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Prove existence of a real root.

If $a_0$+$\frac{a_1}{2}$+$\frac{a_2}{3}+\ldots+\frac{a_n}{n+1}=0$, how to prove ${{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\cdots +{{a}_{n}}{{x}^{n}}=0$ has at least one real root in $(0,1)$.

I know constructor $f(x)={{a}_{0}}x+\frac{{{a}_{1}}}{2}{{x}^{2}}+\frac{{{a}_{2}}}{3}{{x}^{3}}+\cdots +\frac{{{a}_{n}}}{n+1}{{x}^{n+1}}$, and then use the Mean Value Theorem.

I want to know whether we can use mathematical induction to prove, obviously $n = 1$ proposition holds.

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marked as duplicate by lhf, Thomas, Rudy the Reindeer, draks ..., Noah Snyder Oct 10 '12 at 9:25

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"I know constructor"? What do you mean? Anyway, have you tried to prove the induction step? Does it work? –  Siminore Oct 8 '12 at 10:06
    
This is a natural number proposition.So I wanted to try to use mathematical induction.I made some attempts, without success, to seek help.Thank you –  geometryscience Oct 8 '12 at 10:10
    
Perhaps possible.. but not easy.. WHY do you insist on induction? –  Berci Oct 8 '12 at 10:15
    
There is no special intention.Just suddenly thought, because the classical solution of this problem is the intermediate value theorem, I want to try the other solution is also very interesting.Thank you –  geometryscience Oct 8 '12 at 10:18
    
Let me know if you find one.. –  Berci Oct 8 '12 at 10:36

2 Answers 2

Yes.

You just proved it, by the Mean Value Theorem, no? What is an induction needed for?

Let's denote the LHS of the original equation $g(x)$, then your 'constructor' has $f'(x)=g(x)$. So, there is a $\xi\in (0,1)$: $$f'(\xi)=\frac{f(1)-f(0)}{1-0} $$ But now $f(0)=0$ and, by the hypothesis, also $f(1)=0$. We're done.

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Because $f'(x)={{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}}+\cdots +{{a}_{n}}{{x}^{n}}$ and $f(0)=f(1)=0$ then exists point $x_0 \in (0,\,1)$ such that $f'(x_0)=0$

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