Finding Taylor series coefficients for ln(sec(x)) at x=0

[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!

 

[rock.freak667]

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?

 

[rock.freak667]

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.

 

[Defennder]

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.

 

[Dick]

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.

Yes, it is right. If you simplify that godawful mess by cancelling sec's, you just get sec^2(x). You are right. But listen to Defennder, you are working WAY TOO HARD.