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Taylor and Maclaurin Series:

i.e \( f(x)=\sum_{n=0}^{\infty} a{n}(x-c)^{n}\quad ,\quad |x<-c|<r \)Region of Convergence

then \( a n=\frac{f^{n}{(c)}}{n !} \quad ,\quad f(x)=\sum_{n=0}^{\infty} \frac{f^{n}{(c)}}{n !}(x-a)^{n} \)

\( f(x)=f(c)+\frac{f^{\prime}(c)}{1 !}(x-c)+\frac{f^{\prime \prime}(c)}{2 !}(x-c)^{2}+\cdots \)

Taylor series of the function \(F\) at \(C\)

If \( c=0 \)

\( f(x)=\sum_{n=0}^{\infty} a{n} \ x^{n}=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n} \)

\( f(x)=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots \)

Maclawrin Series of \(F\)

 \( f(0)^{(0)}=f(0) \)

② \( 0 !=1 \)

\( \sum_{n=2}^{\infty} f(n)=\sum_{n=1}^{\infty} f(n+1) \)                                \( n=2 \quad n+1=2 \rightarrow n=1 \)

④ \( \sum_{n=1}^{\infty} f(n)=\sum_{n=2}^{\infty} f(n-1) \)                               \( n=1 \quad n-1=1 \quad \rightarrow n=2 \)

Find the Maclaurin Series of the function \( f(x)=e^{x} \) and it's radius of convergence.

Maclaurin series  \( f(x)=\sum_{n=0}^{\infty} \frac{f^{n}(0)}{n !} x^{n}=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\cdots \)

 

 

 

 

 

 

 

Maclaurin Series \( (M \cdot s) \) of \( e^{x} \) is \( \sum_{n=0}^{\infty} \frac{f^{n}{(0)}}{n !} x^{n}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} \)

                                                                          \( =1+\frac{x}{1 !}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots \cdots \)

To check convergence, use Ratio test:

\( \lim _{n \rightarrow \infty} |\frac{u_{n+1}}{u_{n}} |, \sum_{n=0}^{\infty} \frac{x^{n}}{n !} u_{n}=\frac{x^{n}}{n !} \quad, \quad u_{n+1}=\frac{x^{n+1}}{(n+1) !} \)

\( \lim _{n \rightarrow \infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow \infty}\left|\frac{x^{n+1}}{(n+1) !} \cdot \frac{n !}{x^{n}}\right|=\lim _{n \rightarrow \infty} \frac{n !}{(n+1) !}\left|\frac{x^{n+1}}{x^{n}}\right| \)

\( \lim _{n \rightarrow \infty}\left|\frac{u_{n+1}}{u_{n}}\right|=\lim _{n \rightarrow \infty} \frac{n !}{(n+1) n !}|x| =|x| \lim _{n \rightarrow \infty} \frac{1}{n+1}=|x| \lim \frac{1}{\infty} \)

\( \lim _{n \rightarrow \infty}\left|\frac{u_ n+1}{u_ n}\right| =0<1 \rightarrow \sum_{n=0}^{\infty} \frac{x^{4}}{n !} \) is \( A \cdot c \) for all \(x\)

\( e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} \) , region of convergence \( =(-\infty, \infty) \)

                            Radius\( =\infty \)

\( e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !}=1+\frac{x}{1!}+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\cdots \)

① If \( x=1 \Rightarrow e=1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+\cdots =\sum_{n=0}^{\infty} \frac{1}{n !} \)

② If we replace \((x)\) by \((-x)\)

\( e^{-x}=1+\frac{-x}{1!}+\frac{x^{2}}{2 !}-\frac{x^{3}}{31}+\dots=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{n}}{n !} \)

③ \( \frac{e^{x}+e^{-x}}{2} : \) \( e^{x}=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3 !}+\cdots \)

                    \( e^{-x}=1-\frac{x}{1 !}+\frac{x^{2}}{2 !}-\frac{x^{3}}{3 !}+\cdots \)

\( \frac{e^{x}+e^{-x}}{2}= 1+\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}+\cdots= \sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !}=\cos h (x) \)

but \( \cos h (x)=\frac{e^{x}+e^{-x}}{2} \)

\( \lim _{n \rightarrow \infty} | \frac{u_ n+1}{u_ n} |= |\frac{x^{2(n+1)}}{(2(n+1)) !} \cdot \frac{(2 n) !}{x^{2 n}}| =\frac{(2 n) !}{(2 n+2) !} .\left|\frac{{x^{2 n+2}}}{x^{2 n}}\right| \)

\( =\frac{2 n !}{(2 n+2)(2 n+1)(2 n) !} .|\frac{x^{2}}{1}| =x^{2} \lim _{n \rightarrow \infty}|\frac{1}{(2 n+2)(2 n+1)}|=0 \)

\( \sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !} \)is Converngent for all \( x \in R \) 

\( \cos h (x)=\sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !} \) , Interval of Converagence \( =(-\infty, \infty) \)

                                        Radius of Convergence \( =\infty \)

\( \sin h (x)= \sum_{n=0}^{\infty} \frac{x^{2 n+1}}{(2 n+1) !} \) ,  Interval of Converagence \( =(-\infty, \infty) \)

                                            Radius of Convergence \( =\infty \)

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