# Taylor Series Expansion

1. Nov 17, 2005

### Nusc

y=kcosh(x/k) = k(e^(x/w) + e^(-x/k))

y ~ k[1 + x/k + (1/2!)(x^2/k^2) + (1/3!)(x^3/k^3)] +...+ k[1 - x/k +(1/2!)(x^2/k^2) - (1/3!)(x^3/k^3) +...]

All odd terms except 1 cancel out.

So we are left with

y = k [2 + (2/2!)(x^2/k^2) + (2/4!)(x^4/k^4) + (2/6!)(x^6/k^6) +....]

I've been staring at this for hours and I'm supposed to show that:

y= k + (1/2k)(x^2) for k>>x

What did I do wrong??? My error lies somewhere where the twos are bolded.

2. Nov 17, 2005

### shmoe

you lost a 1/2 in your definition of cosh:

$$cosh(x)=\frac{e^x+e^{-x}}{2}$$

3. Nov 17, 2005

### Hurkyl

Staff Emeritus
Well, you obviously are missing a factor of two somewhere... but I suppose you figured that out already. The answer lies here:

$$\cosh x = \frac{e^x + e^{-x}}{2}$$