A theorem to rule them all

Calculus Level 2

True or False?

Let $a_0, a_1 , \ldots , a_n$ be reals such that $\dfrac {a_0}1 + \dfrac{a_1}2 + \cdots + \dfrac{a_n}{n+1} = 0$ , then there exists a real $z\in [0,1]$ such that $a_0 + a_1 z + \cdots + a_n z^n = 0$ .

True False

1 solution

Mohammad Hamdar
Dec 22, 2016

Relevant wiki: Rolle's Theorem

Consider the function $f\left( x \right) ={ a }_{ 0 }x+\frac { { a }_{ 1 }{ x }^{ 2 } }{ 2 } +\frac { { a }_{ 2 }{ x }^{ 3 } }{ 3 } +...+\frac { { a }_{ n }{ x }^{ n+1 } }{ n+1 }$
$f$ is continuous over $\left[ 0,1 \right]$
$f$ is differentiable for $0<x<1\quad$

$f(0)=f(1)=\frac { { a }_{ 0 } }{ 1 } +\frac { { a }_{ 2 } }{ 2 } +...+\frac { { a }_{ n } }{ n+1 } =0$ Then by Rolle's theorem , there exists $z\in \left[ 0,1 \right]$ such that $f^{ ' }\left( z \right) =0$ i.e. ${ a }_{ 0 }+{ a }_{ 1 }z+\cdots+{ a }_{ n }{ z }^{ n }=0$ .

A theorem to rule them all

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