$ from $GL_2(\Bbb R)$ to $\Bbb C$ given by $M = \begin{bmatrix}x &amp; y \\ z &amp; a\end{bmatrix}$ goes to $\frac{y^2}{x^2+ y^2}e^{i a}$. Then clearly $f$ is a continuous surjection from $GL_2(\Bbb R)$ to</span><span style="font-family: &quot;helvetica&quot;; font-size: large;">&nbsp;$\{(x,y) \in \Bbb R^2 : |x|^2 + |y^2| \le 1\}$.</span><br /><div style="color: black;"></div><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><span style="font-size: large;"></span><b style="font-size: x-large;">Share to your groups:</b></div></div><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/hfVft0HF25E" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/05/nbhm-2020-part-c-question-31-solution.htmltag:blogger.com,1999:blog-7629869410369132909.post-21760809559632311882020-04-23T20:38:00.000-07:002020-04-23T20:48:11.959-07:00CSIR JUNE 2011 PART C QUESTION 83 (Singularities of $f(z) = \frac{e^z + 1}{e^z -1}$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">The function $f(z) = \frac{e^z + 1}{e^z -1}$ has what type of singularity at $z=0$?&nbsp;</span><br /><span style="font-size: large;">1. a removable singularity.&nbsp;</span><br /><span style="font-size: large;">2. a pole.</span><br /><span style="font-size: large;">3. an essential singularity.</span><br /><span style="font-size: large;">4. the residue of $f(z)$ at $z = 0$ is $2$.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><b><span style="font-size: large;">Solution:&nbsp;</span></b><br /><span style="font-size: large;"><b>Results:&nbsp;</b></span><br /><span style="font-size: large;">1.<b>&nbsp;</b>If $z_0$ is a removable singularity of $f(z)$ if $\lim_{z \to z_0} (z - z_0)f(z) = 0$.</span><br /><span style="font-size: large;">2.&nbsp;</span><b style="font-size: x-large;">&nbsp;</b><span style="font-size: large;">If $z_0$ is a pole of $f(z)$ if $\lim_{z \to z_0} (z - z_0)f(z)$ exists and non-zero.</span><br /><span style="font-size: large;">3. The residue of the function $f(z)$ at $z = z_0$ is given by the coefficient of $\frac{1}{z}$ in the Laurent series expansion of $f(z)$. Which can be calculated by $\lim_{z \to z_0} \, (z-z_0)f(z)$.</span><br /><span style="font-size: large;"><b>Remark</b>: From the above results, we observe that the limit used to check whether a singularity $z_0$ is removable or pole of $f(z)$ is same the limit used to find the residue of $f(z)$ at $z_0$. Indeed, in both the cases the limit is $\lim_{z \to z_0}(z-z_0)f(z)$. We conclude that,&nbsp;</span><br /><span style="font-size: large;">1. The residue of $f(z)$ at $z = z_0$ is zero if and only if $z_0$ is a removable singularity of f(z). If particular there is no $\frac{1}{z}$ term in the Laurent series expansion of $f(z)$.</span><br /><span style="font-size: large;">2. The residue of $f(z)$ at $z = z_0$ is non-zero if and only if $z_0$ is a pole of f(z). If particular, the coefficient of $\frac{1}{z}$ term in the Laurent series expansion of $f(z)$ is equal to the residue.</span><br /><br /><span style="font-size: large;">We have $e^z - 1 = z + \frac{z^2}{2} + \frac{z^3}{3!} + \cdots$. Then $\frac{e^z -1}{z} = 1 + \frac{z}{2} + \frac{z^2}{3!} + \cdots$. We observe that the series expansion of the function $h(z) = \frac{e^z - 1}{z}$ has constant term $1$ and hence $h(0) = 1$. This shows that the function $g(z) = \frac{e^z+1}{h(z)}$ is analytic at $z = 0$. In particular, $\lim z \to 0 \,g(z)$ exists. We observe that $$f(z) = \frac{g(z)}{z}$$ and $\lim_{z \to 0} (z-0) f(z) = \lim_{z \to 0} g(z) =&nbsp; \lim_{z \to 0} \frac{z(e^z+1)}{e^z - 1}&nbsp; = 2 \ne 0$. This shows that, from the results given at the beginning, $0$ is a simple pole for $f(z)$.</span><br /><span style="font-size: large;"><b>So option 2 is true.</b></span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;">Now,&nbsp;</span><span style="font-size: large;">the residue of $f(z)$ at $z=0$ is equal to $\lim_{z \to 0} \, (z-0)f(z) = \lim_{z \to 0}g(z) = 2$. <b>Hence option 4 is True.</b></span><br /><span style="font-size: large;"><br /></span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/REQJS9-eWXw" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-83.htmltag:blogger.com,1999:blog-7629869410369132909.post-77681180549261825572020-04-22T04:01:00.001-07:002020-04-22T04:01:19.849-07:00NBHM 2020 PART C Question 28 Solution ($\mathcal F = \{f :(0,\infty) \to \Bbb R \mid f(x) = f(2x) \,\text{for all}\, x \in (0,\infty)\}$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $\mathcal F = \{f :(0,\infty) \to \Bbb R \mid f(x) = f(2x) \,\text{for all}\, x \in (0,\infty)\}$. Which of the following&nbsp; statements are true?&nbsp;</span><br /><span style="font-size: large;">1. $f \in \mathcal F$ implies $f$ is bounded,&nbsp;</span><br /><span style="font-size: large;">2</span><span style="font-size: large;">. $f \in \mathcal F$ implies $f$ is uniformly continuous,&nbsp;</span><br /><span style="font-size: large;">3</span><span style="font-size: large;">. $f \in \mathcal F$ implies $f$ is differentiable,&nbsp;</span><br /><span style="font-size: large;">4. Every uniformly bounded sequence in $\mathcal F$ has a uniform convergence subsequence.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><b><span style="font-size: large;">Solution:&nbsp;</span></b><br /><span style="font-size: large;"><b>Option 1: (True)</b> Let $f&nbsp; \in \mathcal F$, then&nbsp;</span><span style="font-size: large;">for all $x \in (0,\infty)$ we have&nbsp;</span><span style="font-size: large;">$$f(x) = f(2x).$$ We note that $f(1) = f(2)$ and $$f([1,2]) = f([2,2^2]) = f([2^2,2^3]) = \cdots$$ In the other direction the given condition can be written as $$f(\frac{x}{2}) = f(x)$$ and&nbsp;</span><br /><span style="font-size: large;">$$f([1,2]) = f([\frac{1}{2},1]) = f([\frac{1}{2^2},\frac{1}{2}]) = \cdots$$</span><br /><span style="font-size: large;">This shows that $f$ is determined by its value in the interval $[1,2]$ with the additional condition that $f(1) = f(2)$. This also shows that $\text{Range}(f) = f([1,2])$, Since $f$ is continuous and $[1,2]$ is compact, <b>$f$ is bounded and option 1 is true.</b>&nbsp;</span><br /><span style="font-size: large;"><b>Option 3:(False)</b> From the above observations, we can start with any function $g : [1,2] \to \Bbb R$ satisfying $g(1) = g(2)$ and can be extended to a function $\widetilde g \in \mathcal F$. Note that $\widetilde g$ restricted to $[1,2]$ is $g$ itself.</span><br /><span style="font-size: large;">Define $g: [1,2] \to \Bbb R$ by $g(x) = |x - \frac{3}{2}|$ then $g(1) = g(2) = \frac{1}{2}$ and $g$ is not differentiable at $x = \frac{3}{2}$ ($|x|$ is not differentiable at $0$ proof works here also). Since $g$ is not differentiable at $\frac{3}{2}$ its extension $\widetilde g$ is also <b>not differentiable</b> at $\frac{3}{2}$ (</span><span style="font-size: large;">$\widetilde g$ restricted to $[1,2]$ is $g$ itself</span><span style="font-size: large;">).<b> So option 3 is false</b></span><br /><span style="font-size: large;"><b>Option 2:(False) </b>Consider the function $g:(0,\infty) \to \Bbb R$ defined by $$g(x) = sin(\frac{2\pi\text{log }x}{\text{log }2}).$$</span><br /><span style="font-size: large;">$$g(2x) = sin(\frac{2\pi\text{log }2x}{\text{log }2}) = sin(\frac{2\pi(\text{log }2+\text{log x})}{\text{log }2}) = sin(\frac{2\pi\text{log }x}{\text{log }2} + 2\pi) \\ = g(x).$$</span><br /><span style="font-size: large;">Therefore $g \in \mathcal F$.&nbsp;</span><br /><b style="font-size: x-large;">Result</b><span style="font-size: large;">: $f$ is uniformly continuous on $(0,\infty)$ if and only if $f$ can be extended to a continuous function on $[0,\infty)$. In other words, $\lim_{x \to 0}f(x)$ exists.</span><br /><span style="font-size: large;">In our example, since $\lim_{x \to 0}g(x)$ doesn't exists (<b>because of log x</b>), $g$ cannot be extended to a continuous function on $[0,\infty)$. Therefore $g$ is <b>not uniformly continuous.</b></span><br /><span style="font-size: large;"><b>option 4:(False)</b>&nbsp;I made a mistake in the argument of this option. I am correcting it. I will update this option shortly. Sorry for the inconvenience.</span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/0GqNDfesTIs" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/nbhm-2020-part-c-question-28-solution.htmltag:blogger.com,1999:blog-7629869410369132909.post-40061489789117368742020-04-22T03:57:00.000-07:002020-04-22T03:57:34.257-07:00CSIR JUNE 2011 PART C QUESTION 81 SOLUTION ($f(z) = \frac{z}{3z+1}$ maps $H^+$ to $H^{+}$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Define $H^{+} = \{z=x+iy \in \Bbb C : y &gt; 0\}$</span><br /><span style="font-size: large;">$H^{-} = \{z=x+iy \in \Bbb C : y &lt; 0\}$</span><br /><span style="font-size: large;">$L^{+} = \{z=x+iy \in \Bbb C : x &gt; 0\}$</span><br /><span style="font-size: large;">$L^{-} = \{z=x+iy \in \Bbb C : x &lt; 0\}$&nbsp;</span><br /><span style="font-size: large;">Then the bilinear transformation $$f(z) = \frac{z}{3z+1}$$ maps&nbsp;</span><br /><span style="font-size: large;">1. $H^{+}$ onto $H^{+}$ and $H^{-}$ onto $H^{-}$.</span><br /><span style="font-size: large;">1. $H^{+}$ onto $H^{-}$ and $H^{-}$ onto $H^{+}$.</span><br /><span style="font-size: large;">1. $H^{+}$ onto $L^{+}$ and $H^{-}$ onto $L^{-}$.</span><br /><span style="font-size: large;">1. $H^{+}$ onto $L^{-}$ and $H^{-}$ onto $L^{+}$.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><span style="font-size: large;"><b>Solution:&nbsp;</b></span><br /><span style="font-size: large;"><b>Option 1: (True)</b> We have $$f(z) = \frac{z}{3z+1} = (3+\frac{1}{z})^{-1}.$$</span><br /><span style="font-size: large;">Let $z=x+iy \in H^{+}$ then $\frac{1}{z} = \frac{x}{x^2+y^2}+i\frac{-y}{x^2+y^2}$ whose imaginary part is negative. Hence $\frac{1}{z}(H^{+}) = H^{-}$. The function $g(z) = z + 3$ preserves all four regions since it is just a translation. Therefore if $z \in H^{+}$ then $\frac{1}{z} \in H^{-}$ and $\frac{1}{z}+3 \in H^{-}$. This shows that, $f(z) = (\frac{1}{z}+3)^{-1} \in H^{+}$. <b>Therefore $f$ sends $H^+$</b></span><br /><span style="font-size: large;"><b>onto $H^{+}$</b>. Similarly if $z \in H^{-}$ then $\frac{1}{z} \in H^{+}$ and $\frac{1}{z}+3 \in H^{+}$. This shows that, $f(z) = (\frac{1}{z}+3)^{-1} \in H^{-}$. <b>Therefore $f$ sends $H^{-}$ onto $H^{-}$</b>.</span><br /><span style="font-size: large;"><br /></span><b style="font-size: x-large;">Option 2:(False)&nbsp;</b><br /><span style="font-size: large;"><b>Option 3:(False)&nbsp;</b></span><br /><b style="font-size: x-large;">Option 4:(False)</b><br /><span style="font-size: large;">We have proved that&nbsp;</span><span style="font-size: large;">$f$ sends $H^+$&nbsp;</span><span style="font-size: large;">onto $H^{+}$ and&nbsp;</span><span style="font-size: large;">$H^{-}$ onto $H^{-}$. Hence they cannot be mapped to any other sets. Hence all the remaining options are false.</span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/plwO4Xhz3Zg" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-81.htmltag:blogger.com,1999:blog-7629869410369132909.post-12875409982140650492020-04-21T03:43:00.003-07:002020-04-23T03:38:17.424-07:00NBHM 2020 PART C Question 26 Solution (Jordan canonical form and Nilpotent matrices)(NBHM interview concept)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $T$ be a nilpotent linear operator on the vector space $\Bbb R^5$ (i.e., $T^k = 0$ for some k). Let $d_i$ denote the dimension of the kernel of $T^{i}$ . Which of the following can possibly occur as a value of $(d_1,d_2,d_3)$?&nbsp; <br />1. $(1,2,3)$, <br />2.$(2,3,5)$, <br />3.$(2,2,4)$, <br />4.$(2,4,5)$.</span><br /><span style="font-size: large;"><b>I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </b></span><br /><span style="font-size: large;"><b>Solution:</b></span><br /><span style="font-size: large;">Let $A$ be a $5 \times 5$ nilpotent matrix. We know that its eigenvalues are $0,0,0,0,0$. By Jordan canonical form theorem, there exists an invertible matrix $P$ such that $J(A) := P A P^{-1}$ will be the Jordan canonical form of $A$. Now, $J(A)$ is a block diagonal matrix in which each block is of size varies between one to five. For example, a block of size five is of the form $$\begin{bmatrix}0&amp;1&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">a block of size four is of the form&nbsp;</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">and so on for the blocks of size $3,2,1$.&nbsp;</span><br /><span style="font-size: large;"><b>Observation1</b>: If $B$ is a size $k$ Jordan block of $A$ then $\text{N}(B) := \text{ Nullity of$B$} = \text{dimension of Kernal of$B$}= 1$ and $\text{R}(B) := \text{rank of$B$} = \text{dimension of Range of$B$} = k-1$.</span><br /><span style="font-size: large;">Proof:&nbsp; $B$ is a $k \times k$ matrix with first $k-1$ rows are linearly independent and the last row is zero. The result follows.</span><br /><span style="font-size: large;"><b>Observation2</b>: Let $B$ be the following size $5$ Jordan block.</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;1&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">Then $B^2, B^3,B^4$ and $B^5$ are given respectively by the following matrices</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">Note that the above matrices are not in their Jordan canonical form. So power of a Jordan block need not be in the Jordan form again.&nbsp;</span><span style="font-size: large;">We observe that the Jordan block $B$ has one zero row&nbsp; and when we take the powers every step the last non-zero row is converted into a new zero row. Hence for $1 \le k \le 5$, $$N(B^k)&nbsp; = k.$$&nbsp;</span><br /><span style="font-size: large;"><b>Observation3</b>: Let $A$ be a $n \times n$ nilpotent matrix, then the number of Jordan blocks in the Jordan decomposition of $A$ is equal to $N(A)$ the dimension of the kernal of $A$.</span><br /><span style="font-size: large;">Proof: Let $J(A)$ equal to&nbsp;</span><br /><span style="font-size: large;">$\begin{bmatrix}B_1&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;B_2&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;B_3&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;B_4&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;B_5\end{bmatrix}$</span><br /><span style="font-size: large;">be the Jordan canonical form of the $n \times n$ matrix $A$. We have $5$ blocks in this decomposition. Let their sizes be $k_1,k_2,k_3,k_4$ and $k_5$ then $k_1 + \cdots +k_5 = n$. In particular the sizes of the blocks form a partition of $n$. For example, if we consider a $5 \times 5$ nilpotent matrix then the possible Jordan block sizes in the Jordan canonical form are $5$&nbsp;</span><br /><span style="font-size: large;">(one single $5$ block), $4+1$ (one $4$ block and one $1$ block ), $3+2$, $3+1+1$, $2+2+1$, $2+1+1+1$, $1+1+1+1+1$ given by all the partitions of $5$. Now, by the observation 1, $N(B_i) = 1$ for all $1 \le i \le 5$. Since $J(A)$ is a block diagonal matrix with blocks $B_1,\dots,B_5$ we have $N(J(A)) = N(B_1)+N(B_1)+\cdots+N(B_5) = 5$. Since $A$ and $J(A)$ are similar matrices we have $N(A) = N(J(A)) = 5 = \text{ number of blocks in }J(A)$. This proves the observation.</span><br /><span style="font-size: large;"><b>Observation4</b>:&nbsp; The $k$th power of $J(A)$&nbsp;</span><br /><span style="font-size: large;">$$\begin{bmatrix}B_1&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;B_2&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;B_3&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;B_4&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;B_5\end{bmatrix}^k$$ is equal to&nbsp;</span><br /><span style="font-size: large;">$$\begin{bmatrix}B_1^k&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;B_2^k&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;B_3^k&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;B_4^k&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;B_5^k\end{bmatrix}.$$</span><br /><span style="font-size: large;">&nbsp;</span><span style="font-size: large;">With these observations we will investigate the options.</span><br /><span style="font-size: large;"><b>option 1</b>:(1,2,3)(<b>Possible</b>) It is given that $d_1 = 1$, $d_2 = 2$ and $d_3 = 3$. By the observation 2, we have to find a $5 \times 5$ matrix $A$ such that $J(A)$ has one Jordan block, $N(A^2) = 2$ and $N(A^3) = 3$.&nbsp;</span><span style="font-size: large;">Let $A$ be the $5\times 5$ matrix given below which itself is in the Jordan canonical form.</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;1&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">Then $A^2, A^3$ are given respectively by the matrices</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;0&amp;1&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">$$\begin{bmatrix}0&amp;0&amp;0&amp;1&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;1 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0 \\ 0&amp;0&amp;0&amp;0&amp;0\end{bmatrix}$$</span><br /><span style="font-size: large;">Clearly $N(A^2) = 2$ and $N(A^3) = 3$.</span><br /><span style="font-size: large;"><b>option 2</b>:<b>(2,3,5)</b>(<b>Not Possible</b>) It is given that $d_1 = 2$, $d_2 = 3$ and $d_3 = 5$. By the observation 2, we have to find a $5 \times 5$ matrix $A$ such that $J(A)$ has two Jordan blocks, $N(A^2) = 3$ and $N(A^3) = 5$. Since $J(A)$ has two Jordan blocks, the possible block sizes are $3+2$ or $4+1$. Suppose $J(A)$ has block decomposition $3+2$. Let $B_1$ be the size $3$ block and $B_2$ be the size $2$ block in $J(A)$. When we find $(J(A))^2$, the last non-zero row of $B_1$ will be changed to zero row and similarly in $B_2$. Hence $N((J(A)^2) = N(J(A))+2 = 2+2 = 4$. But we want this number to be $3$ for option b. Hence the block size $3+2$ is not possible for option b. We continue to calculate $N((J(A))^3)$ which will be helpful for option d.&nbsp;</span><span style="font-size: large;">When we find $(J(A))^3$, again the last non-zero row will be changed to zero row in $B_</span><span style="font-size: large;">1$.&nbsp;</span><span style="font-size: large;">Since $B_2$ is a $1 \times 1$ block it wont have a non-zero row to change.</span><span style="font-size: large;">&nbsp;Hence $N((J(A)^3) = N((J(A))^2)+1 = 4+1 = 5$. <b>Hence $(d_1,d_2,d_3) = (2,4,5)$ is possible which is option d.</b></span><br /><span style="font-size: large;">Suppose $J(A)$ has block decomposition $4+1$. Let $B_1$ be the size $4$ block and $B_2$ be the size $1$ block in $J(A)$. When we find $(J(A))^2$, the last non-zero row of&nbsp;</span><span style="font-size: large;">&nbsp;</span><span style="font-size: large;">$B_1$</span><span style="font-size: large;">&nbsp;will be changed to zero-row. Since $B_2$ is a $1 \times 1$ block it won't have a non-zero row to change. Hence $N((J(A)^2) = 2 + 1 = 3$. Again, w</span><span style="font-size: large;">hen we find $(J(A))^3$, similar argument can be appliead and $N((J(A)^3) =3 + 1 = 4$.</span><span style="font-size: large;">&nbsp;&nbsp;But we want this number to be $5$. Hence the block size $4+1$ is also not possible for option b.&nbsp;</span><span style="font-size: large;">Since $3+2$ and $4+1$ are the only two possible block sizes we have proves that </span><b style="font-size: x-large;">$(d_1,d_2,d_3) = (2,3,5)$ is not possible</b><span style="font-size: large;">.</span><br /><span style="font-size: large;"><b>option c:(2,2,4)(Not Posssible) </b>The argument given in option $2$ shows that if $d_1 = 1$ then the possible $(d_1,d_2,d_3)$ are $(2,3,4)$ and $(2,4,5)$. Hence $(2,2,4)$ is not possible.</span><br /><span style="font-size: large;"><b>option d:(2,4,5)(Possible) </b>This is done as part of option b.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;">A pictorial approach of the same can be seen here:</span><br /><span style="font-size: large;"><a href="https://maksmaths.blogspot.com/2020/04/nbhm-problem.html">https://maksmaths.blogspot.com/2020/04/nbhm-problem.html</a></span><br /><br /></div><span style="font-size: large;"><span style="font-size: small;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems</span><span style="font-size: medium;">.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/4UYSzkGOIZ8" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com4https://nbhmcsirgate.theindianmathematician.com/2020/04/nbhm-2020-part-c-question-26-solution.htmltag:blogger.com,1999:blog-7629869410369132909.post-43524819101236501452020-04-20T20:37:00.001-07:002020-04-22T02:48:47.040-07:00CSIR JUNE 2011 PART C QUESTION 81 SOLUTION (Analytic function with $f(0) = \frac{1}{2}$ and $f(\frac{1}{2}) = 0$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $f$ be an analytic function from the unit disc $\Delta$ to $\Delta$ with $\Delta = \{z \in \Bbb C : |z| &lt; 1\}$ satisfying $f(0) = \frac{1}{2}$ and $f(\frac{1}{2}) = 0$.&nbsp; Which of the following is/are true? <br />1. $|f^{'}(0)| \le \frac{3}{4}$, <br />2. $|f^{'}(\frac{1}{2})| \le \frac{4}{3}$. <br />3. option 1 and 2 both are true. <br />4. $f$ is the identity function.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><span style="font-size: large;">Solution:<br /> Schwarz's pick lemma:&nbsp;</span><span style="font-size: large;">Let $f$ be an analytic function from the unit disc $\Delta$ to $\Delta$. Further if $f(a) = b$ then we have $|f^{'}(a)| \le \frac{1-|f(a)|^2}{1-|a|^2}$.&nbsp;</span><br /><span style="font-size: large;">Given that $f(0) = \frac{1}{2}$. This implies that $$|f^{'}(0)| \le \frac{1 - (\frac{1}{2})^2}{1-(0)^2} = \frac{3}{4}.$$</span><br /><span style="font-size: large;">Given that $f(\frac{1}{2}) = 0$. This implies that $$|f^{'}(\frac{1}{2})| \le \frac{1 - (0)^2}{1-(\frac{1}{2})^2} = \frac{4}{3}.$$ <b>So option 1, 2 and 3 are true.</b></span><br /><span style="font-size: large;"><b>option 4: (False)</b>&nbsp;Let $f(z) =frac{az + b}{cz + d}$. Now, $f(0) = \frac{1}{2}$ implies that $2b = d$ and we have $f(z) = \frac{az+b}{cz+2b}$. Now, $f(\frac{1}{2}) = 0$ implies that $a = -2b$. Let us take $a = 2, b= -1, c= 1$ and $d = 2$, then all these conditions are met and $ad-bc \ne 0$. Therefore the function $$f(z) = \frac{2z - 1}{z - 2}$$ satisfies the given conditions of the problem which is different from the identity function.</span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/N4il54WdXFU" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-80_20.htmltag:blogger.com,1999:blog-7629869410369132909.post-59796668486739942322020-04-20T20:04:00.001-07:002020-04-20T20:37:59.037-07:00CSIR JUNE 2011 PART B QUESTION 52 SOLUTION (Random variable with $E(X) = Var(X)$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $X$ be a random variable with $E(x) = Var(X)$. Then the distribution of $X$ has to be<br />1. Poisson, <br />2. Exponential, <br />3. Normal, <br />4. cannot be identified from the given condition&nbsp;&nbsp;</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><span style="font-size: large;"><b>Solution:&nbsp;</b></span><br /><span style="font-size: large;">Poisson: Let $X$ be a discrete random variable. $X$ has a Poisson distribution with a positive real parameter $\lambda$ if the probability distribution function of $X$ is equal to $$\frac{\lambda^n e^{-\lambda}}{n!}$$ for $n \ge 0$. In this case mean = variance = $\lambda$.</span><br /><span style="font-size: large;">Normal: Let $X$ be a continuous random variable. $X$ has a Normal distribution with parameters $\mu$ and $\sigma$ if the probability distribution of $X$ is equal to $$\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-1}{2}(\frac{x-\mu}{\sigma})^2}.$$ In this case mean = $\mu$ and variance = $\sigma^2$. It is possible to take $\mu = 1$ and $\sigma = 1$ the we get $mean = variance$.</span><br /><span style="font-size: large;">So we conclude that the condition $E(X) = Var(X)$ can occur for both Poisson and Normal. Also, for many other distributions and please comment below if you know any other such distribution. This shows that from the condition $E(X) = Var(X)$ we cannot conclude the distribution uniquely. <b>so option 4 is correct</b>.</span></div><span style="font-size: large;"><span style="font-size: medium;"><b>Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</b></span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/V86iiZapK0U" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-b-question-52.htmltag:blogger.com,1999:blog-7629869410369132909.post-15580039753211875032020-04-20T11:00:00.000-07:002020-04-20T20:43:19.277-07:00CSIR JUNE 2011 PART C QUESTION 80 SOLUTION ($\text{Re}(f(z)$ bounded then $f(z)$ is constant: Three different proofs)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $f$ be an entire function. If $\text{Re} f$ is bounded then <br />1. $\text{Im} f$ is constant, <br />2. $f$ is constant, <br />3. $f \equiv 0$, <br />4. $f^{'}$ is a non-zero constant.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><span style="font-size: large;"><b>Solution: </b><br />First, we prove that option 2 is correct which will imply that option 1 is also true.</span><br /><span style="font-size: large;"><b>Option 2: (True)&nbsp;</b></span><br /><span style="font-size: large;"><b>Proof 1: </b>Given that $f$ is entire. Assume $f$ is not a constant function, then by little Picards's theorem, range of $f$ is either $\Bbb C$ or $\Bbb C \backslash \{a\}$ for some $a \in \Bbb C$. Therefore $\text{Re}(f)$ is unbounded. It is given that $\text{Re}(f)$ is bounded. Hence $f$ has to a constant function.</span><br /><span style="font-size: large;"><b>Proof 2:</b></span><br /><span style="font-size: large;">Assume that the real part of $f$ is bounded then there exists $k$ such that $\text{Re}(f(z)) \le k$. We define $h(z) = \frac{1}{k - f(z) + 1}$. Now $\text{Re}f(z) \le k$ implies that $\text{Re} (k - f(z)) \ge 0$ and $\text{Re}(k-f(z)+1) \ge 1$. This shows that the denominator $k-f(z)+1$ will never zero and hence $h(z)$ is an entire function. Let $h(z) = \frac{1}{u + i v}&nbsp; = \frac{u-iv}{u^2 + v^2} = \frac{u}{u^2+v^2} -i \frac{v}{u^2 + v^2}$. We have $u = \text{Re}(k-f(z)+1)$ and $|h(z)|^2 = (\frac{u}{u^2+v^2})^2+(\frac{v}{u^2+v^2})^2 = \frac{1}{u^2 + v^2} \le 1$ since $u = \text{Re}(k-f(z)+1) \ge 1$. This shows that $|h(z)| \le 1$ and hence h(z) is a bounded entire function which has to be constant by the Liouville's theorem. But $h(z)$ is constant implies that <b>$f(z)$ is constant.&nbsp;</b></span><br /><span style="font-size: large;"><b>proof 3: </b>Let $f(z) = u + iv$. It is given that $u$ is bounded. Consider the function $h(z) = e^{f(z)}$. Now, $|e^z| = |e^{u+iv}| = |e^u| \,|e^{iv}| = e^u$ which is bounded as $u$ is bounded. By Liouville's theorem $h(z)$ is constant and hence $f(z)$ is also constant.</span><br /><span style="font-size: large;"><b>option 4:(False)&nbsp;</b>&nbsp;We have proves that $f$ is constant and hence <b>its derivative $f^{'}$ has to be zero.</b></span><br /><span style="font-size: large;"><b>option 1: (True) </b>We have shown that the function $f$ itsels is constant in the previous option. Hence $\text{Im}(f)$ has to be bounded.</span><br /><span style="font-size: large;"><b>option 3</b>:(<b>False</b>) Consider the constant function $f(z) = 1$ for all $z \in \Bbb C$. Then $\text{Re}(f)$ is bounded but $f$ is not identically zero.</span><br /><br /></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/RZjDdnHReDw" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-80.htmltag:blogger.com,1999:blog-7629869410369132909.post-29396481853938344402020-04-18T11:11:00.001-07:002020-04-20T20:42:16.596-07:00CSIR JUNE 2011 PART C QUESTION 79 SOLUTION (Analytic functions and Identity theorem application)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $\Bbb D = \{z \in \Bbb C : |z| &lt; 1\}$ be the unit disc. Let $f : \Bbb D \to \Bbb C$ be an analytic function satisfying $$f(\frac{1}{n}) = \frac{2n}{3n+1}.$$ Then <br />1. $f(0) = \frac{2}{3}$, <br />2. $f$ has simple pole at $z = -3$, <br />3. $f(3) = \frac{1}{3}$, <br />4. no such $f$ exists.</span><br /><span style="font-size: large;"><b>I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </b></span><br /><span style="font-size: large;"><b>Solution: </b>We will explicitly calculate the function $f$ then the options are easy to verify.</span><br /><span style="font-size: large;"><b>Identity Theorem:&nbsp;</b></span><span style="font-size: large;">Let $\Bbb D = \{z \in \Bbb C : |z| &lt; 1\}$ be the unit disc. Let $f,g : \Bbb D \to \Bbb C$ be two analytic function. If&nbsp; $f(\frac{1}{n}) = g(\frac{1}{n})$ for all $n \in \Bbb N$ then $f \equiv g$ (i.e., $f(z) = g(z)$ for all $z \in \Bbb D$).&nbsp;</span><br /><br /><span style="font-size: large;">Replace $n$ by $\frac{1}{n}$ in $\frac{2n}{3n+1}$ we get $\frac{2}{3 + n}$.&nbsp;</span><span style="font-size: large;">Let $g : \Bbb D \to \Bbb C$ be an analytic function defined by $g(z) = \frac{2}{3+z}$ (Note that $g$ has no singularities in $\Bbb D$). Then it is given in the problem that&nbsp; $f(\frac{1}{n}) = g(\frac{1}{n})$ for all $n \in \Bbb N$. Therefore by the above said identity theorem we have $f(z) = g(z)$ for all $z \in \Bbb D$. Hence $$f(z) = \frac{2}{3+z}.$$</span><br /><span style="font-size: large;"><b>option 1</b>: (<b>True</b>) Clearly $f(0) = \frac{2}{3}$.</span><br /><b style="font-size: x-large;">option 2</b><span style="font-size: large;">: (</span><b style="font-size: x-large;">True</b><span style="font-size: large;">)&nbsp;Clearly $-3$ is a simple pole of $f(z)$.</span><br /><b style="font-size: x-large;">option 3</b><span style="font-size: large;">: (</span><b style="font-size: x-large;">True</b><span style="font-size: large;">) Clearly $f(3) = \frac{1}{3}$</span><br /><b style="font-size: x-large;">option 4</b><span style="font-size: large;">: (</span><b style="font-size: x-large;">False</b><span style="font-size: large;">)&nbsp; The function $f(z) = \frac{2}{3+z}$ is one such function.</span><span style="font-size: large;">&nbsp;</span></div><span style="font-size: large;"><b><span style="font-size: small;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</span>Share to your groups: <!-- Go to www.addthis.com/dashboard to customize your tools --> </b></span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/DJX_TjTgN7M" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-79.htmltag:blogger.com,1999:blog-7629869410369132909.post-55670093014610303022020-04-18T09:56:00.000-07:002020-04-18T09:56:03.248-07:00CSIR JUNE 2011 PART C QUESTION 77 SOLUTION (Diagonalizability of Nilpotent matrices)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $N$ be a $3 \times 3$ non-zero matrix with the property $N^3 = 0$. Which of the following is/are true? <br />1. $N$ is not similar to a diagonal matrix, <br />2. $N$ is similar to a diagonal matrix, <br />3. $N$ has one non-zero eigenvector, <br />4. $N$ has three linearly independent eigenvectors.</span><br /><span style="font-size: large;"><b>I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </b></span><br /><span style="font-size: large;"><b>Solution: </b>An $n \times n$ matrix $N$ is said to be <b>nilpotent</b> if $N^k = 0$ for some $k \in \Bbb N$. A matrix $M$ is said to be <b>diagonalizable over a field $\Bbb F$</b> if there exists an invertible matrix $P$ with entries from the field $\Bbb F$ such that $$P A P^{-1} = \text{diag}(\lambda_1,\lambda_2,\dots,\lambda_n)$$ where $\text{diag}(\lambda_1,\dots,\lambda_n)$ is the diagonal matrix with the diagonal entries eigenvalues of $A$.</span><br /><span style="font-size: large;"><b>Observation1: All the eigenvalues of a nilpotent matrix are zero.&nbsp;</b></span><br /><span style="font-size: large;">Proof: Let $N$ be a nilpotent matrix and $\lambda$ be an eigenvalue of $N$. Then $\lambda^k$ is an eigenvalue of $N^k = 0$. Hence $\lambda^k = 0$ and in turn $\lambda = 0$.</span><br /><span style="font-size: large;"><b>Result1</b>: If $A$ is a non-diagonal matrix whose eigenvalues are equal. Then $A$ is not diagonalizable.&nbsp;</span><br /><span style="font-size: large;">Proof: Suppose $A$ is diagonalizable, then there exists an invertible matrix $P$ such that </span><span style="font-size: large;">$P A P^{-1} = \text{diag}(\lambda,\lambda,\dots,\lambda)$.</span><span style="font-size: large;">&nbsp;This implies that $A = P^{-1}\text{diag}(\lambda,\lambda,\dots,\lambda)P$. But $\text{diag}(\lambda,\dots,\lambda)$ is a scalar matrix and hence it commutes with all the matrices. So $A = P^{-1}\text{diag}(\lambda,\dots,\lambda)P = \text{diag}(\lambda,\dots,\lambda)$ Contradiction to the fact that $A$ is a non-diagonal matrix.</span><br /><span style="font-size: large;"><b>Result2</b>: An $n \times n$ matrix $M$ is diagonalizable if and only if it has $n$ linearly independent eigenvectors.</span><br /><span style="font-size: large;"><b>option 1</b>:(<b>True</b>) Follows from observation 1 and result 1.</span><br /><span style="font-size: large;"><b>option 2</b>:(<b>False</b>) Because option 1 is true.</span><br /><span style="font-size: large;"><b>option 3</b>:(<b>False</b>) We have shown that $0$ is an eigen value of $N$. Hence there exists an eigen vector $v \ne 0$ such that $N v = 0$. Let $\alpha$ be any scalar then $N (\alpha v) = \alpha Nv = 0$. So any scalar multiple of eigenvector is again an eigenvector. So there are more than one eigen vectors.&nbsp;</span><br /><span style="font-size: large;"><b>option 4</b>:(<b>False</b>) We have shown that $N$ is not diagonalizable in option 1. Hence by result 2 it cannot have $n$ linearly independent eigen vectors.</span><br /><br /><br /></div><span style="font-size: large;"><span style="font-size: small;"><b>Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</b></span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/GL0C7w7NT3o" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-77.htmltag:blogger.com,1999:blog-7629869410369132909.post-7263151189103378922020-04-18T06:52:00.001-07:002020-04-18T07:03:14.328-07:00NBHM 2020 PART C Question 24 Solution ($B = \{(z,w) \in \Bbb C^2 : |Re\, z|^2 + |Re\, w|^2 = 1\}$ is not bounded )<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Listed below are four subsets of&nbsp; $\Bbb C^2$. For each of them, write “Bounded” or “Unbounded” in the box as the case may be. ($Re(z)$ denotes the real part of a complex variable $z$).</span><br /><span style="font-size: large;">1. $A = \{(z,w) \in \Bbb C^2 : z^2 + w^2 = 1\}$.</span><br /><span style="font-size: large;">2. $B = \{(z,w) \in \Bbb C^2 : |Re\, z|^2 + |Re\, w|^2 = 1\}$.</span><br /><span style="font-size: large;">3. $C = \{(z,w) \in \Bbb C^2 : |z|^2 + |w|^2 = 1\}$.</span><br /><span style="font-size: large;">4. $D = \{(z,w) \in \Bbb C^2 : |z|^2 - |w|^2 = 1\}$.</span><br /><span style="font-size: large;"><b>I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.</b></span><br /><span style="font-size: large;"><b>Solution: </b>Consider the standard metric in $\Bbb C$ given by $d(z,w) = |z-w|$. Now, the metric on $\Bbb C^2$ is given by $$d((z_1,w_1),(z_2,w_2)) = \sqrt{|z_2 - z_1|^2 + |w_2 - w_1|^2}.$$&nbsp;</span><br /><span style="font-size: large;"><b>Option 1:(Unbounded) </b>We have $\{(it, \sqrt{1-(it)^2}) : t \in \Bbb R\} \subseteq A$. Now, $$d((0,0),(it,\sqrt{1+t^2})) = \sqrt{(t^2)+(1+t^2)} \ge \sqrt{t^2} =&nbsp; t.$$ and can be made arbitrarily large. <b>So $A$ is unbounded.</b></span><br /><span style="font-size: large;"><b>Option 2:(Unbounded)&nbsp;</b></span><span style="font-size: large;">$\{(1, it) : t \in \Bbb R\} \subseteq B$.&nbsp;</span><span style="font-size: large;">Now, $$d((0,0),(1, it)) = \sqrt{(1^2)+(t^2)} = \sqrt{1+t^2} \ge t.$$ which can be made arbitrarily large. <b>So $B$ is unbounded.</b></span><br /><span style="font-size: large;"><b>Option 3:(Bounded) </b></span><span style="font-size: large;">$C = \{(z,w) \in \Bbb C^2 : |z|^2 + |w|^2 = 1\} = \{(z,w) : d((0,0),(z,w)) = 1\}$. This is the unit sphere in $\Bbb C^2$ and <b>hence $C$ is bounded</b>.</span><br /><span style="font-size: large;"><b>Option 4:(Unbounded) </b>We have $cosh^2\,t - sinh^2\,t = 1$ for $t \in \Bbb R$. Hence </span><span style="font-size: large;">$\{(cosh\,t, sinh\,t) : t \in \Bbb R\} \subseteq D$.&nbsp;</span><span style="font-size: large;">Now, $$d((0,0),(cosh\,t, sinh\,t)) = \sqrt{(cosh^2\,t)+(sinh^2\,t)} \\ = \sqrt{(\frac{e^t + e^{-t}}{2})^2+(\frac{e^t-e^{-t}}{2})^2} = \sqrt{\frac{e^{2t}+e^{-2t}}{2}}.$$ which can be made arbitrarily large. So $D$ is unbounded.</span></div><span style="font-size: large;"><span style="font-size: small;"><b>Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</b></span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/S9hm0ghOuZI" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/nbhm-2020-part-c-question-24-solution.htmltag:blogger.com,1999:blog-7629869410369132909.post-50203257008500232562020-04-18T00:15:00.000-07:002020-04-18T07:03:13.229-07:00NBHM 2020 PART C Question 27 Solution ($f(x,y) = x + y$ is a clopen function, preserves dense and discrete sets)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Consider $f : \Bbb R^2 \to \Bbb R$ defined by $f(x,y) = x + y$. For each of the following statements, state whether it is true or false. <br />1. Image under $f$ of any open set if open, <br />2. Image under $f$ of any closed set is closed, <br />3. Image under $f$ of any dense set is dense, <br />4. Image under $f$ of any discrete set is discrete.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you. </span></b><br /><b><span style="font-size: large;">Solution:&nbsp;</span></b><br /><span style="font-size: large;"><b>Option 1</b>: (<b>True</b>) Consider the function $f$. We claim that $f$ is an open map. Let $U$ be an open set in $\Bbb R^2$. We need to show that $f(U)$ is open in $\Bbb R$. Let $a \in f(U)$, we will show that there is a neighborhood of $a$ which is entirely contained in $f(U)$. Let $(x,y) \in U \subseteq \Bbb R^2$ be such that $f(x,y)=x+y = a$. Since $a \in f(U)$ this is possible. Now $(x,y) \in U$ and $U$ is open and so we can find an $\epsilon &gt; 0$ such that the open set $V = (x-\epsilon,x+\epsilon) \times (y-\epsilon,y+\epsilon) \subseteq U$. Now, $f(V) = ((x+y)-2\epsilon, (x+y)+2\epsilon) = (a-2 \epsilon, a+2\epsilon)$ </span><span style="font-size: large;">&nbsp;is an open interval. Since $V \subset U$ we have $f(V) \subseteq f(U)$ is an interval containing $a$. <b>Therefore $f(U)$ is open.</b></span><br /><span style="font-size: large;"><b>option 2:(False) </b>We will construct a closed set $F$ such that $f(F)$ is not closed. Let $F = \Bbb Z \times \sqrt r \Bbb Z$ where $r$ is not a perfect square integer. Then $f(F) = \Bbb Z + \sqrt r \Bbb Z$ which is a dense subset of $\Bbb R$. (<b>Why this is dense? please check my next post</b>). Therefore $f(F)$ is not closed.</span><br /><span style="font-size: large;"><b>option 4</b>:(<b>False</b>) Consider the example given in option 2. $F$ is discrete but $f(F)$ is not.</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>True</b>) Let $D$ be a subset of a metric space $X$, then $D$ is dense in $X$ if given $x$ in $X$ there exists a sequence $d_n$ in $D$ such that $d_n$ converges to $x$. We will use this definition. Consider the function $f$ and let $D$ be a dense subset of $\Bbb R^2$. We claim that $f(D)$ is dense in $\Bbb R$. Let $a \in f(D)$ then there exists a point $(x,y) \in D$ such that $f(x,y) = a$. Now $(x,y) \in D$ and $D$ is dense. So there exists a sequence $(x_n,y_n) \in D$ such that $(x_n,y_n)$ converges to $(x,y)$. Since the function $f$ is continuous, we have $f((x_n,y_n)) = (x_n+y_n)$ converges to $x+y = a$. Since $(x_n,y_n)$ is a sequence in $D$ we have $(x_n+y_n)$ is a sequence in $f(D)$ which converges to $a$. <b>Therefore $f(D)$ is dense in $\Bbb R$.</b></span><br /><span style="font-size: large;"><b>THERE IS A MORE GENERAL RESULT IS TRUE:</b></span><br /><span style="font-size: large;">Every continuous subjective (onto) map between the metric spaces preserves the dense sets. The proof is essentially the same as above. Note that, in this problem given $f$ is a surjective continuous map.&nbsp;</span></div><span style="font-size: large;"><span style="font-size: small;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/wmvXgFcF-Kk" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com4https://nbhmcsirgate.theindianmathematician.com/2020/04/nbhm-2020-part-c-question-27-solution_18.htmltag:blogger.com,1999:blog-7629869410369132909.post-80073291720505783112020-04-17T12:27:00.000-07:002020-04-18T07:03:14.675-07:00NBHM 2020 PART C Question 27 Solution (Structure of the group $\frac{\Bbb Q}{n \Bbb Z}$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">For $n$ a positive integer, let $\frac{\Bbb Q}{n\Bbb Z}$ be the quotient of the group of rational numbers $\Bbb Q$ (under addition) by the subgroup $n\Bbb Z$. For each of the following statements, state whether it is true or false. <br />1) Every element of&nbsp;</span><span style="font-size: large;">$\frac{\Bbb Q}{n\Bbb Z}$</span><span style="font-size: large;">&nbsp;is of finite order,&nbsp;</span><br /><span style="font-size: large;">2) There are only finitely many elements in&nbsp;</span><span style="font-size: large;">$\frac{\Bbb Q}{n\Bbb Z}$</span><span style="font-size: large;">&nbsp;of any given finite order,&nbsp;</span><br /><span style="font-size: large;">3) Every proper subgroup of&nbsp;</span><span style="font-size: large;">$\frac{\Bbb Q}{n\Bbb Z}$</span><span style="font-size: large;">&nbsp;is finite,&nbsp;</span><br /><span style="font-size: large;">4)</span><span style="font-size: large;">$\frac{\Bbb Q}{2\Bbb Z}$&nbsp;</span><span style="font-size: large;">and&nbsp;</span><span style="font-size: large;">$\frac{\Bbb Q}{5\Bbb Z}$</span><span style="font-size: large;">&nbsp;are isomorphic as groups.</span><br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following the blog by email. Also, visit my blog every day and share the solutions with friends. Thank you.</span></b><span style="font-size: large;"></span><br /><b><span style="font-size: large;"><br /></span></b><b><span style="font-size: large;">Solution:</span></b><br /><span style="font-size: large;">The elements of $\frac{\Bbb Q}{ \Bbb Z}$ are cosets of the form $\Bbb Z + s$ where $s = \frac{p}{q}$ is rational (<b>We have gcd(p,q)=1</b>). We have $\Bbb Z + s$ is equal to the identity element $\Bbb Z + 0 = \Bbb Z$ if and only if $s \in \Bbb Z$</span><span style="font-size: large;">.</span><br /><b style="font-size: x-large;">Main Result : Let $n$ be a positive integer then $$\frac{\Bbb Q}{\Bbb Z} \cong \frac{\Bbb Q}{n \Bbb Z}.$$</b><br /><span style="font-size: large;"><b>Proof: </b>The function which takes $\Bbb Z + x$ to $n \Bbb Z + nx$ is the required isomorphism.&nbsp;</span><br /><b style="font-size: x-large;">In view of this result, it is enough to verify the given options to the group $\frac{\Bbb Q}{\Bbb Z}$.</b><br /><b style="font-size: x-large;">option 4</b><span style="font-size: large;">: (</span><b style="font-size: x-large;">True</b><span style="font-size: large;">) By the above result we have the group $\frac{\Bbb Q}{\Bbb Z}$ is isomorphic to the groups $\frac{\Bbb Q}{2 \Bbb Z}$ and $\frac{\Bbb Q}{5 \Bbb Z}$. By transitivity, we have <b>$\frac{\Bbb Q}{2 \Bbb Z}$ is isomorphic to $\frac{\Bbb Q}{5 \Bbb Z}$.</b></span><br /><b style="font-size: x-large;"><br /></b><b style="font-size: x-large;">option 1</b><span style="font-size: large;">: (</span><b style="font-size: x-large;">True</b><span style="font-size: large;">) We claim that every element of $\frac{\Bbb Q}{\Bbb Z}$ has finite order.</span><br /><span style="font-size: large;">Let $x = \Bbb Z + \frac{p}{q}$ be an arbitrary element of $\frac{\Bbb Q}{\Bbb Z}$. Let $m$ be the smallest integer such that <b>$mp = q r$</b> for some $r \in \Bbb N$ (i.e,&nbsp;</span><span style="font-size: large;">$q$ divides $mp$</span><span style="font-size: large;">). Such a $m$ exists because $m = q$ is one possibility such that $q \mid mp$ and so we can consider the minimal one among the $m$ satisfying this condition by the well ordering principle of $\Bbb N$.&nbsp;</span><br /><span style="font-size: large;">We have,</span><span style="font-size: large;">&nbsp;$$x + x =&nbsp; \Bbb Z + \frac{2p}{q}$$ and adding $x$ with itself for $m$ times gives us $$x^{m} = (\Bbb Z + \frac{p}{q})^m =&nbsp; \Bbb Z + \frac{mp}{q} =&nbsp; \Bbb Z + r = \Bbb Z.$$&nbsp;&nbsp;</span><span style="font-size: large;">This shows that $m$ is the smallest power such that $x^m =e$ and so the order of $x = n \Bbb Z + \frac{p}{q}$ is equal to $m$.&nbsp;</span><b style="font-size: x-large;">Hence every element in $\frac{\Bbb Q}{n \Bbb Z}$ is of finite order.</b><br /><b style="font-size: x-large;"><br /></b><b style="font-size: x-large;">option 2:(True)&nbsp;</b><br /><span style="font-size: large;"><b>Observation 1</b>: Let $x = \Bbb Z + \frac{p}{q}$ be an arbitrary element of $\frac{\Bbb Q}{\Bbb Z}$. Let $p = qm + r$ ($0 \le r &lt;q$). Since $gcd(p,q)=1$ we have $1&lt;r&lt;q$ and $\frac{p}{q} = m + \frac{r}{q}$. Now, $$\Bbb Z + \frac{p}{q} = \Bbb Z + (m+\frac{r}{q}) = \Bbb Z + \frac{r}{q}.$$</span><br /><span style="font-size: large;">Hence, whenever we consider an arbitrary element $\Bbb Z + \frac{p}{q}$ of $\frac{\Bbb Q}{\Bbb Z}$<b>we can always assume that $1&lt;p&lt;q$</b>.&nbsp;</span><br /><span style="font-size: large;"><br /></span><b style="font-size: x-large;">Observation 2</b><span style="font-size: large;">: Let $n$ be a positive integer. The set $H_n = \{\Bbb Z + \frac{m}{n}&nbsp; : 0 \le m \le n\}$ (here we are not assuming (m,n)=1) is a cyclic subgroup of $\frac{\Bbb Q}{\Bbb Z}$ order $n$.&nbsp;&nbsp;</span><br /><span style="font-size: large;">Proof: Verify that the element $\Bbb Z + \frac{1}{n}$ is a generator. In general the elements $\Bbb Z + \frac{m}{n}$ with (m,n)=1 in $H_n$ forms the $\phi(n)$ generators of this cyclic group.&nbsp;</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;">Note that, as an element of $\frac{\Bbb Q}{\Bbb Z}$, we always assume $(m,n) = 1$ in $\Bbb Z + \frac{m}{n}$. <b>Hence $\Bbb Z + \frac{m}{n}$ of $\frac{\Bbb Q}{\Bbb Z}$ has order $n$.&nbsp;</b></span><br /><span style="font-size: large;"><b><br /></b></span><span style="font-size: large;">This shows that the set of elements of order $n$ in $\frac{\Bbb Q}{\Bbb Z}$ are given by $\Bbb Z + \frac{p}{n}$ with $1 \le p \le n$, $(p,n) = 1$ and there are $\phi(n)$ of them. <b>Hence there are only finitely many elements of order $n$ in $\frac{\Bbb Q}{\Bbb Q}$ for every positive integer $n$.</b></span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>option 3:</b>(<b>False</b>) Let $m$ be a positive integer. Define $$H(m) = \{\Bbb Z + \frac{p}{q} : q \text{ divides } m\}.$$&nbsp;</span><br /><span style="font-size: large;">It is straight forward to verify that $H(m)$ is indeed an infinite subgroup of $\frac{\Bbb Q}{\Bbb Z}$.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</b></span><span style="font-size: large;">&nbsp;&nbsp;</span><b style="font-size: x-large;">Share to your groups:</b></div><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/-ySq29WaFnY" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com4https://nbhmcsirgate.theindianmathematician.com/2020/04/nbhm-2020-part-c-question-27-solution.htmltag:blogger.com,1999:blog-7629869410369132909.post-4272252843812860462020-04-16T08:28:00.002-07:002020-04-18T07:03:12.918-07:00NBHM 2020 PART B Question 22 Solution ( Distinct homeomorphism classes of topologies on X with exactly n open subsets)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><div style="text-align: center;"><span style="font-size: large;"><b>YOU WANT ME DO NBHM OR CSIR QUESTIONS?. PLEASE COMMENT BELOW.</b></span></div><span style="font-size: large;">Let X be a three-element set. For each of the following numbers n, determine the number of distinct homeomorphism classes of topologies on X with exactly n open subsets (including the empty set and the whole set). Write that number in the box. <br />1) 3, <br />2) 4, <br />3) 5, <br />d) 7.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>I have given a detailed solution below so that the same idea can be used in similar problems. Do read completely.</b></span><br /><span style="font-size: large;"><br /></span><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me by following</span></b><b><span style="font-size: large;">&nbsp;the blog by email.</span></b><b><span style="font-size: large;">&nbsp;Visit my blog every day and share the solutions with friends.&nbsp; Thank you.</span></b><br /><span style="font-size: large;"><b>Solution: </b>Let $(X,\tau_1)$ and $(X,\tau_2)$ be topological spaces on a set $X$. A bijection $f: X \to X$ is said to be homeomorphism from $(X,\tau_1)$ to $(X,\tau_2)$ if $f$ is a continuous open map. In particular, the image and inverse image of an open set is open.&nbsp;</span><br /><span style="font-size: large;">Given $X = \{1,2,3\}$. Note that, <b>in any topology on $X$ a proper open set has cardinality either $1$ to $2$.</b></span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>option 1</b>: (<b>Number of topologies with three open sets</b>) Let $(X,\tau)$ be a topology on $X = \{1,2,3\}$ with three open sets. Then $\tau = \{X,\phi, U\}$ where $U$ is a subset of cardinality either $1$ or $2$. There are three subsets of $X$ with cardinality $1$ and three subsets with cardinality $2$. Any of these $6$ subsets can play the role of $U$ in $\tau$ to form a topology with three open sets. Hence there are $6$ topologies on $X$ with $3$ open sets.&nbsp;</span><span style="font-size: large;">If $|U| = 1$ then we call such topologies of type I and if $|U| = 2$ then we call them as type $II$.&nbsp;</span><br /><span style="font-size: large;">Claim: Any two topologies of type $I$ are isomorphic. </span><span style="font-size: large;">Let $\tau_1 = \{X,\phi,U_1\}$ and $\tau_2 = \{X, \phi, U_2\}$ be two such topologies $U_1 = \{a\}$ and $U_2 = \{b\}$ where $a,b \in \{1,2,3\}$. Now, any permutation of $X= \{1,2,3\}$ which maps $a$ to $b$ is a homeomorphism between $(X,\tau_1)$ to $(X,\tau_2)$.</span><br /><span style="font-size: large;">Claim:&nbsp;</span><span style="font-size: large;">Any two topologies of type $II$ are isomorphic.&nbsp;</span><span style="font-size: large;">Let $\tau_1 = \{X,\phi,U_1\}$ and $\tau_2 = \{X, \phi, U_2\}$ be two such topologies $U_1 = \{a,b\}$ and $U_2 = \{c,d\}$ where $a,b,c,d \in \{1,2,3\}$. Now, any permutation of $X= \{1,2,3\}$ which maps $\{a,b\}$ to $\{c,d\}$ is a homeomorphism between $(X,\tau_1)$ to $(X,\tau_2)$.</span><br /><span style="font-size: large;">Claim: A topology of type $I$ and a topology of type $II$ cannot be homeomorphic. This follows from the following observation.</span><br /><b style="font-size: x-large;">observation:&nbsp;</b><span style="font-size: large;">Since $f$ is a homeomorphism, for an openset $U$ of $X$, we have $|f(U)| = |U|$ and $|f^{-1}(U)| = |U|$.&nbsp;</span><br /><span style="font-size: large;"><b>Therefore there are two homeomorphism classes of topologies on $X$ with three open sets and these classes are given by type $I$ and type $II$ defined above.</b></span><br /><span style="font-size: large;"><b><br /></b></span><span style="font-size: large;"><b>option 2:(Number of topologies with four open sets)</b></span><br /><span style="font-size: large;">We use the same ideas as above. Let $(X,\tau = \{X,\phi,U_1,U_2\})$ be a topology on $X$ with four open sets. Assume $U_1 = \{a\}$ then $U_2$ cannot be an open set with a single point because then their union will also be in $\tau$ which will give us the fifth open set. So $U_2$ has to be a set with two elements. Similarly, if $|U_1| = 2$ then $|U_2| = 1$. Without loss of generality, we assume that $|U_1| = 1$.&nbsp;</span><span style="font-size: large;">Now there are two cases. $U_1 \subseteq U_2$ we call such topologies type $I$ and $U_1 \nsubseteq U_2$ we call such topologies type $II$. There are <b>six</b> topologies of type $I$ and there are <b>three</b> topologies of type $II$. So there are <b>nine</b> topologies on $X$ with four open sets. Among these topologies, like in the previous option, one can show that any two topologies of the same type (type I or type II) are homeomorphic and two topologies of different types are not homeomorphic. <b>Hence there are two homeomorphism classes of topologies with four open sets in $X$.</b></span><br /><span style="font-size: large;"><b><br /></b></span><span style="font-size: large;"><b><br /></b></span><span style="font-size: large;"><b>option 3:(Number of topologies with five open sets)</b></span><br /><div><span style="font-size: large;">Let $(X,\tau = \{X,\phi,U_1,U_2,U_3\})$ be a topology on $X$ with five open sets.&nbsp;</span></div><div><span style="font-size: large;">Assume $U_1 = \{a\}$ then $U_2$ can be an open set with a single point or two points.&nbsp;</span></div><div><span style="font-size: large;"><b>Assume $U_2$ is singleton</b> then $U_3$ has to be $U_1 \cup U_2$ as we have only five open sets in $\tau$. We call such topologies type $I$. There are three topologies of this type.</span></div><div><span style="font-size: large;"><b>Assume $U_2$ has two elements.</b></span></div><div><span style="font-size: large;"><b>Claim: $a \in U_2$. </b>If not, $U_2 = \{b,c\}$ with $b \ne a$</span></div><div><span style="font-size: large;">and $c \ne a$. Lets check the choices for $U_3$&nbsp;</span></div><div><span style="font-size: large;">suppose $|U_3| =1$. If $a \notin U_3$ then $U_3 \cap U_2$ is a sixth open set in $\tau$. So $U_3 = U_1$. This shows that $|U_3| =2$. By our assumption $U_2$ doesn't contain $a$ which means that $U_3$ should contains $a$ otherwise $U_2 = U_3$. If $a \in U_3$ then $U_3 \cup U_2$ is a sixth open set in $\tau$. Contradiction. So if $a \notin U_2$ there is no choice for $U_3$. Hence we have $a \in U_2$.&nbsp;</span></div><div><span style="font-size: large;"><b>Claim: $a \in U_3$.</b> Same argument as above.&nbsp;</span></div><div><span style="font-size: large;"><b>This shows that $\tau = \{X,\phi,\{a\},\{a,b\},\{a,c\}\}$. </b>We call such topologies of type $II$ and there are three of them (a=1 or 2 or 3). Hence, including type $I$, there are a total of&nbsp;<b>six</b> topologies on $X$ with five open sets.&nbsp;</span><span style="font-size: large;">Among these topologies, like in the previous option, one can show that any two topologies of the same type are homeomorphic and two topologies of different types are not homeomorphic.&nbsp;</span><b style="font-size: x-large;">Hence there are two homeomorphism classes of topologies with four open sets in $X$.</b></div><br /><span style="font-size: large;"><b>option 4:(Number of topologies with seven open sets)</b></span><br /><span style="font-size: large;">We claim that there is no topology on $X$ with $7$ open sets.&nbsp;</span><br /><span style="font-size: large;">Let $\tau$ be a topology on $X = \{1,2,3\}$.</span><br /><span style="font-size: large;">Claim: <b>If there are at least seven open sets in $\tau$ then $\tau$ is the discrete topology on $X$</b> (which is nothing but the power set of $X$ where all the subsets are open).</span><br /><span style="font-size: large;"><b>This shows that there are no topology on $X$ with $7$ open sets.</b></span><br /><span style="font-size: large;">proof: It is enough to prove that all the singleton subsets of $X$ are in $\tau$. Since $\tau$ is closed under arbitrary union, $\tau$ will contain all the subsets if it contains all the singleton subsets. We have $|\tau| = 7$ and hence either all the singleton subsets of $X$ are in $\tau$ or all the two elements subsets of $X$ are in $\tau$. In the first case we are done and if all the two-element subsets of $X$ are in $\tau$ then by taking intersection among them we can generate all the singletons and again we are done.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><br /></span><br /><div><span style="font-size: large;"><b><br /></b></span></div></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/RB8StHXQsD4" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com5https://nbhmcsirgate.theindianmathematician.com/2020/04/nbhm-2020-part-b-question-21-solution.htmltag:blogger.com,1999:blog-7629869410369132909.post-52970311841082914232020-04-15T00:45:00.001-07:002020-04-18T07:03:13.387-07:00CSIR JUNE 2011 PART C QUESTION 76 SOLUTION (dimension = degree of the minimal polynomial) <div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $W = \{p(B) : p \text{ is a polynomial with real coefficients}\},$&nbsp; where $$B = \begin{bmatrix}0&amp;1&amp;0\\ 0&amp;0&amp;1\\ 1&amp;0&amp;0 \end{bmatrix}.$$ Then the dimension $d$ of the vector space $W$ satisfies <br />1) $4 \le d \le 6$, <br />2) $6 \le d \le 9$, <br />3)$3 \le d \le 8$, <br />4)$3 \le d \le 4$. <br /><b><span style="font-size: large;">I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me. Visit my blog every day and share the solutions with friends. Follow the blog by email. Thank you.</span></b></span><br /><span style="font-size: large;"><b>Solution</b>:</span><br /><span style="font-size: large;"><b>We will calculate the dimension $d$ explicitly.</b>&nbsp; We start with the following simple observation. Consider the polynomial $p(x) = x^3 + 3x^2 -2x +12$ then $p(A) = A^3 + 3A^2 - 2 A + 12 I$. In particular $p(A)$ is a linear combination of the matrices $A^0 = I, A^1, A^2, \dots$. So the set of all non-negative powers of $A$&nbsp; spans $W$. We will find a basis inside this spanning set whose cardinality will be our required $d$.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;">Let $A$ be an $n \times n$ real matrix and consider its powers $A^0=I, A^1, A^2,\dots,A^m$ <b>for some $m&gt;0$</b>. These matrices are vectors in the vector space $M_n(\Bbb R)$. So it is possible to talk about the linear independence of these matrices. We say these matrices are linearly dependent if, as usual, there exist real scalars $\alpha_0,\alpha_1,\dots,\alpha_m$ <b>not all zero</b> such that $\alpha_0 A^0 + \alpha_1 A^1 + \cdots + \alpha_m A^m = 0$. Equivalently, there exists a non-zero polynomial $p(x) = \alpha_0+\alpha_1 x+ \alpha_2 x^2 + \cdots + \alpha_m x^m$ with real coefficients <b>of degree atmost $m$</b>&nbsp; (some coefficient is non-zero and this coefficient need not be the leading coefficient) such that $p(A) = 0$. Such polynomials are known as the annihilating polynomials of $A$.&nbsp;</span><br /><span style="font-size: large;"><b>Hence, $A^0,A^1,A^2,\dots, A^m$ are linearly dependent if and only if there is a polynomial of degree almost $m$ satisfied by $A$.&nbsp;</b></span><br /><br /><span style="font-size: large;">The minimal spanning set is the basis and hence we need to find the smallest degree polynomial satisfied by $A$. This is nothing but the minimal polynomial of $A$. Hence $$\text{dimension d of W} = \text{degree of the minimal polynomial }m_A(x).$$</span><br /><br /><span style="font-size: large;">Caylay-Hamilton Theorem: Every $n \times n$ matrix satisfies its characteristic polynomial $p_A(x)$ which is a monic polynomial of degree $n$.</span><br /><span style="font-size: large;">This shows that, from the above discussion, the matrices, $A^0, A^1, A^2, \dots, A^n$ are linearly dependent. Therefore $d \le n$. In our problem, the given matrix is of order $3$ and hence we have $$1 \le d \le 3.$$ Next we will prove that $d$ is actually equal to $3$. We need to show that it's minimal polynomial $m_A(x)$ also has degree $3$.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>Result</b>: Every eigenvalue of $A$ is a root of the minimal polynomial $m_A(x)$.</span><br /><span style="font-size: large;"><b>Proof</b>: We will show that the minimal polynomial divides all the annihilating polynomials. In particular, it will divide the characteristic polynomial and hence the result follows.</span><br /><span style="font-size: large;">Let $p(x)$ be an annihilating polynomial of $A$. We have $p(x) = q(x) m_A(x) + r(x)$ with degree of $r(x)$ strictly less than the degree of $m_A(x)$ by the reminder theorem. This implies that $p(A) = q(A)m_A(A) + r(A)$ and we have $r(A) = 0$. Hence $r(x)$ is also an annihilating polynomial of $A$. This is possible only if $r(x)$ is the zero polynomial because the degree of $r(x)$ is strictly smaller then the degree of the minimal polynomial. Therefore $$p(x) = q(x) m_A(x).$$ In particular, degree of the minimal polynomial is less than or equal to $n$.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;">The characteristic polynomial of the given matrix $A$ is equal to $x^3 - 1$ which has three distinct roots. All these roots have to be roots of the minimal polynomial by the above result. So the degree of the minimal polynomial is equal to at least $3$ and hence by the previous paragraph, it should be equal to $3$.&nbsp; This shows that $$d = 3.$$</span><br /><span style="font-size: large;"><b>only option (3) and (4) are correct.</b></span><br /><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b></span></div><div class="addthis_inline_share_toolbox"></div><b><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></b></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/971BCbbQs_s" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-76.htmltag:blogger.com,1999:blog-7629869410369132909.post-59064828041722001412020-04-14T20:20:00.000-07:002020-04-18T07:03:13.733-07:00CSIR JUNE 2011 PART C QUESTION 75 SOLUTION (Which are subspaces ? $\{X \in M: \text{trace}(AX)=0\}$, $\{X \in M: \text{det}(AX)=0\}$.)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $M$ be the vector space of all $3 \times 3$ real matrices and let $$A = \begin{bmatrix}2&amp;1&amp;0\\ 0&amp;2&amp;0\\ 0&amp;0&amp;3\end{bmatrix}.$$ Which of the following are subspaces of $M$? <br />1) $W_1 = \{X \in M: XA = AX\}$, <br />2) $W_2 = \{X \in M: X+A = A + X\}$, <br />3) $W_3 = \{X \in M: \text{trace}(AX)=0\}$, <br />4) $W_4 = \{X \in M: \text{det}(AX)=0\}$.&nbsp;</span><br /><span style="font-size: large;"><b>I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me. Visit my blog every day and share the solutions with friends. Follow the blog by email. Thank you.</b><br /><b>Solution</b>:</span><br /><span style="font-size: large;"><b>Observation</b>: Let $\alpha$ be a scalar then $\alpha A = (\alpha I) A$. That is multiplying every element of $A$ is the same as pre multiplying the matrix $A$ be the scalar matrix $\alpha I$ (diagonal matrix with all the diagonal entries are $\alpha$). Also, these scalar matrices commute with all the matrices in $M$ (Center of the ring $M$).</span><br /><span style="font-size: large;"><b>option 1</b>:(<b>True</b>) Let $X_1,X_2 \in W_1$ then $X_1 A = A X_1$ and $X_2 A = A X_2$. Now, $$(X_1+X_2)A = X_1 A + X_2 A = A X_1 + A X_2 = A (X_1 + X_2).$$&nbsp;</span><b style="font-size: x-large;">Therefore $X_1+X_2 \in W_1$.</b><br /><span style="font-size: large;">Also, since $\alpha I$ commutes with all the matrices, we have $$(\alpha X_1)A = (\alpha I) (X_1 A) = (\alpha I) (A X_1) = A (\alpha I)X_1 =&nbsp; A (\alpha X_1).$$ Therefore <b>$\alpha X_1 \in W_1$</b> and hence $W_1$ is a subspace of $M$.</span><br /><span style="font-size: large;"><b>option 2</b>: (<b>True</b>) Matrix addition is commutative. Therefore every matrix in $M$ commutes with $A$ and we have $W_2 = M$. Hence it is a subspace.</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>True</b>)&nbsp;</span><span style="font-size: large;">Let $X_1,X_2 \in W_3$ then $\text{trace}(A X_1) = 0$ and $\text{trace}(A X_2)=0$. Now, $$\text{trace}(A(X_1+X_2)) = \text{trace}(AX_1)&nbsp; + \text{trace}(AX_2) = 0.$$&nbsp;</span><b style="font-size: x-large;">Therefore $X_1+X_2 \in W_3$.</b><br /><span style="font-size: large;">Also, since $\alpha I$ commutes with all the matrices and $\text{trace}(\alpha A) = \text{trace }((\alpha I)(A)) = \alpha \cdot \text{trace }A$, we have $$\text{trace}(A (\alpha X_1)) = \text{trace}(A ((\alpha I) (X_1)) ) = \text{trace}((\alpha I) (A X_1))\\ = \alpha \cdot \text{trace}(A X_1) =&nbsp; 0.$$ Therefore&nbsp;</span><b style="font-size: x-large;">$\alpha X_1 \in W_3$ </b><span style="font-size: large;">and hence&nbsp;</span><span style="font-size: large;">$W_3$ is a subspace of $M$.</span><br /><span style="font-size: large;"><b>option 4</b>:(<b>False</b>) We have $\text{det}A = 12 \ne 0$. Therefore $\text{det}(AX) = \text{det }A \cdot \text{det}X = 0$ if and only if $\text{det} X =0$. This shows that $W_4 = \{X \in M: \text{det}X = 0\}$. <b>The set of all singular matrices is not a vector subspace of $M$</b>. For example, consider the matrices which are of determinant zero $\begin{bmatrix}1&amp;0&amp;0\\ 0&amp;1&amp;0\\ 0&amp;0&amp;0\end{bmatrix}$ and $\begin{bmatrix}0&amp;0&amp;0\\ 0&amp;0&amp;0\\ 0&amp;0&amp;1\end{bmatrix}$. Clearly their sum is the identity matrix whose determinant is non-zero. Therefore the set of all matrices with determinant zero is not even closed under addition.</span><br /><br /></div><span style="font-size: large;"><span style="font-size: small;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</span>Share to your groups: </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/E6UySZh6chY" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-74_5.htmltag:blogger.com,1999:blog-7629869410369132909.post-33238012778512073182020-04-14T14:51:00.000-07:002020-04-18T07:03:12.605-07:00CSIR JUNE 2011 PART C QUESTION 74 SOLUTION (Let $T: \Bbb R^n \to \Bbb R^n$ be a linear transformation such that $T^2 = \lambda T$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $T: \Bbb R^n \to \Bbb R^n$ be a linear transformation such that $T^2 = \lambda T$ for some $\lambda \in \Bbb R$. Then <br />1)$||T(x)|| = |\lambda| ||x||$ for all $x \in \Bbb R^n$,</span><br /><span style="font-size: large;">2)If $||T(x)|| = ||x||$ for some non-zero vector $x \in \Bbb R^n$ then $\lambda = \pm 1$,</span><br /><span style="font-size: large;">3) $T = \lambda I$ where $I$ is the $n \times n$ identity matrix,</span><br /><span style="font-size: large;">4) If $||T(x)|| &gt; ||x||$ for some non-zero vector $x \in \Bbb R^n$ then $T$ is necessarily singular.&nbsp;</span><br /><span style="font-size: large;"><b>I am spending my crucial Ph.D. time in writing this blog to help others to achieve in Mathematics. So please encourage me. Visit my blog every day and share the solutions with friends. Follow the blog by email. Thank you. </b></span><br /><span style="font-size: large;"><b><br /></b></span><span style="font-size: large;"><b>Solution</b>:<br />&nbsp;Consider the $2 \times 2$ nilpotent matrix $$\begin{bmatrix}0&amp;1\\0&amp;0\end{bmatrix}.$$ We have $A^2 = 0$ and hence this matrix satisfies the given condition $A^2 = \lambda T$ with $\lambda = 0$. Since $A$ is upper diagonal its eigen values are its diagonal entries. Hence the eigen values of $A$ are $0,0$.</span><br /><span style="font-size: large;"><b>option 1</b>: (<b>False</b>) We use the above matrix $A$ as a counter example. Let $x = \begin{bmatrix}0\\1\end{bmatrix}$ then $Ax = y$ where $y = \begin{bmatrix}1\\0\end{bmatrix}$. Now, $$||Ax|| = ||y|| = \sqrt{1^2 + 0^2} = 1.$$ But $|\lambda| ||x|| = 0$. This shows that, for our matrix $A$, $$||Ax|| \ne |\lambda| ||x||.$$</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>False</b>) We again use the above matrix $A$. We ave seen that this matrix satisfies $A^2 = \lambda A$ with $\lambda =0$. We observe that $A$ is a non-zero matrix whereas $\lambda I = 0I$ is the zero matrix. Hence $A \ne \lambda I$ in this case.</span><br /><span style="font-size: large;"><b>option 4</b>: (<b>False</b>) Consider the matrix $$A = \begin{bmatrix}2&amp;0\\ 0&amp;2\end{bmatrix}.$$ Let $x = \begin{bmatrix}0 \\ 1\end{bmatrix}$ then $$||Ax|| = ||\begin{bmatrix}0\\ 2\end{bmatrix}|| = \sqrt{0^2+2^2} = 2&gt;0$$ but $A$ is clearly invertible (non-singular).</span><br /><span style="font-size: large;"><b>option 2</b>: (<b>False</b>) Consider the matrix $$\begin{bmatrix}\sqrt 2 &amp; 0 \\ 0 &amp; 0\end{bmatrix},$$ then $A^2 = \sqrt 2 A$. Hence this matrix satisfies $A^2 = \lambda A$ with $\lambda = \sqrt 2$. Let $x = \begin{bmatrix}1\\ 1\end{bmatrix}$ then $$||x|| = \sqrt{1^2+1^2} = \sqrt 2 = ||\begin{bmatrix}\sqrt 2 \\ 0\end{bmatrix}|| = ||Ax||.$$ But $\lambda = \sqrt 2 \ne \pm 1$.</span><br /><span style="font-size: large;"><b>All the options are false. This question was wrong and the full mark was given to everybody.</b></span><br /><span style="font-size: large;"><br /></span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/LDD7EGUyhbY" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-74_14.htmltag:blogger.com,1999:blog-7629869410369132909.post-31347282741140895252020-04-14T07:47:00.000-07:002020-04-18T07:03:12.574-07:00CSIR JUNE 2011 PART C QUESTION 73 SOLUTION (which are positive definite? 1) $A+B$, 2) $ABA^{*}$, 3) $A^2+I$, 4) $AB$.)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); padding: 10px;"><div style="color: black;"><span style="font-size: large;">Suppose $A$ and $B$ are $n \times n$ positive definite matrices and $I$ be the $n \times n$ identity matrix. Then which of the following matrices are positive definite? </span></div><div style="color: black;"><span style="font-size: large;">1) $A+B$, </span></div><div style="color: black;"><span style="font-size: large;">2) $ABA^{*}$,</span></div><div style="color: black;"><span style="font-size: large;">3) $A^2+I$,</span></div><div style="color: black;"><span style="font-size: large;">4) $AB$. </span></div><div style="color: black;"><span style="font-size: large;"><b>Solution</b>:</span></div><div style="color: black;"><span style="font-size: large;">Let $A$ be a real <b>symmetric</b> matrix, then $A$ is said to positive definite if it satisfies any of the following equivalent conditions.</span></div><div style="color: black;"><span style="font-size: large;">i) all its eigenvalues are positive.</span></div><div style="color: black;"><span style="font-size: large;">ii) $x^t A x &gt; 0$ for all vectors $x \ne 0$.</span></div><div style="color: black;"><span style="font-size: large;">iii)$&lt;x,Ax&gt; &gt; 0$ for all vectors $x \ne 0$.</span></div><div style="color: black;"><span style="font-size: large;">iii) A is positive definite <b>if and only if</b> it can be written as $A = R^tR$, where $R$ is possibly a rectangular matrix, with independent columns.</span></div><div style="color: black;"><span style="font-size: large;">iv) All the principal minors of $A$ are positive. (<b>Please share if you know any other equivalent conditions in the comment below</b>)</span></div><div style="color: black;"><span style="font-size: large;">Let $A$ and $B$ be two $n \times n$ positive definite matrices. We have $x^t A x&gt;0$ and $x^t B x&gt;0$ for $x \ne 0$. <b>We will solve each given option by each of the above given definition of positive definiteness in order to understand them clearly.</b></span></div><div style="color: black;"><span style="font-size: large;"><b>option 1</b>. (<b>True</b>)We have for $x \ne 0$, $$x^t (A+B) x = x^t A x + x^t B x &gt; 0.$$</span></div><div style="color: black;"><span style="font-size: large;">Therefore option 1 is true.</span></div><div style="color: black;"><span style="font-size: large;"><b>option 2</b>. (<b>True</b>) We have for $x \ne 0$, $Ax \ne 0$ since $A$ is invertible.</span></div><div style="color: black;"><span style="font-size: large;">$$&lt;x,ABA^*x&gt; = &lt;A^*x,BA^*x&gt; = &lt;B^*A^*x,A^*x&gt;&nbsp; \\ = &lt;BAx,Ax&gt; &gt; 0 .$$</span></div><div style="color: black;"><span style="font-size: large;">Therefore option 2 is true. Since $A$ and $B$ are real symmetric, we have $A^* = A$ and $B^* = B$.</span></div><div style="color: black;"><span style="font-size: large;"><b>option 3</b>. (<b>True</b>) Let the eigen values of $A$ be $\lambda_1,\lambda_2,\dots,\lambda_n$. Since $A$ is positive definite we have all these eigen values are positive. We observe that the eigen values of $A^2+1$ are $\lambda_1^2+1,\lambda_2^2+1,\dots,\lambda_n^2+1$ which are all positive. Hence $A^2+1$ is positive definite.&nbsp;</span></div><div style="color: black;"><span style="font-size: large;"><b>option 4</b>. (<b>False</b>)</span></div><div style="color: black;"><span style="font-size: large;"><b>The product&nbsp;of two positive definite matrices need not be even symmetric</b>.&nbsp; In particular, we have $AB$ is symmetric if and only if $A$ and $B$ commutes with each other. Because $$(AB)^* = B^*A^* = A^*B^* = AB.$$ Note that, if $A$ and $B$ commutes with each other, then $A^*$ and $B^*$ commutes with each other.&nbsp;</span></div><div style="color: black;"><span style="font-size: large;"><br /></span></div><div style="color: black;"><span style="font-size: large;">To illustrate this, consider the positive definite matrices $$A = \begin{bmatrix}11 &amp; 10 \\ 10 &amp; 10\end{bmatrix}$$ and $$B</span><span style="font-size: large;">&nbsp;= \begin{bmatrix}11 &amp; 5 \\ 5 &amp; 10\end{bmatrix}</span><span style="font-size: large;">$$&nbsp;</span></div><div style="color: black;"><span style="font-size: large;">text{ Then their product } $$AB = \begin{bmatrix}171&amp;155 \\ 160&amp;150\end{bmatrix}$$ which is not symmetric.</span></div><div style="color: black;"><span style="font-size: large;"><br /></span></div></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/Hyh70v1FTCs" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-72_14.htmltag:blogger.com,1999:blog-7629869410369132909.post-85193768147690466062020-04-06T07:15:00.000-07:002020-04-18T07:03:14.895-07:00CSIR JUNE 2011 PART C QUESTION 72 SOLUTION ($A = (a_{ij}), a_{ij} = a_ia_j$ for $1 \le i,j \le n$ is positive semi-definite)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $a_{ij} = a_ia_j$ for $1 \le i,j \le n$ and $a_1,a_2,\dots,a_n$ are real numbers. Let $A = (a_{ij})$ be the $n \times n$ matrix. Then <br />1) It is possible to choose $a_1,a_2,\dots a_n$ so as to make the matrix $A$ non-singular. <br />2) The matrix $A$ is positive definite if $(a_1,a_2,\dots,a_n)$ is a non-zero vector. <br />3) The matrix $A$ is positive definite for all $(a_1,a_2,\dots,a_n)$. <br />4) For all $a_1,a_2,\dots,a_n$ we have zero is an eigen value of $A$.&nbsp;</span><br /><span style="font-size: large;"><b>(We assume $n&gt;1$. otherwise it is stright forward.)</b><br /><b>Solution</b>:&nbsp;</span><br /><span style="font-size: large;"><b>option 1</b>: (<b>False</b>)Let $v = (a_1,a_2,\dots,a_n)$, then the first row of the matrix $A$ is given by the vector $(a_1a_1,a_1a_2,a_1a_3,\dots,a_1a_n) = a_1 v$ ($v$ scalar multiplied by the scalar $a_1$). Similarly the $i$th row of $A$ is given by $a_i v$. This shows that, all the rows of $A$ are some scalar multiples $v$. Therefore the dimension of the row space of $A$ = row rank of $A$ = rank of $A \le 1$. Therefore $A$ is singular for all choices of $v$. (unless n=1 the matrix $A$ is an $1 \times 1$ matrix).</span><br /><span style="font-size: large;"><b>option 2</b>: (<b>False</b>) If $A$ is positive definite then all its eigen values are positive. Hence $\text{det }A \ne 0$. We have shown that $A$ has determinant zero for all $(a_1,a_2,\dots,a_n)$.</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>True</b>) If $A$ is symmetric then $A$ is positive semi-definite if and only if&nbsp; all its principal minors are non-negative. We will prove that, all the principal minors of $A$ are non-negative. Let $1 \le k \le n$ and consider the principal $k$ minor $A(k)$ given by $A(k)_{ij} = a_{ij}$ for $1\le i,j \le k$.&nbsp;</span><span style="font-size: large;">Let $v(k) = (a_1,a_2,\dots,a_k)$, then the first row of the matrix $A(k)$ is given by the vector $(a_1a_1,a_1a_2,a_1a_3,\dots,a_1a_k) = a_1 v(k)$ ($v(k)$ scalar multiplied by the scalar $a_1$). Similarly the $i$th row of $A(k)$ is given by $a_i v(k)$. This shows that, all the rows of $A(k)$ are some scalar multiples $v(k)$. This shows that dimension of the row space of $A(k)$ = row rank of $A(k)$ = rank of $A(k) \le 1$. Therefore $A(k)$ is singular. (unless k=1, this case $A(1)$ can be invertible). This shows that $A$ is positive semi definite.</span><br /><span style="font-size: large;"><b>option 4</b>:(<b>True</b>) We have shown that $A$ is not invertible and hence zero is an eigen value of $A$.</span><br /><span style="font-size: large;"><br /></span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span>&nbsp;</span><br /><span style="font-size: large;"><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/GNqBgTtVLI8" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-72.htmltag:blogger.com,1999:blog-7629869410369132909.post-11534060105519527872020-04-06T04:37:00.001-07:002020-04-18T07:03:11.494-07:00CSIR JUNE 2011 PART C QUESTION 70 SOLUTION ($\mathcal{F}= \{f:\Bbb{R} \to \Bbb{R}: |f(x)-f(y)| ≤ K|(x-y)|^a \}$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><div style="background-attachment: initial; background-clip: initial; background-image: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); background-origin: initial; background-position: initial; background-repeat: initial; background-size: initial; padding: 10px;"><span style="font-size: large;">Consider the set $\mathcal{F}= \{f:\Bbb{R} \to \Bbb{R}: |f(x)-f(y)| ≤ K|(x-y)|^a \}$ for some $a&gt;0$ and $K&gt;0$, then which of the following statements are true?</span><br /><span style="font-size: large;">1. Every $f \in \mathcal F$ is continuous.</span><br /><span style="font-size: large;">2. Every $f \in \mathcal F$ is uniformly continuous.</span><br /><span style="font-size: large;">3. Every differentiable function $f : \Bbb{R} \to \Bbb{R}$ is in $\mathcal{F}$.</span><br /><span style="font-size: large;">4. Every function $f \in \mathcal{F}$ is differentiable. (This is converse of (3))</span><br /><br /><span style="font-size: large;"><b>Solution</b>:</span><br /><span style="font-size: large;"><b>option 3</b>. (<b>false</b>) Consider the function $f(x) = x^2$, then $f$ is a differentiable function from $\mathbb{R}$ to $\mathbb{R}$. We claim that $f \notin \mathcal{F}$. We have for any $x,y \in \mathbb{R}$, $$|x^2 - y^2| = |x-y| |x + y|$$</span><br /><span style="font-size: large;">Since $x,y \in \mathbb{R}$ are arbitrary $|x+y|$ can also be arbirtarily large. So there cannot exist a required $K&gt;0$ and $f \notin \mathcal{F}$.</span><br /><span style="font-size: large;"><br /></span><b><span style="font-size: large;">Remark:</span></b><br /><span style="font-size: large;">If $f(x) = x^2$ is consider as function on $(a,b)$ a bounded interval (open, closed, semi open any kind), then the term $|x+y|$ can be bounded by&nbsp; $2b$ since $x,y \in (a,b)$ and we can take $k = 2b$ and $a =1$ which will make $f \in \mathcal F$.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>option 4</b>: (<b>false</b>) Consider the function $f(x) = |x|$ and this function satisfies $$||x| - |y|| \le |x - y|.$$ Hence for $k=1$ and $a=1$ this function is in $\mathcal F$. But $|x|$ is not differentiable at the origin.</span><br /><span style="font-size: large;"><br /></span><span style="font-size: large;"><b>options 1 and 2</b>. (<b>True</b>)&nbsp; Let $f$ be a function from $\mathcal{F}$, then $f$ satisfies $$|f(x)-f(y)| ≤ K|(x-y)|^a$$ for some $K&gt;0$ and $a&gt;0$. We claim that $f$ is uniformly continuous. Let $\epsilon &gt; 0$, we find a $\delta$ which depends only on $\epsilon$ and hence the function will be uniformly continuous.&nbsp; We take $\delta = (\frac{\epsilon}{K})^{\frac{1}{a}}$ then $|x-y| &lt; \delta$ implies that $|f(x) - f(y)| \leq K |x-y|^ a &lt; K \delta^a =&nbsp; K ((\frac{\epsilon}{K})^{\frac{1}{a}})^a = \epsilon$. This completes the proof.</span><br /><br /></div></div><span style="font-size: large;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> <br /><div class="addthis_inline_share_toolbox"></div>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/a-DOb0S1c6Y" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-70_6.htmltag:blogger.com,1999:blog-7629869410369132909.post-64462318357777247502020-04-06T03:39:00.002-07:002020-04-18T07:03:13.921-07:00CSIR JUNE 2011 PART C QUESTION 70 SOLUTION (Dense subsets of $\Bbb R^2$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Which of the following sets are dense in $\Bbb R^2$ with respect to the usual topology. <br />1) $\{(x,y) \in \Bbb R^2 : x \in \Bbb N\}$, <br />2) $\{(x,y)\in \Bbb R^2 : x+y \text{ is rational}\}$, <br />3) $\{(x,y)\in \Bbb R^2 : x+y^2 = 5\}$, <br />4) $\{(x,y) \in \Bbb R^2 : xy \ne 0\}$. <br /><b>Solution</b>:</span><br /><span style="font-size: large;">A subset $A$ of a metric space $(X,d)$ is said to be dense if $A$ intersects all the non-empty open balls in $X$. In particular, to show a subset is not dense in $X$ enough to construct a non-empty open ball $U$ which is disjoint from $A$. i.e., $A \cap U = \Phi$.</span><br /><span style="font-size: large;"><b>option 1</b>:(<b>Not dense</b>) We claim that the given set&nbsp;</span><span style="font-size: large;">$A := \{(x,y) \in \Bbb R^2 : x \in \Bbb N\}$ is closed. Hence it cannot be dense.&nbsp;</span><span style="font-size: large;">Let $(x,y)$ be a limit point of this set. Then there exists a sequence $(x_n,y_n) \in A$ such that $(x_n,y_n)$ converges to $(x,y)$. This means that $x_n \to x$ and $y_n \to y$. Note that $(x_n,y_n) \in A$ implies that $x_n$ is a converging sequence of natural numbers. This is possible, only if, $x_n = k \in \Bbb N$ for sufficiently large $n$. In particular, this sequence is eventually constant and hence is converging to $k$. This shows that $x=k$ as the limit of a sequence is unique. Therefore $(x,y) \in A$ and $A$ is closed.</span><br /><span style="font-size: large;"><b>option 2</b>:(<b>True</b>) From the definition, we observe that, if $A \subseteq B$ and $A$ is dense in $X$, then $B$ is also dense in $X$.&nbsp;</span><br /><span style="font-size: large;">Proof: Let $U$ be a non-empty open set in $X$. Since $A$ is dense in $X$, $A \cap U \ne \Phi$. This implies that $B \cap U \ne \Phi$ as $A \subseteq B$.&nbsp;</span><br /><span style="font-size: large;">We have $\Bbb Q \times \Bbb Q$ is dense in $\Bbb R^2$ since the product of dense sets is dense. This is immediate using the projection maps. We have&nbsp;</span><span style="font-size: large;">$\Bbb Q \times \Bbb Q \subseteq \{(x,y)\in \Bbb R^2 : x+y \text{ is rational}\}$ and hence the given set is dense in $\Bbb R^2$ by the above observation.</span><br /><span style="font-size: large;"><b>option 3</b>:(<b>False</b>) Again, we will prove that the set&nbsp;</span><span style="font-size: large;">$C:= \{(x,y)\in \Bbb R^2 : x+y^2 = 5\}$ is closed in $\Bbb R^2$. Let $f: \Bbb R^2 \to \Bbb R$ be a function defined by $f(x,y) = x+y^2-5$, then being the polynomial function $f$ is continuous. Now $f^{-1}(0) = C$. Since $\{0\}$ is closed and $f$ is continuous, we have $C$ is closed.</span><br /><span style="font-size: large;"><b>option 4</b>:(<b>True</b>) Let&nbsp;</span><span style="font-size: large;">$D: = \{(x,y) \in \Bbb R^2 : xy \ne 0\}$. We will prove that this set is dense in $\Bbb R^2$.&nbsp;</span><span style="font-size: large;">Let $B(x,r)$ be a open ball in $\Bbb R^2$. We claim that $B((a,b),r) \cap D \ne \Phi$.</span><br /><span style="font-size: large;">Case 1: $(a,b) \in D$, then clearly&nbsp;</span><span style="font-size: large;">$B(x,r) \cap D \ne \Phi$.</span><br /><span style="font-size: large;">Case 2: $(a,b) \notin D$. then $a = 0$ or $b=0$.&nbsp;</span><br /><span style="font-size: large;"><b>Subcase 1</b>: $a = 0$ and $b \ne 0$, then $(a,b)$ lies in the y axis. Consider the point $(a+\frac{r}{2},b) \in D$, then $d((a+\frac{r}{2},b),(a,b)) = \sqrt {|(a+\frac{r}{2}-a)^2 + (b-b)^2| } = \sqrt{\frac{r^2}{4}} = \frac{r}{2} &lt; r$. Hence the point $(a+\frac{r}{2},b) \in B(x,r)$.&nbsp;</span><br /><span style="font-size: large;"><b>Subcase 2</b>: $a \ne 0$ and $b = 0$. Similar to the previous subcase consider the point $(a,b+\frac{r}{2}) \in D$.&nbsp;</span><br /><span style="font-size: large;"><b>Subcase 3</b>: $a=0$ and $b=0$. Consider the point&nbsp;</span><span style="font-size: large;">$(a+\frac{r}{2},b+\frac{r}{2}) \in D$, then&nbsp;</span><span style="font-size: large;">$d((a+\frac{r}{2},b+\frac{r}{2}),(a,b)) = \sqrt {|(a+\frac{r}{2}-a)^2 + (b+\frac{r}{2}-b)^2| } \\ = \sqrt{\frac{r^2}{4}+\frac{r^2}{4}} = \frac{\sqrt 2 r}{2} &lt; r$</span><br /><span style="font-size: large;">Hence the point $(a+\frac{r}{2},b+\frac{r}{2}) \in B(x,r)$.</span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/vuCArMk4ARE" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-70.htmltag:blogger.com,1999:blog-7629869410369132909.post-40023629566989386242020-04-06T02:33:00.002-07:002020-04-18T07:03:12.886-07:00CSIR JUNE 2011 PART C QUESTION 69 SOLUTION (The derivative of $f(x,y) = (3x - 2y + x^2, 4x +5y + y^2)$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Consider the function $f: \Bbb R^2 \to \Bbb R^2$ defined by $f(x,y) = (3x - 2y + x^2, 4x +5y + y^2)$. Then <br />1) f is discontinuous at $(0,0)$, <br />2) $f$ is continuous at $(0,0)$ and all the directional derivatitives exist at $(0,0)$, <br />3) $f$ is differentaible at $(0,0)$ but the derivative $Df(0,0)$ is not invertible, <br />4) $f$ is differentaible at $(0,0)$ and the derivative $Df(0,0)$ is invertible. <br /><b>Solution</b>: Let $f_1(x,y) = 3x-2y+x^2$ and $f_2 = 4x+5y+y^2$ be the coordinate functions of $f$.&nbsp;</span><span style="font-size: large;">The partial derivatives of $f$ are $\frac{\partial f_1}{\partial x} = 3 + 2x$, $\frac{\partial f_1}{\partial y}= -2$, $\frac{\partial f_2}{\partial x} = 4$ and $\frac{\partial f_2}{\partial y} = 5 + 2y$.</span><br /><span style="font-size: large;"><b>option 1</b>:(<b>True</b>) The coordinate functions $f_1$ and $f_2$ of $f$ are polynomials and hence $f$ is continuous. This can also be seen from the continuity of partial derivativatives of the coordinate functions of $f$.</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>False</b>) The derivative of a function $f: \Bbb R^n$ to $\Bbb R^m$ at a point $x \in \Bbb R^n$, <b>if it exists</b>, is the unique linear transformation $D(f(x)) \in \mathcal{L}(\Bbb R^n,\Bbb R^m)$ such that $$\text{lim}_{h \to 0} \frac{||f(x+h) - f(h) - D(f(x))h||}{||h||} \to 0.$$ where&nbsp;</span><br /><span style="font-size: large;">$$Df(x) = \begin{bmatrix} \frac{\partial f_1}{\partial x}&amp;\frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x}&amp;\frac{\partial f_2}{\partial y}\end{bmatrix} =\begin{bmatrix}3+2x&amp;-2\\4&amp;5+2y\end{bmatrix}$$ and </span><span style="font-size: large;">$Df((0,0)) = \begin{bmatrix}3&amp;-2\\4&amp;5\end{bmatrix} .$</span><br /><br /><span style="font-size: large;">Let $x = \begin{bmatrix}0 \\ 0 \end{bmatrix}$ and&nbsp;</span><span style="font-size: large;">$h = \begin{bmatrix}h_1 \\ h_1 \end{bmatrix}$</span><span style="font-size: large;">&nbsp;then $f(x+h) = f(h)$. The above limit becomes&nbsp;</span><span style="font-size: large;">$$\text{lim}_{h \to 0} \frac{||- D(f((0,0)))\cdot h||}{||h||} = \text{lim }_{h \to 0}\frac{||\begin{bmatrix}3h_1-2h_2 \\ 4h_1+5h_2 \end{bmatrix}||}{||\begin{bmatrix}h_1 \\ h_2 \end{bmatrix}||}$$</span><br /><span style="font-size: large;">By calculating the norm explicitly, we notice that, the above limit exists. Hence&nbsp;</span><span style="font-size: large;">$Df((0,0)) = \begin{bmatrix}3&amp;-2\\4&amp;5\end{bmatrix}$ is the required derivative of $f$ at $(0,0)$ which is clearly invertible and <b>option (4) is true</b> and in turn <b>option (3) is false</b> . Since $f$ is differential at $(0,0)$, we have <b>option (2) is also correct</b>.</span><br /><span style="font-size: large;"><span style="font-size: small;"><br /></span></span> <span style="font-size: large;"><span style="font-size: small;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank">here</a> for more problems.</span>&nbsp;</span></div><span style="font-size: large;">Share to your groups: <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;"><b>FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</b></span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/47oH4dUgj4A" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-66.htmltag:blogger.com,1999:blog-7629869410369132909.post-57351897866665410352020-04-06T00:06:00.000-07:002020-04-18T07:03:13.669-07:00CSIR JUNE 2011 PART C QUESTION 68 SOLUTION ($||x||_1 \le d \,||x||_{\infty}$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">For $(x_1,x_2,\dots,x_d) \in \Bbb R^d$, and $p \ge 1$ define $$||x||_p = (\sum\limits_{i=1}^d|x_i|^p)^{\frac{1}{p}}$$ and $$||x||_{\infty} = \text{max }\{x_i : 1 \le i&nbsp; \le d\}.$$ Which of the following inequalities hold for all $x \in \Bbb R^d$?&nbsp;</span><br /><span style="font-size: large;">1)$||x||_1 \ge ||x||_2 \ge ||x||_{\infty}$,</span><br /><span style="font-size: large;">2)$||x||_1 \le d \,||x||_{\infty}$,</span><br /><span style="font-size: large;">3)$||x||_1 \le \sqrt d \,||x||_{\infty}$,</span><br /><span style="font-size: large;">4)$||x||_1 \le \sqrt d \,||x||_2$.<br /><b>Solution</b>: Assume $p \ge 1$.</span><br /><span style="font-size: large;"><b>option 1</b>: (<b>True</b>)We have, for $1 \le i \le d$,&nbsp;</span><span style="font-size: large;">$|x_i|^p \le (||x||_p)^p.$ Hence $$||x||_{\infty} \le ||x||_p \,\,\,\,\,\,\,\,\,\,(\text{ for any }p \ge 1).$$</span><br /><b style="font-size: x-large;">Claim</b><span style="font-size: large;">:&nbsp; </span><span style="font-size: large;">If $p \le q$ then $||x||_q \le ||x||_p$.</span><br /><span style="font-size: large;"><b>Proof:&nbsp;</b></span><br /><span style="font-size: large;">Assume $0&lt;a\le 1$, then $$(\sum_{i=1}^d |x_i|)^a \le \sum_{i=1}^n |x_i|^a.$$&nbsp;</span><span style="font-size: large;">Because, $\frac{\sum_{i=1}^d |x_i|^a}{(\sum_{j=1}^d |x_j|)^d} = \sum_{i=1}^d \frac{|x_i|^a}{(\sum_{j=1}^d |x_j|)^a} = \sum_{i=1}^d (\frac{|x_i|}{\sum_{j=1}^d |x_j|})^a \ge \sum_{i=1}^d \frac{|x_i|}{\sum_{j=1}^d|x_j|} = 1$.</span><br /><span style="font-size: large;">Note that, in the last step, we have used the fact that $x^a &gt; x$ if $x, a \in (0,1]$.</span><br /><span style="font-size: large;">Now, assume $p \le q$ then $\frac{p}{q} \le 1$. By using the above result, We have $$||x||_q = \big(\sum_{i=1}^d|x_i|^q\big)^{\frac{1}{q}} = \big(\big(\sum_{i=1}^d |x_i|^q\big)^{\frac{p}{q}}\big)^{\frac{1}{p}} \le (\sum_{i=1}^d (|x_i|^{q})^{\frac{p}{q}})^{\frac{1}{p}} = \sum_{i=1}^d |x_i|^p)^{\frac{1}{p}} \\ = ||x||_p.$$</span><br /><span style="font-size: large;">In particular we have, $$||x||_2 \le ||x_1||.$$</span><br /><span style="font-size: large;"><b>option 4</b>: (<b>True</b>)The Cauchy-Schwarz inequality in $\Bbb R^d$:</span><br /><span style="font-size: large;">Let $x = (x_1,x_2,\dots,x_d), y=(y_1,y_2,\dots,y_d)$. Then</span><br /><span style="font-size: large;">$$(\sum_{i=1}^d x_iy_i)^2 \le (\sum_{i=1}^d x_i^2)(\sum_{i=1}^d y_i^2).$$&nbsp;</span><span style="font-size: large;">Substitute $x = (1,1,\dots,1) \in \Bbb R^d$ in this equation we get the required result.</span><br /><span style="font-size: large;"><b>option 2</b>:(<b>True</b>) Let $x=(x_1,x_2,\dots,x_d) \in \Bbb R^d$ and $a = \text{max }_{1 \le i \le n}|x_i| = ||x||_{\infty}$. First, we claim that $||x||_2 \le \sqrt d \,||x||_{\infty}$. We have $(||x||_2)^2 = \sum_{i=1}^d |x_i|^2 \le \sum_{i=1}^d a^2 = (||x||_{\infty})^2 (\sum_{i=1}^d 1)&nbsp; = d (||x||_{\infty})^2.$</span><br /><span style="font-size: large;">This proves the claim.</span><br /><span style="font-size: large;">Now, using this result in option 4 we get $$||x||_1 \le \sqrt d\, ||x||_{2} \le d \,||x||_{\infty}.$$</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>False</b>) Take $x = (1,1) \in \Bbb R^2$ and calculate the LHS and RHS. We get $2 \le \sqrt 2$.&nbsp;</span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><b>Share to your groups:</b> <!-- Go to www.addthis.com/dashboard to customize your tools --> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/GqS8wFrF45g" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-68.htmltag:blogger.com,1999:blog-7629869410369132909.post-14001469593387326042020-04-05T03:48:00.000-07:002020-04-18T07:04:10.065-07:00CSIR JUNE 2011 PART C QUESTION 67 SOLUTION (If $f_n(x) \to 0$ (a.e) then $\int_a^b f_n(x) dx \to 0$)<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Let $f_n$ be a sequence of integrable functions defined on an interval $[a,b]$. Then, </span><br /><span style="font-size: large;">1) If $f_n(x) \to 0$ (a.e) then $\int_a^b f_n(x) dx \to 0$, </span><br /><span style="font-size: large;">2) $\int_a^b f_n(x) \to 0$ then $f_n(x) \to 0$, </span><br /><span style="font-size: large;">3) If $f_n(x) \to 0$ and each $f_n$ is a bounded function then $\int_a^b f_n(x) \to 0$, </span><br /><span style="font-size: large;">4) If $f_n(x)\to 0$ and $f_n$'s are uniformly bounded then $\int_a^b f_n(x) dx \to 0$. </span><br /><span style="font-size: large;"><b>Solution</b>: <b>option 1:</b>&nbsp;(<b>False</b>) Consider the sequence of functions on $[0,1]$ defined by $$f_n(x) = \begin{cases}n \text{ if } x \in [0,\frac{1}{n}] \\ 0 \text { if } x \in (\frac{1}{n}, 1] \end{cases}.$$&nbsp;</span><br /><span style="font-size: large;">Claim: $f_n(x) \to 0$. Let $x \in (0,1]$ then there exists $n_0 \in \Bbb N$ such that $\frac{1}{n_0} &lt; x$ and in particular $x \in (\frac{1}{n_0},1]$ (by using the archimedian property). For $n &gt; n_0$ we have $\frac{1}{n}&lt;\frac{1}{n+0}$ and so $x \in (\frac{1}{n},1]$. Hence $f_n(x) = 0$. This shows that the sequence $f_1(x),f_2(x),\dots$ is eventually zero. <b>Hence $f_n(x) \to 0$</b>.&nbsp;</span><br /><span style="font-size: large;">Now $\int_0^1 f_n(x) dx = \int_0^{\frac{1}{n}}n dx = n \int_0^{\frac{1}{n}}dx = 1$ for all $n$.&nbsp;</span><span style="font-size: large;">Hence $\int_0^1 f_n(x)dx \to 1 \ne 0$.</span><br /><span style="font-size: large;"><b>option 2</b>: (<b>False</b>) Define $f_n(x) = \begin{cases} \frac{1}{2} \text{ if } x = \frac{1}{2} \\ 0 \text{ otherwise} \end{cases}$ , then $f_n$ converge to the function $$f(x) = \begin{cases} \frac{1}{2}&nbsp; \text{ if } x = \frac{1}{2} \\ 0 \text{ otherwise }\end{cases}.$$</span><br /><span style="font-size: large;">Clearly $\int_0^1 f_n(x)dx = 0$ for all $n$ and hence $\int_0^1f_n(x) \to 0$. But $f(x) \not\equiv 0$.</span><br /><span style="font-size: large;"><b>option 3</b>: (<b>False</b>) Each of the function $f_n$ given in option (1) are bounded. So the same example works for this option also.</span><br /><span style="font-size: large;"><b>option 4</b>: (<b>True</b>) (Because all the other options are false, but we will prove the result which is the correct way to do Mathematics)</span><br /><span style="font-size: large;">Claim: Suppose $f_n$s are uniformly bounded, then there exists $k$ such that $|f_n(x)| \le k$ for all $n \in \Bbb N$ and $x \in [a,b]$. Let $g(x)$ be the constant function $k$ defined on $[a,b]$ then we have $|f_n(x)| \le g(x)$. Now, by the dominated convergence theorem, we can interschange the limit and the integral. Hence, we have $$\lim_{n \to \infty} \int_{a}^b f_n(x)dx = \int_a^b \lim_{n \to \infty}f_n(x) dx = \int_a^b 0 dx = 0.$$</span></div><span style="font-size: large;"><span style="font-size: medium;">Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span>&nbsp;</span><br /><span style="font-size: large;"><b>Share to your groups:</b> </span><br /><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/_EFx0k5Ooic" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-67.htmltag:blogger.com,1999:blog-7629869410369132909.post-56134015583649782182020-04-04T03:56:00.002-07:002020-04-18T07:03:13.637-07:00CSIR JUNE 2011 PART C QUESTION 65 SOLUTION ($\text{lim}_{n \to \infty} n\,\text{log}(\frac{1}{n+1})$))<div dir="ltr" style="text-align: left;" trbidi="on"><div style="background-size: cover; background: url(&quot;https://4.bp.blogspot.com/-M2t8f00ifu0/Xn7pUXtuW5I/AAAAAAAAADU/12qrgAHndf8UteEW4cztLmO50xl84YoSQCLcBGAsYHQ/s320/watermark5.png&quot;); color: black; padding: 10px;"><span style="font-size: large;">Which of the following is/are correct? <br />1) $n\,log(1+\frac{1}{n+1}) \to 1$ as $n \to \infty$, <br />2) $(n+1)\,log(1+\frac{1}{n}) \to 1$ as $n \to \infty$, <br />3) $n^2\,log(1+\frac{1}{n}) \to 1$ as $n \to \infty$, <br />4) $n\,log(1+\frac{1}{n^2}) \to 1$ as $n \to \infty$. <br /><b>Solution</b>:&nbsp;</span><br /><span style="font-size: large;"><b>option a</b>:(<b>True</b>) $n\,log(1+\frac{1}{n+1}) = log((1+\frac{1}{n+1})^n) =log(\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n+1})})$.</span><br /><span style="font-size: large;">$lim_{n \to \infty} log(\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n+1})}) = log(\lim_{n \to \infty} \frac{(1+\frac{1}{n+1})^{n+1}}{(\frac{1}{n+1})} )$ [Since log is a continuous function]. Now, the numerator converges to $e$ as $\lim_{n \to \infty}(1+\frac{1}{n})^n = e$ and the denominator converges to $1$, so the ratio converge to $e$. <b>So the limit is $log(e) = 1$.</b></span><br /><span style="font-size: large;"><b>option b</b>:<b>(True) </b>$(n+1)(log(1+\frac{1}{n})) = log(1+\frac{1}{n})^{n+1} = log((1+\frac{1}{n})^n (1+\frac{1}{n}))$. Taking limit $n \to \infty$, we see that the inside limit is $e \cdot 1 = e$ and the required limit is $log(e) = 1$.</span><br /><span style="font-size: large;"><b>option c</b>: (<b>False</b>) $n^2\,log(1+\frac{1}{n}) = \frac{log(1+\frac{1}{n})}{\frac{1}{n^2}} = \frac{0}{0} \text{form}$. We can use L'Hospital's rule. Differentiate the numerator and the deminator we get $\frac{(\frac{n}{n+1})(\frac{-1}{n^2})}{\frac{-2}{n^3}} = \frac{n^2}{2(n+1)}$ which is divergent as $n \to \infty$ and hence <b>the required limit doesn't exists</b>.&nbsp;</span><br /><span style="font-size: large;"><b>option d</b>: (<b>False</b>) $n log(1+\frac{1}{n^2}) = \frac{log(1+\frac{1}{n^2})}{\frac{1}{n}} = \frac{0}{0} \text{ form}$. Again, we can use the L'Hospital's rule. Differentiate the numerator and the denominator we get $\frac{(\frac{n^2}{1+n^2})(\frac{-2}{n^3})}{\frac{-1}{n^2}} = (\frac{n^2}{1+n^2})(\frac{2}{n})$ which converges to $0$ as $n \to \infty$. Hence the required limit is $log(0)$ which doesn't exist.</span><br /><span style="font-size: medium;">&nbsp;Click <a href="https://nbhmcsirgate.theindianmathematician.com/" target="_blank"><b>here</b></a> for more problems.</span><span style="font-size: large;">&nbsp;</span><br /><b style="font-size: x-large;">Share to your groups:</b></div><div class="addthis_inline_share_toolbox"></div><span style="font-size: large;">FOLLOW BY EMAIL TO GET NOTIFICATION OF NEW PROBLEMS. SHARE YOUR DOUBTS IN THE COMMENTS BELOW. ALSO, YOU CAN SUGGEST PROBLEMS TO SOLVE WHICH WILL BE SOLVED IMMEDIATELY.</span></div><img src="http://feeds.feedburner.com/~r/TheIndianMathematician/~4/6cSFC07pj4I" height="1" width="1" alt=""/>THE INDIAN MATHEMATICIANhttp://www.blogger.com/profile/10264936586823809199noreply@blogger.com0https://nbhmcsirgate.theindianmathematician.com/2020/04/csir-june-2011-part-c-question-65.html