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[MUSIC]. 

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

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

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

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I should try to antidifference n to the 
4th. 

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

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

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

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

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

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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 

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

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

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

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

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

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781. 
And the difference between 1,024 and 

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

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

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this. 
So I'll write the differences, between 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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somehting with a difference of n to the 
4th. 

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

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

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

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

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these numbers by five. 
And since all these coefficients except 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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term by adding up on the differences of n 
cubed. 

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

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

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

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get rid of this n squared term. 
Right? 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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adding up the first k numbers on the 
list. 

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

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

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between these numbers give you the 4th 
power. 

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

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Now I just have to worry about that 
constant. 

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But then constant is zero, we can check 
it. 

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

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

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

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

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

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

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

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

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

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

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

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0. 
This whole process is really analogous to 

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

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

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

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

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This analogy runs really deep.