# Thread: S_4=1^4+2^4+...+n^4 (linear algebra method)

1. ## S_4=1^4+2^4+...+n^4 (linear algebra method)

This problem provides a linear algebra method for computing sums of the form $\displaystyle S_k=1^k+2^k+3^4+\ldots+n^k$ with $\displaystyle k$ positive integer.

1.
Let $\displaystyle \mathbb{R}_5[x]$ the real vector space of all polynomials of degree $\displaystyle \leq 5$. Consider the map $\displaystyle T:\mathbb{R}_5[x]\rightarrow{\mathbb{R}_5[x]}$ defined by $\displaystyle \forall p(x) \in \mathbb{R}_5[x],\;T(p(x))=p(x+1)-p(x)$ . Prove that $\displaystyle T$ is a linear map.

2.
Prove that $\displaystyle p\in \ker T\Leftrightarrow\;p$ is constant.

3.
Find the image by $\displaystyle T$ of the elements of the canonical basis of $\displaystyle \mathbb{R}_5[x]$ and determine $\displaystyle T^{-1}(x^4)$ .

4.
Using 3. , find an expression for $\displaystyle S_4=1^4+2^4+3^4+\ldots+n^4$ .

2. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

This is what I have for 1.
We have to prove that:
$\displaystyle \forall a(x),b(x) \in \mathbb{R}_5[x]: T(a(x)+b(x))=T(a(x))+T(b(x))$ (1)
$\displaystyle \forall a(x)\in \mathbb{R}_5[x], \forall \lambda \in \mathbb{R}: T(\lambda a(x))=\lambda T(a(x))$ (2)

Proof (1):
$\displaystyle T(a(x)+b(x))=T((a+b)(x))=(a+b)(x+1)-(a+b)(x)=a(x+1)+b(x+1)-a(x)-b(x)=[a(x+1)-a(x)]+[b(x+1)-b(x)]=T(a(x))+T(b(x))$

Proof (2):
$\displaystyle T(\lambda a(x))=\lambda a(x+1)-\lambda a(x)=\lambda[a(x+1)-a(x)]=\lambda T(a(x))$

Therefore $\displaystyle T$ is a linear map.

3. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

2. suppose T(p(x)) = 0, with $\displaystyle p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$. then

$\displaystyle T(p(x)) = a_0 + a_1(x+1) + a_2(x+1)^2 + a_3(x+1)^3 + a_4(x+1)^4 + a_5(x+1)^5\newline - a_0 - a_1x - a_2x^2 - a_3x^3 - a_4x^4 - a_5x^5$

$\displaystyle = (a_1+a_2+a_3+a_4+a_5) + (2a_2+3a_3+4a_4+5a_5)x + (3a_3+6a_4+10a_5)x^2\newline+(4a_4+10a_5)x^3+5a_5x ^5$

since this is by supposition the 0-polynomial, we have immediately that:

$\displaystyle 5a_5 = 0 \implies a_5 = 0 \implies 4a_4 =0 \implies\newline a_4 = 0 \dots \implies a_3 = a_2 = a_1 = 0$

hence $\displaystyle p(x) = a_0$, a constant polynomial. the reverse implication is trivial.

3. $\displaystyle T(1) = 0$ (see above).
$\displaystyle T(x) = 1+x-x = 1$
$\displaystyle T(x^2) = 1+2x$ (just take my word for it,ok?).
$\displaystyle T(x^3) = 1+3x+3x^2$ (think: truncated pascal's triangle).
$\displaystyle T(x^4) = 1+4x+6x^2+4x^3$(*whistling*)
$\displaystyle T(x^5) = 1+5x+10x^2+10x^3+5x^4$.

so what is the pre-image of $\displaystyle x^4$? this amounts to solving:

$\displaystyle \begin{bmatrix}0&1&1&1&1&1\\0&0&2&3&4&5\\0&0&0&3&6 &10\\0&0&0&0&4&10\\0&0&0&0&0&5\\0&0&0&0&0&0 \end{bmatrix} \begin{bmatrix}a_0\\a_1\\a_2\\a_3\\a_4\\a_5 \end{bmatrix} = \begin{bmatrix}0\\0\\0\\0\\1\\0\end{bmatrix}$

so $\displaystyle a_5 = \dfrac{1}{5}$
$\displaystyle 4a_4 + 10a_5 = 0 \implies 4a_4 + 2 = 0 \implies a_4 = -\dfrac{1}{2}$
$\displaystyle 3a_3 + 6a_4 + 10a_5 = 0 \implies 3a_3 - 3 + 2 = 0 \implies a_3 = \dfrac{1}{3}$
$\displaystyle 2a_2 + 3a_3 + 4a_4 + 5a_5 = 0 \implies 2a_2 = 0 \implies a_2 = 0$
$\displaystyle a_1 + a_2 + a_3 + a_4 + a_5 = 0 \implies a_1 + \dfrac{1}{3} - \dfrac{1}{2} + \dfrac{1}{5} = 0 \implies a_1 = -\dfrac{1}{30}$

as we have seen T annihilates constants, so:

$\displaystyle T^{-1}(x) = \{c - \dfrac{1}{30}x + \dfrac{1}{3}x^3 - \dfrac{1}{2}x^4 + \dfrac{1}{5}x^5 \in \mathbb{R}_5[x] : c \in \mathbb{R}\}$

(short break for some etoufee for dinner)

4. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

ok, back. dinner was tasty!

4. let $\displaystyle q(x) = -\dfrac{1}{30}x + \dfrac{1}{3}x^3 - \dfrac{1}{2}x^4 + \dfrac{1}{5}x^5$

then, as we have seen, $\displaystyle T(q(x)) = x^4$.

this means that $\displaystyle q(x+1) - q(x) = x^4$ for all real x. so, in particular:

$\displaystyle 1^4 = q(2) - q(1)$
$\displaystyle 2^4 = q(3) - q(2)$
$\displaystyle \vdots$
$\displaystyle n^4 = q(n+1) - q(n)$

therefore, $\displaystyle 1^4 + 2^4 + 3^4 + \dots + n^4 = q(n+1) - q(1)$

but $\displaystyle q(n+1) - q(1) =$

$\displaystyle -\dfrac{1}{30}(n+1) + \dfrac{1}{3}(n+1)^3 - \dfrac{1}{2}(n+1)^4 + \dfrac{1}{5}(n+1)^5 + \dfrac{1}{30} - \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{5}$

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

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

hence $\displaystyle 1^4 + 2^4 + 3^4 + \dots + n^4 = \dfrac{6n^5 + 15n^4 + 10n^3 - 1}{30}$

which is the desired formula for $\displaystyle S_4$

(i'd like to add that this was a very entertaining problem, and that (in theory) Sn could be calculated from this method without much trouble, as the difficulty involved grows linearly with n. we still wind up with a upper triangular matrix for T, of rank n-1, so the resulting system of linear equations we get are easy to solve, although for large n, the repetition could get mind-numbing).

5. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

Beautiful! Can I ask, what is the source of the problem?

6. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

Originally Posted by Bourbaki
Beautiful! Can I ask, what is the source of the problem?
That is a problem proposed in an exam of Linear Algebra (Industrial Engineering School, UPM, Madrid)

7. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

(Off-topic) , there is a way to solve the general case using derivatives

Theorem
$\displaystyle s(n) = \sum^{n}_{k=0} k^p =\sum^{p+1}_{k=1}a_k n^k$
where

$\displaystyle a_{t+1} =\frac{1}{t+1}\sum^{p+1}_{k=t+2}a_k{ k \choose t} (-1)^{k-t}\;\; \mb{and}\;\; a_{p+1}=\frac{1}{p+1}.$

Proof

From $\displaystyle s(n) = \sum^{n}_{k=0} k^p$, we have $\displaystyle s(n) -s(n-1)=n^p$.

suppose $\displaystyle s(n)= \sum^{p+1}_{k=1}a_k n^k$

with independent term $\displaystyle a_0=0$ because $\displaystyle s(0) =0$.

Apply the $\displaystyle p$-th derivative on $\displaystyle s(n) -s(n-1)=n^p$ the result is

$\displaystyle s^{(p)}(n) -s^{(p)}(n-1) =p!$

we have $\displaystyle s^{(p)}(n) = a_p.p! + a_{p+1}(p+1)!n$ so $s^{(p)}(n) -s^{(p)}(n-1) (n)=a_{p+1}(p+1)!=p!$ then

$\displaystyle a_{p+1}=\frac{1}{(p+1)} .$

We gonna find the other coefficients . We take the $\displaystyle t$-th derivative on $\displaystyle s(n) -s(n-1)=n^p$, with $\displaystyle 0 \leq t<p$, so

$\displaystyle s^{(t)}(n) -s^{(t)}(n-1) =t!{p \choose t} n^{p-t}$, taking $\displaystyle n=0$ we have

$\displaystyle s^{(t)}(0) =s^{t}(-1)$

Using $\displaystyle s(n)= \sum^{p+1}_{k=1}a_k x^k$ and applying again the $\displaystyle t$-th derivative

$\displaystyle s^{(t)} (n)= \sum^{p+1}_{k=t}a_k(t!){ k \choose t} n^{k-t}$

from $\displaystyle s^{(t)}(0) =s^{t}(-1)$ follows

$\displaystyle \sum^{p+1}_{k=t}a_k(t!){ k \choose t} 0^{k-t} = \sum^{p+1}_{k=t}a_k(t!){ k \choose t} (-1)^{k-t} \Rightarrow$

$\displaystyle a_t.t! = t!\sum^{p+1}_{k=t+2}a_k{ k \choose t} (-1)^{k-t} + a_t(t!)-a_{t+1}(t+1)! \Rightarrow$

$\displaystyle a_{t+1} =\frac{1}{t+1}\sum^{p+1}_{k=t+2}a_k{ k \choose t} (-1)^{k-t}$

8. ## Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

and this should not be surprising, because:

1) the derivative is just a difference quotient. in fact, you don't need derivatives, you could do an induction argument on successive differences.
2) the rref of the matrix for T, and the differential operator D, are the same, so this is essentially the "same process".
3)pascal's triangle rulez (binomial coefficients are great with a little cream and sugar. oh wait, that's biscotti. my bad).