# Series expansion of logarithmic function ln(cosx)

## Homework Statement

A question asks me to find the first three non-zero terms of ln(cosx)

## 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?

NascentOxygen
Staff Emeritus
[A question asks me to find the first three non-zero terms of ln(cosx)

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?

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 ?

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

dynamicsolo
Homework Helper
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.

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

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

A question asks me to find the first three non-zero terms of ln(cosx)

## 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?

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

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

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

## Homework Statement

A question asks me to find the first three non-zero terms of ln(cosx)

## 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?

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
The solution suggested by dynamicsolo seems easy enough to me.

dynamicsolo
Homework Helper
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?

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)!