Finding Polynomial Equations for $1^4 + 2^4 + 3^4 + \ldots n^4$

Question

Find a polynomial expression for

$$ 1^4 + 2^4 + 3^4 + ... + n^4 $$

I know you have to use the big theorem but I can't figure out how you would start to compute the differences. Suggestions?

Answer 1

If you know that a quadratic captures $1+2+3+\cdots n$ and that a cubic captures $1^2+2+3+3^3 + \cdots + n^2$ it's not a stretch to guess that a $5th$ degree equation $ax^5 +bx^4+cx^3+dx^2+ex+f$ captures your sum. You can plug in $1, 2, 3, 4, 5, 6$ for $x$, and you will have $6$ linear equations with $6$ variables which you could solve (easily but tediously) via matrices. Then you could just prove your formula works by induction.

Answer 2

Note that \begin{aligned} \sum_{i=1}^{n}i^5 &= \sum_{i=0}^{n-1}(i+1)^5 \\ & = \sum_{i=0}^{n-1}\left(i^5 + 5i^4 + 10i^3 + 10i^2+5i+1 \right) \\ &= \sum_{i=0}^{n-1}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\ &= -n^5 + \sum_{\bbox[yellow]{i=1}}^{\bbox[yellow]{n}}i^5 + \sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right) \\ \end{aligned}

After canceling out the expression $\sum i^5$ from the sides, we'll get $$n^5=\sum_{i=0}^{n-1}\left(5i^4 + 10i^3 + 10i^2+5i+1 \right)$$

Which can be rewritten as \begin{aligned} 5\sum_{i=0}^{n-1}i^4 &= n^5 - \sum_{i=0}^{n-1}\left(10i^3 + 10i^2+5i+1 \right)\\ \sum_{i=0}^{n-1}i^4 &=\frac15\left(n^5- 10\sum_{i=0}^{n-1}i^3 -10\sum_{i=0}^{n-1}i^2 -5\sum_{i=0}^{n-1}i -n \right) \end{aligned}

By defining $S_k=\sum_{i=0}^{n-1}{i^k}$, this is equivalent to $$S_4 =\frac15\left(n^5- 10S_3 -10S_2 -5S_1 -n \right) $$ or in the form below to resemble the similarity with the binomial expansion ($S_0=n$) $$ 5S_4 +10S_3+10S_2+5S_1 + S_0 =n^5 $$

So, to find $S_4$, you just need to replace the expressions for $S_3$, $S_2$ and $S_1$, or repeat the above process to manually derive them.

Answer 3

Ignoring big theorems, $$ f(n) = 1^4+2^4+\ldots+n^4 $$ is quite blatantly a polynomial in $n$ with degree $5$. It follows that it is enough to compute $f(0),f(1),\ldots,f(5)$ to recover the coefficients of such a polynomial by Lagrange interpolation:

$$ f(x) = \sum_{k=0}^{5}\,f(k)\!\!\!\!\prod_{\substack{j\in\{0,\ldots,5\}\\ j\neq k}}\frac{x-j}{k-j}.$$ Alternative approach: since $$ \sum_{n\geq 0} n^4 e^{-nx} = \frac{d^4}{dx^4}\sum_{n\geq 0}e^{-nx} = \frac{d^4}{dx^4}\left(\frac{1}{1-e^{-x}}\right) =\frac{e^x(e^x+1)(e^{2x}+10 e^{x}+1)}{(e^x-1)^5}$$ it follows that for any $x\in(0,1)$ we have $$ \sum_{n\geq 0}n^4 x^n = \frac{x \left(1+11 x+11 x^2+x^3\right)}{(1-x)^5} $$ and $f(N)$ is the coefficient of $x^N$ in $\frac{x \left(1+11 x+11 x^2+x^3\right)}{(1-x)^6}$. By stars and bars, $$\boxed{\,f(N) = 1^4+2^4+\ldots+N^4= \color{red}{\frac{N(N+1)(2N+1)(3N^2+3N-1)}{30}} }$$

Answer 4

Even though this might be too much for simple task you want (you can assume the polynomial of fifth degree and calculate coefficients) I think it is worth mentioning that there is a general formula for finding polynomial of sum of $p$-th powers $1^p+2^p+\dots+n^p$, and it is called Faulhaber's formula.

In short

$$1^p+2^p+\dots+n^p = \frac{1}{p+1} \sum_{j=0}^{p}(-1)^j\binom{p+1}{j}B_j n^{p+1-j}$$

where $B_j$ are called Bernoulli numbers, which you can compute for example as

$$ B_n = \sum_{k=0}^{n}\frac{1}{k+1}\sum_{r=0}^{k}(-1)^r \binom{k}{r} r^n $$

giving first couple values $1,-1/2$,$1/6$,$0$,$-1/30$,$\dots$.

So in your case this becomes

\begin{align} 1^4+2^4+\dots+n^4 &= \frac{1}{5} \sum_{j=0}^{4}(-1)^j\binom{5}{j}B_j n^{5-j}\\ &= \frac{1}{5} ((-1)^0\binom{5}{0}B_j n^{5}+(-1)^1\binom{5}{1}B_1 n^{5-1}\\ &+(-1)^2\binom{5}{2}B_2 n^{5-2}+(-1)^3\binom{5}{3}B_3 n^{5-3}+(-1)^4\binom{5}{4}B_4 n^{5-4})\\ &= \frac{1}{5} (B_0 n^{5}-5 B_1 n^{4}+10 B_2 n^{3}-10 B_3 n^{2}+5 B_4 n)\\ &= \frac{1}{5} (n^{5}+\frac{5}{2}n^{4}+\frac{10}{6} n^{3}-\frac{5}{30} n)\\ \end{align}

Answer 5

If you know that $f(n) = a_5n^5 + \cdots + a_0$, you can calculate the coefficients using limits: $$ a_5 = \lim_{n\to\infty}\frac{f(n)}{n^5} = \lim_{n\to\infty}\frac{1^4+\cdots+n^4}{n^5} = \lim_{n\to\infty}\frac{(n+1)^4}{(n+1)^5-n^5} = \cdots = \frac15. $$ (Cesàro-Stolz used in the third $=$).

Continue with $$a_4 = \lim_{n\to\infty}\frac{f(n)-a_5n^5}{n^4} = \cdots$$ $$\cdots$$

Answer 6

There is a simpler formula without using the Bernoulli numbers,

$$(n+1)((n+1)^p-1) = \sum_{k=1}^{p} {p+1 \choose k }S_k $$

where $S_p = 1^p + 2^p + \dots + n^p$

For example, $p=1$

$$ (n+1)(n+1- 1)= {2 \choose 1} S_1 = 2S_1$$ $$ S_1 = \frac{n(n+1)}2 $$

Similarly, for $p=4$

$$ (n+1)((n+1)^4 - 1) = 5S_1 + 10S_2 + 10S_3 +5S_4$$

from where $S_4$ can be easily computed in terms of lower sums, but this representation is easier to remember.

Note this is a generalization of @Babak's answer.

Answer 7

Bernoulli found the following expression for what you describe:

$\sum_{i=1}^{n} i^4 = \frac{6n^5+15n^4+10n^3-n}{30}$

[ derived from p.90 of David Eugene Smith's A Source Book in Mathematics, Dover ed. ISBN 0-486-64690-4 ]