Natural Logarithm Taylor series: f(x) = ln(1 + x^2)

[Gchem45]

The question I'm working on is:
a. Find the first four non-zero terms of the Maclaurin series for f(x) = ln(1 + x^2). Use it to find f(0.3).

Now I dont have a great deal of experience doing Taylor/Maclaurin series but I've differentiated the first through fifth derivatives and can't seem to fit the polynomial on the numerator to fit in any sort of pattern.

f^n(x)=(((-1)^n)*(polynomial)(n-1)!)/(1+x)^n

I think I'm on the right track, but I could be writing gibberish. I've looked online for examples that would fit something similiar to this but have only found ln(1+x) which doesnt follow the same sort of pattern as this one. I'm not looking for a complete answer but a push in the right direction would help so much. Hope I arranged this coherantly.

 

[Unco]

It seems you should review http://mathworld.wolfram.com/TaylorSeries.html .

Find f(0), f'(0), f''(0), etc. and substitute into the formula until you have 4 nonzero terms (the odd derivatives are zero).

 

[Gchem45]

Thanks a lot thats exactly the push I needed thank you.

 

[tkhunny]

Are we losing our minds? Don't let the easy ones get away from you.

\(\displaystyle \frac{d}{dx}ln(1+x^{2})\;=\;\frac{2x}{1+x^{2}}\)

\(\displaystyle \frac{1}{1+x}\;=\;1-x+x^{2}-x^{3}+...\)

\(\displaystyle \frac{1}{1+x^{2}}\;=\;1-x^{2}+x^{4}-x^{6}+...\)

\(\displaystyle \frac{2x}{1+x^{2}}\;=\;2x-2x^{3}+2x^{5}-2x^{7}+...\)

\(\displaystyle ln(1+x^{2})\;=\;x^{2}-\frac{x^{4}}{2}+\frac{x^{6}}{3}-\frac{x^{8}}{4}+...\)

What is required for this sort of thing, Uniform Convergence? I forget.