Finding a3 in the Taylor Series for x^3ln(1+x^2)

lmannoia
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Homework Statement


Let f(x) = x3ln(1+x2), and let the summation (from n=0 to infinity) anxn be the Taylor series for f about 0. Then what is a3?

Homework Equations





The Attempt at a Solution


What?! I definitely don't expect the answer, but does anyone know how I could go about finding this out from the vague information given about the summation?
 
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lmannoia said:

Homework Statement


Let f(x) = x3ln(1+x2), and let the summation (from n=0 to infinity) anxn be the Taylor series for f about 0. Then what is a3?

Homework Equations





The Attempt at a Solution


What?! I definitely don't expect the answer, but does anyone know how I could go about finding this out from the vague information given about the summation?
How do you normally go about finding the coefficients of the terms in a Taylor's series? In this case, it's a Maclaurin series.
 
It would be f(0)+f1(0)x+f2(0)x^2/2! +f3(0)x^3/3!.. what I'm not getting is, how can I do this when I'd be plugging in zero? Wouldn't all of the coefficients just be zero? Or am I thinking about this in the wrong way?
 
What do you mean by f1(0)? f2(0)? f3(0)? Do you mean f'(0), f''(0), and f'''(0)?

Clearly f(0) = 0, and it turns out that f'(0) = 0. Have you determined what f''(x) and f'''(x) are?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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