Noori Selina

Assignment 14

Taylor Series

\(f(x)=\frac{1}{1-x}\)

This function is not defined when \(x = 1\).

Derivatives and evaluation at \(x=0\):

• \(f(x)=\frac{1}{1-x} \rightarrow f(0)=1\)
• \(f'(x)=\frac{1}{(1-x)^2} \rightarrow f'(0)=1\)
• \(f''(x)=\frac{2}{(1-x)^3} \rightarrow f''(0)=2\)
• \(f'''(x)=\frac{6}{(1-x)^4} \rightarrow f'''(0)=6\)
• \(f^{(4)}(x)=\frac{24}{(1-x)^5} \rightarrow f^{(4)}(0)=24\)
• \(f^{(5)}(x)=\frac{120}{(1-x)^6} \rightarrow f^{(5)}(0)=120\)

Use McLaurin Series formula:

\[1 + \frac{1}{1!}x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \frac{120}{5!}x^5 + \ldots\]

This is simplified to:

\[1+x+x^2+x^3+x^4+x^5+\ldots+x^n\]

Presented in summation form:

\[\sum_{n=0}^\infty x^n\]

\(f(x)=e^x\)

Derivatives and evaluation at \(x=0\):

• \(f(x)=e^x \rightarrow f(0)=1\)
• \(f'(x)=e^x \rightarrow f'(0)=1\)
• \(f''(x)=e^x \rightarrow f''(0)=1\)
• \(f'''(x)=e^x \rightarrow f'''(0)=1\)
• \(f^{(4)}(x)=e^x \rightarrow f^{(4)}(0)=1\)
• \(f^{(5)}(x)=e^x \rightarrow f^{(5)}(0)=1\)

Using the McLaurin Series formula:

\[1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \ldots\]

This simplifies to:

\[1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+\ldots+\frac{x^n}{n!}\]

In summation form:

\[\sum_{n=0}^\infty \frac{x^n}{n!}\]

\(f(x)=\ln(1+x)\)

Derivatives and evaluation at \(x=0\):

• \(f(x)=\ln(1+x) \rightarrow f(0)=0\)
• \(f'(x)=\frac{1}{x+1} \rightarrow f'(0)=1\)
• \(f''(x)=-\frac{1}{(x+1)^2} \rightarrow f''(0)=-1\)
• \(f'''(x)=\frac{2}{(x+1)^3} \rightarrow f'''(0)=2\)
• \(f^{(4)}(x)=-\frac{6}{(x+1)^4} \rightarrow f^{(4)}(0)=-6\)
• \(f^{(5)}(x)=\frac{24}{(x+1)^5} \rightarrow f^{(5)}(0)=-24\)

Using the McLaurin Series formula:

\[0 + \frac{1}{1!}x - \frac{1}{2!}x^2 + \frac{1}{3!}x^3 - \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \ldots\]

Simplifies to:

\[x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 + \ldots + (-1)^{n+1}\frac{x^n}{n}\]

In summation form:

\[\sum_{n=0}^\infty (-1)^{n+1}\frac{x^n}{n}\]