Today, we will be expanding some taylor series for a handful of functions. http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx
This week, we’ll work out some Taylor Series expansions of popular functions.
The taylor series expansion is given by the following:
\[ f(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} } \\ =f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\frac{f'''(a)}{2!}(x-a)^{3}+... \] 1) \[ f(x)=\frac{1}{(1-x)}=(1-x)^{-1} \] for our purposes, we will evaluate the Maclaurin series where a=0. This alters the formla for the taylor series. We will compute the first few terms of our series and see if there exists a visible pattern. I can tell that all the derivatives will be positive. If we apply out chain rule, there will always be two negatives that will cancel each other out.
when a=0 \[ f(a=0)=(1-x)^{-1}\\ f'(a=0)=(1-x)^{-2}=1\\ f''(a=0)=2(1-x)^{-3}=2\\ f'''(a=0)=(2)(3)(1-x)^{-4}=6\\ f''''(a=0)=(2)(3)(4)(1-x)^{-5}\\ ...\\ ...\\ ...\\ f^{n}(x)=n!(1-x)^{-n}=\frac{n!}{(1-x)^{n}} \] Build the taylor series expansion
\[ T(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} }=1+x+x^{2}+x^{3}+x^{4}+... \]
Follow similar procedure to problem 1 \[ f(a=0)=e^{x}=1\\ f'(a=0)=e^{x}=1\\ f''(a=0)=e^{x}=1\\ f'''(a=0)=e^{x}=1\\ f''''(a=0)=e^{x}=1\\ ...\\ ...\\ ...\\ f^{n}(x)=e^{n} \]
Build the taylor series expansion
\[ T(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} }=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\frac{x^{5}}{5!}+...+ \]
Compute a few derivatives when x=0 or a=0
\[ f(x)=ln(1+x)=ln(1)=0\\ f'(x)=\frac{1}{(x+1)}=1\\ f''(x)=-\frac{1}{(x+1)^{2}}=-1\\ f'''(a=0)=\frac{2}{(x+1)^{3}}=2\\ f''''(a=0)=-\frac{(2)(3)}{(x+1)^{4}}=-6\\ f'''''(a=0)=\frac{(2)(3)(4)}{(x+1)^{5}}=24 \] We can build out our Taylor series expansion for this function. It should be noted that we actually have an alternating series since each term changes from positive to negative.
\[ T(x)=\sum _{ n=0 }^{ \infty }{ (-1)^{n}\frac { { f }^{ n }(a) }{ n! } (x-a)^{n} }=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}.... \]