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This week we will work on some of Taylor Series expansion of popular function
For each function only consider its valid range as indicated in the notes when you are computing Taylor series fuction.
\[\begin{aligned} &f(x)=\frac{1}{(1-x)} \\ &f(x)=e^{x} \\ &f(x)=\ln (1+x) \end{aligned}\]\[ f(x)=\frac{1}{(1-x)} \]
This function is not defined when \(x=1\). Derivaties and evaluation at \(x=0\).
\[ \begin{aligned} &f(x)=\frac{1}{(1-x)} \rightarrow f(0)=1 \\ &f^{\prime}(x)=\frac{1}{(1-x)^{2}} \rightarrow f^{\prime}(0)=1 \\ &f^{\prime \prime}(x)=\frac{2}{(1-x)^{3}} \rightarrow f^{\prime \prime}(0)=2 \\ &f^{\prime \prime \prime}(x)=\frac{6}{(1-x)^{4}} \rightarrow f^{\prime \prime \prime}(0)=6 \\ &f^{4}(x)=\frac{24}{(1-x)^{5}} \rightarrow f^{4}(0)=24 \\ &f^{5}(x)=\frac{120}{(1-x)^{6}} \rightarrow f^{5}(0)=120 \end{aligned} \]
Use McClaurin Series formula:
\[ 1+\frac{1}{1 !} x^{1}+\frac{2}{2 !} x^{2}+\frac{6}{3 !} x^{3}+\frac{24}{4 !} x^{4}+\frac{120}{5 !} x^{5}+\ldots \]
Simplifies to:
\[ 1+x+x^{2}+x^{3}+x^{4}+x^{5}+\ldots+x^{n} \]
In summation form:
\[ \sum_{n=0}^{\infty} x^{n} \]
\(f(x)=e^{x}\)
Derivatives and evaluation at x=0:
\[ \begin{aligned} &f(x)=e^{x} \rightarrow f(0)=1 \\ &f^{\prime}(x)=e^{x} \rightarrow f^{\prime}(0)=1 \\ &f^{\prime \prime}(x)=e^{x} \rightarrow f^{\prime \prime}(0)=1 \\ &f^{\prime \prime \prime}(x)=e^{x} \rightarrow f^{\prime \prime}(0)=1 \\ &f^{4}(x)=e^{x} \rightarrow f^{4}(0)=1 \\ &f^{5}(x)=e^{x} \rightarrow f^{5}(0)=1 \end{aligned} \]
Use McClaurin Series formula:
\[ 1+\frac{1}{1 !} x^{1}+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\ldots \]
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 \(x=0\).
\[ \begin{aligned} &f(x)=\ln (1+x) \rightarrow f(0)=0 \\ &f^{\prime}(x)=\frac{1}{x+1} \rightarrow f^{\prime}(0)=1 \\ &f^{\prime \prime}(x)=\frac{-1}{(x+1)^{2}} \rightarrow f^{\prime \prime}(0)=-1 \\ &f^{\prime \prime \prime}(x)=\frac{2}{(x+1)^{3}} \rightarrow f^{\prime \prime \prime}(0)=2 \\ &f^{4}(x)=\frac{-6}{(x+1)^{4}} \rightarrow f^{4}(0)=-6 \\ &f^{5}(x)=\frac{24}{(x+1)^{5}} \rightarrow f^{5}(0)=-24 \end{aligned} \]
Use McClaurin Series formula:
\[ 0+\frac{1}{1 !} x^{1}-\frac{1}{2 !} x^{2}+\frac{2}{3 !} x^{3}-\frac{6}{4 !} x^{4}+\frac{24}{5 !} x^{5}+\ldots \]
Simplifies to:
\[ x-\frac{1}{2} x^{2}+\frac{1}{3} x^{3}-\frac{1}{4} x^{4}+\frac{1}{5} x^{5}+\ldots(-1)^{n+1} \frac{1}{n} x^{n} \]
In summation form:
\[ \sum_{n=0}^{\infty}(-1)^{n+1} \frac{1}{n} x^{n} \]