# Maclaurin series and general calculus question

1. Apr 4, 2016

### NihalRi

1. The problem statement, all variables and given/known data
This question has four parts which may follow up from each other so I incuded all the parts. The real problem I'm having is with d

Consider the function f ang g given by f (x)=( e^x+[e^-x])/2 & g (x) =( [e]^x]-[e^-x])/2
a) show f'(x) = g (x) and g'(x) = f (x)

b) find the first three non zero terms in the Maclaurin expansion of f (x)

c) hence find the value of lim (as x approaches 0) (1-f (x))/[x^2]

d) find the value of the improper integral ∫(0 to ∞) g (x)/[f (x)^2]dx

2. Relevant equations

3. The attempt at a solution
a) was pretty straight forward using the quotient rule
b) using the formula I got
1 + [x^2]/2! + [x^4]/4!
c) I replaced c for f (x) and simplifying gave me -1/2
d) I replaced g (x) with its differential from a) becoming
∫ (0 to ∞) f'(x)/[f(x)^2]
= lim (b approaches ∞) ∫ (0 to b) f'(x)/[f(x)^2]
I thought maybe I'd try substitution now but that's not working

Last edited: Apr 4, 2016
2. Apr 4, 2016

### Ray Vickson

What substitution did you try? There is one particular substitution that makes this problem very easy, indeed.

3. Apr 4, 2016

### NihalRi

Letting u = f (x) I think is ideal but my upper and lower limits would have to change to f (b) and -1 respectively. I'm getting,
∫(-1 to f (b)) u^-2 du
Which is [-1×u^-1](from -1 to f (b))
Since f (x) = u
We've got [-1×f(x)^-1)](from -1 to f (b))
now when I started replacing things started to get messy (I'd type it out but it's just really messy and probably won't end up making much sense) which makes me think that perhaps there is another way, that relates to part c) somehow.

I just tried making b approach infinity and I think the upper part of the definite integral vanishes and I'm left with just the lower part which equals 0.65. I'm not sure if this is right and I'm still suspicious that there could be another method:)

Last edited: Apr 4, 2016
4. Apr 4, 2016

### SammyS

Staff Emeritus
When you switch from u back to ƒ(x), the limits of integration go back to 0 and b .