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Math Help - Integrals problem 1

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    MHF Contributor Also sprach Zarathustra's Avatar
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    Integrals problem 1

    Let p(x) be a polynomial so that int(a-->a+1){p(x)dx} = 0 for all a in R.
    Prove p(x)=0 for all x in R.
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let p(x) be a polynomial so that int(a-->a+1){p(x)dx} = 0 for all a in R.
    Prove p(x)=0 for all x in R.

    Suppose p(x)\neq 0 and let a\in\mathbb{R} be the largest real root of p(x) (if the pol. has no real roots then the contradiction is immediate since then the pol's values are all positive or all negative), and get a contradiction evaluating the integral from a to a+1.

    Tonio
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    MHF Contributor Also sprach Zarathustra's Avatar
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    "...and get a contradiction evaluating the integral from a to a+1."

    How I get this kind of contradiction(for the maximality of a as root of p(x))?

    My try:

    let deg(p(x))=n
    so int(a-->a+1){p(x)dx} = q(a+1) - q(a) = 0 , when deg(q(x))=n+1
    so, deg{q(x+1) - q(x)}=n
    q(a+1) - q(a)= m(a+1)=0
    contradiction? (m(x)!=p(x))
    Thanks!
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let p(x) be a polynomial so that int(a-->a+1){p(x)dx} = 0 for all a in R.
    Prove p(x)=0 for all x in R.
    So, I_n=\sum_{j=1}^{n}\int_{j}^{j+1}p(x)dx=\int_{1}^{n  +1}p(x)=0. Thus, \lim_{n\to\infty}I_n=0. Thus, if p(x)=a_0+\cdots+a_nx^n then \int p(x)=a_0x+\cdots+ a_nx^{n+1} and so \lim_{n\to\infty}\left(a_0x+\cdots+a_nx^{n+1}\righ  t)=0. Do your stuff.
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    WHY the coefficients p(x) and int(p(x)) are identical?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    WHY the coefficients p(x) and int(p(x)) are identical?
    Oops. Stupid typo. Won't affect anything.
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    Quote Originally Posted by Also sprach Zarathustra View Post
    "...and get a contradiction evaluating the integral from a to a+1."

    How I get this kind of contradiction(for the maximality of a as root of p(x))?


    Because after the largest real root the polynomial will be either all positive or all negative, so its integral cannot be zero, of course.

    Tonio



    My try:

    let deg(p(x))=n
    so int(a-->a+1){p(x)dx} = q(a+1) - q(a) = 0 , when deg(q(x))=n+1
    so, deg{q(x+1) - q(x)}=n
    q(a+1) - q(a)= m(a+1)=0
    contradiction? (m(x)!=p(x))
    Thanks!
    .
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