That and Why

When I was a kid, my parents had two ways of justifying rules. In the first class there was a pretty understandable reason:

Me: Why do I have to brush my teeth?
Mom: Because it will give you a beautiful smile.
Dad: And because if you don’t the germs from your gums will spread to your nervous system and rot the area of your brain related to inhibitions.
Mom: Oh, please don’t tell him things like that, honey.
Dad: Don’t tell me what to do, woman.
Mom: Are you off your meds again?
Dad: What did I just say? Everyone’s a critic. [to the pet turtle] What are you staring at you retractable hockey puck?
Mom: Mark, dear, see, this is what happens. Your father didn’t brush his teeth when he was a little boy.
Dad: That turtle is a demon. Somebody get me a soldering iron and some holy water.

These days I have some doubts as to the authenticity of those little performances, but they were certainly effective. On the other hand, sometimes my parents’ justifications could be a little obscure:

Me: Why do I have to take out the trash?
Mom: Because I say so.
Dad: And because if you don’t, I will tell you in detail what sex really is, and remember in my Navy days I did two tours on a submarine.

Both are devastatingly convincing – either way I am completely sure I need to do my chores. But in only one case do I feel like there’s a real reason why.

I recently saw this mathematical relation somewhere (I forget where, but it’s pretty well-known):

1^3 + 2^3 + \ldots + n^3 = (1 + 2 + \ldots + n)^2

For example, if n = 5, then

1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 1 + 8 + 27 + 64 + 125 = 225

and

(1 + 2 + 3 + 4 + 5)^2 = 15^2 = 225

This will be true no matter how big n gets. Obviously no one has checked all the way up to n = 935467568777043682111, for example. Even with a computer it would not be possible, and if you did check up to that number, how do you know it would still work for that, plus one? We’ll come back to this.

A simpler example is this one:

2 + 4 + 6 + \ldots + 2n = n(n+1).

Again you can check it out for as many numbers as you want. For n = 7 it says

2 + 4 + 6 + 8 + 10 + 12 + 14 = 7*(7+1)

And that’s right. The left hand side adds up to 56 and the right hand side is 7*8 = 56.

The idea in math, though, is to show that it’s always true, even for n equal to the number of stars in the universe.

Here is proof based on dominoes with dots on them. We’ll lay the dominoes out on a table so that both the sum 2 + 4 + \ldots + n and the multiplication problem n(n+1) count the number of dots on the dominoes.

Our dominoes will have two dots, one on each half, like this:

It's like Wall-E has fallen on his side.  And also lost his body.

One domino. Two dots. This will be the 2 from 2 + 4 + ... + 2n

That domino represents 2, the first number in our sum 2 + 4 + \ldots + 2n.

Next we add 4, which means 2 more dominoes:

You know that story about how if you hold your nose while you sneeze, you'll blow a hole in the back of your head? That's basically what happened to whales.

Two more dominoes with four more dots. Now we have 2 + 4 dots.

But we can rearrange the dominoes however we want and still keep the number of dots the same, so for the hell of it, let’s put them like this:

If there is a deity, it is probably pretty confused about how to live up to everyone's expectations at once.

The same dominoes rearranged.

Now we add three more dominoes to get this:

I bet Carmen Sandiego and Waldo are doing it right now.  I just don't know where.

Adding three more dominoes. Now we have 2 + 4 + 6 dots.

Can you see the pattern? Each time we want to add a another number in the series, we add another layer of dominoes around the edge of the rectangle. It’ll be more obvious if we gray out the middle layer a little.

I will always stand by you. Not because I support you, but because it makes me look good by comparison.

Same image with the layers highlighted. Each layer is one number from the sum (2 + 4 + ... + 2n) and n is the number of layers.

We can keep doing that until we have n layers of dominoes. So, for n=5, our finished picture would look like this:

I am so tired of these stupid domino pictures.  Hey, stop filling in the alt-text on my images, Dad.

For n = 5, we have (2 + 4 + 6 + 8 + 10) dots.

We know that we’ve added the numbers 2 + 4 + 6 + 8 + 10 because each layer has one more domino, hence two more dots, than the previous one. But we can also use a shortcut to count the dots. Ignoring the edges of the dominoes and just focusing on the dots, we see that the dots form a rectangle. The rectangle is 5 dots high and 6 dots long, so the total number of dots is 5*6 = 30. This checks out. 2 + 4 + 6 + 8 + 10 = 30.

The same pattern will hold for as many layer as we please. It’s clear from the picture that each new layer adds two dots, so n layers will have 2 + 4 + \ldots + n dots. But it’s also clear that each new layer makes the rectangle one dot higher and one dot longer, so that in all the rectangle is n dots high and n+1 dots long. That means the total number of dots can be counted two ways, and since the number of dots is the same either way,

2 + 4 + \ldots + 2n = n(n+1)

That’s one sort of proof. We might say, after seeing this proof, that now we understand not only that the equation

2 + 4 + \ldots + 2n = n(n+1)

is true, but also why it is true.

This is a subjective thing. This particular proof makes a lot of sense to me, but to someone else it might not. The proof is very informal. What if there’s an error I just don’t see?

Let’s look at a different type of proof – a more formal type based on symbols rather than pictures. This time we’ll prove the more difficult equality

1^3 + 2^3 + \ldots + n^3 = (1 + 2 + \ldots + n)^2

from the beginning of the post.

Let’s start with the right hand side.

We already know

2 + 4 + \ldots + 2n = n(n+1)

so dividing both sides by 2

1 + 2 + \ldots + n = \frac{n(n+1)}{2}

Squaring,

(1 + 2 + \ldots + n)^2 = \frac{n^2(n+1)^2}{4} = \frac{n^4 + 2n^3 + n^2}{4}

That’s the right hand side. Now for the left. By algebra

k^4 - (k-1)^4 = 4k^3 - 6k^2 + 4k - 1

so

\sum_{k=1}^n k^4 - (k-1)^4 = 4\sum_{k=1}^n k^3 - 6\sum_{k=1}^n k^2 + 4\sum_{k=1}^n k - \sum_{k=1}^n 1

But also

\sum_{k=1}^n k^4 - (k-1)^4 = \sum_{k=1}^n k^4 - \sum_{k = 1}^n (k-1)^4 = \sum_{k=1}^n k^4 - \sum_{k=0}^{n-1} k^4  = n^4

So

n^4 = 4\sum_{k=1}^n k^3 - 6\sum_{k=1}^n k^2 + 4\sum_{k=1}^n k - \sum_{k=1}^n 1

or

\sum_1^n k^3 = \frac{n^4 + 6\sum_1^n k^2 + 4 \sum_1^n k - \sum_1^n 1}{4}.

In general, we can find the sum \sum_{k=1}^n k^p for any p based on the binomial coefficients and the sums for lesser powers. Simplifying out the algebra in this case gives

\sum_{k=1}^n k^3 = \frac{n^4 + 2n^3 + n^2}{4},

which is the same as the result from before, so

1^3 + 2^3 + \ldots + n^3 = (1 + 2 + \ldots + n)^2.

This proof is pretty much solid. You could make it formal and rigorous if you wanted to. But unlike the first proof, I don’t get from it much sense of the “why”. Sometimes I feel like numbers are just telling me what to do and threatening me with horrible consequences if I don’t.

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2 Responses to “That and Why”

  1. More Why « Arcsecond Says:

    […] Arcsecond « That and Why […]

  2. What formula would be used to for the summation of (the sum of the integers from 1 to n) from 1 to n? - Quora Says:

    […] curious, I wrote a blog post describing one possible method for a very similar problem a while ago.https://arcsecond.wordpress.com/2…This answer .Please specify the necessary improvements. Edit Link Text Show answer summary […]

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