Educational Mathy Quirks

Are There Better Ways To Find GCF and LCM?

2011-08-24T07:30:00.003-04:00

Some of you may have been exposed to this in school. Some of you, like me, were not directly taught this method. Just a heads up.

When I was in fourth or fifth grade, I learned the terms greatest common factor and least common multiple.

For those unfamiliar, the greatest common factor (GCF) refers to the largest number that will cleanly divide (leave no remainder) two or more specific numbers. For example, the biggest number that goes into 2 and 3 is 1, so 1 is the GCF of 2 and 3.

The least common multiple (LCM) refers to the smallest number that can be cleanly divided (leave no remainder) by two or more specific numbers. For example, the smallest number divisible by 2 and 3 is 6, so 6 is the LCM of 2 and 3.

The way I learned to find the GCF of two numbers is as follows.

First, take two numbers. Our numbers will be 8 and 10.

Next, list the factors of each number. For 8, we have factors 1 and 8 (1*8=8) and 2 and 4 (2*4=8). For 10, we have factors 1 and 10 (1*10=10) and 2 and 5 (2*5=10). For each number, we write the factors in increasing order.

Now we circle the factors that 8 and 10 have in common. We notice that there are two factors, 1 and 2, which are factors of both 8 and 10.

But only one can be the GCF. As you may have guessed, we want the greatest number, and that happens to be 2. This means that the GCF of 8 and 10 is 2.

What about the LCM? We start again by writing our numbers 8 and 10.

Since we're considering multiples instead of factors, we want to list the multiples of each number. For 8, we get 1*8=8, 2*8=16, 3*8=24, 4*8=32, 5*8=40, 6*8=48 and so on. For 10, we get 1*10=10, 2*10=20, 3*10=30, 4*10=40, 5*10=50 and so on.

Now if we look through the multiples, we find that they have 40 in common. In reality, the two numbers have infinitely many multiples in common, but 40 is the lowest in value, making it the least common multiple.

Those ways are effective, but once you get to large numbers, they become terribly inefficient. Is there a better way? Let's start with 8 and 10 and figure out the GCF a new way.

We need to factor each number into its primes. We can make 8 into 2*4, and 2 is prime. We can make 4 into 2*2, both of which are prime. So 8 is made of 2*2*2 or 2^3.

We can factor 10 into 2*5, both of which are prime.

Now we need to compare quantities of numbers. For 8 (2^3), we have three 2s. For 10 (2*5), we have one 2 and one 5.

What do both have in common? Each has a single 2. Even though 8 has two extra 2s and 10 has one extra 5, we don't include those. We only care about what they have in common.

Let's repeat for LCM. First, prime factorization.

Now we count quantities.

Now we have to compare those quantities.

We include one 2 (because both have one two). We include two extra 2s (because only the 8 had two extra 2s). We include one extra 5 (because only the 10 had one extra 5). This gives 2*4*5, or 40. Notice that if we had included the 2 they had in common twice (once for each), we'd have 80, or 8*10. Sometimes we can just multiply the two numbers and get the LCM, but that is only true if the two numbers are co-prime (sharing no factors).

Let's try two bigger numbers, say 48 and 64. As always, we complete a prime factorization.

Now we compare the quantities and take only what they have in common (four 2s). This gives us a GCF of 16.

We repeat for LCM.

Again, by comparing quantities, we get four 2s in common, one 3 specific to the first and two 2s specific to the second.

Multiplying these out, we get an LCM of 192.

These methods are perhaps most useful in algebra. When adding fractions, like 1/2 and 1/3, you must first find a common denominator (the least common denominator). In that case, we would get 6 to add 3/6 and 2/6. If you're looking at 1/x and 1/(x+1), the problem still remains. Let's look at two factored expressions below.

Now, like always, we count the factors.

Now we take what they have in common, what is specific to the first and what is specific to the second.

Multiplying them out, we get the factored expression below.

To finish the fraction addition, we would simply have to multiply the top of each fraction by what its denominator was missing from the least common denominator. For example, the first expression did not have an (x+3), so the numerator and denominator would be multiplied by (x+3). For 1/2, what is the bottom missing to get to 6? Since it is missing a 3, we multiply 1*3 and 2*3 to get 3 and 6 to give us 3/6, just as we would expect.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I don't exactly recall learning this explicitly. I saw the types of problems in Algebra I and Algebra II to an extent, but I more figured out the method than I was taught it.

The images used here were created by me using Noteshelf for iPad.

Can Algebra II Help Me With Finances?

2011-08-22T07:30:00.000-04:00

If you have money in a savings account (or owe money on a loan or credit card), you're probably familiar with interest. Simple interest is calculated by multiplying the amount you have in the bank (principal) by the interest rate by the amount of time.

So if you have $100 in the bank at 5% interest per year, after one year you'll have $5 more. In banking, we don't usually worry about simple interest.

When we calculate interest, we need to take account of the interest your money has already earned. We also have to know how many times the interest is taken.

For example, suppose your principal, P, is still $100, and the interest rate, r, is still 5% per year. We refer to calculating interest as compounding interest. We let n equal the number of times interest is compounded in a year. If we compound interest once per year, then we have $105 after one year, like simple interest. But what if we compound twice? Then we would have $105.06 at the end of one year.

The formula (with the time in years equal to t) for compound interest is:

How does changing the number of times interest is compounded affect the amount of money we earn? To make things simple, we'll start with $1 and assume 100% interest per year. We'll look at the account for one year (t=1). This gives us a simple formula for A.

We'll consider compounding with the following frequency:

1) Yearly
2) Monthly

3) Daily

4) Hourly

5) Per Minute

6) Per Second

7) Per Tenth-of-Second

Though the amount we earn at the end of one year doesn't really change after compounding hourly $1, notice something about the numbers. They appear to be stabilizing. The larger number we use for n, the closer A appears to get to another number. What happens if we let n get really big? If we compound interest continuously (that is, we are ALWAYS calculating interest), then we can say that n goes to infinity. By taking a limit, we can see how A looks if we let n get really big.

The expression below is read as: the limit as n goes to infinity of the quantity one plus one over n quantity to the n power.

This is one of the original definitions for the constant e, which has a value of about 2.718. If we look at our table of values, we can see that as n gets bigger, A gets closer and closer to e.

What if the interest rate wasn't 100%? Would we still get e? As it turns out, we don't. We get something slightly different. If interest rate is a random number, a, then our limit doesn't yield e; it gives us e to the a, as seen below.

What happens if we continuously compound interest? We need to take the limit as n goes to infinity of the compound interest formula. Since P does not depend on n, we can bring it outside the limit.

Since t doesn't depend on n either, we can move the limit inside of the t-power. If we look at what we're taking the limit of now, it simplifies to e to the r, just like e to the a earlier. Using our rules for exponents, we know that raising an exponent to a power requires us to multiply the exponents, leaving us with e to the r t power, all multiplied by P.

So our formula for continuously compounded interest is:

This is a little surprising, as nothing about interest necessarily indicates the presence of e. The fact is, e has the habit of showing up whenever exponentials are involved, so keep an eye out for it.

Personally, I would never teach this in an Algebra II setting. I would teach the compound interest formula when learning exponential functions, but I would not delve into the e-problem. In my experience, Algebra II students don't really care about interest, they don't know the definitions of e and they usually don't know limits.

That being said, this would be good enrichment for the advanced students in an Algebra II class, and this would be a valuable lesson for a Precalculus or Calculus class reviewing limits.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first saw this definition for e from my Calculus Teacher Jan Haynes. I saw the table derivation for e in compound interest from my mentor teacher Jennifer Davis when I was completing my student teaching at Somerville High School and finished the rest of the derivation on my own.

The images used here were created by me using Noteshelf for iPad.

Are All Repeating Decimals Fractions? Or What Is One? (Baby, Don't Round Me)

2011-08-19T07:30:00.001-04:00

In math, we often categorize numbers. All numbers fall under the term complex numbers, numbers which have both real and imaginary components (though at least one of which may be zero). The real numbers can be further divided into the irrational numbers and the rational numbers. The terms rational and irrational do not refer to how well the numbers can reason; rather, they come from the word ratio. Irrational numbers cannot be written as a ratio, while rational number can. In other words, rational numbers are numbers that can be rewritten as fractions.

What makes a number rational? To determine whether or not a number is rational, all we have to consider is the number after the decimal. If the decimal terminates (or repeats), then the number is rational. Other wise, it is irrational.

Which decimals terminate? One example is 4. Since there are no numbers after the decimal (other than zero), we can say it terminates. The number 1.2345 also terminates, but the number 1.243531723... does not terminate. Neither does 3.14159... or 2.71828... But what about a number like 3.333333.... or 0.123123123... or 2.45321321... or 7.9999....? They don't terminate, but they do repeat.

I could easily say 4/1 is the fraction of 4, or that 1 and 2345/10000 is fraction of 1.2345. Those repeating decimals are not so easy. We've already said that repeating decimals indicate that the number is rational, so there must be a way to convert a repeating decimal into a fraction. But how?

When I was in 7th grade, I had a two-page "cheat sheet" of mathematical formulas to memorize for my scholars' bowl team. Two of the formulas on there related to turning a repeating decimal into a fraction. When I was in 10th grade, though, my Algebra II/Trig and Math Team teacher, Suzanne Ingram, taught me a method for finding these fractions. The best part is that the method requires only Algebra I level skills.

Since most of you reading know that 0.333... is 1/3, we'll start with 3.33... so there's some familiarity to build upon. We start by setting our decimal, 3.33..., equal to x.

Now we want to multiply x by 10 enough times so that the number after the decimal doesn't change (but at least once). Here, we can multiply by 10 one time and the number after the decimal is the same.

Now we have two equations. We want to subtract the smaller from the larger.

This leaves us with 9x on the left and 30 on the right.

Since we have an equation, we just divide both sides by 9. This gives us x on the left and 30/9 on the right, which reduces to 10/3. If we work that out, we find that 10/3 does equal 3.333...

Let's try 0.123123... for x. How many 10s will it take? Since we have three numbers before the repeat, we will have to multiply by 10 three times, or 1,000.

Again, we subtract the smaller from the larger to get 999x on the left and 123 on the right.

Dividing by 999 gives us 123/999, which reduces to 41/333. If we were to work that out, we would find that it does equal 0.123123123...

In the next case, the decimal doesn't repeat right away. We have 2.45321321... Since this problem is different, we should treat it slightly different. The first step is to multiply by 10 enough times so that all of the decimal repeats. In other words, we need to get rid of that 45. To do so, we will multiply by 100 (10 two times). Now, we have to multiply by 10 enough times so that the number after the decimal doesn't change like we did before. Since there are three numbers, 3, 2 and 1, we will have to multiply by 1000 (10 three times). But we're not multiplying the original x; we're multiplying the new 100x to get 100000x.

We have to take the 100000x and subtract 100x. This will give us 99900x on the left and 245076 on the right.

Dividing both sides by 99900 gives us 245076/99900, which reduces to 20423/8325. If we work it out, we find that does equal 2.45321321...

What about 7.999...? This one looks simple like the first case, and it is. However, it is much more interesting. Again, we only have to multiply by 10 one time to maintain our repeating decimal.

Subtracting the two yields 9x on the left and 72 on the right.

Dividing both sides gives us 72/9, which reduces to 8.

Hold on, now. We just showed that 7.999.... = 8. If we subtract 7 from both sides, we are effectively saying: 0.9999.... = 1. Could that be true?

Let's walk through the divisions by nine to convince ourselves.

1/9 = 0.111....

2/9 = 0.222...

3/9 = 0.333... = 1/3

4/9 = 0.444...

5/9 = 0.555...

6/9 = 0.666... = 2/3

7/9 = 0.777...

8/9 = 0.888...

Even though we know that 9/9 is equal to 1, the pattern implies that 9/9 = 0.999... The existence of the pattern does not prove that 0.999... = 1. It does, however, provide us some insight as to whether or not the solution we came up with makes sense.

Some people prefer formulas; I'm okay with them, but I'd much rather see a process than memorize a formula.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first saw this method from by Algebra II/Trig, Precalculus and Math Team teacher Suzanne Ingram. I first saw the 0.999... = 1 problem online my senior year of high school and asked my Calculus teacher Jan Haynes about it.

The images here were produced by me using Noteshelf for iPad.

Are Soup And Cereal Companies Evil?

2011-08-17T07:30:00.005-04:00

Unfortunately, exposing evil empires will require some knowledge of calculus. If you understand first derivatives, you should be fine. If you don't, I'll try my best to explain what you need to know, but it may prove to be difficult.

To preface, this is one of my favorite things to do in math, because it's a real world problem. Many, many students want to stick to equations and not bother with word problems. Unfortunately, you don't see many equations in real life; you have to make them yourself from information given. While you have all the information you need to solve these problems, they do present a way to apply mathematics to problems we face when grocery shopping.

There are two properties of a container that we care about for this exercise. The first is the container's volume. This is important because we want to know how much soup or cereal our container will hold. The second is the container's surface area. This is important because we want to know how much the container will cost to make, since that depends on how much material we need to make the container.

We'll start with soup companies, since the math is easier.

We know that soup comes in cylindrical cans. The volume of a cylinder is given by the formula pi times radius (r) squared times height (h) (the first formula below). The surface area is a little more complicated. That formula is two times pi times radius (r) squared plus two times pi times radius (r) times height (h) (the second formula below).

Now, we're going to assume that volume is constant. By using a constant volume, we can figure out what gives us the most volume for a given surface area. If it sounds backwards, don't worry. It'll work out, I promise. Now we can begin work.

Our first problem is that in the surface area formula, we have two variables, r and h. Since we said volume was constant, we can solve the volume formula for h. We find that h = V/pi r squared. We can then plug h into the area equation. So far, we've only used algebra to get our new formula for area.

Now for the calculus. We want to take the derivative of A (area) with respect to r (radius). If that sounds messy, here's what you need to know. We want to know how the area changes if the radius changes. You can imagine that if radius increases, area increases. The derivative just gives us a formula that tells us exactly how the radius affects the surface area. We write the derivative as dA/dr (area to radius). We usually use derivatives to find minimum and maximum values of functions. In our case, we want to find when the surface area is at a minimum. This happens whenever the derivative is zero (don't worry about why this is if you don't know for now; you can always look it up later). On the first line, I've found the derivative; on the last line, I set it equal to zero.

We can now solve for V in terms of r.

This means that when surface area is minimized, volume equals two times pi times radius cubed.

Using this and the original equation for volume, we can find out which value of h gives us the smallest surface area for a given volume.

This means that when the can is as wide (2r = diameter) as it is tall, we get the lowest surface area for a given volume. In other words, we get the most volume for a given surface area. Unfortunately, I haven't noticed a lot of cans this shape. Most of the cans I see are tall and skinny, even though that shape isn't optimized.

Let's try cereal boxes. This math will get a little heavier, but the idea is the same.

The volume of a box (any box) is given by length times width times height. The formula for surface area is a little more complicated. Just know that l is length, w is width and h is height, and the second formula below will give you surface area.

For volume, we have three variables, which is two too many. Again, let's let volume be constant and solve for l (this will eliminate one variable). We get that l = V/wh. Now we plug that into our area equation.

Again, we've done no calculus yet, but here it comes. We're going to take two different derivatives. Sounds messy, but I'll explain. First we want to find the derivative of surface area with respect to height (how does area change with height?) (dA/dh). Second, we want to find the derivative of surface area with respect to width (how does area change with width?) (dA/dw). You might notice the funny symbol I used for d. Don't worry too much; it just means I have more than one variable. Just like before, though, I want to set them equal to zero to find the minimum value for surface area.

If you know how to take derivatives, know that these are partial derivatives. And if you know how to take derivatives, then know in the first case, I assume w is constant, and in the second case, I assume h is constant.

When I set dA/dh equal to zero, I solve it and get that V = wh^2.

When I do the same for dA/dw, I get V = hw^2.

Since volume is constant, I can set them equal to each other. I find out that h = w, so I make them both equal to s, a new variable to represent either h or w.

With s, I get the two equations for volume and surface area below. Again, in surface I have two variables and want to get rid of l by solving my volume equation. This gives me l = V/s^2. If I plug this into the area equation, I get the result at the bottom.

I want to know how surface area changes with s, so I need to take the derivative (regular this time) to find dA/ds. Like always, I want to set that equal to 0.

When I work that out, I find that surface area is minimized for a given volume (or volume is maximized for a given surface area) when volume equals s-cubed, which means all three sides are the same length.

The shape that optimizes this problem, then, is a cube. Again, I don't see many cereal boxes this shape. Most I see are tall and skinny, just like the soup cans.

So what gives?

Which can did you think would hold more: tall/skinny or short/fat?

What about the box: cube or tall rectangle?

If you're the average person, you picked the tall one both times. It's a psychological thing that makes us think the taller container holds more; we completely ignore the other two dimensions!

Companies know this. That's why they don't bother optimizing their containers. They know they sell less in a box that costs the same to make while making you think it's more than it is.

So are soup and cereal companies evil? Not necessarily, but they do know how to play with your mind.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first learned optimization problems from my high school Calculus teacher Jan Haynes.

The images here were produced by me using Noteshelf for iPad.

Where Does The Quadratic Formula Come From?

2011-08-15T07:30:00.002-04:00

If you've ever went through some form of math class, you've likely heard the term linear equation. You'll probably recognize the slope-y-intercept formula y = mx + b. If you completed a whole Algebra I curriculum, you likely recall going beyond these linear equations and studying quadratic equations. They were the ones with an x-squared and looked like this:

All long as a does not equal zero. Look familiar? You might also remember solving these (somewhat ugly-looking) problems. The first way was factoring, and life was easy if you could factor. But what if you couldn't factor? Most of us turned to the universal quadratic formula. You remember that, right? It had that 4ac, the over 2a, the b-squared, the plus or minus... Yes, it was messy, but it worked! But that wasn't the only other way. Some of you, like me, preferred the method called completing the square. Whichever you preferred, the order of learning usually went like this:

1) Factoring

2) Completing the Square

3) Quadratic Formula

Now, I understood learning factoring before the quadratic formula, but I never really knew why I needed to know how to complete the square. Yes, it becomes useful when you study integral calculus, but was there a reason to teach it to Algebra I students? Well, I don't recall going too far in depth on this, but the only reason to learn completing the square in Algebra I is to derive the quadratic formula. So how do you get the quadratic formula by completing the square? Well, we know the quadratic formula works for all quadratic equations, so we start with our generic equation. But there's an important rule for completing the square. Do you remember it? The number in front of x-squared has to be one (1). The problem here is that a can be any number (except 0). Fortunately, since a cannot be 0, we can divide everything in the problem by a.

This makes the number in front of x-squared one (1). And zero divided by a will simplify to zero (0).

The next step is to subtract the constant term (the one without an x). Here, that happens to be c/a. Remember, since this is an equation, we have to do everything to both sides equally.

We now have the problem below.

This is the trickiest part of completing the square. We now have to add a number to both sides, but not just any number. We want to take the coefficient (the number in front) of x, divide it by 2 and then square it. In symbols, it would look like this:

In this case, the coefficient of x is b/a. So we put b/a in the empty space.

We can rewrite the inside as b/2a. We can then square it.

We now have the term we want to add to our equation.

Since we're dealing with the equation, we add our new term to both sides.

To make everything easier later, we want to combine both terms in the right-hand side. To do so, we must first find the common denominator. Here, it's 4-a-squared. This makes c/a into 4ca/4-a-squared.

We then combine the two terms into one fraction.

Now to deal with the left-hand side. This method is called completing the square, and that's what we're trying to do. Notice the x added to a blank below? That whole term will be squared, and it's the square we're trying to complete.

But what number goes there? Well, remember the number we squared earlier? We want to use the number before it was squared (b/2a). Notice I'm not adding anything to the right-hand side. This is because we are not adding b/2a; we are only rewriting the left-hand side.

The reason we wanted to complete the square is so we could take a square root. Since we're taking the square root (and changing the left-hand side), we have to do it to both sides. This is important: make sure to include ONE plus/minus sign. Since we are introducing a square root to the problem, we have to remember to include the plus/minus sign. It makes the most sense to put it on the right, so that's where it'll go.

We simplify the left-hand side.

On the right-hand side, we can simplify our radical by taking the square root of 4-a-squared to get 2a on bottom.

Now we want to get x alone (since we're solving for x). We have to subtract b/2a from both sides.

This gives us x by itself on the left. Now, we combine our two terms on the right-hand side. This leaves us with the quadratic formula, in all it's messy glory!

I don't remember ever learning this in Algebra I, but I think it would've helped if we did. I like to know why we learn what we do (completing the square) and where they come from (the quadratic formula), and I'm sure know both sides would help some students learn. If a student can understand completing the square AND can work with variables, he or she can follow a lesson on this. This derivation is simple enough that it can be given as enrichment work for students who finish early with class work or a test.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first saw quadratic equations during my eight grade year from my Algebra I and Math Team teacher Tiffany Brooks.

The images used here were produced by me using Noteshelf for iPad.

Can I Solve An Infinite Root?

2011-08-12T07:30:00.003-04:00

For the purposes of this entry, and infinite root looks like this:

The infinite roots in this entry (like the one above) repeat within themselves. The first one we'll consider is the repeating "root-2." Is there a way to evaluate this? For now, we'll set the repeating infinite root equal to x, the quantity for which we want to solve.

This next part is probably the most important to understand for these problems. If we cover up the biggest square root sign, the first 2 and the first addition sign, what do we see? It's another repeating infinite root. This repeating infinite root is special, because it is exactly what we defined x to be at the beginning.

Using the rule of substitution, we can let everything after that first addition sign be replaced by x. This gives us the far simpler equation below.

In order to solve this problem, we need to get rid of the square root sign. To do that, we square the right-hand sign. Since this is an equation, whatever we do to one side, we must equally do to the other, giving us a new equation.

We now set the left-hand side equal to 0 by subtracting x and 2 from both sides.

Using our rules from Algebra I, we can factor the left-hand side of the equation into (x - 2) and (x + 1). Since we have two things multiplied equal zero, we know one of the two things has to be zero. We then set each factor equal to zero.

On the left problem, we add 2 to both sides. On the right problem, we subtract 1 from both sides. This gives us two answers, 2 and -1.

Let's look at the original repeating infinite root. Notice how no sign is written in front of the root? This means we're looking for a positive root. So now we have to check our answers.

The -1 solution is what is called an extraneous root. This basically means it shows up as an answer from us changing the problem, but it does not solve the original problem. Extraneous roots usually appear whenever you start raising things to powers (e.g. when we squared both sides) or in other cases. We didn't do anything wrong, it's just a part of working with mathematics.

I chose the case with 2 because that's how the problem was introduced to me. My Algebra II/Trig and Math Team Teacher Suzanne Ingram told us that the first time a student saw this, the answer given was 2 because "there were a lot of twos." If you notice, two also was the answer. But is this always the case? Let's look at another example using 3 instead of 2.

Just as before, notice how x is contained under the root after the first 3 and first addition sign.

Now we square both sides.

After subtracting x and 3 from both sides, we notice that this won't factor. I prefer completing the square, but we'll use the quadratic formula since it's faster.

When we work everything out, it simplifies to one plus or minus the square root of thirteen all over two. We separate that out into two solutions, each with a square root of thirteen. The larger is about 2.3 while the smaller is about -1.3.

Notice how one was positive and the other negative? Our original problem did not have a sign, so we want the positive answer. Since the solution that was about -1.3 is negative, we reject it as an extraneous root. This means that our one solution is 1 plus the square root of 13 all over 2 (or about 2.3).

So, yes, it is possible to solve these infinite root problems. Some of them can get quite tricky, but the ones with just one number repeating (with addition) can all be solved like these. And the answer is not always as simple as the number that repeats over and over.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first saw this problem during my sophomore of high school from my Algebra II/Trig, Precalculus and Math Team teacher Suzanne Ingram.

The images used here were produced by me using Noteshelf for iPad.

How Many Sides Does A Circle Have?

2011-08-11T07:30:00.005-04:00

One? Zero? A bunch?

To answer this question, we're going to use some trigonometry, some precalculus and some calculus. Don't worry if you don't know it yet. I'll explain what you need to know.

First, the trig. The angle around the center of a polygon is 360 degrees (i.e. a circle). In our math, we'll be using radians (not degrees). All we need to know is that 360 degree = 2π. In a right triangle, the longest side (opposite the right angle) is known as the hypotenuse. The sine of an angle (sin θ) is the opposite side divided by the hypotenuse. The cosine of an angle (cos θ) is the adjacent side divided by the hypotenuse. The only other thing we need to know is:

2(sin(θ))(cos(θ)) = sin(2θ). Don't worry about why, just know that it is (we may cover this another time).

Next, the precalculus. The limit of a function is a value. If I say "The limit of y as x approaches infinity," I want to know what y looks like as x gets really big.

Here's an example. y = 5x. What is the limit of y as x approaches 2? In some cases, we can plug in. If we do that here, we get y = 10. So the limit of y as x approaches 2 is 10.

Now, y = 1/x. What's the limit of y as x approaches zero? As x approaches infinity?

If something is divided by zero, we say it approaches infinity. If something is divided by infinity, we say it approaches zero.

Now for the calculus. The two words I'll use are L'Hopital and derivative. If I get a limit that becomes infinity/infinity or 0/0, I can use L'Hopital's rule to calculate the limit. The works by using derivatives. To follow my work, you don't need to know what a derivative is. In fact, if you'd like, just substitute the word "ice cream" for derivative. After all, who doesn't love ice cream?

That covers all you need to know. Let's get started.

First, recall from last time that the area of a regular convex polygon is given by (1/2)(a)(p) and that p is equal to (n)(s), where a is the apothem, p is perimeter, n is the number of sides and s is the side length.

When thinking about circles, we don't usually come across the term "apothem," so we'll want a way to write our equations without it. Unfortunately, we also don't hear side lengths in reference to circles, we need to get rid of that as well. To do this, we'll use the other side of our little triangles running from the center to a corner. We'll call this longest side (the hypotenuse) r, for reasons that will become obvious later. Here, the hypotenuse is r, the opposite side is s/2 and the adjacent side is a. Using the definitions for sine (opposite/hypotenuse) and cosine (adjacent/hypotenuse) for our angle θ, we get the following:

Now that we have expressions for s and a, we need a way to write θ in terms of other things we're using. We know the central angle is 2π radians and that there are 12 little triangles that are all the same. This means the top angle of a little triangle is 2π/12 (or π/6).

For an n-gon, the central angle is still 2π, but we now have 2n little triangles (from yesterday's post), with means our angle is going to be π/n for each one.

Using the the equations for s and a found earlier, we can rewrite the equation for area.

Remember the other trig equation from earlier? We can now use it to make sine and cosine into one function, making the math that comes later much easier.

Now we substitute in our π/n.

We take r to be a constant, so our only variable is n. This means area is a function of n and can be written as A(n) (Read: A of n). Now, we know that if there are no sides, there can be no enclosed area, so that case isn't very interesting. But what if there are infinitely many sides? To find out, let's take the limit of A(n) as n goes to infinity.

We ended up with infinity times zero, which is no directly solvable using limits. Since r and 2 aren't dependent upon n, I can bring them outside to cleanup my function before I mess with it.

Remember when I mentioned L'Hopital? I said it would let me work with infinity/infinity or 0/0. Well, now I have infinity times 0. Fortunately, there's a trick I can use to make it 0/0!

Think of it like this. Consider 2. Now, consider 1/2. Finally, consider 1/(1/2). The last is the same as 2, but it is written differently. In our case, we have n (which is becoming infinity). If I rewrite it as 1/n, I've got 0 (on the top), but my expression is different. To fix that, I have to bring 1/n into the denominator. It's like earlier. I started with 2. I flipped it, but it was different. To make it the same, I brought it to the bottom of the fraction. My old expression and my new expression are the same thing, but they're written in different languages. Now, I have 0/0.

L'Hopital's rule works like this, but only if I'm dealing with 0/0 or infinity/infinity. It says that the limit of a fraction of two functions equals the limit of a fraction of two function's derivatives (or ice creams). If our expression is f(n)/g(n) (Read: f of n over g of n), then the derivatives (ice creams) are f'(n) and g'(n) (Read: f prime of n and g prime of n). I've calculated the derivatives (ice creams) below. Don't get caught up in how they are what they are (for today), just keep up with which is which.

Since f(n)/g(n) gave me 0/0, let's see my new expression using f'(n) and g'(n).

There's a -1/n^2 term in both the top and bottom, so it simplifies away.

Now I plug my new expression back into my limit. Don't forget the r^2/2 outside of the limit.

Since 2π is a constant, I can bring it outside the limit and simplify.

Since cosine is a function, I can move my limit inside the parentheses. So now we're taking the limit of 2π/n as n approaches infinity. What did we say about this earlier?

By plugging in and dividing by infinity, we create a fraction that becomes 0. We now take the cosine of 0, which has a value of 1. Since we had πr^2 outside the limit, we multiply this by 1 to get the expression πr^2.

So we can say that the limit of the area of a regular convex polygon as the number of sides approaches infinity is equal to πr^2 (Read: pi r squared), which is formula for the area of a circle. This implies that a circle is like a polygon with infinitely many sides!

This is one of my favorite problems to work out and show people. It's one of those things that doesn't really have a practical purposes, but it's kinda cool to see.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I've been doing this problem for a couple of years now, but I'm not sure when I started. At the earliest, it was during my senior year of high school (after I learned L'Hopital's rule from Mrs. J. Haynes in Calculus). At the latest, it was this past December.

The images used here were created by me using Noteshelf for iPad.

What's The Area Of A Regular Polygon?

2011-08-10T07:30:00.000-04:00

Yes, squares and rectangles are easy things for which to calculate area. Just use base times height. And triangles are almost the same. Just multiply base times height times one-half. But what about other shapes?

First, a note on vocabulary and notation. In this entry, I will be showing how to get the formula for the area of any regular convex polygon. A polygon is a shape with at least three sides. A regular polygon has all sides of the same length and all angles of the same measure. A convex polygon has each angle measuring less than 180 degrees.

The symbol s always refers to the length of a side.

The symbol s/2 always refers to half of the length of a side.

The symbol p always refers to the perimeter of a figure.

The symbol n always refers to the number of sides.

The symbol A always refers to an area.

The symbol a always refers to the apothem. The apothem is a line drawn from the center of the figure to a side. The apothem cuts the side into equal halves (bisects) at two 90 degree angles (perpendicular).

Finally, a note on the writing. As I said, I'm talking about regular polygons, so all sides within a polygon are the same length (even if they do not appear to be so). The drawings may look a little rough, but they are legible and should drive the point across.

Let's begin by looking at the simplest polygon: the triangle.

This triangle has three equal sides, all of length s. Each dotted line represents an apothem. The details are similar for the square and hexagon below, except the square has four sides, and the hexagon has six sides.

In general, we refer to a polygon that has n sides as an n-gon.

From the images above, we can notice a pattern in the relationship between the number of sides in a polygon and the number of little triangles created with lines drawn through each angle to the center and lines drawn from the center to the sides. Please note: in general, it is not safe to assume a trend from a few examples. For the purposes of this exercise, we just need to convince ourselves that the trend below is true (an n-gon has 2n triangles).

In the case of the square, we can calculate the area of the shaded triangle below using the formula for the area of a triangle, where the base is s/a and the height is the apothem.

The same is true for the hexagon.

In fact, the area of a little triangle of an n-gon is also A = sa/4. Again, we cannot, in general, make such assumptions from a few examples. If we can convince ourselves this is true, it will suffice for our purposes.

Now to find the area of the entire square, we just need to multiply the area of one little triangle by the number of little triangles. Since these are regular convex polygons, each little triangle of the shape contains the same area. As we said in the case of the square, there are 8 little triangles. So the total area is 2sa.

In the case of the hexagon, there are 12 little triangles, so we get a total area of 3sa.

For an n-gon, as we said earlier, we have 2n little triangles. This gives us a total area of (2n)(sa/4). This simplifies to nsa/2.

Perimeter is simply the sum of the lengths of the sides. For the square, there are four sides of length s, so the perimeter is 4s. In other words, one side is one-fourth of the perimeter.

For the hexagon, there are six sides of length s, so the perimeter is 6s. In other words, one side is one-sixth of the perimeter.

For an n-gon, there are n sides of length s, so the perimeter is ns. In other words, one side is one-nth (1/n) of the perimeter. Or, s = p/n.

We can combine our equations for area with the equation relating side length and perimeter. Since, in the case of the square, s = p/4, we can plug p/4 into the Area = 2sa equation.

The process is similar for the hexagon.

We repeat the process one final time for the n-gon.

In all three cases, we came up with the formula Area = (1/2)(a)(p). While this was not a rigorous proof, it was a way to see where the formula comes from. And while not very interesting on its own merits, it will be quite useful in answering tomorrow's question. I just wanted to devote an entry to this problem because it is worth seeing.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first saw this formula in my freshman year of high school in Mrs. J. Plunkett's Geometry I class. I didn't see any sort of derivation until I worked one out myself the summer following my sophomore year of high school.

The images here were produced using Noteshelf for iPad.

What's The Square Root Of Five?

2011-08-09T07:30:00.000-04:00

It's about 2.24.

That's the quick answer. I just typed 5^0.5 into Google, and I got the result I wanted. I could have just as easily typed the problem into my calculator, or my phone, and found the same result.

According to Wikipedia, the first scientific calculator came out in 1968 and cost $32,000 in 2011 US Dollars. Without tables or a slide rule handy, in some cases, how did people calculate square roots?

Some were easy. The square root of 1 is 1, 4 is 2, 9 is 3 and so on. These numbers (1, 4, 9, ...) are called perfect squares because their square roots (1, 2, 3, ...) are integers.

Unfortunately, not every number is a perfect square. The first obvious example is 2. Another is 3. In fact, it seems obvious that there would be more non-perfect squares than there are perfect squares.

So how can we calculate the square root of a non-perfect square? The short answer is long division.

Yes, the same techniques learned way back in third grade (or so) to find quotients and remainders can be applied to finding square roots of numbers (albeit with slight modification).

The numbers 2 and 3 are between 1 and 4, so this means their square roots are between 1 and 2. Since we don't want to make any statements that aren't true, we'll use 5 so we don't run into any issues with the number 1 (since 1 is its own square and square root).

Where do we begin?

At the beginning, of course.

Now, we want to find an answer to two decimal places. Remember from long division that however many decimal places you want to find, you have to calculate one more. If you want one decimal place, find two. And so on. We want two decimal places, so we need to calculate our answer to three places.

For every digit after the decimal, we need to add two zeroes after the five's decimal point. Since we want two decimals, we need to calculate three. Since we need to calculate three, we'll need six zeroes.

Now we can start calculating. What's the biggest perfect square that goes into five? One goes into five, four goes into five, but nine does not go into five. So our biggest perfect square is 4.

What's the square root of four? Since the square root of 4 is 2, we write 2 above the five (like we'd do in long division).

Like long division, we have to subtract what we've already taken account of. In this case, we took care of 4 using that 2, so we're going to subtract away 4 from 5, giving us a difference of 1.

In long division, you would bring down a zero. In finding square roots, we're going to bring down two zeroes. Now he's where it starts to get tricky.

Take the number on top (2) and double it (4). Bring the number (4) down beside the difference (1 with two zeroes, 100). Write a blank next to the new number (4) and put both that number (4) and the blank in parentheses. Your work should look like this:

This is the hardest part to think about. You need to put a number in that blank, so that if you multiply the number (4_) by the number you choose, it is less than the other number (100). Confused? Here's how it works. If you choose 1, you're going to multiply 41x1. If you choose 4, you're going to multiply 44x4. If you choose 9, you'll multiply 49x9. Make sense now? You're going to choose a number between 0 and 9, and you want the biggest product less than that 100. If we choose 1, we get 41x1 = 41. If we choose 2, we get 42x2 = 84. If we choose 3, we get 43x3=129. So the biggest one that works is 2. Now we write 2 in that blank from earlier.

We also write a 2 on top after the decimal place over our first two zeroes. As we said, 42x2 = 84, so we subtract that from 100, giving us a difference of 16.

Like earlier, we carry down the next two zeroes. We also take the top number (22), double it (44), and write it next to a blank space in parentheses like we did earlier.

We run through blanks again. If we choose 1, 441x1 = 441. If we choose 2, 442x2 = 884. If we choose 3, 443x3 = 1329. If we choose 4, 444x4 = 1776, which is too big. So we choose 3. As before, the 3 goes in the blank and above the next two zeroes.

As we said, 443x3 is 1329. So again we subtract the product and carry down the next two zeroes. We then take number on top (223), double it (446) and write it in parentheses with a blank (446_).

We try numbers again. If we choose 1, 4461x1 = 4461. If we choose 2, 4462x2 = 8924. If we choose 4, 4464x4 = 17856. If we choose 8, 4468x8 = 35744. If we choose 9, 4469x9 = 40221, which is too big. So our rounding digit is 8.

In reality, we don't care about this digit. It only matters for rounding (since we want two places). All we need to check is the case of 5. In that case, 4465x5 = 22325. Since the number is less than 37100, we know to round up (since the digit is 5 or greater). If 4465x5 had been more than 37100, we would know to round down (since our digit would have to be 4 or less). Since the number we calculated is 2.238, we know the square root of 5, to two decimal places, is 2.24, which is what I calculated at the beginning of this entry.

Since we can easily calculate this with tools, this method's use is somewhat outdated, but it's a cool thing to know. Some other numbers to practice on include 2, 3, 6, 7 and 8. Once you've mastered those, try your hand at 10 and the double digits.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I learned this from Suzanne Ingram, who was my teacher for three years of math team, one year of Algebra II/Trig and one year of Precalculus in high school.

The images used here were produced by me using Noteshelf for iPad.

How Big Is Earth? Or, Is Three Close Enough?

2011-08-08T07:30:00.001-04:00

As an engineer, I absolutely love estimation. Early in high school, I hated it, though. I hated the uncertainty and craved exactitude. By my sophomore year of college, I'd come to appreciate the elegant simplicity of estimation. One of my favorite demonstrations of this elegance is the calculation of Earth's volume.

Just how big is Earth?

Perhaps the best part of estimation is that it can be used to get anything close enough. In elementary and middle school, we were taught to use 3.14 or 22/7 for pi, even though those aren't exact values. For most applications (even beyond grade school math), they're close enough. Ask people what "pi" is. A mathematician will tell you it's the ratio of a circle's circumference to its diameter. A champion pi-digit memorizer will tell you it's 3.14159265... and so on. But an engineer will tell you it's about 3.

How close is that?

With a little estimation (and less knowledge), both questions are easily answered.

According to Wikipedia, the volume of Earth is 1,083,210,000,000 km^3 (or 259,875,899,000 miles^3).

Also according to Wikipedia, the value of pi is 3.14159... to infinity. Also according to Wikipedia, pi has been calculated to around 5,000,000,000,000 digits.

How can we easily calculate Earth's volume?

My favorite way is to use the meter. If you look it up, a meter is defined as the distance light travels, in a vacuum, in 0.00000000333564095 seconds.

While interesting (an precise), this definition is of no use to us. What was the meter originally defined as?

One meter was one ten-millionth of the distance from the North Pole to the Equator. So from the North Pole to the Equator was to measure 10,000,000 meters. We're already estimating. Even though the original definition and the new definition aren't the same, we know the values to be very close (close enough for our purposes).

This is one-fourth of the distance to make a complete circle around the world. This means that the distance around Earth (travelling through both the North and South Poles) is 40,000,000 meters. Here again is an estimation. We know that Earth is not a perfect sphere. For our calculations, though, it is very close.

So now we have Earth's circumference to be 40,000,000 meters. We want to get to volume, so we'll need the radius (or half the diameter). First, we'll use the formula below.

C = 2

πr

Circumference equals two times pi times the radius. Plugging in, we get this:

40,000,000 m =

πr

Simplifying results in:

(20,000,000 m)/

π = r

To make things easier, we'll use the fact that 1,000 m = 1 km. This gives us:

(20,000 km)/

π = r

We're going to estimate pi to be 3 to see how close we can get.

This gives us:

6,666.6666... km = r

We'll estimate this number a few different ways.

First, we'll try r = 7,000 km.

Second, we'll try r = 6,700 km.

Third, we'll try r = 6,670 km.

Finally, we'll try r = 6,667 km.

How do we calculate volume of a sphere?

We have the following formula:

V = (4/3)(

π)(r^3)

Since we're letting pi be 3, the formula simplifies a little.

V = 4(r^3)

So we'll cube our radius in each case.

(1) r^3 = 343,000,000,000 km^3

(2) r^3 = 300,763,000,000 km^3

(3) r^3 = 296,740,963,000 km^3

(4) r^3 = 296,340,742,963 km^3

To find the volume, just multiply each by 4.

(1) V = 1,372,000,000,000 km^3 (or 329,160,305,000 miles^3)

(2) V = 1,203,052,000,000 km^3 (or 288,627,524,000 miles^3)

(3) V = 1,186,963,852,000 km^3 (or 284,767,772,000 miles^3)

(4) V = 1,185,362,971,852 km^3 (or 284,383,700,000 miles^3)

Recall the original volume we found was 1,083,210,000,000 km^3 (or 259,875,899,000 miles^3).

Our estimates were off by between 102,152,971,852 and 288,790,000,000 km^3! That's enormous!

Or is it?

We can find out how "wrong" our guesses are by using the percent error formula.

Using our accepted value from Wikipedia, we obtain the following percent errors:

(1) % Error = 26.66%

(2) % Error = 11.06%

(3) % Error = 9.58%

(4) % Error = 9.43%

Our best estimate was off by less than 9.5%, while our worst was only off by about 27%. What if we'd used a more accurate value for pi? Using 3.14159 gives us radii of 6,000 km, 6,400 km, 6,370 km and 6,366 km. These values result in percent errors of:

(1) % Error = 16.47%

(2) % Error = 1.37%

(3) % Error = 0.05%

(4) % Error = 0.24%

When we used a radius of 6,370 km, we got our best approximation for the volume.

What's Earth's actual radius?

According to Wikipedia, it's 6,371 km, making our radius percent error 0.01%

In the end, our biggest error was letting pi equal three. Even our biggest estimation for the radius gave us a volume off by only about 16.5%. But the less we rounded, the better our guess (when pi was 3).

Is it okay to let pi equal three? Well, sometimes. If a bicycle wheel is two feet wide, the circumference is about 6 feet. In reality, it's closer to 6.28318 feet, but our percent error is 4.5%. In our worst guess about Earth's volume, we were about 74% right. If you want a rough estimate, then by all means use three. If you need a better answer, then use 3.14159.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

The image for percent error is from http://www.pstcc.edu.

Most of the information used in my calculation can be found at http://www.wikipedia.org.

When Does The Imaginary Become Real?

2011-08-05T07:30:00.002-04:00

The square root of 4 is 2 (or -2), and the square root of 100 is 10 (or -10). But what's the square root of -4? -100? To take the square root of negative numbers, we have to introduce the idea of √(-1) (Read: the square root of negative one). We call this i, the imaginary number.

The number i has a few interesting properties.

It cannot be combined with a real number like other real numbers. To combine i with the real number 2, you would get the result 2 + i. This is a complex number because it has both real and imaginary parts.

Its exponential powers cycle. So i to the first power is i, i to the second power (i squared) is -1, i to the third power (i cubed) is -i and i to the fourth power is 1. The pattern repeats to infinity (1-4 = 5-8 = 9-12, etc.).

Even though the imaginary number is difficult to consider in many applications, it is even possible to take its square root (this may be covered later).

One of the most interesting problems involving i that I've come across is i to the i-th power (i^i).

Two to the second power (2^2) is 4, and five to the fifth power (5^5) is 3,125, but what is i to the i-th power? Is it real, imaginary or complex? Can we even calculate it?

The answer, surprisingly, is yes, we can calculate i^i.

To do so, let's recall Euler's Formula from an earlier post.

e^(ix) = cos(x) + i sin(x)

We want to find a value for x so that we get the following:

e^(ix) = i

Since cos(x) is entirely real, we want that to equal 0.

Since sin(x) is going to be the imaginary coefficient, we want it to equal 1.

cos(x) = 0 whenever x = π/2, 3π/2, 5π/2, ...

sin(x) = 1 whenever x = π/2, 5π/2, 9π/2, ...

Any value of x = π/2, 5π/2, 9π/2, ... will work. Since π/2 is the easiest, we'll use it.

e^(iπ/2) = cos(π/2) + i sin(π/2)

e^(iπ/2) = 0 + (i)(1)

e^(iπ/2) = i

We can now rewrite i^i as [e^(iπ/2)]^i.

Recalling rules of exponents, whenever a power is raised to a power, the powers multiply.

Our two powers are iπ/2 and i. When these multiply, we get (i^2)(π/2).

As we said earlier, i^2 is -1. So our new exponent is (-π/2).

This makes our new equation:

i^i = e^(-π/2)

We no longer have an imaginary number in our exponent! We can easily calculate this and find:

e^(-π/2) = 0.207879576...

What if we'd used another value like 5π/2?

A quick calculation gives us:

e^(-5π/2) = 0.000388203204...

We got a different value! Let's try one more value. Even though -3π/2 is a negative number, it is a solution for e^(ix) = i.

The only difference is that the exponent will be positive, instead of negative, this time.

e^(3π/2) = 111.317778...

Three different values and three different answers. If we tried every value of x that solves e^(ix) = i, we'd get a different solution every time. So does this mean i^i has no solution?

In one of those peculiarities of mathematics, we've arrived at a multi-valued function. This means that i^i doesn't have one value; it has infinitely many values, and they're all positive real numbers.

In most cases, π/2 will be the value mentioned (since it is the principal value to make sin(x) = 1).
As such, the principal value for i^i usually given will be 0.207879576...

Just remember that this is one of many, many values.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I came across this mathy quirk while reading the book An Imaginary Tale: The Story of √-1. (ISBN: 978-0-691-14600-3).

How Are The Most Important Numbers Related?

2011-08-04T15:00:00.001-04:00

To get started, what are the most important numbers in math?

To be sure, there are quite a few. For the purposes of this problem, we'll focus on five numbers. The first three should be familiar to you, but the other two may not.

The first number is 0.

The second number is 1.

These should be familiar to pretty much everyone.

The third number is π (pi) (3.14159..., or about 3.14).

Pi is the ratio of a circle's circumference to its diameter. Pi should be familiar to most people who took math in elementary and junior high school.

The fourth number is e (2.71828...).

The number e is found most often in Algebra II-level and above math classes. It pops up in exponential and logarithmic problems, and is frequently used to calculate interest and growth on a bank account or loan.

The fifth and final number is i, the imaginary number (or the square root of negative one).

The number i, like e, often appears beginning in Algebra II curricula.

Other numbers are also important, such as 2, the square root of two and phi (the golden ratio), but we're going to focus on these five.

So how are they related?

To figure that out, we'll start with the two which are least familiar to most people, e and i.

Heads up, the next section can get REALLY mathy. I'll be going through a derivation. If you don't care to know it (though you should) or if you get lost, you can skip ahead and just accept the equation marked (1) as fact.

When e is used in exponential functions, it is usually seen as f(x) = e^x.

Using calculus, you can create a Taylor series for e^x. Don't worry if this is new to you. All you need to know is that a Taylor series basically takes a complicated function (like e^x) and makes it into a polynomial (like x x^2 + x^3...) that can go on forever.

The Taylor series for e^x is e^x = 1 + x + (x^2/2) + (x^3/6) + (x^4/24) + (x^5/120) + ...

If you're familiar with trigonometry, you probably know the words sine (sin) and cosine (cos). These are trigonometric functions that also have Taylor series.

The Taylor series for sin(x) is sin(x) = x - (x^3/6) + (x^5/120) - ... (notice the signs alternate).

The Taylor series for cos(x) is cos(x) = 1 - (x^2/2) + (x^4/24) - ... (notice the signs alternate).

If you're noticing a pattern, you're among famous company. Leonard Euler (and many other famous mathematicians) noticed it as well. Could there be a way to make e^x equal sin(x) and cos(x)?

This is where the imaginary number comes in. What if we had e^(ix) instead of e^x?

Our Taylor series would be e^(ix) = 1 + ix + ((ix)^2/2) + ((ix)^3/6) + ((ix)^4/24) + ((ix)^5/120) + ...

The powers of i repeat, so i to the first is i, i to the second is -1, i to the third is -i and i to the fourth is 1.

This also means i to fifth is still i, and i to the sixth is still -1.

So we can simplify to get e^(ix) = 1 + ix + -(x^2/2) + -i(x^3/6) + (x^4/24) + i(x^5/120) + ...

By rearranging the terms, we see that e^(ix) = [1 - (x^2/2) + (x^4/24) - ...] + [ix - i(x^3/6) + i(x^5/120) - ...].

The left brackets should look familiar. That's the Taylor series for cos(x)!

So now we have e^(ix) = cos(x) + [ix - i(x^3/6) + i(x^5/120) - ...].

Everything in the right side has an i, so we can factor that out.

Now we have e^(ix) = cos(x) + i[x - (x^3/6) + (x^5/120) - ...].

Now the right brackets are exactly sin(x)!

Our final equation is:

(1) e^(ix) = cos(x) + i sin(x) (1)

This is known as Euler's Formula, and it will be very useful in later discussions.

If that seems like a lot of math, it is. But that was the hard part. The rest is smooth sailing.

Now we introduce the third number, π. If we let x=π, we get the following equation:

e^(iπ) = cos(π) + i sin(π).

Pi is a radian measure (π radians = 180 degrees). If we know (or use a calculator to find) that cos(π) (or cos(180 degrees)) is -1 and that sin(π) (or sin(180 degrees)) is 0, our equation simplifies.

e(^iπ) = -1 + (i)(0)

Since we know anything times 0 is 0, we end up with the following equation.

e(^iπ) = -1

Now we want to introduce the second number, 1, by adding it to both sides of the equation.

e^(iπ) + 1 = -1 + 1

Our last step is to simplify the right hand side.

In our final step, the first number, 0, fell out of the equation!

So there it is, the relationship between the most important numbers. Some have called this equation the most beautiful and elegant in all of mathematics, and, so far, I agree. This one is my favorite, and I hope it is now one of yours.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I first came across this fact during my junior year of high school in my math team class. I later saw explanations in my senior year calculus class, and I saw the explanation given here in my freshman year of college in 18.01A Single Variable Calculus.

The image of the final step was found at http://www.informath.org.

How Many Degrees Are In A Triangle?

2011-08-03T15:00:00.000-04:00

In other words, what is the sum of the interior angles of a triangle?

After thinking way back to geometry (further for some than for others), we can recall (or look up) that there are 180 degrees in any triangle. This idea is the basis of many proofs high school students attempt in their mathematical careers, so many are familiar with it.

But is it a lie?

If you're most people (and most of you are), then you probably deal almost exclusively with Euclidean geometry. You can look it up on Wikipedia for a thorough description, but for our purposes it basically means that your shapes are drawn on flat, smooth surfaces.

So when we look at triangles in Euclidean geometry, all of the interior angles of a triangle add up to 180 degrees, just as we were taught.

But what if the triangle is drawn on a surface that isn't flat? And where on Earth would I find such a triangle?

There's the beauty. You find that kind of triangle everywhere on Earth.

Take a look at this globe.

Let's imagine drawing a triangle. Side A runs along the equator. Side B runs from the North Pole to the equator (at 20E, or 20 degrees to the right). Side C runs from the North Pole to the equator (at 40E, or 40 degrees to the right).

Now that we know our three sides, let's get the three angles.

The angle between Side B and Side C, Angle A, is 20 degrees (the difference between 40 and 20 is 20).

Now for Angle B and Angle C. Side B and Side C are lines of longitude and Side A is a line of latitude. Lines of latitude and lines of longitude are perpendicular (another geometry word). This means they make right angles (or angles of 90 degrees).

Because of this, we know the angle between Side A and Side B, Angle C, is 90 degrees.

We also know the angle between Side A and Side C, Angle B, is 90 degrees.

So to recap:

Angle A = 20 degrees

Angle B = 90 degrees

Angle C = 90 degrees

Add them up and we get 200 degrees! These triangles exist in what is referred to as elliptical geometry, not Euclidean.

A slightly more complicated example shows a triangle with less than 180 degrees in what is known as hyperbolic geometry (you can read more on that on your own).

So there we have it. Sometimes a triangle has 180 degrees, but sometimes it doesn't. And you don't have to look beyond Earth to find one that doesn't.

Anyway, I hope you got something out of this. If not, well, there's always next time.

If you have any questions, or if anything isn't clear, just let me know in the comments.

I stumbled across this in the book The Big Questions: Mathematics. (ISBN: 978-1-4351-1131-8).

The globe image was found on http://kaffee.50webs.com/Science/index.html.

Welcome!

2011-08-03T13:27:00.001-04:00

Welcome! This blog is my attempt to gather resources for educators. These resources, as one may expect, will be mathy. Now, that is not to say every post will deal with math. Every post will deal with something mathy. Don't expect many posts (or any for that matter) to be deeply profound mathematical statements. Most of what I will post here will be interesting tid bits or ideas I come across (ideally those that may be discussed in a high school classroom). I will try to frame the title of each entry as a question (when possible) and dedicate the body of each post to elaborate upon the problem (if needed) and how to arrive at an answer. This is mostly for me (a nice place to conveniently gather mathy things I'd find valuable to teach). However, anyone who stumbles across this information depository is more than welcome to use, reference or link to it. I will cite my sources for clarity (URLs for websites and ISBNs for books). With that being said, welcome to this blog, and I hope you get something out of it.