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Let $b_0, b_1,...,b_n$ be real numbers with the property that

$$ b_0 + \frac{b_1}{2} + \frac{b_2}{3}+...+\frac{b_n}{n+1}=0 $$ Prove that the equation $$ b_0 + b_1x + b_2x^2+...+b_nx^n=0 $$ Has at least one solution in the interval $(0,1)$

How can I prove this? I really don't know where to start...

james42
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2 Answers2

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Hint: Let $f(x)=b_0x+\frac{b_1}{2}x^2+\cdots +\frac{b_n}{n+1}x^{n+1}$. Then $f(0)=f(1)=0$. Now use the Mean Value Theorem.

Remark: I think the following argument is more intuitive. Let $g(t)=a_0+a_1t +\cdots +a_nt^n$. We are told that $\int_0^1 g(t)\,dt=0$. If follows that either $g$ is identically $0$ or must change sign on the interval. I used the less intuitive Mean Value Theorem approach because it fits in better with the usual order in calculus courses.

André Nicolas
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    Question: What if the only roots are at $0$ and $1$? – pancini Dec 01 '15 at 01:07
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    The only roots of what? I imagine you mean $f(x)$. Makes no difference, for by the MVT there is a $c$ between $0$ and $1$ such that $\frac{f(1)-f(0)}{1-0}=f'(c)$. And the problem asks us in effect to show that $f'(x)=0$ has a solution between $0$ and $1$. We are not really interested in the roots of $f(x)$, it is a tool to get at $a_0+a_1x+\cdots+a_nx^n$. – André Nicolas Dec 01 '15 at 01:14
  • It follows more directly from Rolle's theorem, but it's just a special case of the Mean Value Theorem. – user236182 Dec 01 '15 at 01:18
  • Recall that the MVT says, under suitable differentiability conditions, that there is a $c$ such that $a\lt c\lt b$ and $\frac{f(b)-f(a)}{b-a}=f'(c)$. – André Nicolas Dec 01 '15 at 01:19
  • @user236182: Thanks, I has a typo in the remark. – André Nicolas Dec 01 '15 at 02:41
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Let $f(x)=b_0+b_1x+b_2x^2+........b_{n-1}x^{n-1}+b_nx^n$ We have $$\int _0^1 f(x)dt =\left[b_0x+\frac{b_1}{2}x^2+\frac{b_2}{3}x^3+........+\frac{b_{n-1}}{n}x^{n}+\frac{b_n}{n+1}x^{n+1}\right]_0^1=0$$ $$\int _0^1 f(x)dt =0$$ Hence, because the continuous function $f(x)$ is not identically zero on $[0,1]$, the equation $f(x)=0$ must have at least a root in $[0,1]$ (in order the area be zero, it must have regions of area negative and positive).

Piquito
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