# If $a_0, a_1,\ldots,a_n \in R$ and $\frac{a_0}{1}+\frac{a_1}{2}+\cdots+\frac{a_n}{n+1}=0$, then exists $x\in(0,1)$ s.t. $a_0+a_1x+\cdots+a_nx^n=0$ [duplicate]

Prove or disprove: If $a_0, a_1,\ldots,a_n \in R$ satisfy $\frac{a_0}{1}+\frac{a_1}{2}+\cdots+\frac{a_n}{n+1}=0$, then exists $x\in(0,1)$ such that $a_0+a_1x+\cdots+a_nx^n=0$

Let's define f(x)= $a_0+a_1x+a_2x^2+\cdots+a_nx^n$

If I check the limit of f(x) as $x\rightarrow 0$ I get $a_0$. The limit of f(x) as $x\rightarrow 1$ is $a_0+a_1+a_2+\cdots+a_n$

I'm not sure if I can use the IVT to claim there exists a solution since I'm not sure if $a_0\gt0$ or $a_0+a_1+a_2+\cdots+a_n \gt 0$

Assuming I do need to use the IVT, can I split it up into different cases? i.e. four cases, one when $a_0\gt0$ and $a_0+a_1+a_2+\cdots+a_n \gt 0$, one with $a_0\lt 0$ etc.

Is there another concrete way to prove this?

## marked as duplicate by Arnaud D., Paul Enta, Ethan Bolker, Holo, Namaste calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 6 '18 at 10:58

• Consideer using the MVT rather than the IVT. – servabat May 28 '16 at 18:16

Let $F(x) = \sum_{i=0}^n \frac{a_i}{i+1} x^{i+1}$.

$F$ is continuous on $[0,1]$ and differentiable on $(0, 1)$, and $f(x) = F'(x) = \sum_{i=0}^n a_i x^{i}$.

By the mean value theorem : $$\exists c \in (0,1) \text{ s.t. } f(c) = \frac{F(1) - F(0)}{1 - 0} = 0$$ (That is actually Rolle's Theorem as $F(0) = F(1)$)

• Rolles theorem requires that f(0)=f(1). $f(0)=a_0$, whereas f(1)= $a_0+a_1+a_2...$ How do we prove that f(0)=f(1)? – user300011 May 28 '16 at 18:28
• @RonaldB : yes, but consideer using $F$ rather than $f$. The $F$ function I defined is such that $F(0) = F(1)$ and such that $F' = f$ (but actually, any antiderivative of $f$ would work). – servabat May 28 '16 at 18:29
• I'm not quite sure of what exactly F is. Are we assuming that that the function at x=0 and x=1 are equal? – user300011 May 28 '16 at 18:32
• Just look how I defined $F$ : $F(x) = \sum_{i=0}^n \frac{a_i}{i+1} x^{i+1}$. $F$ is an antiderivative of $f$ (i.e. $F' = f$). So $F(0) = 0$ and $F(1) = \sum_{i=0}^n \frac{a_i}{i+1} = 0 = F(0)$. Which mean you can apply Rolle's Theorem on $F$. Because $F$ it is an antiderative of $f$, Rolle's will give you a property on $f$. – servabat May 28 '16 at 18:34

Denote by $P(x)$ the polynomial $a_{0}+a_{1}x+...a_{n}x^{n}.$$\\$

By Mean Value Theorem, $\exists c\in (0,1)$ such that -$\\$

$\int_{0}^{1}P(x)dx = P(c)$.

Observe that $P(c) = 0.$