For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.

In order to find Taylor Series we have to take the next derivative, divide by n! and multiply by \((x-a)^n\)

\(f(x)=f(a) + \frac{f'(a)}{1!}(x-a)+ \frac{f''(a)}{2!}(x-a)^2+ \frac{f'''(a)}{3!}(x-a)^3+...\)

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

first derivitive: \(f'(x)= \frac{1}{(1-x)^2}\)

second derivitive: \(f''(x)= 2\frac{1}{(1-x)^3}\)

third derivitive: \(f'''(x)= 6*\frac{1}{(1-x)^4}\)

forth derivitive: \(f^4(x)= 24*\frac{1}{(1-x)^5}\)

fifth derivitive: \(f^5(x)= 120*\frac{1}{(1-x)^6}\)

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n-th derivitive: \(f^n(x)= n!*\frac{1}{(1-x)^{n+1}}\)

\(f(x)=\frac{1}{1-a} + \frac{\frac{1}{(1-a)^2}}{1!}(x-a)+ \frac{2\frac{1}{(1-a)^3}}{2!}(x-a)^2+ \frac{-6*\frac{1}{(1-a)^4}}{3!}(x-a)^3+ \frac{24*\frac{1}{(1-x)^5}}{4!}(x-a)^4+\frac{-120*\frac{1}{(1-x)^6}}{5!}(x-a)^5 +...=\frac{1}{1-a} + \frac{(x-a)}{(1-a)^2}+ \frac{(x-a)^2}{(1-a)^3}+ \frac{(x-a)^3}{(1-a)^4}+ \frac{(x-a)^4}{(1-a)^5} +...=\sum _{ n=0 }^{ \infty }{ {\frac{\left(x - a \right)^{n}}{ \left(1 - a \right)^{\left( n + 1\right)} }}}\)

Now put a=0

\(\frac{1}{1-x}=\sum _{ n=0 }^{ \infty }{ {\frac{\left(x - 0 \right)^{n}}{ \left(1 - 0 \right)^{\left( n + 1\right)} }}}=\sum _{ n=0 }^{ \infty }x^n\)

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

first derivitive: \(f'(x)= e^x\)

second derivitive: \(f''(x)= e^x\)

third derivitive: \(f'''(x)= e^x\)

forth derivitive: \(f^4(x)= e^x\)

fifth derivitive: \(f^5(x)= e^x\)

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n-th derivitive: \(f^n(x)= e^x\)

\(f(x)=e^a + \frac{e^a}{1!}(x-a)+ \frac{e^a}{2!}(x-a)^2+ \frac{e^a}{3!}(x-a)^3+ \frac{e^a}{4!}(x-a)^4+\frac{e^a}{5!}(x-a)^5 +...=\sum _{ n=0 }^{ \infty }\frac{e^a(x-a)^n}{n!}\)

Now put a=0

\(e^x=\sum _{ n=0 }^{ \infty }\frac{e^0(x-0)^n}{n!}=\sum _{ n=0 }^{ \infty }\frac{x^n}{n!}\)

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

first derivitive: \(f'(x)= \frac{1}{1+x}\)

second derivitive: \(f''(x)= -\frac{1}{(1+x)^2}\)

third derivitive: \(f'''(x)= \frac{2}{(1+x)^3}\)

forth derivitive: \(f^4(x)= -\frac{6}{(1+x)^4}\)

fifth derivitive: \(f^5(x)= \frac{24}{(1+x)^5}\)

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n-th derivitive: \(f^n(x)= \frac{(n-1)!}{(1+x)^n}\)

\(f(x)=ln(1 + a) + \frac{\frac{1}{1+a}}{1!}(x-a)+ \frac{-\frac{1}{(1+a)^2}}{2!}(x-a)^2+ \frac{\frac{2}{(1+a)^3}}{3!}(x-a)^3+ \frac{-\frac{6}{(1+a)^4}}{4!}(x-a)^4+\frac{\frac{24}{(1+a)^5}}{5!}(x-a)^5 +...=ln(1 + a) + \frac{(x-a)}{1+a}- \frac{(x-a)^2}{2(1+a)^2}+ \frac{(x-a)^3}{3(1+a)^3}-\frac{(x-a)^4}{4(1+a)^4}+\frac{(x-a)^5}{5(1+a)^5} +...=ln(1 + a)+\sum _{ n=0 }^{ \infty }(-1)^{n+1}\frac{(x-a)^n}{n(1+a)^n}\)

Now put a=0

\(ln(1 + x)=ln(1 + 0)+\sum _{ n=0 }^{ \infty }(-1)^{n+1}\frac{(x-0)^n}{n(1+0)^n}=\sum _{ n=0 }^{ \infty }(-1)^{n+1}\frac{x^n}{n}\)