Question

Taylor series of f(x)=1/x a=1
please

Answer 1

whats your question?

Answer 2

what is the taylor series of f(x)=1/x a=1

Answer 3

Well I'm not sure what Taylor Series actually is, but from searching it up.

I can try to solve it.
So first you need the derivative of f(x)
which would be f'(x)=-1/x^2
and then plug 1 into f'(x)
f'(1)=-1/(-1)^2
=-1

Answer 4

Not sure if its right though, cuz never learned the Taylor Series.

Answer 5

its probably wrong lol

Answer 6

yes, your answer is wrong...

Answer 7

\[f(x)=\dfrac{1}{x}=x^{-1}\\~\\f'(x) = -\dfrac{1}{x^2} = -x^{-2} \\~\\f''(x) = \dfrac{2}{x^3} = 2x^{-3}\\~\\f'''(x) = -\dfrac{6}{x^4}=-6x^{-4}\\~\\f''''(x) = \dfrac{24}{x^5}=24x^{-5}\]So from pattern, we can see that \(f^{(n)}(x) = (-1)^n~n!\dfrac{1}{x^{n+1}}\)

So since we need to find Taylor series at \(a=1\), we have \[f^{(n)}(1) = (-1)^n~n!\dfrac{1}{1^{n+1}} = (-1)^n~n!\]

That should help you. Use this formula:

\[f(x) = \sum_{n\to0}^{\infty} \dfrac{f^{(n)}(a)}{n!}(x-a)^n ~\text{ at }~x=a\]

Answer 8

Does that help though? @daniel9511 I don't know what you are struggling with, so...

Answer 9

Oops \(n\to0\) below \(\sum\) is supposed to be \(n=0\)

Answer 10

thank you

Answer 11

ok glad I helped.