# Homework Help: Series expansion of logarithmic function ln(cosx)

1. Aug 15, 2011

### seboastien

1. The problem statement, all variables and given/known data
A question asks me to find the first three non-zero terms of ln(cosx)

2. Relevant equations

3. The attempt at a solution

I wrote cos x as 1+(1-cos x), used the power series of ln function, expanded cosx and simplified, here is my answer: 1/2x^2 - 1/6x^4 + 1/16x^6

Is that right?

2. Aug 15, 2011

### Staff: Mentor

You can easily check for yourself. Try some suitable data, and use your calculator. Say, x=0.7

ln(cos(0.7)) = -0.282

How close is your approximation to this, for x=0.7 ?

3. Aug 15, 2011

### seboastien

okay, so It's wrong... where exactly have I messed up then?

4. Aug 15, 2011

### dynamicsolo

They're being cute here. What is ln(cos x) the integral of? (If you don't recall, differentiate this function.) What is the series expansion of that function? Now integrate that term-by-term.

5. Aug 15, 2011

### seboastien

thanks, i'm not entirely certain, but is the differential -tanx?

6. Aug 15, 2011

### Ray Vickson

If all you want are a few terms, the easiest method is to expand ln(1-z) in powers of z, then substitute the first few terms of 1-cos(x) in place of z, that is, to use z = x^2/2 - x^4/4! + x^6/6! - ... . As to where you messed up: do you *really* think that cos(x) = 1 + (1 - cos(x))?

RGV

7. Aug 15, 2011

### dynamicsolo

It is: you have either worked out its series expansion in your course, or you can get the first few terms yourself.

8. Aug 15, 2011

### jackmell

What's wrong with just brute-force calculation of the first six derivatives of that function at zero, then form the Taylor series at zero? Think it would have been easier than your persuit otherwise.

9. Aug 15, 2011

### SammyS

Staff Emeritus
The solution suggested by dynamicsolo seems easy enough to me.

10. Aug 15, 2011

### dynamicsolo

There's nothing wrong with the direct approach. However, if a function can be recognized as being related to another one through algebraic manipulation, differentiation, or integration, you could save yourself a lot of work (sometimes work you already did to get the other Taylor/Maclaurin series).

And you get a bonus: the radius of convergence of a power series is not altered by differentiation or integration (and can be re-evaluated easily in the case of algebraic manipulation)!