# Could someone help me get my head around this Taylor Series stuff

the7joker7

## Homework Statement

The Taylor series for f(x) = ln(sec(x)) at a = 0 is sum_(n=0to infinity) c(sub n) (x)^n.

Find the first few coefficients.

## The Attempt at a Solution

I've been trying to figure out where to start by looking it up...I've seen instructions that each coefficient is just the nth derivative of the function, other's saying it's the nth derivative divided by n!, others say it's the nth derivative times n...and nothing comes out properly.

Help!

Homework Helper
The Taylor series of a function,f(x) about a point x=a is given by:

$$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)(x-a)^2}{2!}+\frac{f'''(a)(x-a)^3}{3!}+\frac{f''''(a)(x-a)^4}{4!}+...+\frac{f^n(a)(x-a)^n}{n!}+...$$

as you can see, the nth term of a Taylor series is $\frac{f^n(a)(x-a)^n}{n!}$.
But what you need to find are the first few terms, so find f(a), then f'(a) and so forth and just put those values into the equation above.

the7joker7
Okay, so I did...

ln(sec(0)) = 0, first term, this is correct...

(sec(1) * tan(1) / sec(1)) * (1 - 0) = 1.5574, but the correct answer is 0...ugh!

Help?

Homework Helper
Okay, so I did...

ln(sec(0)) = 0, first term, this is correct...

(sec(1) * tan(1) / sec(1)) * (1 - 0) = 1.5574, but the correct answer is 0...ugh!

Help?

Your f(x)=ln(secx). You chose to expand about the point x=0. So the 'a' in the equation is 0

f'(x)=secxtanx/secx, f'(0)=sec0tan0/sec0=0.

the7joker7
Is (sec(x)*(sec(x)*sec(x)^2+tan(x)*sec(x)*tan(x))-sec(x)*tan(x)*sec(x)*tan(x))/sec(x)^2 f''(x)?

EDIT: Jeez, these derivatives get huge...I must be working too hard somehow.

Last edited:
Homework Helper
No it does not.

f'(x) is given above by rock.freak667. Just differentiate it to get the answer. Note that you can cancel out sec x in the given expression to make your life much, much easier.