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This week we are working on taylor series.
The taylor series expansion is given by the following:
\[ f(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} } \\ =f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2}+\frac{f'''(a)}{2!}(x-a)^{3}+... \]
Problem Statement
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 \[
T(x)=f(x)=\sum _{ n=0 }^{ \infty }{ \frac { { f }^{ n }(a) }{ n! } (x-a)^{n} }\\
=\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{n}(n-1)! }{ n! } (x)^{n} }\\
=\sum _{ n=0 }^{ \infty }{ \frac { (-1)^{n}(n-1)! }{ n(n-1)! } (x)^{n} }\\
=\sum _{ n=0 }^{ \infty } (-1)^{n} \frac{x^{n}}{n}
\]