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f(x)= 1/(1-x)

\[ f'(x)= \frac{1}{(1-x)^2}\] \[ f''(x)= \frac{2}{(1-x)^3}\]

\[ f'''(x)= \frac{6}{(1-x)^4}\]

\[ f''''(x)= \frac{24}{(1-x)^5}\]

• Evaluate at x=0 \(f'(0)=1\) \(f''(0)=2\) \(f'''(0)=6\) \(f''''(0)=24\)

• generalized \[ f^n(x)= \frac{n!}{(1-x)^{n+1}}\]

• therefore as our coefficents are always n! \[\displaystyle\sum_{n=0}^{\infty} \frac{f^n(x)}{n!}*x^n== \displaystyle\sum_{n=0}^{\infty} x^n \]

f(x) = e^x

\[f^n(x)= e^x\]

\[e^n(0)=1\]

• Therefore our coefficents are always one leaving us \[\displaystyle\sum_{n=0}^{\infty} \frac{e^0}{n!}*x^n==\frac{x^n}{n!}\]

f(x)=ln(1+x)

\[ f'(x)= \frac{1}{(x+1)}\]

\[ f''(x)= \frac{-1}{(x+1)^2}\]

\[ f'''(x)= \frac{2}{(x+1)^3}\] \[ f''''(x)= -\frac{6}{(x+1)^4}\]

• evaluate at x=0 \(f'(0)=1\) \(f''(0)=-1\) \(f'''(0)=2\) \(f''''(0)=-6\)

• series looks like

\[ 0 + \frac{1}{1}x^{1}-\frac{1}{2}x^{2} + \frac{2}{6}x^{3} - \frac{6}{24}x^{4} \]

• We can see that coefficents are alternating signs and simplifies to \(x^n/n\) , but starts at n=1 becuase it is undefined at n=0 \[ \sum_{n=1}^\infty=(-1)^{n+1} \frac{x^n}{n} \]