Question

Need help with a Taylor series.

f(x)=1/x around a=-3
I just need help writing it out as a summation

Answer 1

the general formula for the taylor series of f(x) about x=a is\[\sum_{n=0}^{\infty}{f^{(n)}(a)\over n!}(x-a)^n \]for your formula a=-3, and the nth derivative of f(x) is\[(-1)^nn!x^{-(n+1)}\]so our formula becomes\[\sum_{n=0}^{\infty}(-1)^n(-3)^{-(n+1)}(x+3)^n\]

Answer 2

Oh I see now, thank you!

Answer 3

I think we can actually simplify this a little further, but I'm a bit unsure how.

I think we can say
\[(-1)^n(-3)^{-(n+1)}=(-1)^n(-1)^{-(n+1)}3^{-(n+1)} =(-1)(3)^{-(n+1)}\]but I'm not positive about that...

Answer 4

Hmmm....okay. I'm not sure either. I would have to ask when I go for tutoring tomorrow. Thanks again for trying :)