Find the Maclaurin series for ln(1-x^2) and its interval of validity.

[freshcoast]

Homework Statement

find the Maclaurin series and find the interval on which the expansion is valid.

f(x) = ln(1-x² )

Homework Equations

The Attempt at a Solution

I'm pretty confident in my skill at problems like these, except for this one I am getting an answer different from the book and I can not see why. I have done similar problems using similar steps and have gotten the correct answers but this one I keep getting something different.

the answer in the book is

Ʃ 5^Nx^N / N!

 

[Dick]

Homework Statement

find the Maclaurin series and find the interval on which the expansion is valid.

f(x) = ln(1-x² )

Homework Equations

The Attempt at a Solution

I'm pretty confident in my skill at problems like these, except for this one I am getting an answer different from the book and I can not see why. I have done similar problems using similar steps and have gotten the correct answers but this one I keep getting something different.

the answer in the book is

Ʃ 5^Nx^N / N!

The books answer isn't right. But yours isn't either. There's a sign problem. The derivative of ln(1-5x) is -5/(1-5x), isn't it?

 

[Dick]

And so why does your problem say f(x)=ln(1-x^2) and your working say f(x)=ln(1-5x)? Are you just showing the working for a 'similar problem'?

 

[freshcoast]

Oh yeah that's true. and damn these books, I've spent a considerably large amount of time trying to figure out why I was wrong!

 

[freshcoast]

Oops, I meant to solve for ln(1 - 5x),

But I am also having a problem with ln(1-x^2) aswell, because the book shows differ from my calculations.

the book shows

Ʃ x^2N / N

 

[Dick]

Oops, I meant to solve for ln(1 - 5x),

But I am also having a problem with ln(1-x^2) aswell, because the book shows differ from my calculations.

the book shows

Ʃ x^2N / N

I don't know how you are getting those derivatives, but don't you need to use the quotient rule to find the derivatives higher than the first? The simpler way is just find the series for ln(1-x) and then substitute x^2 for x.

 

[freshcoast]

Oh alright, I used that way and managed to get the answer, and I'm getting my derivatives using chain rule.

ex.

(1 / x) is equal to x^-1, so using chain rule I get -1(x)^-2 ---> -1 / x^2

 

[freshcoast]

Can you also take at a look at this?

They are asking for the Taylor series centered f(x) = 1 / X at c = 1

the answer in the book is

Ʃ (-1)^n (x - 1)^n

which I am finding hard to understand

 

[Dick]

Can you also take at a look at this?

They are asking for the Taylor series centered f(x) = 1 / X at c = 1


the answer in the book is

Ʃ (-1)^n (x - 1)^n

which I am finding hard to understand

Why don't you just try that again. You are matching up the wrong derivatives with the wrong factorials. E.g. 2! should go with the second derivative 2/x^3.