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Math Help - roll theorem question..

  1. #1
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    roll theorem question..

    i am given that
    <br />
 \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n  = 0 \\<br />
    prove that there is a solution for 0<x<1
    using roll theorem
     <br />
 a_0 x^n  + a_1 x^{n - 1}  + .. + a_n  = 0 \\ <br />
    Last edited by transgalactic; February 5th 2009 at 09:45 PM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    i am given that
    <br />
 \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n  = 0 \\<br />
    prove that there is a solution for 0<x<1
    using roll theorem
     <br />
 a_0 x^n  + a_1 x^{n - 1}  + .. + a_n  = 0 \\ <br />
    Consider the function f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x .
    Notice that f(0) = f(1) = 0.
    Now apply Rolle's theorem and get f'(x_0) = 0 for some 0 < x_0 < 1.
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  3. #3
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    if i put x=1 on your function
    f(1) doesnt equal zero

    how you get f(1)=0
    ??
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  4. #4
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    Quote Originally Posted by transgalactic View Post
    how you get f(1)=0
    ??
    That is the hypothesis you were given.
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  5. #5
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    i cant see the equation that if i will put x=1
    i will get 0

    ?
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  6. #6
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    there is no such hypothesis in the posted question
    ??
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  7. #7
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    Quote Originally Posted by transgalactic View Post
    there is no such hypothesis in the posted question
    ??
    f(1) = \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0
    That is the hypothesis.
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  8. #8
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    Quote Originally Posted by ThePerfectHacker View Post
    f(1) = \frac{{a_0 }}{{n + 1}} + \frac{{a_1 }}{n} + \frac{{a_2 }}{{n - 1}} + .. + a_n = 0
    That is the hypothesis.
    so its only an assumption
    its not a fact??
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  9. #9
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    Quote Originally Posted by transgalactic View Post
    so its only an assumption
    its not a fact??
    You are given this assumption part of the problem.
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  10. #10
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    ok if f(0)=f(1) then there is a point c for which
    f'(c)=0
    0<c<1
    but its not a solution for f(x)

    its an extreme point
    ??
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  11. #11
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    Quote Originally Posted by transgalactic View Post
    ok if f(0)=f(1) then there is a point c for which
    f'(c)=0
    0<c<1
    but its not a solution for f(x)

    its an extreme point
    It is a solution to f'(x) = 0 (with x=c).
    But f'(x) = a_0x^n + ... + a_n.
    Thus, you found a solution to the polynomial equation a_0x^n + ... + a_n = 0.
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  12. #12
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    so you take a function
    <br />
f(x) = \tfrac{a_0}{n+1}x^{n+1} + \tfrac{a_1}{n}x^n + ... + a_n x\\<br />
    which is an anti derivative of my given equation that i need find solution to.
    <br />
f'(x)=a_0 x^n + a_1 x^{n - 1} + .. + a_n <br />

    so the extreme point of f(x) is the solution of f'(x)

    thanks
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  13. #13
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    Yes, finally you understand.
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