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Math Help - problem

  1. #1
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    problem

    problem attached.
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  2. #2
    Member mohammadfawaz's Avatar
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    Let f(x)=a_0x+\frac{a_1}{2}x^2+...+\frac{a_n}{n+1}x^{n  +1}
    Clearly, this function is differentiable in the interval [0,1].
    Let's apply the mean value theorem on that interval: f(1)-f(0)=f'(c)(1-0) where c\in(0,1).
    Hence: f(x)=a_0+\frac{a_1}{2}+...+\frac{a_n}{n+1}-0=a_0+a_1c+a_2c^2+...+a_nc^n (check this!)
    Refer to the given to get: 0=a_0+a_1c+a_2c^2+...+a_nc^n
    Hence, we have found a constant c\in(0,1) satisfying the given equation. Thus, given equation has at least one solution in (0,1)

    Hope this helps
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  3. #3
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    Quote Originally Posted by mohammadfawaz View Post
    Let f(x)=a_0x+\frac{a_1}{2}x^2+...+\frac{a_n}{n+1}x^{n  +1}
    Clearly, this function is differentiable in the interval [0,1].
    Let's apply the mean value theorem on that interval: f(1)-f(0)=f'(c)(1-0) where c\in(0,1).
    Hence: f(x)=a_0+\frac{a_1}{2}+...+\frac{a_n}{n+1}-0=a_0+a_1c+a_2c^2+...+a_nc^n (check this!)
    Refer to the given to get: 0=a_0+a_1c+a_2c^2+...+a_nc^n
    Hence, we have found a constant c\in(0,1) satisfying the given equation. Thus, given equation has at least one solution in (0,1)

    Hope this helps
    nice.
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