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Math Help - Roots and the IVT

  1. #1
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    Roots and the IVT

    Here is the question:

    a_0 + a_1/2 + ... + a_n/(n+1) = 0, all a's are real.
    Show that at least one real root betw. 0 and 1 exists for a_0 + a_1*x + ... + a_n*x^n = 0.

    I decided to go at this with induction. The n=1 step is easy; just solve a 2X2 system of equations. But I am stuck on the inductive step. I asked a classmate, and he suggested integrating the polynomial and then using the intermediate value theorem. But integrating the polynomial with respect to x doesn't produce anything that seems relevant or usable. What should I do?
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  2. #2
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    Re: Roots and the IVT

    The integral of the polynomial between 0 and 1 is zero.
    That means that either the polynomial is identically zero or the area lying above the x axis is equal to the area below the x axis.
    Which means that .... .
    Thanks from phys251 and topsquark
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Re: Roots and the IVT

    Quote Originally Posted by phys251 View Post
    Here is the question:

    a_0 + a_1/2 + ... + a_n/(n+1) = 0, all a's are real.
    Show that at least one real root betw. 0 and 1 exists for a_0 + a_1*x + ... + a_n*x^n = 0.

    I decided to go at this with induction. The n=1 step is easy; just solve a 2X2 system of equations. But I am stuck on the inductive step. I asked a classmate, and he suggested integrating the polynomial and then using the intermediate value theorem. But integrating the polynomial with respect to x doesn't produce anything that seems relevant or usable. What should I do?
    Hint: Let \displaystyle F(x)=a_0x+\frac{a_1}{2}x^2+\cdots+\frac{a_n}{n+1}x  ^{n+1}. Note that F(0)=F(1)=0. What does Rolle's theorem tell you?


    Quote Originally Posted by BobP View Post
    The integral of the polynomial between 0 and 1 is zero.
    That means that either the polynomial is identically zero or the area lying above the x axis is equal to the area below the x axis.
    Which means that .... .
    Unfortunately this does not actually work. Consider, for example, the fact that \displaystyle \int_{-1}^{1}x^3=0 yet x^3 is not simultaneously zero on that interval (you can use this to cook up an example for [0,1]. If you know that p(x) was nonnegative on [0,1] this would work (since p(x) would need to be zero a.e. and thus zero everywhere since it's continuous).

    Best,
    Alex
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    Re: Roots and the IVT

    Hi Drexel28

    Could you expand on your objection to the statement I made in my earlier post ?

    The example you give, \int^{1}_{-1}x^{3}dx=0, doesn't contradict my statement.

    My statement would say that the integral is equal to zero \bold{either} because the integrand is identically zero, (which clearly it isn't), \bold{or} because the part of the area lying above the x-axis is equal to the part of the area lying below the x-axis, (which is the case here).

    The implication of that is that f(x)=x^{3} is zero at at least one point in the interval, (in order that the curve crosses the x-axis).
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    MHF Contributor Drexel28's Avatar
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    Re: Roots and the IVT

    Quote Originally Posted by BobP View Post
    Hi Drexel28

    Could you expand on your objection to the statement I made in my earlier post ?

    The example you give, \int^{1}_{-1}x^{3}dx=0, doesn't contradict my statement.

    My statement would say that the integral is equal to zero \bold{either} because the integrand is identically zero, (which clearly it isn't), \bold{or} because the part of the area lying above the x-axis is equal to the part of the area lying below the x-axis, (which is the case here).

    The implication of that is that f(x)=x^{3} is zero at at least one point in the interval, (in order that the curve crosses the x-axis).
    BobP,

    I apologize for my previous statement. I too hastily read your argument. Everything seems correct with that--very clever!


    Best,
    Alex
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  6. #6
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    Re: Roots and the IVT

    Quote Originally Posted by BobP View Post
    The integral of the polynomial between 0 and 1 is zero.
    That means that either the polynomial is identically zero or the area lying above the x axis is equal to the area below the x axis.
    Which means that .... .
    Ohhhh, DEFINITE integration. Yep, that did the trick. Thanks.
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