Proving Inequality in Rudin 8.10

[ehrenfest]

[SOLVED] rudin 8.10

Homework Statement

Prove that (1-x)^{-1} \leq \exp 2x[/tex] when 0 \leq x \leq 1/2.<br /> <br /> <h2>Homework Equations</h2><br /> e^x = \sum_{i=0}^{\infty}\frac{x^n}{n!}<br /> <br /> 1/(1-x) = 1+x+x^2+\cdots<h2>The Attempt at a Solution</h2><br /> I tried working with the series and that failed miserable. Maybe I need to use calculus and find out whether the function is increasing or decreasing but I started that but I tried that a little and did not see how it would help.

 

[HallsofIvy]

1. What are their values when x= 0?


2. How do their derivatives compare?

 

[ehrenfest]

1. What are their values when x= 0?

They are both 1.

2. How do their derivatives compare?

Let f(x) = 1/(1-x) and g(x) = e^{2x}.

Then f&#039;(x) = 1/(1-x)^2 and g&#039;(x) = 2 e^{2x}.

Is there an obvious inequality between f'(x) and g'(x) when 0 \leq x \leq 1/2? I do not see it.

 

[MathematicalPhysicist]

well I think it's best to evaluate:
(e^(2x)(1-x))=f(x)

and to find where is its minimum point in the interval: [0,1/2]

 

[ehrenfest]

Yay, I got it thanks!