Finding Taylor Expansion of f(x) and ln(1+x^2)

[Quincy]

Homework Statement

a) Using a geometric series, find the Taylor expansion of the function f(x) = x/(1+x^2)

b) Use the series found in (a) to obtain the Taylor expansion of ln(1 + x^2)

Homework Equations

The Attempt at a Solution

I really don't know where to start; I can't find anything about it in the book, and I don't remember the teacher ever talking about it. I know what Taylor/Mclaurin series are, but what does it mean by the "Taylor expansion"?

 

[hgfalling]

The Taylor expansion is just the written-out Taylor series for a function. So:

$$e^x = \sum_{k=1}^{\infty} \frac{x^k}{k!}$$

The expression on the right is the Taylor expansion of $e^x$ around 0.

 

[Mark44]

1/(1 - x) = 1 + x + x^2 + ... + x^n + ... is a geometric series, and is also a Maclaurin series. A Maclaurin series is a Taylor series in powers of x.

Can you come up with a series for 1/(1 + x^2)? If so, then you should be able to come up with a series for x/(1 + x^2).

For part b, think about the relationship between x/(1 + x^2) and ln(1 + x^2).

 

[Quincy]

(a) x/(1+x^2) = a/(1-r) --> a = x, r = -x^2

expansion: Sum from n = 1 to infinity of (-1)^(n+1) * X^(2n-1)?

For part b, think about the relationship between x/(1 + x^2) and ln(1 + x^2).

The derivative of ln(1+x^2) is 2x/(1+x^2), which is x/(1+x^2) * 2, but I don't understand what it means when it says "use the series in (a) to obtain the Taylor expansion".

 

[Mark44]

(a) x/(1+x^2) = a/(1-r) --> a = x, r = -x^2

expansion: Sum from n = 1 to infinity of (-1)^(n+1) * X^(2n-1)?

It's probably more useful to write the series in expanded form, rather than in closed form (the form with the summation symbol).

The derivative of ln(1+x^2) is 2x/(1+x^2), which is x/(1+x^2) * 2, but I don't understand what it means when it says "use the series in (a) to obtain the Taylor expansion".

So d/dx(ln(1 + x^2) = 2x/(1 + x^2) ==> ln(1 + x^2) = ?? What can you replace 2x/(1 + x^2) with on the right side?