Series expansion of logarithmic function ln(cosx)

[seboastien]

Homework Statement

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

Homework Equations

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]

[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 ?

 

[seboastien]

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

 

[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.

 

[seboastien]

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

 

[Ray Vickson]

Homework Statement

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

Homework Equations

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]

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.

 

[jackmell]

Homework Statement

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

Homework Equations

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]

The solution suggested by dynamicsolo seems easy enough to me.

 

[dynamicsolo]

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)!