My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum

LinkBack Thread Tools Display Modes
November 28th, 2017, 07:14 AM   #1
Joined: Jan 2016
From: Blackpool

Posts: 95
Thanks: 2

Descarte's rule of signs proof

Prove by induction on n that if p(x) is a polynomial of degree n then the function

\[e^{x}-p(x)\] has at least n+1 zeros.

By zero I mean that f(x) for some x=0

This is my solution however I feel the proof is incomplete and missing some theorems.
Jaket1 is offline  
November 28th, 2017, 03:27 PM   #2
Senior Member
Joined: Aug 2017
From: United Kingdom

Posts: 203
Thanks: 60

Math Focus: Algebraic Number Theory, Arithmetic Geometry
I'm not really following your argument. How have you used the assumption that it's true for polynomials of degree $k$ to prove it's true for polynomials of degree $k+1$?

A possible approach: suppose the statement is true for all polynomials of degree k. Let $p(x)$ be any polynomial of degree $k+1$ and write $f(x) = e^x - p(x)$. If you differentiate this with respect to $x$ you get $f'(x) = e^x - p'(x)$. Now $p'(x)$ is a polynomial of degree $k$, so by the inductive hypothesis, $f'(x)$ has at most $k$ zeros. But the zeros of $f'(x)$ are precisely the stationary points of $f(x)$.

Can you see how to use this to bound the number of zeros of $f(x)$?
cjem is offline  

  My Math Forum > College Math Forum > Real Analysis

descarte, proof, rule, signs

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Quotient Rule Proof hyperbola Calculus 7 April 21st, 2015 02:19 AM
Descarte's Rule of Signs and Imaginary Roots mepomero Applied Math 4 January 24th, 2014 08:36 AM
Descartes's Rule of Signs Dan154 Algebra 1 October 4th, 2012 07:30 AM
rule of signs proof dryajov Applied Math 2 January 21st, 2009 07:49 PM
Quotient rule proof courteous Calculus 6 October 11th, 2008 06:10 AM

Copyright © 2018 My Math Forum. All rights reserved.