problem

**ilikeapples** · Jun 10th 2010, 08:14 AM

problem attached.

**mohammadfawaz** · Jun 10th 2010, 10:02 AM

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

**ilikeapples** · Jun 10th 2010, 10:48 AM

Originally Posted by mohammadfawaz

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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