# Rudin 8.10

1. May 7, 2008

### ehrenfest

[SOLVED] rudin 8.10

1. The problem statement, all variables and given/known data
Prove that $(1-x)^{-1} \leq \exp 2x[/tex] when [itex]0 \leq x \leq 1/2$.

2. Relevant equations
$$e^x = \sum_{i=0}^{\infty}\frac{x^n}{n!}$$

$$1/(1-x) = 1+x+x^2+\cdots$$

3. The attempt at a solution
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.

2. May 8, 2008

### HallsofIvy

Staff Emeritus
1. What are their values when x= 0?

2. How do their derivatives compare?

3. May 8, 2008

### ehrenfest

They are both 1.

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

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

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

4. May 8, 2008

### 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]

5. May 8, 2008

### ehrenfest

Yay, I got it thanks!