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

  1. #1
    MHF Contributor FernandoRevilla's Avatar
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    S_4=1^4+2^4+...+n^4 (linear algebra method)

    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 .
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

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

    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)
    Last edited by Deveno; November 18th 2011 at 04:53 PM.
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  4. #4
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    Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

    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).
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  5. #5
    Newbie Bourbaki's Avatar
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    Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

    Beautiful! Can I ask, what is the source of the problem?
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Re: S_4=1^4+2^4+...+n^4 (linear algebra method)

    Quote Originally Posted by Bourbaki View Post
    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)
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  7. #7
    Junior Member Renji Rodrigo's Avatar
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    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
    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}
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  8. #8
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    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).
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