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  1. #1
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    problem

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

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

    Hope this helps
    nice.
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