# It's true because pictures

May 26, 2015 6:19 PM Subscribe

Wow, yeah, this is really good proof that I don't learn visually. I mean, I get it, but I don't get it from looking at the picture. If you showed me that in school and told me, a-ha, here's the proof!, I think I'd be pretty frustrated.

posted by uncleozzy at 6:33 PM on May 26, 2015 [7 favorites]

posted by uncleozzy at 6:33 PM on May 26, 2015 [7 favorites]

....What is it supposed to be trying to prove?....

posted by EmpressCallipygos at 6:39 PM on May 26, 2015

posted by EmpressCallipygos at 6:39 PM on May 26, 2015

Nice! The inductive argument that I've seen for that is...less obvious. (I'm sure it's the same, but when I was teaching it this spring, the algebra ended up feeling like "and then a miracle occurred".)

posted by leahwrenn at 6:39 PM on May 26, 2015 [1 favorite]

posted by leahwrenn at 6:39 PM on May 26, 2015 [1 favorite]

(Actually the non-animated one is a bit more clear for me, I think. Either way, it's neat that it can be shown this way.)

posted by uncleozzy at 6:47 PM on May 26, 2015

posted by uncleozzy at 6:47 PM on May 26, 2015

It's proving that for any positive integer n, (1 + 2 + ... + n)^2 = 1^3 + 2^3 + ... + n^3.

For example, if n = 2, (1+2)^2 = 3^2 = 9 and 1^3 + 2^3 = 1 + 8 = 9.

If n = 3, (1+2+3)^2 = 6^2 = 36 and 1^3 + 2^3 + 3^3 = 1+8+27 =36.

So, it starts out with (1 + 2 + ... + n)^2, where in the picture, n = 5. (blah)^2 is (the area of) a blah x blah square (which is why we sometimes talk about blah-squared), but since we're working with (1 + 2 + ... + n)^2, they chunked up the pieces of the square into cleverly chosen rectangles. They then colored the rectangles so that you could see which ones they were stacking up: you had stair-step stacks of red, on top of a blue square, combined with a stair-step stack of green, chosen so that you get cubes of increasing size. Now (blah)^3 is (the volume of) a blah x blah x blah cube (hence, we might say blah-cubed), and if you look at the cubes they ended up with, we have 1^3 + 2^3 + ... + n^3 (where again, n = 5).

An interesting thing I learned this spring is that one of Rene Descartes' huge innovations was that he was willing to allow working with x^2, x^3, x^4 as just numbers that could represent, say, length, rather than being constrained that x^3 could *only* represent a volume, or x^2 could *only* represent area. Earlier algebraists had to go through elaborate contortions in their computations so that the `dimensions' of the quantity that they were working with would match (i.e., they required homogeneity)---since of course, since antiquity, real math meant geometry.

posted by leahwrenn at 6:49 PM on May 26, 2015 [17 favorites]

For example, if n = 2, (1+2)^2 = 3^2 = 9 and 1^3 + 2^3 = 1 + 8 = 9.

If n = 3, (1+2+3)^2 = 6^2 = 36 and 1^3 + 2^3 + 3^3 = 1+8+27 =36.

So, it starts out with (1 + 2 + ... + n)^2, where in the picture, n = 5. (blah)^2 is (the area of) a blah x blah square (which is why we sometimes talk about blah-squared), but since we're working with (1 + 2 + ... + n)^2, they chunked up the pieces of the square into cleverly chosen rectangles. They then colored the rectangles so that you could see which ones they were stacking up: you had stair-step stacks of red, on top of a blue square, combined with a stair-step stack of green, chosen so that you get cubes of increasing size. Now (blah)^3 is (the volume of) a blah x blah x blah cube (hence, we might say blah-cubed), and if you look at the cubes they ended up with, we have 1^3 + 2^3 + ... + n^3 (where again, n = 5).

An interesting thing I learned this spring is that one of Rene Descartes' huge innovations was that he was willing to allow working with x^2, x^3, x^4 as just numbers that could represent, say, length, rather than being constrained that x^3 could *only* represent a volume, or x^2 could *only* represent area. Earlier algebraists had to go through elaborate contortions in their computations so that the `dimensions' of the quantity that they were working with would match (i.e., they required homogeneity)---since of course, since antiquity, real math meant geometry.

posted by leahwrenn at 6:49 PM on May 26, 2015 [17 favorites]

that is fascinating! I had heard that this was how ancient Greek mathematicians thought - their concept of number was constrained to a geometric model - but I had no idea that this persisted into the 16th century!

posted by thelonius at 6:54 PM on May 26, 2015

posted by thelonius at 6:54 PM on May 26, 2015

*It's proving that for any positive integer n, (1 + 2 + ... + n)^2 = 1^3 + 2^3 + ... + n^3.*

For example, if n = 2, (1+2)^2 = 3^2 = 9 and 1^3 + 2^3 = 1 + 8 = 9.

If n = 3, (1+2+3)^2 = 6^2 = 36 and 1^3 + 2^3 + 3^3 = 1+8+27 =36

For example, if n = 2, (1+2)^2 = 3^2 = 9 and 1^3 + 2^3 = 1 + 8 = 9.

If n = 3, (1+2+3)^2 = 6^2 = 36 and 1^3 + 2^3 + 3^3 = 1+8+27 =36

In English please? And why are you folks are awed by it? For a lot of us, maths isn't a second, third, or even fourth language.

Anywho, the possible reason the animation doesn't click with a lot of people is that it's, basically, a cartoon. And, since you can make anything happen in a cartoon, we are apt to simply not believe what we see.

posted by Thorzdad at 7:06 PM on May 26, 2015 [2 favorites]

If you like color coded geometry, check out this 1847 version of The First Six Books of The Elements of Euclid where color is used instead of A/B/C for angle names. (Warning, use of the old-style long s may make the text a little tougher to read.)

posted by fings at 7:11 PM on May 26, 2015 [3 favorites]

posted by fings at 7:11 PM on May 26, 2015 [3 favorites]

Are there any other exponents (a , b) such that sum(1,2,3,...)^a == sum(1^b, 2^b, 3^b, ...)?

posted by CheeseDigestsAll at 7:23 PM on May 26, 2015 [1 favorite]

posted by CheeseDigestsAll at 7:23 PM on May 26, 2015 [1 favorite]

Really the meta-cool thing is the variety of ways people creatively explain the same thing.

This is exactly why the wailing about common core math distresses me. It's about a vocabulary of ways to do things, not The One Way that you happened to be taught (like 'long' division). It's flexibility of thought, and pattern recognition, not carrying a digit.

posted by Dashy at 7:24 PM on May 26, 2015 [9 favorites]

This is exactly why the wailing about common core math distresses me. It's about a vocabulary of ways to do things, not The One Way that you happened to be taught (like 'long' division). It's flexibility of thought, and pattern recognition, not carrying a digit.

posted by Dashy at 7:24 PM on May 26, 2015 [9 favorites]

I've heard the Machine Elves are delighted to show you one of these for the Riemann hypothesis, but when you try to remember it projected onto 2+1 dimensions it's just a jumble of random, staticky Benson clips.

posted by mubba at 7:35 PM on May 26, 2015 [5 favorites]

posted by mubba at 7:35 PM on May 26, 2015 [5 favorites]

*Wow, yeah, this is really good proof that I don't learn visually. I mean, I get it, but I don't get it from looking at the picture*

I thought, that's super cool. Then I went to find the proof by induction so I could really get it.

posted by jeather at 7:46 PM on May 26, 2015 [5 favorites]

*Are there any other exponents (a , b) such that sum(1,2,3,...)^a == sum(1^b, 2^b, 3^b, ...)?*

No. For this to be true, you'd need

3^a = 1 + 2^b.

This holds when a=2 and b=3, but beyond that case, a power of 2 can never be one less than a power of 3. If you like, this is a case of the Catalan conjecture, which was proved by Preda Mihaelscu in 2002, but this case is much simpler; in fact, it was proven by Levi ben Gershon about 700 years earlier! Here's a description of his proof from MathOverflow.

posted by escabeche at 7:49 PM on May 26, 2015 [7 favorites]

*Rabbi*Levi ben Gershon (Gersonides, Ralbag) was quite a brilliant person. He actually invalidated the Ptolemaic model of the solar system through observation well before Copernicus; sadly, he didn't advance the heliocentric model.

posted by Joe in Australia at 8:02 PM on May 26, 2015 [5 favorites]

If'n you like math visualizations be sure to check out LucasVB's wikipedia contributions.

posted by Jpfed at 8:08 PM on May 26, 2015

posted by Jpfed at 8:08 PM on May 26, 2015

*I thought, that's super cool. Then I went to find the proof by induction so I could really get it.*

I'm glad that I'm not the only person who needed to do that. Visual math proofs, although fun to look at, don't actually help me understand what's happening without me working out the math on the side.

posted by tealNoise at 8:10 PM on May 26, 2015 [1 favorite]

This visual proof does translate pretty directly to a symbolic one, without using induction.

(1 + 2 + ... + n)^2 = 1^2 + 2*2*(1) + 2^2 + 2*3*(1 + 2) + 3^2 + ... + 2*n*(1 + 2 + ... + (n-1)) + n^2

= 1^2 + 2*2 + 2^2 + 3*2*3 + 3^2 + ... + n*(n-1)*n

= 1^2*(1+0) + 2^2*(1+1) + 3^2*(1+2) + ... + n^2*(1 + (n-1))

= 1^3 + 2^3 + 3^3 + ... + n^3

So it is in some sense really just a bunch of applications for Gauss' trick for computing 1+2+...+n, extruded into the third dimension.

(Another proof is to note that both sides are polynomials of degree 4, so if they agree for four values of n, they must be identical. So check the easiest four values.)

posted by eruonna at 8:51 PM on May 26, 2015 [6 favorites]

(1 + 2 + ... + n)^2 = 1^2 + 2*2*(1) + 2^2 + 2*3*(1 + 2) + 3^2 + ... + 2*n*(1 + 2 + ... + (n-1)) + n^2

= 1^2 + 2*2 + 2^2 + 3*2*3 + 3^2 + ... + n*(n-1)*n

*(by Gauss' trick)*= 1^2*(1+0) + 2^2*(1+1) + 3^2*(1+2) + ... + n^2*(1 + (n-1))

*(group like terms)*= 1^3 + 2^3 + 3^3 + ... + n^3

So it is in some sense really just a bunch of applications for Gauss' trick for computing 1+2+...+n, extruded into the third dimension.

(Another proof is to note that both sides are polynomials of degree 4, so if they agree for four values of n, they must be identical. So check the easiest four values.)

posted by eruonna at 8:51 PM on May 26, 2015 [6 favorites]

I always assumed I had an average IQ but I'm starting to realize that I'm not bringing up the average.

posted by bonobothegreat at 8:58 PM on May 26, 2015 [2 favorites]

posted by bonobothegreat at 8:58 PM on May 26, 2015 [2 favorites]

*In English please?*

An inductive proof is three steps: "Assuming that f(x) = g(x), show that f(x+n) = g(x+n)". By analogy, this amounts to: "show that the rungs on this ladder are evenly spaced."

Let's say for example that we postulate that (x+1)*(x-1) = (x

^{2}- 1) then we could as a next step say OK, so, therefore (x+1+1)*(x+1-1) should equal (x+1)

^{2}-1 and hey, it does.

The next step is to say, ok, show that it's true for some fixed X. In this example we can just pick x=1 as an easy bucket, giving us 0=0. Finally, by induction, we say that since we know that if it's true at some integer it must be true for that number plus N, and we know it's true for x=1, it's true for all 1+N for any arbitrarily large integer N.

We prove that every rung in this ladder is evenly spaced, and that if you can reach the second rung from the first, that means you can eventually reach every rung, no matter how many there are.

In this case we can do that by noting that if (1 + 2 + ....+ x)

^{2}= 1

^{3}+ 2

^{3}+ ... + x

^{3}is true, then since (a + b)

^{2}= a

^{2}+ 2ab + b

^{2}, we see that:

(1 + 2 + ....+ x + x +1 )

^{2}= ( (1 + 2 + ....+ x) + ( x +1 ) )

^{2}= (1

^{3}+ 2

^{3}+ ... + x

^{3}) + 2(1 + 2 + ....+ x) (x+1) + (x+1)

^{2}

and since we know the sum of 1 to X is x

^{2}/2, we can quickly say that all that equals

(1

^{3}+ 2

^{3}+ ... + x

^{3}) + 2(1 + 2 + ....+ x) (x+1) + (x+1)

^{2}= (1

^{3}+ 2

^{3}+ ... + x

^{3}) + x

^{2}(x+1) + (x+1)

^{2}

= (1

^{3}+ 2

^{3}+ ... + x

^{3}) + (x+1)

^{3}

And the base case is easy.

posted by mhoye at 9:02 PM on May 26, 2015 [1 favorite]

This is bizarrely soothing and something about it makes my brain think all is right in the world even if I don't totally understand it.

posted by Hermione Granger at 9:48 PM on May 26, 2015

posted by Hermione Granger at 9:48 PM on May 26, 2015

This seems a bit like a higher-order version of the somewhat more well-known fact (at least, to me) that

1 + 3 + 5 + ... + 2n - 1 = n²

mainly due to the similar but far simpler picture proof where you make a square by "wrapping up" consecutive smaller numbers. ASCII isn't great for this, but basically in the following you start from the upper left and the numbers correspond to the terms on the left side of the above equation.

Anyway I'm wondering if the fact in OP is related to this, and if so, whether there are further identities involving higher powers?

posted by valrus at 10:07 PM on May 26, 2015

1 + 3 + 5 + ... + 2n - 1 = n²

mainly due to the similar but far simpler picture proof where you make a square by "wrapping up" consecutive smaller numbers. ASCII isn't great for this, but basically in the following you start from the upper left and the numbers correspond to the terms on the left side of the above equation.

13579 33579 55579 77779 99999So there are three 3s, five 5s, etc. and you can see that if you added eleven 11s along the right and bottom you'd get a bigger square, and so on ad infinitum.

Anyway I'm wondering if the fact in OP is related to this, and if so, whether there are further identities involving higher powers?

posted by valrus at 10:07 PM on May 26, 2015

According to Wikipedia (This is a nontrivial result), sums of odd exponents of consecutive integers:

1^t + 2^t+...+n^t for odd positive t,

are always polynomial in the nth triangular number a_n, where

a_n=1+2+...+n

.

..

...

....

(See, it makes a triangle when you add them up).

There are also various identities for sums of 4th powers, etc---also I think due originally to Levi Ben Gerson, but maybe not. Certainly the sums of consecutive 4th powers were known pretty early in the development of algebra. If not Levi Ben Gerson, then maybe one of the medieval/Renaissance Islamic mathematicians? I can't remember, and I'm not being able to find the right search string on Wikipedia.

posted by leahwrenn at 10:39 PM on May 26, 2015

1^t + 2^t+...+n^t for odd positive t,

are always polynomial in the nth triangular number a_n, where

a_n=1+2+...+n

.

..

...

....

(See, it makes a triangle when you add them up).

There are also various identities for sums of 4th powers, etc---also I think due originally to Levi Ben Gerson, but maybe not. Certainly the sums of consecutive 4th powers were known pretty early in the development of algebra. If not Levi Ben Gerson, then maybe one of the medieval/Renaissance Islamic mathematicians? I can't remember, and I'm not being able to find the right search string on Wikipedia.

posted by leahwrenn at 10:39 PM on May 26, 2015

Damn, I thought this was proving "the break a square of chocolate out of a bar and the bar still has the original number of squares in it" thingy I've seen. Seen in my dreams.

posted by maxwelton at 10:41 PM on May 26, 2015

posted by maxwelton at 10:41 PM on May 26, 2015

I like this because unlike a lot of "visual proofs" this really does fairly trivially imply a fully rigorous argument for the truth of the identity.

posted by Proofs and Refutations at 11:01 PM on May 26, 2015

posted by Proofs and Refutations at 11:01 PM on May 26, 2015

In English please?

The square of sums equals the sum of cubes.

posted by XMLicious at 11:29 PM on May 26, 2015 [3 favorites]

*I had heard that this was how ancient Greek mathematicians thought - their concept of number was constrained to a geometric model - but I had no idea that this persisted into the 16th century!*

TBQH I still sometimes reflect upon the non-Euclidean geometry class I took, and I feel a deep unease.

"Just what the fuck are radians, anyway?", I wonder. "And what do they want? What's their angle?"

posted by nom de poop at 12:14 AM on May 27, 2015 [5 favorites]

mhoye, the sum of 1 to x is x(x+1)/2. The rest of your math works out though.

Visually, I prefer to have a rectangle that's n^2 long and n wide, but there's many ways to view it. To grasp the inductive step in the given visual example, I'd show that all the pieces which have one side that's the greatest length, in this case 5, can be matched to form groups of 5, which would make a square. So then you'd match to 4*5 to the 1*5, the 3*5 to the 2*5, etc. The 5*5 square doesn't need to be matched. This would net you the cube it shows.

posted by halifix at 12:22 AM on May 27, 2015

Visually, I prefer to have a rectangle that's n^2 long and n wide, but there's many ways to view it. To grasp the inductive step in the given visual example, I'd show that all the pieces which have one side that's the greatest length, in this case 5, can be matched to form groups of 5, which would make a square. So then you'd match to 4*5 to the 1*5, the 3*5 to the 2*5, etc. The 5*5 square doesn't need to be matched. This would net you the cube it shows.

posted by halifix at 12:22 AM on May 27, 2015

pi: *pops up in an equation*

me: We meet again, you little fuck.

posted by nom de poop at 12:26 AM on May 27, 2015 [2 favorites]

me: We meet again, you little fuck.

posted by nom de poop at 12:26 AM on May 27, 2015 [2 favorites]

*I thought, that's super cool. Then I went to find the proof by induction so I could really get it.*

Same here. I think I have unintentionally trained my brain to STOP LOOKING AT THE PICTURE when trying to grasp counterintuitive or non-visualizable stuff.

posted by Dr Dracator at 1:24 AM on May 27, 2015

One reason I particularly liked this is that, in elementary school, we encourage students to associate a rectangle ("rectangular array", if you must) with every multiplication fact. So they should actually be primed to look at their "times table" exactly like this:

posted by Wolfdog at 1:50 AM on May 27, 2015 [2 favorites]

1 2 3 ___________________ 1| * ** *** | 2| * ** *** | * ** *** | 3| * ** *** | * ** *** | * ** *** (and so on)So I think I'm going to (1) see if students recognize their times table when it looks like that, and (2) practice doing the assembly with snap cubes because I know some students who will

*love*learning how to form the rectangles into a family of cubes in the manner of the GIF.posted by Wolfdog at 1:50 AM on May 27, 2015 [2 favorites]

In case it hasn't been explained well enough, that "times table of rectangles" is exactly what you get when you expand (1 + 2 + 3)(1 + 2 + 3), or more generally, (1 + ... + n)(1 + .... + n). Each number in the first set of parens gets paired up with each number in the second pair of parens, and you get every possible product of two numbers, in both orders, just like your schoolbook multiplication table has.

I've have 5th graders playing with snap cubes and graph paper discover the rule for factoring a

posted by Wolfdog at 1:59 AM on May 27, 2015 [2 favorites]

I've have 5th graders playing with snap cubes and graph paper discover the rule for factoring a

^{2}-b^{2}(although they didn't express it formulaically like that), and I bet they'll recognize it, and find it meaningful, when it shows up later in algebra. It's never too early to learn that you can see - or count - things more than one way.posted by Wolfdog at 1:59 AM on May 27, 2015 [2 favorites]

I think when people ask for explanations "in English" they aren't really talking about the kind where you just casually drop mathematical notation inline whenever something would be hard to write in English.

posted by LogicalDash at 3:49 AM on May 27, 2015 [1 favorite]

posted by LogicalDash at 3:49 AM on May 27, 2015 [1 favorite]

Let me put this one other way.

Take a multiplication table of any size - n by n in general, or for concreteness, say the old 10 by 10 table that I learned from.

On each entry in the table, place the corresponding number of snap cubes (little cubes that link together).

Then you'll be able to assemble all those little cubes into a family of cubes of all different sizes: for me, ten cubes from the 1x1x1 up to the big 10x10; for you,

posted by Wolfdog at 4:02 AM on May 27, 2015 [1 favorite]

Take a multiplication table of any size - n by n in general, or for concreteness, say the old 10 by 10 table that I learned from.

On each entry in the table, place the corresponding number of snap cubes (little cubes that link together).

Then you'll be able to assemble all those little cubes into a family of cubes of all different sizes: for me, ten cubes from the 1x1x1 up to the big 10x10; for you,

*n*cubes from a little 1x1x1 up to a big nxnxn. The picture shows you exactly how to do it. You have exactly the right number of little cubes! That's neat.posted by Wolfdog at 4:02 AM on May 27, 2015 [1 favorite]

Or, if you make the multiplication table out of little snap cubes then you can pick any entry

posted by TreeRooster at 5:07 AM on May 27, 2015

*nxn*on the diagonal, plus all the entries above*nxn*it its column, plus those before*nxn*in its row, and make a perfect*nxnxn*Rubix cube out of all those little cubes.posted by TreeRooster at 5:07 AM on May 27, 2015

Exactly! I think the animation would be better, actually, if it assembled the cubes one at a time. You get the free bonus identity n

I have just been doing this with actual snap cubes, and it's as neat and simple as I hoped. Once you have the 4x4 multiplication table laid out, you can make and unmake the large cubes in a few seconds. I don't have enough cubes to do the 5's. GRANT PLEASE!

posted by Wolfdog at 5:11 AM on May 27, 2015 [2 favorites]

^{3}= 2n(1 + ... + n-1) + n^{2}.I have just been doing this with actual snap cubes, and it's as neat and simple as I hoped. Once you have the 4x4 multiplication table laid out, you can make and unmake the large cubes in a few seconds. I don't have enough cubes to do the 5's. GRANT PLEASE!

posted by Wolfdog at 5:11 AM on May 27, 2015 [2 favorites]

There's an animation for the sum of squares too!

posted by Wolfdog at 5:24 AM on May 27, 2015 [2 favorites]

posted by Wolfdog at 5:24 AM on May 27, 2015 [2 favorites]

By the way, if you wanted to do the 10x10 times table, you would need 3025 snap cubes. 3025 is one of my favorite numbers, because 3025 = (30 + 25)

posted by Wolfdog at 7:45 AM on May 27, 2015

^{2}posted by Wolfdog at 7:45 AM on May 27, 2015

*I think when people ask for explanations "in English" they aren't really talking about the kind where you just casually drop mathematical notation inline whenever something would be hard to write in English.*

Well, we could all go back to rhetorical algebra...

(Although, the previous comment that what we're proving is that the square of the sum (of the first n consecutive integers) is equal to to the sum of the cubes (of the first n consecutive integers) was pretty good!)

posted by leahwrenn at 9:56 AM on May 27, 2015

*Are there any other exponents (a , b) such that sum(1,2,3,...)^a == sum(1^b, 2^b, 3^b, ...)?*

No. For this to be true, you'd need

3^a = 1 + 2^b.

No. For this to be true, you'd need

3^a = 1 + 2^b.

Huh? Why?

posted by polymodus at 1:27 PM on May 27, 2015

That is just looking at the special case (1+2)^a = (1^b + 2^b), not even worrying about the longer sum for arbitrary n's.

posted by Wolfdog at 1:38 PM on May 27, 2015

posted by Wolfdog at 1:38 PM on May 27, 2015

I see now. It is implied by the statement.

Hmm. What if the claim was relaxed to "for all n > N, for some sufficiently large N".

posted by polymodus at 2:04 PM on May 27, 2015

Hmm. What if the claim was relaxed to "for all n > N, for some sufficiently large N".

posted by polymodus at 2:04 PM on May 27, 2015

It is equivalent because both sides are polynomial in n, so if they agree for infinitely many n, they agree for all n.

posted by eruonna at 4:12 PM on May 27, 2015

posted by eruonna at 4:12 PM on May 27, 2015

*It is equivalent because both sides are polynomial in n, so if they agree for infinitely many n, they agree for all n.*

So actually a different question I also had was, consider:

F(n) = ∑

_{i ∊ 1 … n}f(

*i*)

G(n) = ∑

_{i ∊ 1 … 2n}g(

*i*)

Which to me shows that conceivably you could have summations that don't have their number of terms to be linear in the input size. To me it is not immediately apparent that the sides have the same "degree", since both sides are technically series.

My previous wondering about sufficiently large N was just this:

{(a, b) | Exist (N): For all (n > N): (∑

_{i = 1…n}i)

^{a}≈ ∑

_{i=1…n}i

^{b}}

For some definition of ≈ (approximate, converges, proportional…), and asking if this set is empty.

posted by polymodus at 5:15 PM on May 27, 2015

*What if the claim was relaxed to "for all n > N, for some sufficiently large N".*

I can prove it for sufficiently

*small*N; does that help?

posted by Joe in Australia at 5:38 PM on May 27, 2015

I couldn't quite follow the animation because it went too fast, so I added some delays and annotations. Maybe it'll help someone else:

http://imgur.com/q9YHmFR

posted by moonmilk at 7:15 PM on May 27, 2015 [3 favorites]

http://imgur.com/q9YHmFR

posted by moonmilk at 7:15 PM on May 27, 2015 [3 favorites]

« Older Now Something for the Ladies: Feminine... | FIFA officials arrested on corruption charges Newer »

This thread has been archived and is closed to new comments

posted by Songdog at 6:26 PM on May 26, 2015 [1 favorite]