# Taylor series

1. Jul 23, 2007

### ehrenfest

1. The problem statement, all variables and given/known data

For which real numbers c is $$(e^x+e^{-x})/2 \leq e^{cx^2}$$ for all real x?

2. Relevant equations

3. The attempt at a solution
I think you can expand both sides into series and term by term compare them. The left side is
$$\sum_{n=0}^{\infty}\frac{x^{2n}}{2n!}$$
Can someone help me with the right side?

2. Jul 23, 2007

### christianjb

Try x=iy. LHS = cos(y), RHS=exp(-cy^2)

Hmmm... maybe not. x is real.

cosh(x)=1+x^2/2!+x^4/4!+x^6/6!+...

exp(cx^2)=1+cx^2+c^2x^4/2!+c^3x^6/3!+...

exp(cx^2)-cosh(x)=(c-1/2!)x^2+(c^2/2!-1/4!)x^4+(c^3/3!-1/6!)x^6+...

c>0.5 means that each term is +ve.

Last edited: Jul 24, 2007
3. Jul 24, 2007

### nrqed

I guess you mean $$\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!}$$

( 2n! really means 2 (n!))

Why don't you expand the RHS also to get
$$\sum_{n=0}^{\infty}\frac{(cx^2)^{n}}{n!} = \sum_{n=0}^{\infty}\frac{c^n \times x^{2n}}{n!}$$

and then simply compare

4. Jul 24, 2007

### Kummer

5. Jul 24, 2007

### ehrenfest

Sorry--my calculus is a little fuzzy.

Are you allowed to use the same Taylor series expansion of exp(x) with exp(cx^2)? Because you generate the Taylor series expansion by taking the derivatives of exp(x) at 0, and the derivatives of exp(x) at 0 are different from the derivatives of exp(cx^2) at 0, it seems like you might need some variant of the chain rule to make that work. I mean instead of just plugging in cx^2 for x.

6. Jul 24, 2007

### christianjb

Nah, you can just substitute x'=cx^2 on both LHS and RHS.