Today, we will be working with Taylor Series http://tutorial.math.lamar.edu/Classes/CalcII/TaylorSeries.aspx
The taylor series expansion is given by the following:
\[ f(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} } \\ =f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\frac{f'''(a)}{2!}(x-a)^{3}+... \]
8.8.10 find a formula for the nth term of the Taylor series of f(x), centered at c, by finding the coefficients of the first few powers of x and looking for a pattern
By definition, we will be evaluating the Maclaurin series where a=0
\[ f(x)=ln(1+x)\\ c=0 \] We want to try and find the nth derivative by finding a few derivatives first and observing a pattern that may emerge
\[ f(x)=ln(1+x)\\ f'(x)=(x+1)^{-1}=\frac{1}{(x+1)}\\ f''(x)=-(x+1)^{-2}=-\frac{1}{(x+1)^{2}}\\ f'''(x)=2(x+1)^{-3}=\frac{2}{(x+1)^{3}}\\ f''''(x)=(2)(-3)(x+1)^{-4}=-\frac{(2)(3)}{(x+1)^{4}}\\ f'''''(x)=(2)(-3)(-4)(x+1)^{-5}=\frac{(2)(3)(4)}{(x+1)^{5}}\\ ...\\ ...\\ ...\\ f^{n}(x)=(-1)^{n}(n-1)!(x+1)^{-n}=(-1)^{n}\frac{(n-1)!}{(x+1)^{n}} \]
what is the value of our derivatives when x=0?
\[ f(0)=ln(1+x)=ln(1+0)=0\\ f'(0)=\frac{1}{(0+1)}=1\\ f''(0)=-\frac{1}{(0+1)^{2}}=-1\\ f'''(0)=\frac{2}{(0+1)^{3}}=2\\ f''''(0)=-\frac{(2)(3)}{(0+1)^{4}}=-(2)(3)\\ f'''''(0)=\frac{(2)(3)(4)}{(x+1)^{5}}=(2)(3)(4)\\ ...\\ ...\\ ...\\ f^{n}(0)=(-1)^{n}\frac{(n-1)!}{(0+1)^{n}}=(-1)^{n}(n-1)! \]
Lets compute our nth Taylor Series
\[ T(x)=f(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} }\\ =\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{n}(n-1)! }{ n! } (x)^{n} }\\ =\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{n}(n-1)! }{ n(n-1)! } (x)^{n} }\\ =\sum _{ n=0 }^{ \infty } (-1)^{n} \frac{x^{n}}{n} \]