Root in (0,1) (0,1)

Let a0,a1,,an a_0, a_1, \ldots, a_n be real numbers which satisfy

ann+1+an1n++a23+a12+a0=0. \frac{a_n} { n+1} + \frac{a_{n-1} } {n} + \ldots + \frac{a_2}{3} + \frac{a_1} { 2} + a_0 = 0 .

Show that the equation

anxn+an1xn1++a1x+a0=0 a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0

has at least one solution in the interval (0,1) (0,1) .


This is a list of Calculus proof based problems that I like. Please avoid posting complete solutions, so that others can work on it.

Note by Calvin Lin
5 years, 7 months ago

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Consider the polynomial function : f(x)=anxn+1n+1+an1xnn++a1x22+a0x.f(x)=\frac{a_n x^{n+1}}{n+1}+\frac{a_{n-1}x^n}{n} +\dots +\frac{a_1x^2}{2}+a_0 x. Clearly f(0)=0f(0)=0, and f(1)=0f(1)=0 by the first given equality. By Rolle's theorem there is some x(0,1)x\in (0,1), such that f(x)=0f'(x)=0 which can be written as : anxn+an1xn1+a1x+a0=0.a_n x^n +a_{n-1}x^{n-1}+\dots a_1 x+a_0=0.

Haroun Meghaichi - 5 years, 7 months ago

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Good work! I did the same, but with due respect "please avoid posting complete solutions, so that others can work on it"

Piyal De - 5 years, 7 months ago

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It appears to b a problem based on ROLLE's THEOREM........ On integrating the given equation we get a new equation whose roots are 0 &1. Now the new equation is a polynomial so it is continuos as well as differentiable in (0,1) . Now apply ROLLE'S THEOREM which states that between any 2 roots of a continuous and differentiable function there must b atleast one root of its differentiable function.

Abhishek Pal - 5 years, 7 months ago

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Rolle's theorem for interval (0,1). :)

Vishal Sharma - 5 years, 7 months ago

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JEE PROBLEM

U Z - 4 years, 9 months ago

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