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

**FernandoRevilla** · August 11th 2011, 01:26 AM

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

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

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

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

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

**Siron** · November 18th 2011, 01:35 PM

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

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

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

Therefore $T$ is a linear map.

**Deveno** · November 18th 2011, 04:09 PM

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

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

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

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

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

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

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

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

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

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

as we have seen T annihilates constants, so:

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

(short break for some etoufee for dinner)

**Deveno** · November 18th 2011, 05:49 PM

ok, back. dinner was tasty!

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

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

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

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

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

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

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

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

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

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

which is the desired formula for $S_4$

(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).

**Bourbaki** · November 20th 2011, 09:23 PM

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

**FernandoRevilla** · November 20th 2011, 10:07 PM

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)

**Renji Rodrigo** · November 21st 2011, 02:06 AM

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

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

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

Proof

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**Deveno** · November 21st 2011, 02:44 AM

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).

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

LinkBack

Thread Tools

Display

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

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

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

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

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

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

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

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

Similar Math Help Forum Discussions

Linear Algebra: Finding a specific linear transformation issue

Linear Algebra: Linear Independence question

Is linear algebra considered to be a type of abstract algebra?

[SOLVED] Elementary Linear Algebra, Systems of Linear Equations

introduction To Linear Transformations (sec. 1.8 Linear Algebra book by Lay)

Search Tags