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Let's find the sum of the 4th powers. In particular, I want to calculate the

sum, n goes from 1 to k, of n to the 4th. Playing with this analogy, I could think

about this problem the same way that I'd approach an integration problem.

I should try to antidifference n to the 4th.

I should try to find a list of numbers, whose successive differences are n to the

fourth, just try to find a list of numbers whose differences are n to the

4th. Let's just try to find some lists of

numbers and their differences. So as a first example let's take a look

at say n to the 5th, and let's look at the 5th power.

So 0 the 5th is 0, 1 to the 5th is 1, 2 to the 5th is 32, 3 to the 5th is 243, 4

to the 5th is the same as 2 to the tenth that's 1024.

Five to the 5th is 3125 and so on. So that's just lists of of 5th powers.

Now, let's try to find the differences between subsequent numbers in this list.

So the difference between 1 and 0 is 1. The difference between 32 and 1 is 31.

The difference between 243 and 32 is 211. The difference between 1,024 and 243 is

781. And the difference between 1,024 and

3,125 is 2101 and so forth. Now it's maybe not so helpful just to see

specific numbers. Let's try to write down a formula for

this. So I'll write the differences, between

the 5th powers. Well that's n to the 5th minus the

previous 5th power, so minus n minus 1 to the 4th.

And then I could expand this out. I could take n to 5th minus this thing

expanded and I'd get 5 n to the 4th is the n to the 5th will cancel minus 10 n

cubed plus 10 n squared minus 5 n plus 1. Right, and the plus 1 comes from

subtracting a minus 1 to the 5th. Okay, so that formula there gives these

green numbers, it's the differences in the 5th powers.

I'm going to play the same kind of game for a different list of numbers.

Let's try to find the differences in the list of say 4th powers, well that will be

the 4th power of n, minus the fourth power of the pervious number of n minus

1. And again I can expand this out and n to

the 4th minus, there's an n to the fourth that cancels.

So the highest power that survives is an n cubed term with a coefficient of 4

minus 6 n squared plus 4 n minus 1. So this formula tells me the differences

between subsequent 4th powers. Now I want to combine those, to get

somehting with a difference of n to the 4th.

Okay, but how am I going to do that? Well, here's something I could do to make

it a little bit easier to see what's going on.

Instead of looking at differences between 5th powers, I can look at the differences

between 1 5th of 5th power. So I, that is the effect of dividing all

these numbers by five. And since all these coefficients except

for the last one are multiples of five, that'll make that formula look a little

bit nicer. And I could do the same thing with with n

to the 4th. Instead of looking at 4th powers, I could

look at 1 quarter of fourth powers, and I'd have the effect of dividing all of

these coefficients by by four. Let me just write down you know, what,

what we've got here, kind of summarize the resulting formulas.

So, the differences between n to the 5th over 5, well, according to this that will

be n to the 4th minus 2 n cubed plus 2 n squared minus n plus a 5th.

And I get a similar kind of formula for looking at differences of n to the 4th

over 4. It's this divided by 4 and that will give

me an n, cubed minus 3 halves and squared plus n minus a quarter.

Well, let's just try to combine those two.

But how, exactly, do I want to combine them?

Well, the deal is that I've got an n to he 4th here, and that's really what I

want at the end of this process. I want something with the differences n

to the 4th, because I'm trying to antidifference n to the 4th.

And I've got a minus 2n cubed. And I've just got an n cubed here.

So if I took this and added two copies of this, I'd be in pretty good shape.

So what would I get in that case? I'll be looking at the differences of n

to the 5th over 5 plus two copies of n to the 4th over 4.

And that would give me an n to the 4th. The n cubed term would go away.

I'd get a minus n squared, because I've got a 2n squared minus two copies of

three halves n squared. And then I get a plus n, because I've got

minus n plus two copies of n. And then a 5th minus 2 copies of a 4th

will give me minus 3 tenths. I could try to get rid of that n squared

term by adding up on the differences of n cubed.

Why does that work? Well, if you calculate the differences

for the list of numbers n cubed over 3. you end up getting, n squared minus n,

plus a third. So, if I take this and add this, that'll

get rid of this n squared term. Right?

So I'm going to look at the differences of n to the 5th over 5 plus two copies of

n to the 4th over 4 plus n cubed over 3, and I'm left with an n to the fourth.

No more n squared anymore. The plus n minus n also cancels, and then

I've got minus three tenths plus a third, and that ends up being plus 1 30th.

Now I can get rid of the 1 over 30th term.

Now how do I do that? Well, look at the differences of the list

of numbers n over 30, and that's just 1 30th, right.

This list of numbers, each number in that list differs by 1 30th compared to the

previous number in the list. So if I take this and subtract n over 30,

then the differences in that list are exactly n to the 4th, right?

By which I mean that I take d of n to the 5th over 5 plus 2 n to the 4th over 4

plus n to the 3rd over 3 minus n over 30. And then I got n to the 4th plus a 30th

minus a 30th and what I'm left with is just n to the 4th.

So in light of all of this, What do I know right know?

Taking differences between a list of numbers, and this accumulation function,

adding up the first k numbers on the list.

Those are inverse operations, right? This is the discreet version of the

fundamental theorem of calculus. In this particular case, the differences

between these numbers give you the 4th power.

So if I sum the 4th powers I get this, at least up to some constant.

Now I just have to worry about that constant.

But then constant is zero, we can check it.

Here's a sum for n equals 1 just to 1 of n to the 4th.

And I can just plug in any value of K that I like, since this constant C

doesn't depend upon K. I can figure out what that constant is by

looking at when K equals 1. So, this is just 1.

But on the other side, I've got what I get when I plug in k equals 1.

Which is, 1 5th plus 2 times the 4th, plus, you know, 1 to the 3rd over 3.

So plus a 3rd minus a 30th, plus that constant c that doesn't depend upon k at

all. But a 5th plus a half plus a third minus

a 30th is 1. So I've got 1 equals 1 plus C.

So that means C equals 0. And that tells me what the formula then,

for the sum of the 4fourh powers is. It's just this, and that constant is just

0. This whole process is really analogous to

the fundamental theorem of calculus and how we use it, right.

In this example I'm antidifferencing n to the 4th in order to sum n to the 4th.

In the same way that if I were to antidifferentiate x to the fourth, that

would help me to integrate x to the 4th, right.

This analogy runs really deep.