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What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?

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There is a pictorial derivation and the formula is here: math.stackexchange.com/questions/241760/…. This can be proved with induction, by studying the vector space of polynomials using $\binom{x}{n}$ as basis vectors instead of $x^n$, or in several interesting other ways. I seek a geometric explanation in that post. –  alex.jordan Mar 19 at 21:52
    
    
See my answer. –  Mhenni Benghorbal Mar 19 at 23:42

5 Answers 5

up vote 11 down vote accepted

If you don't want to look up Faulhaber's formula you could try $$(n+1)^5-n^5=5n^4+10n^3+10n^2+5n+1$$

If you sum both sides of this, and know how to compute the sums of $n^3, n^2, n \text{ and } 1$ you should be able to work out the sum of $n^4$.

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One way is by using the Euler-Maclaurin summation formula, something you strangely don't find in most standard calculus books.

http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula

Let $f(x) = x^4$.

Then

$$f'(x) = 4x^3, f'''(x) = 24 x, f^{(5)}(x) = f^{(7)}(x) = ... = 0 $$

And according to the formula,

$$ \sum_{k=1}^{n} k^{4} = \sum_{k=0}^{n} k^{4} = \int_{0}^{n} x^{4} \ dx - B_{1} \left( f(n) - f(0) \right) + \frac{B_{2}}{2!} \left(f'(n) - f'(0) \right) + \frac{B_{4}}{4!} \left(f'''(n) - f'''(0) \right)$$

where $B_{i}$ is a Bernoulli number.

http://en.wikipedia.org/wiki/Bernoulli_number

So

$$\sum_{k=1}^{n} k^{4} = \frac{n^{5}}{5} + \frac{1}{2} (n^{4}) + \frac{1}{2} \left( \frac{1}{6} \right) (4n^{3}) + \frac{1}{24} \left(-\frac{1}{30}\right)(24n) $$

$$ = \frac{n^{5}}{5} + \frac{n^{4}}{2} + \frac{n^{3}}{3} - \frac{n}{30}$$

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Every rational polynomial $P$ that takes integer values $P[n]$ for $n\in\Bbb N$ can be written uniquely as an integer linear combination of the polynomials $\binom nk$ (for fixed $k$; it has degree $k$ in $n$): $$P[n]=\sum_{k=0}^{\deg P}c_k\binom nk \quad\text{for all $n\in\Bbb N$.} $$ These individual polynomials satisfy $\sum_{i=0}^{n-1}\binom ik=\binom n{k+1}$, and correspondingly $\binom{n+1}k-\binom nk=\binom n{k-1}$ (when $k>0$). Since each $\binom nk$ starts with $k$ values $\binom 0k=\binom1k=\cdots\binom{k-1}k=0$ followed by $\binom kk=1$, we can find the coefficient $c_0$ of $\binom n0$ as $P[0]$, and the remaining coefficients by forming $P'[n]=P[n+1]-P[n]=\sum_{k=1}^{\deg P}c_k\binom n{k-1}$ (this is not really the derivative of $P$ of course), and continuing recursively with$~P'$. In practice one need not compute $P'$ as a polynomial, just as a list of values at various$~n$, obtained by taking differences from the sequence of $P$. For $P[n]=n^4$ one gets $$ \begin{matrix} P: & 0 & & 1 & & 16 & & 81 & & 256 & & 625 & & 1296\\ & P': & 1 & & 15 & &65 & & 175 & & 369 & & 671\\ & & P'': & 14 & & 50 & & 110 & & 194 & & 302 \\ & & & & 36 & & 60 & & 84 & & 108\\ & & & & & 24 & & 24 & & 24 \\ \end{matrix} $$ so $(c_0,c_1,\ldots,c_4)=(0,1,14,36,24)$. Then $n^4=\binom n1+14\binom n2+36\binom n3+24\binom n4$, and $$ \sum_{i=0}^{n-1}n^4=\binom n2+14\binom n3+36\binom n4+24\binom n5. $$ You can easily expand that into a quintic polynomial expression in $n$. Please note the summation is up to $n-1$; this works better in this setting.

I think this is called the discrete version of Taylor expansion (or something close to that).

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The sum can be derived via combinatorial argument.

Count the number of quintuples $(x_1,x_2,x_3, x_4,x_5)$ with the property $x_5> \max\limits_{i\in S}$ from the set $S=\{1,2,...,n,n+1\}$,

then as $x_5$ varies from $2$ to $n+1$, the positions $(x_1,x_2,x_3, x_4)$ the positions can be filled in $\sum\limits_{i=1}^n i^4$ ways.

Counting it according to the cases: $5 = 4 + 1 = 2 + 2 + 1 = 3 + 1 + 1 = 2+1+1+1=1+1+1+1+1$ (where the numbers in partitions indicate the number of equal components in that partition, i.e. for example $5=4+1$ corresponds to the case $x_1=x_2=x_3=x_4<x_5$; and $2+2+1$ corresponds to the case two pairs have equal value but the value of the pair are different ,.. and so on). Count the number of ways in which each of the partitioning can happen: $1,14,36,24$ respectively (in the same order as I have presented the partitions).

You get the number of ways to be: ${n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5}$

Therefore, $\sum\limits_{i=1}^n i^4 = {n+1\choose 2}+14{n+1\choose 3}+36{n+1 \choose 4}+24{n+1\choose 5} = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n$.

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Here is my favorite method which works for any polynomial summand and you only need to remember two basic facts, one from calculus one and one about polynomials. First, since summations are analogous to integration, we have that

$$\int x^k \approx x^{k+1} \Rightarrow \sum x^k \approx x^{k+1}.$$

For your problem, let us define

$$f(n)=\sum_{i=1}^n i^4$$

and since the summand is a polynomial of degree four, the sum $f(n)$ must be a polynomial of degree five. If you don't believe me then just compute at least the first six values of $f(n)$ and compute the sixth differences and all of the terms will be zero (analogous to the sixth derivative of a fifth degree polynomial being zero).

Then using the (second) fact that a polynomial of degree five can be uniquely determined by six points, use points $$(1,f(1)),(2,f(2)),(3,f(3)),(4,f(4)),(5,f(5)),(6,f(6))$$ and compute the unique interpolating polynomial and you get

$$f(n)=\frac{6n^5+15n^4+10n^3-n}{30}.$$

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