Y!A: What is the Taylor series of ln((x^2)+1) centered at c=0?

this is a question i answered on Y!Answers and can be found here:

http://answers.yahoo.com/question/index;_ylt=AmsLM3qKuCKKGf8_wa161Zfty6IX;_ylv=3?qid=20090426105041AA1f5lh&show=7#profile-info-2AVjODBNaa

there are two ways you can do this:

the first is by deriving the taylor series directly. however that involves a lot of messy product rules for derivatives in this case (but i will start it for you anyway).

the second is to use substitution and integration.

THE FIRST METHOD:
you would write normally, as if you were deriving a taylor series:
f(x) = a ₀ + a₁x + a₂x² + ... = ln(x² + 1)

but you can see that when we differentiate both sides, ln(x² +1) yields confusing derivatives. so, i will use a different, smarter method.

THE SECOND METHOD:
the first thing i will do is take a known taylor series, and later manipulate it:
the taylor series for f(x) = 1/(1-x) is known to be 1 + x + x² + x³ + x⁴ + ...
so i will take f(-x) = 1 / (1 - (-x) ) = 1 / (1+x)
for which the taylor series becomes: 1 - x + x² - x³ + x⁴ - ... (plugging in -x everywhere for x)

next, i integrate 1 / (1+x) and its taylor series:
ln(1+x) = C + x - (x²)/2 + (x³)/3 - (x⁴)/4 + (x⁵)/5 - ...
(and we can eliminate C now by plugging in 0 for x and getting ln(1) = C = 0)

now, using some substitution, take our new function g(x) = ln(1+x) and its taylor series, and substitute x² for x. that is, g(x²) = ln(1 + x²), and its taylor series becomes (x²) - ((x²)²)/2 + ((x²)³)/3 - ((x²)⁴)/4 + ((x²)⁵)/5 - ... (placing x²everywhere that we see an x), which equals
x² - (x⁴)/2 + (x⁶)/3 - (x⁸)/4 + ...

thus ln( 1 + x² ) = x² - (x⁴)/2 + (x^6)/3 - (x⁸)/4 + (x^10)/5 - ...
and in general form:
∑ (-1)ⁿ⁻¹ * x²ⁿ/n

[from n=1 to ∞]

though this method seems very confusing, it will get easier and more natural with practice. the trick to picking the correct initial function will also come with practice and your familiarity with taylor series.

i hope this wasn't too confusing!