Question

Find the Taylor series for f(x)=1/(sqrt(x)) about x=9

Answer 1

\[\sum_{n=0}^{\infty}{f^{(n)}(a)\over n!}(x-a)^n\]\[f(x)=x^{-\frac12}\implies f^{(n)}(x)=(-\frac12)^nx^{\frac12-n}\]\[\therefore \sum_{n=0}^{\infty}{f^{(n)}(9)\over n!}(x-9)^n={(-\frac12)^n9^{\frac12-n}\over n!}(x-9)^n\]

Answer 2

I could write it in some slightly different ways of course
is there a part you don't understand?

Answer 3

slight typo in the exponent*\[\sum_{n=0}^{\infty}{f^{(n)}(a)\over n!}(x-a)^n\]\[f(x)=x^{-\frac12}\implies f^{(n)}(x)=(-\frac12)^nx^{-\frac12-n}\]\[\therefore \sum_{n=0}^{\infty}{f^{(n)}(9)\over n!}(x-9)^n={(-\frac12)^n9^{-\frac12-n}\over n!}(x-9)^n\]

Answer 4

No not at all this is perfect I just had an answer and you have confirmed it. I've forgotten most things about Taylor series and I was just getting confirmation I was dong it correctly . Thanks very much

Answer 5

very welcome :D