Given $a_0, a_1, .., a_n$ are the real numbers satisfying $$\dfrac {a_0}{n+1} + \dfrac {a_1}{n} +......+\dfrac {a_{n-1}}{2}+a_n=0$$ then prove that there exists at least one real root of the equation $a_0 x^n + a_1 x^{n-1} +....+a_n=0$ such that $x\in (0,1)$.

My teacher told me to integrate the left hand side of the given equation and call it as $g(x)$, then after verifying the requirements of Rolle's Theorem again differentaite $g(x)$ to get the required equation and this completes the proof. However, I don't understand that first integrating and then differentiating the same thing works here.

  • $\begingroup$ Rolle's Theorem allows you to conclude something about a value of $f'(x)$ given some information about some values of $f(x)$. You want to conclude something about $a_0x^n+a_1x^{n-1}+\cdots +a_n$, so you need to view this as $f'(x)$ for some $f(x)$. Hence you need to integrate to figure out what $f(x)$ is. $\endgroup$ – kccu Mar 31 at 17:39

Let $$p(x) = a_0x^n + a_1x^{n-1}+\ldots + a_n$$

and let $g$ be an antiderivative of $p$ which satisfies $g(0)=0$, that is

$$g(x) = \frac{a_0x^{n+1}}{n+1}+\frac{a_1x^n}{n}+\ldots +a_nx$$

Check that $g(1)=g(0)=0$, now apply Rolle's theorem to conclude about root of $p$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.