   Chapter 4.5, Problem 103E

Chapter
Section
Textbook Problem

PUTNAM EXAM CHALLENGEIf a 0 , a 1 , … , a n are real numbers satisfying a 0 1 + a 1 2 + ⋯ + a n n + 1 = 0 show that the equation a 0 + a 1 x + a 2 x 2 + ⋯ + a n x n = 0 has at least one real root.

To determine

To Prove: If a0,a1,,an are real then, a0+a1x++anxn=0 has at least one real root.

Explanation

Given:

The numbers a0,a1,...,an are real and satisfy,

a01+a12++ann+1=0

Formula Used: The mean value theorem isto be applied as per whichif f is continuous on the closed interval [a,b], there exists a number c in the closed interval [a,b] such that, abf(x)dx=f(c)(ba)

Proof: Let f(x)=a0+a1x++anxn.

On integration with limits 0, and 1 to get,

01f(x)dx=01(a0+a1x++anxn)dx=[a0x+a1x22+

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