\[f(x) = \frac{1}{(1-x)}\]
First we want to determine a pattern when taking the derivative
\[f'(x) = \frac{1}{(1-x)^2}\] \[f''(x) = \frac{2}{(1-x)^3}\] \[f'''(x) = \frac{6}{(1-x)^3}\] \[f^{(4)}(x) = \frac{18}{(1-x)^4}\]
So now that we can see a parttern we can figure out f^n(x)
\[f^{(n)}(x) = \frac{n!}{(1-x)^n}\]
Plugging that in the the Taylor Series gives us
\[\sum_{n=0}^{\infty} \frac{n!}{(1-c)^nn!}(x-c)^n\]
Reducing this down gives us:
\[\sum_{n=0}^{\infty} (\frac{x-c}{1-c})^n\]
If we take it one step further and solve for the Maclaurin Series (where c=0) we get:
\[\sum_{n=0}^{\infty} x^n\]
Which makes sense!
\[f(x)=e^x\]
Lets solve some derivatives:
\[f'(x)=e^x\] \[f''(x)=e^x\] \[f^{(3)}(x)=e^x\]
In this case itβs easy to find f^n(x):
\[f^{(n)}(x)=e^x\]
Plugging in to the taylor series:
\[\sum_{n=0}^{\infty} \frac{e^c}{n!} (x-c)^n\]
Lets solve for the Maclaurin Series as well:
\[\sum_{n=0}^{\infty} \frac{x^n}{n!}\]
This is also the answer we are expecting
\[f(x)=ln(1+x)\] \[f'(x)=\frac{1}{1+x}\] \[f''(x)=\frac{-1}{(1+x)^2}\] \[f'''(x)=\frac{2}{(1+x)^3}\] \[f^{(4)}(x)=\frac{-6}{(1+x)^4}\] \[f^{(5)}(x)=\frac{24}{(1+x)^5}\]
Now solving for f^n:
\[f^{(n)}(x)=\frac{(-1)^{n+1}(n-1)!}{(1+x)^n}\]
Plugging into Taylor series:
\[\sum_{n=0}^{\infty} \frac{(-1)^{n+1}(n-1)!}{(1+c)^nn!}(x-c)^n\] \[\frac{(n-1)!}{n!} = \frac{1}{n}\] \[\sum_{n=0}^{\infty} \frac{(-1)^{n+1}}{(1+c)^nn}(x-c)^n\]
Maclaurin Series
\[\sum_{n=0}^{\infty} (-1)^{n+1}\frac{x^n}{n}\]