Here's an interesting mental math challenge. Given two different positive real numbers, which we'll call a and b, which is greater, a^b or b^a? If you're able to calculate both exponents mentally, such as 2³ vs 3², then that's probably the simplest way to go. What happens if one or both exponential expressions are too hard to mentally calculate?

To solve this, we'll need to find a general rule. Some advanced calculation will be required to find it. However, once we do the work to find the rule, you'll see that almost no math will be needed to solve these types of problems!

SEARCH FOR THE RULE: Consider that 2³ < 3², but 3⁴ > 4³. Just looking at the placement of smaller and bigger numbers, there doesn't seem to hard and fast rule which applies. The best place to start is just by assuming what we wish to discover:



From that point, let's see if we can separate a and b somehow. The quickest way is to raise both sides to the power of 1/ab:



In English, then, when the ath root of a is greater than the bth root of b, then a^b > b^a. Don't worry, though. You won't have to do any roots in your head. Instead, just look at this Desmos graph where y=xth root of x. If you click in the box marked 1 at the upper left, Desmos highlights two points, the point with the minimum y value, which is (0, 0), and the point with the maximum value, (2.718, 1.445).

Wait a minute! The value of 2.718 sounds familiar. Could the maximum value of xth root of x really be e? Sure enough, Wolfram|Alpha verifies that the maximum is e!

Look at both sides of the graph, then. From 0 to e, the graph increases. From e on upwards, the value of the xth root of x will steadily decrease. In fact, the value will just keep getting closer and closer to 1.

This means we've found the start of our rule. When you have two numbers, both of which are equal to or greater than e, the smaller number x will always yield a greater xth root of x than the higher number. This tells us that the ath root of a is greater than the bth root of b when a is the smaller number. Working backwards to our original assumption, this means that:

When a and b are both equal to or greater than e, a^b is always greater than b^a when a is the smaller number and b is the larger number.

There's our rule!

WORKING WITH INTEGERS: What about when one or both of a and b are less than e? In the case of positive integers, this means we only have to consider the cases involving 1 and 2.

The case of 1 is easy. Assume that a equals 1 and b has a value of 2 or more. 1^b will always be 1, and b¹ will always be b, which is 2 or more. That covers every case involving 1.

What about 2? Let's work through each case individually, always assigning a to a value of 2, and b a value other than 2. If b is 1, we just covered that case. b can't be 2, as it would be the same as a. What about when b is 3? We already covered the fact that 2³ < 3², so we remember that special case. What happens when b is 4? This is a very unique case, as 2⁴ is exactly equal to 4²! In fact, this is a well-known special case, as it's the only time when a^b equals b^a when a and b are different integers. When b is 5 ore greater, as you can see in this graph, we can fall back on the rule we set above.

A CLASSIC CHALLENGE: Which is greater, π^e or e^π? There's a problem you probably never thought you'd solve in your head!

Let's go see if our primary rule applies. Are a and b both equal to or greater than e? e, of course, is exactly equal to e (2.71828 and so on). π, as any Grey Matters readers should already know, is roughly 3.1415, and therefore also greater than e.

This means we can apply our primary rule! e is smaller than π, which means that e^π must be greater than π^e. A quick verification with Wolfram|Alpha shows that this is correct.

About 4 years ago, Presh Talwalkar wrote up several approaches to this classic problem. Interestingly, this approach wasn't included. The 4th method does work with the xth root of x as we did, but it uses a deeper approach involving calculus.

For this post, I'd like to turn to a variation of a classic Henry Dudeney puzzle, from his book Amusements in Mathematics. It can also be found in Martin Gardner's October 1960 Scientific American column, or his book, New Mathematical Diversions, as the 5th puzzle ,“Bisecting Yin and Yang”, in chapter 12, “Nine Problems”.

As you've no doubt guessed, this puzzle involved the yin (dark) yang (light) symbol. For this puzzle, I've drawn it in a very mathematically precise way over at Desmos. The outside is a unit circle (so, the radius is 1 unit), the main semicircular divisions of the design have a radius of ¹⁄₂ unit, and the opposite-color dots have a radius of ¹⁄₆ unit. Here's the challenge: What's the equation of the line that divides the design so that each side of the line contains exactly equal amounts of the dark and light portions?



As with many puzzles, this one seems hard, until you break it down into simpler steps. Let's start with a much easier puzzle: If the top half of the puzzle were dark, and the bottom half light, as in this rendering, where would you draw the line? The answer is easy. It should be a vertical line, so the equation would be x=0.



Next, we change the design a bit, so as to get closer to the yin yang symbol. Starting with the previous design, we cut a half circle (¹⁄₂ unit radius, remember?) of the dark portion from the right side, and add that to the left side, giving us the design below. The vertical line obviously won't work anymore, and we'll need to rotate that line by some amount to compensate, but how much?



Again, the secret is to take steps slowly. If you remember your high school geometry, you remember that the formula for a circle is A=πr², and that our design as a whole, being a unit circle, has an area equal to π.

The formula for the semicircle we've moved, then, is A=¹⁄₂πr². Plugging in ¹⁄₂ for the radius, we get ^π⁄₈ units. So, to compensate for ^π⁄₈ units out of a full circle with an area of π-units, we simply rotate the formerly-vertical line counterclockwise by ¹⁄₈ of the circle, or 45°! The upper left quadrant completely dark, so that makes this adjustment simple. The blue line is the dividing line for this design:



This, in fact, is the answer to the original puzzle as posed by Dudeney and Gardner. This is NOT the answer to the problem I posed above. When I first ran across this puzzle, it annoyed me that it wasn't done with the full yin yang symbol. The dots are a symbol of how, in nature, nothing is purely one thing or the other, and are a very important part of the design.

It's time to go back to the full design. Compared to the previous step, we're going to be removing some of the dark area from the right side to the left side. This means that we'll end up rotating our dividing line some distance clockwise this time, and we need to figure out by how much. Yes, once again, we'll be using our area formulas to work out the adjustment. We even know that the result should be easy to interpret, since the result will π over something, and this comes out of a π-unit circle.



The dots, of course, are full circles, so we use the formula for the area of a full circle once again. The dots have a radius of ¹⁄₆ of a unit, and plugging that into the formula, we get ^π⁄₃₆. In other words, the line needs to be moved back clockwise ¹⁄₃₆ of a full circle, or 10°. That brings the line to being 35° off of the original vertical line.

A 35° line must be our answer, right? Wrong. Go back and look at the original question: Here's the challenge: What's the equation of the line that divides the design so that each side of the line contains exactly equal amounts of the dark and light portions? We need to work out enough details for the line equation y = mx + b, where m is the slope, b is where the line crosses the y-axis, and y and x remain as variables. The line, of course, crosses the y-axis at 0, so b = 0. That reduces the equation to just y = mx, so we need to figure out the slope.

First, angles are usually measured in relation to the positive x axis, so we're actually talking about a 125° (35° + 90°), or ^25π⁄₃₆ radians (Confused? Read Intuitive Guide to Angles, Degrees and Radians). In geometry, we'd say we were trying to calculate rise over run (rise ÷ run). In trigonometry, we're trying to calculate the opposite site of the angle by the adjacent side (Confused? Read How To Learn Trigonometry Intuitively), and that means we need to use the tangent!

So, the equation is y = tan(^25π⁄₃₆)x, or y = tan(125°)x, if you prefer. The actual slope is an irrational number which is roughly equal to -1.428148. There you have it, the equation to a line which divides the design into equal parts of light and dark, as shown below.

Ever since I started this blog, I've been waiting for this day. I started Grey Matters on 3/14/05, specifically with the goal of having its 10th blogiversary on the ultimate Pi Day: 3/14/15!

Yes, it's also Einstein's birthday, but since it's a special blogiversary for me, this post will be all about my favorite posts from over the past 10 years. Quick side note: This also happens to be my 1,000th published post on the Grey Matters blog!

Keep in mind that the web is always changing, so if you go back and find a link that no longer works, you might be able to find it by either searching for a new place, or at least copying the link and finding whether it's archived over at The Wayback Machine.

2005

My most read posts in 2005 were 25 Years of Rubik's Cube (at #2), and Free Software for Memory Training (at #1). It was here I started to get an idea of what people would want from a blog about memory feats.

2006

In the first full January to December year of Grey Matters, reviews seemed to be the big thing. My reviews of Mathematical Wizardry, Secrets of Mental Math, and Mind Performance Hacks all grabbed the top spots.

2007

This year, I began connecting my posts with the interest of the reader, and it worked well. My series of “Visualizing” posts, Visualizing Pi, Visualizing Math, and Visualizing Scale were the biggest collectively-read posts of the year.

Fun and free mental improvement posts also proved popular in 2007. Unusual Lists to Memorize, my introduction to The Prisoner's Dilemma, and my look at Calculators: Past, Present, and Future (consider Wolfram|Alpha was still 2 years away) were well received! 10 Online Memory Tools...For Free! back-to-back with my Memorizing Poetry post also caught plenty of attention.

2008

I gave an extra nod to Pi this year, on the day when Grey Matters turned Pi years old on May 5th. The most popular feature of the year was my regularly update list of How Many Xs Can You Name in Y Minutes? quizzes, which I had to stop updating.

Lists did seem to be the big thing that year, with free flashcard programs, memorizing the elements, and tools for memorizing playing card decks grabbed much of the attention in 2008.

2009

Techniques took precedence over lists this year, although my series on memorizing the amendments of the US Constitution (Part I, Part II, Part III) was still popular. My web app for memorizing poetry, Verbatim, first appeared (it's since been updated). Among other techniques that caught many eyes were memorizing basic blackjack strategy, the Gilbreath Principle, and Mental Division with Decimal Precision.

2010

This year opened with the sad news of the passing of Kim Peek, the original inspiration for the movie Rain Main. On a more positive note, my posts about the game Nim, which developed into a longer running series than even I expected, started its run.

As a matter of fact, magic tricks, such as Bob Hummer's 3-Object Divination, and puzzles, such as the 15 Puzzle and Instant Insanity, were the hot posts this year.

Besides Kim Peek, 2010 also saw the passing of Martin Gardner and Benoît Mandelbrot, both giants in mathematics.

2011

The current design you see didn't make its first appearance until 2011. Not only was the blog itself redesigned, the current structure, with Mental Gym, the Presentation section, the Videos section, and the Grey Matters Store, was added. This seemed to be a smart move, as Grey Matters begin to attract more people than ever before.

The new additions to each section that year drew plenty of attention, but the blog has its own moments, as well. My list of 7 Online Puzzle Sites, my update to the Verbatim web app, and the Wolfram|Alpha Trick and Wolfram|Alpha Factorial Trick proved most popular in 2011.

My own personal favorite series of posts in 2011, however, was the Iteration, Feedback, and Change series of posts: Artificial Life, Real Life, Prisoner's Dilemma, Fractals, and Chaos Theory. These posts really gave me the chance to think about an analyze some of the disparate concepts I'd learned over the years when dealing with various math concepts.

2012

In 2012, I developed somewhat of a fascination with Wolfram|Alpha, as its features and strength really began to develop. I kicked the year off with a devilish 15-style calendar puzzle, which requires knowing both how to solve the 15 puzzle and how to work out the day of the week for any date in your head! Yeah, I'm mean like that. I did, however, release Day One, my own original approach to simplifying the day of the week for any date feat.

Estimating Square Roots, along with the associated tips and tricks was the big feat that year. The bizarre combination of controversy over a claim in a Scam School episode about a 2-card bet and my approach to hiding short messages in an equation and Robert Neale's genius were also widely read.

2013

After we lost Neil Armstrong in 2012, I was inspired to add the new Moon Phase For Any Date tutorial to the Mental Gym. A completely different type of nostalgia, though, drove me to post about how to program mazes. Admittedly, this was a weird way to kick off 2013.

Posts about the Last Digit Trick, John Conway's Rational Tangles, and Mel Stover were the first half of 2013's biggest hits on Grey Matters.

I also took the unusual approach of teaching Grey Matters readers certain math shortcuts without initially revealing WHY I was teaching these shortcuts. First, I taught a weird way of multiplying by 63, then a weird way of multiplying by 72, finally revealing the mystery skill in the 3rd part of the series.

2014

Memory posts were still around, but mental math posts began taking over in 2014. A card trick classically known as Mutus Nomen Dedit Cocis proved to have several fans. The math posts on exponents, the nature of the Mandelbrot set, and the Soma cube were the stars of 2014. Together, the posts Calculate Powers of e In Your Head! and Calculate Powers of π In Your Head! also grabbed plenty of attention.

Wrap-up

With 999 posts before this one, this barely even scratches the surface of what's available at this blog, so if you'd made it this far, I encourage you to explore on your own. If you find some of your own favorites, I'd love to hear what you enjoyed at this blog over the years in the comments below!

Earlier this year, I posted about calculating powers of e in your head, as well as powers of Pi.

This time around, I thought I'd pass on a method for calculating powers of a much more humble number: 2. It sounds difficult, but it's much easier than you may think!

BASICS: For 2⁰ up to 2¹⁰, you'll memorize precise answers. For answers to 2¹¹ and higher integer powers, you'll be estimating the numbers in a simple way that comes very close.

First, you must memorize the powers of 2 from 0 to 10 by heart. Here they are, along with some simply ways to memorize each of them:

Problem   Answer    Notes 
  20    =     1     Anything to the 0th power is 1
  21    =     2     Anything to the 1st power is itself
  22    =     4     22 = 2 × 2 = 2 + 2
  23    =     8     3 looks like the right half of an 8
  24    =    16     24 = 42
  25    =    32     5 = 3 + 2
  26    =    64     26 begins with a 6
  27    =   128     26 × 21
  28    =   256     Important in computers
  29    =   512     28 × 21
  210   =  1024     210 begins with a 10

Take a close look at 2¹⁰, which is 1024. It's very close to 1,000, so we're going to take advantage of the fact that 2¹⁰ ≈ 10³!

When multiplying 2^x × 2^y, remember that you simply add the exponents together. For example, 2³ (8) × 2⁷ (128) = 2^{7 + 3} = 2¹⁰ (1024). Similarly, you can break up a single power of 2 into two powers which add up to the original power, such as 2⁹ (512) = 2^{6 + 3} = 2⁶ (64) × 2³ (8).

TECHNIQUE: We'll start with 2¹⁵ as an example.

Start by breaking up the given power of 2 into the largest multiple of 10 which is equal to or less than the given power, and multiply it by whatever amount is leftover, which will be a number from 0 to 9.

Using this step, 2¹⁵ becomes 2^{5 + 10}, which becomes the problem 2⁵ × 2¹⁰.

For an powers from 0 to 9, you should know by heart, so you can convert these almost instantly. In the example problem we've been doing, we know that 2⁵ is 32, so the problem is now 32 × 2¹⁰.

Now we deal with the multiple of 10. For every multiple of 10 involved, you can replace 2¹⁰ with 10³. With our problem which is now 32 × 2¹⁰, there's only a single multiple of 10 in the power, so we can replace that with 10³. This turns our current problem into 32 × 10³.

At this point, it's best to represent the number in scientific notation. In this feat, that simply refers to moving the decimal point to the left, so that the left number is between 0 and 10, and then adding 1 to the power of 10 for each space you moved the decimal. Converting to scientific notation, 32 × 10³ becomes 3.2 × 10⁴.

That's all there is to getting our approximation!

How close did we come? 2¹⁵ = 32,768, while 3.2 × 10⁴ = 32,000. I'd say that's pretty good for a mental estimate!

EXAMPLES: Over 6 years ago, I related the story of Dr. Solomon Golomb. While in college, he took a freshman biology class. The teacher was explaining that human DNA has 24 chromosomes (as was believed at the time), so the number of possible cells was 2²⁴. The instructor jokingly added that everyone in the class knew what number that was. Dr. Golomb immediate gave the exact right answer.

Can you estimate Dr. Golomb's answer? Let's work through the above process with 2²⁴.

First, we break the problem up, so 2²⁴ = 2^{4 + 20} = 2⁴ × 2²⁰.

Next, replace the smaller side with an exact amount. In this step, 2⁴ × 2²⁰ becomes 16 × 2²⁰.

Replace 2^10x with 10^3x, which turns 16 × 2²⁰ into 16 × 10⁶.

Finally, adjust into scientific notation, so 16 × 10⁶ becomes 1.6 × 10⁷.

If you know your scientific notation, that means your estimated answer is 16 million. Dr. Golomb, as it happened, had memorized the 1st through 10th powers of all the integers from 1 to 10, and new that 2²⁴ was the same as 8⁸, so he was able to give the exact answer off the top of his head: 16,777,216. 16 million is a pretty good estimate, isn't it?

Below is the classic Legend of the Chessboard, which emphasizes the powers of 2. In the video, the first square has one (2⁰) grain of wheat placed on it, the second square has 2 (2¹) grains of wheat on it, with each square doubling the previous number of grains.



The 64th square, then, would have 2⁶³ grains of wheat on it. About how many is that? I'm going to run through the process a little quicker this time.

Step 1: 2⁶³ = 2^{3 + 60} = 2³ × 2⁶⁰

Step 2: 2³ × 2⁶⁰ = 8 × 2⁶⁰

Step 3: 8 × 2⁶⁰ ≈ 8 × 10¹⁸

While 2⁶³ is 9,223,372,036,854,775,808, our estimate of 8,000,000,000,000,000,000 works.

TIPS: If you're really worried about the error, there is a way to improve your estimate. Percentage-wise, the difference between 1,000 (10³) and 1,024 (2¹⁰) is only 2.4%. So, for every multiple of 10 to which you take the power of 2 (or every power of 3 to which you take 10), you can multiply that by 2.4% to get a percentage difference. You can then multiply that percentage difference by your estimate to improve it.

Just above, we converted 2⁶³ into 8 × 10¹⁸. Since we started with six 10s, our percentage difference would be 6 × 2.4%, or 14.4%. In other words, our estimate of 8 × 10¹⁸ could be made closer by adding 14.4% to 8.

Assuming your comfortable with doing percentages like this in your head, 8 increased by 14.4% is 8 + 1.152 = 9.152, so our improved estimate would be 9.152 × 10¹⁸. Considering the actual answer is roughly 9.223 × 10¹⁸, that's quite close!

Practice this, and you'll have an impressive skill with which to impress family, friends, and computer geeks!

This week's post is a day early because today is Tau Day!

Tau, for those not familiar with it, is the mathematical constant equal to 2 × π (Pi), or roughly 6.28, and is represent by the greek letter Tau: τ. Today (6/28), we're going to look at the internet battle that's erupted over π versus τ since 2010.

The opening salvos in this year's battle were already launched back on Pi Day by DNews, with their Is Tau Better Than Pi? video:



3 days ago, Scientific American continued with their post, “Why Tau Trumps Pi”. Just 1 day later, prooffreader.com jumped in with Pi vs. tau: Ultimate Smackdown.

Starting at about 2:40 into the DNews video above, they come very close to a good answer. Yes, geometry and trigonometry rely heavily on 2π, and in those cases τ makes more sense, especially when it comes to concepts like τ radians in a circle. However, π has plenty of uses beyond those subjects on its own.

I'm in favor of adopting τ as a commonly-used constant, but not as a replacement for π. Talking about τ versus π to me is like getting in a heated argument over degrees versus radians. Which one is better? The answer can be degrees OR radians, depending on the context of the problem at hand. Should we use base 10, base 2, or base e in logarithms and exponents? Again, the answer depends on the context.

Kalid Azad of BetterExplained.com has 3 questions that can make even the toughest math concepts understandable: What relationship does this model represent? What real-world items share this relationship? Does that relationship make sense to me? In fact, as James Sedgwick points out in his essay, “The Meaning of Life”, you only have meaning if you have a relationship set in a context.

Yes, the endless internet battles over Pi versus Tau can be fun, but when it comes down to the important aspects, I believe we should focus on solving the problem at hand, and using the most effective tools. Besides, if we only keep one or the other, that's one less geeky holiday to celebrate in the year.

Happy Tau Day!

Today, I'm going to take you on a journey. It's a sort of imaginary journey, but not in the way that you may think.

This is a mathematical journey, and if you can grasp where we start, I think you'll enjoy where we wind up.

There are two important concepts with which you'll need to be familiar before we begin. The first is the idea of imaginary and complex numbers, which are explained intuitively in A Visual, Intuitive Guide to Imaginary Numbers. The other concept is the number e, as explained in An Intuitive Guide To Exponential Functions & e. Take your time, if needed, to understand these concepts, and the rest of this post will be well worth the time spent.

We'll start simply, with a right triangle that's half as high as it is long (via Desmos.com). The coordinates of the vertices in Cartesian coordinates are (0,0), (1,0), and (1, 0.5). Converting these to complex numbers, they respectively become as 0 (0 + 0i), 1 (1 + 0i), and 1 + 0.5i. As you can see, converting between complex numbers and Cartesian coordinates requires little effort.

Just for fun, let's see what happens if we take this 1 + 0.5i right triangle, and take it the integer powers from 1 to 10. Wolfram|Alpha calculates the answers to (1 + 0.5i)ⁿ quickly, for values of n from 1 to 10, but the answers are just complex numbers, with no real apparent meaning.

Desmos.com can't handle complex numbers, but can handle tables of Cartesian coordinates, so if we use those answers, and draw the corresponding triangles, we get a very visual interpretation of these complex numbers!



What we get is a series of larger and larger right triangles! Think about why this is, using our old friend, the Pythagorean theorem. The horizontal leg is 1 unit, the vertical leg is 0.5 units, so the Pythagorean theorem tells us that the hypotenuse of that first triangle is about 1.11803 units long. That's also the base of the next right triangle.

The distance formula can be used to work out the height of the new triangle. It starts at (1, 0.5) and ends at (0.75, 1), so we plug the numbers in to get d = √(0.75 - 1)² + (1 - 0.5)² = √(-0.25)² + (0.5)² = √0.0625 + 0.25 = √0.3125 ≈ 0.559017.

Notice that, if you double 0.559017, you get 1.11803. That's not a coincidence. What you're getting is a series of triangles with progressively longer legs, and whose heights will always be half of that length. In short, (1 + 0.5i)ⁿ builds n right triangles with a 1:0.5 (or 2:1) ratio for the legs, and hypotenuses who lengths grow in accordance with the Pythagorean theorem!

Now, the formula (1 + 0.5i)ⁿ looks a little like the part of the formula for e. What would happen if we changed it to (1 + ^0.5i⁄_n)ⁿ? Let's look at this step by step.

With just 1 power, and therefore just 1 triangle, the formula gives us 1 + 0.5i, not surprisingly. This gives us our original right triangle.

Next, let's change the power to 2, and step through the formula twice to get 2 triangles. Wolfram|Alpha returns 1+0.25i and 0.9375+0.5i. We still get 2 triangles, but now they occupy about the same height as the previous single triangle!



Sure enough, if we take 3 steps through the 3rd power, we get 3 right triangles with about the same total height as the others.



This should make sense, especially as this approach was inspired by e. Remember how the change in e gets smaller and smaller as bigger numbers are used for n? Just like e packs smaller and smaller numbers in the same numeric space, our right triangles are packing more and more triangles in the same space on the graph!

Yes, if we take this approach to the 10th power and take 10 steps through it, we get 10 right triangles packed into the same space.



Let's take a closer comparison of this formula to e. First, the actual definition of e is:



We didn't take that limit into consideration in our triangle demonstration. What happens as our number of triangles get closer and closer to infinity? There will be more and more, so the height will get closer and closer to 0 units. If the height gets closer to 0, the Pythagorean theorem tells use that the hypotenuse will get closer and closer to the length of the original leg, which is 1 unit.

If you can picture infinitely many triangles packed into that same space, you can see that it would almost be like the lengths of the long legs would all be 1 unit, so it would effectively be a perfect arc!

The more general definition of e defines what happens when e is taken to a power:



So, if we apply that limit to our original complex equation, (1 + ^0.5i⁄_n)ⁿ, that means the infinitely many right triangles yields the same result as e^0.5i! Remember that taking our formula to the 10th power ended in 0.888809+0.485079i, so you shouldn't be surprised to see that e^0.5i roughly equals 0.877583 + 0.479426i.

Wolfram|Alpha's Arg[] command takes a complex number and returns the equivalent in radians. Entering Arg[0.877583 + 0.479426i] gives us the rather interesting result of 0.5 radians!

Let's think about this. Plotting out an infinite number of triangles of the form (1 + ^0.5i⁄_n)ⁿ, the equivalent of e^0.5i, results in an arc that's 0.5 radians long.

Yep, we're actually getting visual and mathematical proof that e^ix will result in an arc that's x radians long! If you understand trigonometry, this means you can use sine and cosine to work out the same point calculated by e^ix, which is exactly what Euler's formula says!

Yep, since Pi radians is half a circle, then our formula becomes e^iπ, which is Euler's famous identity! Since the cosine of π = -1 and the sine of π = 0, it works out -1 + 0i, or simply -1.

Did you ever think that playing around with a few triangles would ever lead you to an understanding of e^iπ = -1?

As a follow-up to last week's tutorial on calculating powers of e in your head, I'm going to teach you how to do the same for our old friend Pi!

As an added bonus, calculating powers of Pi can be slightly easier than powers of e, so even if you passed up last week's tutorial, you should still give this one a look.

BASICS: In a manner similar to the previous tutorial, we'll request a number x, and solve for y in the equation π^x = 10^y.

This method, just like the previous one, is also based on turning this problem into a logarithm. It becomes log₁₀(π^x) = y, which simplifies to x × log₁₀(π) = y. This works out to about x × 0.49714987... = y.

This time around, we can take advantage of the fact that this number is close to 0.5!

TECHNIQUE: We'll use π¹⁶ as our example requested power.

The first step this time involves setting up a subtraction problem with 2 numbers, both of which start with the given number (16, in our example).

BUILDING THE SUBTRACTION PROBLEM: To begin, take the given number an multiply it by 500. This can be made simpler, if you prefer, by simply tacking “,000” on the end of your number, and then dividing by 2. This is the first number we need.

Applying this step to our example number 16, we add a comma and 3 zeroes to it (16,000), and then divide by 2 to get 8,000. We have our first number.

Note: If you're given an odd number, you will always end up with a “,500” at the end. For example, if the given number was 15, this step would result in 7,500. Knowing this is a handy way to make sure you didn't misplace the comma.

To get the second number for the subtraction problem, simply multiply the given number by 3. This should be easy enough to do without any special tips.

Doing this, our second number is 16 × 3 = 48.

SUBTRACTION:Having set up the 2 numbers for the subtraction problem the next step, not surprisingly, is to perform the subtraction by subtracting the smaller number from the larger number.

We've worked out the numbers 8,000 and 48 in our 16 example, so the subtraction problem is 8,000 - 48.

If you're like most people, though, you remember writing down subtraction problems with lots of zeroes in school, and having to borrow over multiple places. That being the case, you're probably wondering how to deal with all this in your head! The following video teaches you how to deal with problems like these without ANY borrowing:



To work out 8,000 - 48 using the above technique, it's probably better if you think of the problem as 8,000 - 048. The first step, as in the video, is to round the leftmost digit up, from 0 to 1 in this case, and seeing that 80 - 1 = 79. We already know the answer begins with 79!

How far up would you have to go from 48 cents to get to a whole dollar? Getting the answer of 52 cents shouldn't be a problem here. That's the other half of the answer.

Your running total, at this point, is 7,952. After a little practice, subtracting from zeroes in your head will seem not only less scary, but nearly effortless.

ADJUSTING FOR APPROXIMATION: We're going to add a little now to improve the accuracy of our answer. How do we do that?

Take the number you just subtracted, and throw away the ones digit. Divide the remaining digits by 2. If that ends in a .5, just throw the .5 away, as well. This is the number you add to your running total.

We just subtracted 48 to get 7,952. We take 48 and throw away the ones digit, leaving 4. Dividing that by 2, we get 2. Finally, we add 2 to 7,952 to get 7,954 as our new running total.

DIVIDING BY 1,000: To divide by 1,000, I could tell you to move the decimal point three places to the left, but there's an even simpler technique this time. All you have to do is replace the comma in the total with a decimal point!

With this approach, 7,954 instantly becomes 7.954 with very little effort.

At this point, you're done! As you can verify on Wolfram|Alpha, π¹⁶ ≈ 10^7.954.

THE FULL PROCESS ALL AT ONCE: To run through this at once, and to better acquaint you with the full range of situations you'll run across, let's try to work out π³³ = 10^y.

• Multiply 33 × 1,000 to get 33,000, and divide by 2, getting 16,500.
• Multiply 33 × 3 to get 99.
• Subtract 16,500 - 99 = 16,401.
• Throw away the 9 (the ones place of 99), leaving 9, and divide that by 2 (4.5), throwing away the .5 to leave 4.
• Add 16,401 + 4 = 16,405.
• Replace the comma with a decimal point, resulting in 16.405.

Once again, check for yourself on Wolfram|Alpha to see that π³³ ≈ 10^16.405

TIPS: Most of the tips I gave for e apply for π, as well. I'll repost the relevant ones below for convenience, modified for our Pi examples.

• By looking at the whole number part of the answer (the significand) and adding 1 to that, you can state the number of digits the full answer would have. In our 7.954 example, we take the whole number part, the 7, and add 1 to get 8, so we can state that the answer is a 8-digit number. Having worked out π³³ to be about 10^16.405, we can state that the answer is a 17-digit (16 + 1) number!

• You can handle exponents with a decimal in them by working them out as if they were a whole number, and then adjusting for an appropriate number of additional decimal points when you're moving the decimal point. For example, π^1.6 is the same as π¹⁶, but with the decimal moved one place more to the left. Since π¹⁶ ≈ 10^7.954, it's easy to see that π^1.6 ≈ 10^0.7954.

• If you want to take this a step further, and be able to say that π¹⁶ is roughly equal to 9 × 10⁷, check out Nerd Paradise's Calculating Base 10 Logarithms in Your Head, the video Calculating logarithms in your head, and the PDF How to Quickly Calculate Logarithms to Three Decimal Places in Your Head.

135 years ago today, Albert Einstein was born. 9 years ago today, Grey Matters was born. Of course, today will always be Pi Day!

Welcome to a geeky, yet still special, day of the year that is close to my heart!

Vi Hart kicked things off with an anti-Pi rant. This might seem like a strange action to take on Pi Day, but she does it in her own inimitable style, and really does make some good points about Pi in the process:



Numberphile followed closely behind with some Pi-sinpired music, and then enlisted James Grime to talk about river lengths, and their amazing connection to Pi:



Here's the 1996 paper that inspired this claim, but there is a bit of controversy over this point. It's still an interesting concept to ponder, however.

The good people over at Plus Magazine have some great thoughts, posts, and even artwork all about Pi!

Speaking of Pi artwork, mathematical animator 1ucasvb created a new sine and cosine animation especially for Pi Day that is simple, yet informative. Lucasvb has more animations on that site (this one being a particular favorite), and even more over in his Wikimedia Commons gallery. Even if you don't understand all the math behind them, enjoy and explore the animations.

I'm off to celebrate Pi Day for now, but feel free to enjoy and post your Pi Day wishes in the comments!

Trigonometry is a subject that can strike fear into the heart of almost any high school student.

It's actually quite understandable and useful if taught clearly enough so you grasp it. In this post, I'll show you just where to find those resources.

BASICS: As with any topics, you'll want to make sure you have a few basics down. When getting started with trigonometry, that means being clear on the concepts of similarity, and the good old Pythagorean Theorem.

The series Project Mathematics! is of great help here. Their amazing computer-animated lessons make the concepts of similarity clear and even simple:

Part 1:



Part 2:



Next, make sure you're up to speed on the good old Pythagorean Theorem, and then you'll be ready to proceed to the trigonometry of the unit circle:



THE UNIT CIRCLE: I've posted my own tutorials on the unit circle and trigonometric functions, but I truly have to tip my hat to a recently posted tutorial that far exceeds both of those.

Kalid at BetterExplained.com just posted an awesome article titled, How To Learn Trigonometry Intuitively, including a video to help you along.

What makes this post so great? He slowly introduces each concept and makes each step concrete and understandable. Kalid starts by likening trigonometry to learning anatomy, as if you were learning the anatomy of a circle. After his introduction to sine and cosine with a dome analogy, he points out that the numbers you're seeing are percentages. How much bigger or smaller is this part or that part compared to the radius of the circle? Cleverly, he even goes back to his anatomy metaphor to make this more understandable.

With the idea of a wall next to the dome, he then introduces the tangent and the secant, and then uses a ceiling built over the dome to help drive home the ideas of the cotangent and the cosecant. Probably the most startling moment in the whole post, however, is when Kalid gets you to see the connections between the 3 types of triangles he's been explaining. When you see that they're simply scaled versions of each other, everything begins to fall in place!

Once you have the scaled triangles in your mind, your knowledge of the Pythagorean Theorem and similar triangles make the relationships almost trivial to work out in your head! Instead of memorizing formulas you'll quickly forget after a trig test, you simply grasp the relationships, and can work them out anytime you need them!

Even as good as Kalid's explanations are, he points out that you shouldn't get too attached to the static diagrams. Taking that advice, I used the online graphing calculator as Desmos.com to create some models I could play around with to grasp the concepts for myself, and I've linked to them using the corresponding section names from Kalid's article:

Sine/Cosine: The Dome

Tangent/Secant: The Wall

Cotangent/Cosecant: The Ceiling

Visualize The Connections

FURTHER READING: If you enjoyed learning about trigonometry this way, there are a few other of Kalid's post I highly recommend.

First, read Surprising Uses of the Pythagorean Theorem to help you get out of the mindset that the Pythagorean Theorem is only about triangles. Next, check out How To Measure Any Distance With The Pythagorean Theorem and learn how you can use it to bring problems with a mind-boggling array of factors down to a size you can manage.

Finally, since radians are so important to the unit circle, but come across as more confusing than they should be, Kalid's Intuitive Guide to Angles, Degrees and Radians is a definite must-read.

I sincerely hope you take the time to explore most, if not all, of these resources, as they gave me a new respect and understanding for the tools of trigonometry, and I simply want to share that joy of discovery.

No, I'm not insulting Danica McKellar's math ability, I swear! I even sell her math books over in the Grey Matters Store!

Math Bites is actually the name of Danica McKellar's new weekly video series from Nerdist, which started last month.

Math Bites can be found on YouTube, and you can subscribe to the Nerdist channel for regular updates, as well.

Each episode is only about 5-6 minutes long, and focuses on one math topic. The style is somewhat like that of Square One TV, in that there are short comedy skits and musical segments that help reinforce the concepts or give you breathing time to let the concepts sink in. I especially like the takeaway tips at the end of each episode, which summarize the important points of each episode.

The first episode focused on a topic near and dear to my heart, the constant Pi:



While many Grey Matters readers have memorized Pi to 400 digits, some might prefer the full Dance of the Sugar Pi Fairy to help them memorize Pi to 139 digits.

The second episode talked about the importance of being able to do math in your head. No, this episode doesn't talk about extraordinary mental math feats like those taught in the Mental Gym, but rather knowing how to do basic arithmetic and estimation well enough so that you don't get ripped off:



You can already get an idea of why I like this series so much. It touches on the basics of topics I've posted about many times before. I've posted about binary numbers both on their own, and their importance in Nim, so the 3rd episode on binary numbers was also helpful and entertaining:



The most recent episode as of this posting is about percentages, and focuses on how they can mislead you. Sometimes you can get taken advantage of when businesses can't do math, and also when businesses can do math, so percentages can be very important to understand:



Even if you missed the initial announcement of Math Bites, you're now completely caught up with the series. Here's hoping that Danica McKellar keeps this series up for a long time to come!

Note: In honor of Tau Day, I'm re-posting and updating my original 2011 Tau Day post. Enjoy!

Happy Tau Day! Tau Day is today, but what exactly is it?

If you've been to this site before, you should be well acquainted with Pi. Like Pi (π), Tau (τ) is a a mathematical constant. The value is simple enough, in that it's 2π (2 × Pi).

Since Pi Day is 3/14, that makes Tau Day 6/28. So, why should we make such a big deal over Tau? That's the topic of today's Tau Day post!

Bob Palais started the ball rolling in 2001, with his editorial Pi Is Wrong! He wasn't talking about the actual number being a wrong quantity. When you divide a circle's circumference by it's diameter, you do get 3.1415... and so on. Palais' point was that it's the relationship to the diameter that's the basic problem.

Think about how you draw a circle with a compass. You place the spike at the center of your circle, and rotate the drawing implement around that point. You're taking advantage of the fact that every point on the circle is the same distance from the center of the circle.

In other words, you're taking advantage of the circle's radius, not it's diameter. Of course, if want the formula for a circle's circumference in relation to a radius (r), then you have 2πr. When you start working with the unit circle and radians (distance of the radius around the circumference), you run into 2π again, as the number of radians around a circle.

In formula after formula, 2π pops up again and again. To use Dr. C. Douglas Haessig's analogy, if Pi is a Hollywood star, then 2 is a constant groupie. Since Tau is 2 × Pi, Tau is basically the idea that these two have been dating for long enough, and it's time for them to get married.

What happens when we replace 2π with τ? I'll let Vihart, whose videos I recommended in an earlier post, explain further:



I tried to make radian conversion easy with a mnemonic (indeed, using this same video), but in the long run, you can see that Tau is a better way to go. Here are some handy links for further exploring the concepts in the above video:

• Intuitive Understanding of Sine Waves
• Intuitive Understanding Of Euler’s Formula
• The Tau Manifesto

Robert Dixon voices his support here, and even suggests replacing Pi with τ/2:



For a better understanding of how the area is being used here, check out The Story of Pi video, and A Gentle Introduction To Learning Calculus.

Back in November 2012, Numberphile discussed the advantages of this new approach in its Tau Replaces Pi video:



Numberphile fans being passionate in the way they think about numbers, this naturally led to an intense debate just over a month later:



Ultimately, Tau may offer a better future by offering a better understanding of mathematics, and that's definitely something worthy of celebration, and thus worthy of its own day.

This only leaves one problem. Tau is twice as much as Pi, but Tau's Greek letter has only half as many legs. Just for visual consistency, shouldn't π be 2τ, and τ be .5π? Just kidding, Tau Day fans.

Can you believe it? Grey Matters is 8 years old today, March 14th! How long is 8 years? When I started this blog, YouTube had been formed as a company, but it would be another month before they would publicly unveil their website.

Besides being this blog's 8th blogiversary, it's also Pi Day (3/14) and Albert Einstein's birthday, so let's have a little fun, shall we?

Mental_Floss.com helps gets the party started by sharing 11 unserious photos of Einstein. Yes, of course the famous tongue picture is there, but there are more with which you may not be familiar.

For a Pi Day party, we need food, and what better food than pies? Matt Parker shows us how to calculate Pi using pies:



If you're concerned about food being used in this way, Matt Parker assures:

For the record: we were given the pies, they were no longer fit for consumption (thrown away after) and we donated £314 to a food charity.

— Matt Parker (@standupmaths) March 11, 2013

Your next concern might be about the accuracy of the measurement, which Wolfram|Alpha gives here to 10 places. At first glance, 3.138 doesn't seem as impressive as it should be.

However, if you remember last month's post on bringing pi digits to life, you'll recall that it only takes 38 digits to measure a universe-sized circle with an accuracy to the nearest hydrogen atom. Considering that, measuring a circle in terms of pies to 3.138 is less surprising, and is a considerably good result.

When I was doing research for my continued fractions post, I was thrilled to discover L. J. Lange's continued fraction of Pi, which he developed in his May 1999 paper An Elegant Continued Fraction for π:



As much as you hear about the randomness and unpredictability of Pi, this continued fraction has an astonishingly simple and regular pattern. The denominators, of course, are all 6. The numerators are the squares of all the odd numbers starting with one. In fact the numerator at any level n can be calculated with the formula (2n - 1)². For example, the 6th numerator is calculated as (2 × 6 - 1)² = (12 - 1)² = 11² = 121. Using Gauss' Kettenbruch notation, we can then write this formula for Pi as:



How fast does this get us to the 38 digits for our universe-sized circle which measured to the nearest hydrogen atom?

We can use Wolfram|Alpha to get an idea. The first 10 levels of this fraction give us Pi to 4 digits (the integer part plus 3 decimal places). We get Pi accurate to 7 digits by the 100th level, and to 10 digits by the 1000th level.

Assuming this logarithmic rate of 3 digits for every order of magnitude continues, we would need to go tens of trillions of levels deep to get universe-level accuracy!

Alas, it seems the beauty of this formula's pattern is at the cost of slow convergence to Pi. Since the original formula takes the process to infinite levels, however, at least it gives Pi in the long run. If you're wondering how someone like Archimedes worked out Pi 2200 years ago, without textbooks, calculators, or even calculus, it's actually due to this ingenious approach described at BetterExplained.com. Note that after taking his own approach 96 levels deep, Archimedes also calculated Pi accurate to only 4 digits.

Naturally, many others are celebrating Pi Day today. Check out Ben Vitale's Some Musings on Pi, both part 1 and part 2. The Math Dude podcast also took some time to celebrate the world's best known mathematical constant. One of the more amusing moments in Pi history was the time that the Indiana almost legislated the value of Pi to be exactly 3.2, and James Grime tells the story well.

Thanks to all my readers for reading Grey Matters and keeping this blog going for 8 wonderful years! Now, it's time for you to keep an eye out for what I have in store for my 9th year.

In my previous Quick Feats post, I briefly made use of continued fractions.

As a concept, they aren't well known, yet they are well worth exploring. When you start learning about continued fractions, there are many seemingly-endless surprises!

Let's start with a simple division problem, such as 47 ÷ 17. If we run that problem through Wolfram|Alpha, its' pretty much what we expect. ⁴⁷⁄₁₇ simplifies to 2¹³⁄₁₇, the decimal goes on forever, and that the first 16 digits after the decimal point repeat.

Let's learn something new by changing our point of view. Go through this desmos.com demo, which uses a 47 by 17 rectangle. At each stage, the rectangle is divided up into the largest squares possible. The same is then done with any remaining area, until the entire area is divided up into squares of various sizes.

Assuming you've gone through the process visually and geometrically, we're now going to repeat the process arithmetically. We're going to be dealing with fractional division and reciprocals, so here's a video refresher course, should you need it.

Instead of starting with a 47 by 17 rectangle, we'll just start with the problem ⁴⁷⁄₁₇. We've already seen that this simplifies to 2¹³⁄₁₇, and isn't hard to see how this relates to dividing up our rectangle into 2 perfect squares, with a 13 by 17 rectangle left over.

The next step was dividing up the 17 by 13 rectangle into a one 13 by 13 square, leaving a 4 by 13 rectangle. We need to keep our fraction as the same, but somehow redefine it in terms of 13s at this point. Starting from the fact that a fraction multiplied by its reciprocal equals one, we can work out the following:



Yes, it's a rather strange-looking result, but at least we have the 13 on the bottom, where we need it, and the value of the fraction remains the same. Putting the 2 back into the equation, 2¹³⁄₁₇ becomes:



Remember how we took that 13 by 17 rectangle and divided it up into a single 13 by 13 square with a 4 by 13 rectangle left over? Simplifying ¹⁷⁄₁₃ into 1 + ⁴⁄₁₃ is the same thing. Not surprisingly, we can repeat this process of flipping and simplifying the fractions until we get down to our 1 by 1 squares:



Due to the way in which we flipped the fractions, it's not hard to understand why all the numerators (top numbers of the fractions) are 1. In fact, this is the standard way in which continued fractions are written, with all the numerators as 1 (there are exceptions, of course).

Ignoring the numerators for the moment, look at the sequence of the other numbers - 2, 1, 3, 4. If you walked through the desmos.com demo I linked above, you'll recognize this right away! The geometric process resulted in TWO 17 by 17 squares, ONE 13 by 13 square, THREE 4 by 4 squares, and FOUR 1 by 1 squares, just as our continued fraction resulted in 2, 1, 3, and 4!

That's basically what continued fractions do. They show you how to break up a number so as to better understand its structure, and can often help you discover useful patterns in the process.

To get a better understanding of continued fractions in a very clear manner, there's a wonderful series of father-and-son video series called MathForKids that explains them to any beginner very well. The following is their first continued fraction video:



Towards the end of that video, there's another surprise; the continued fractions help solve quadratics equations with far less difficulty than you probably remember from your days in school!

The second video in the series starts with the simplest continued fraction (all 1s), and yet another surprise develops from this simple pattern. The third video in the series shows you a wonderful shortcut for evaluating continued fractions that automatically generates approximate fractions for any number! The fourth video focuses on working out the square root of 2, and the final video focuses on generating Pi approximations.

For a detailed understanding of the amazing power of continued fractions, R. Knott's course, complete with homework assignments, is tough to beat. It even begins with a similar rectangular division explanation with which you're already familiar.

Plus magazine's Chaos In Numberland article goes on to show you some of the amazing uses to which continued fractions have contributed.

As I mentioned the beginning, the surprises you get as you understand more and more about continue fractions are a consistent treat. Take the time to explore them, and the treasures you'll discover will be well worth the time.

It's one thing to be able to memorize 400 digits of Pi. It's another thing altogether to be able to make the feat memorable and interesting for you audience.

Sure, 400 is an impressive number of Pi digits to memorize, but what do 400 digits really mean? What kind of detail are we talking about?

Numberphile, who has done numerous Pi videos already, has recently released a new video about Pi and the size of the universe, that starts to give you an idea of the sense of Pi's scale:



It almost seems strange that so few digits should be able to take us from the width of a hydrogen atom all the way up to the diameter of the universe. Each place in Pi (as in the tenths, the hundredths, the thousandths, and so on) represents a place that's smaller than the previous one by a factor of 10. There's a classic 1977 film called Powers of 10 that does a wonderful job of dramatizing just how few power of 10 are needed to cover the entire scale of the universe:



This film may look familiar, either because you may have seen it in school, or you've seen one of its parodies, such as in Men In Black, Contact, or The Simpsons.

A good way of keep thing in mind is Tim Rowett's poem, Space:

Seven steps each ten million to one
Describe the whole space dimension
The Atom, Cell’s girth
Our bodies, the Earth
Sun’s System, our Galaxy – done!

– Tim Rowett, Three Limericks – On Space, Time and Speed

BetterExplained.com's article on how the digits of Pi were determined in ancient times gives you a sense of scale in a different manner, and is done so with their usual startling clarity.

As I've mentioned before, memorization and understanding together give you a more complete picture than either could ever do separately.

The last snippets of 2012 are ready!

December's snippets switch back and forth between the nitty-gritty side and the cultured side.

• Picking up from the last entry from November's snippets, it seems that the Tau vs. Pi fight has intensified. This month, NumberPhile treats us to a Tau vs. Pi smackdown:



• Instead of having your math wrestling-style, you might prefer it a bit more cultured. If so, you're in luck, as the Museum of Mathematics, or MoMath for short, had its grand opening in New York on December 12, 2012! Vi Hart's father, George Hart, designed many of the exhibits, and takes us on an opening day tour:



• It was while I was putting together my 2nd integer lattice post that I first took notice of the MyWhyU YouTube channel. They seem to have the same knack for clear explanations as BetterExplained.com, but everything is explained through cartoons. Even their homepage is set up like a notebook, to give it a friendly feel.

Compare BetterExplained's introduction to number systems, to WhyU's introduction (mainly comparing the modern approach to Roman Numerals):



• Since we had the better cultured version of mathematics videos above, how about a better cultured version of an educational video website? You're probably familiar with TED.com, and their amazing selection of free videos from knowledgeable lecturers.

The people behind TED.com have added a new website called TED-Ed, and a corresponding YouTube channel. The TED-Ed site goes beyond videos to include quizzes and other resources that make it possible to get an idea of just how much you or your students are learning. It even allows you to create your own video/quiz/resource combinations! You can get a better idea of TED-Ed's features from their website tour: