MSDS Spring 2018

DATA 605 Fundamentals of Computational Mathematics

Jiadi Li

HW #14 - Taylor Series

For each function, only consider its valid ranges as indicated in the notes when computing the Taylor Series expansion.

\(f(x)\) = \(\frac{1}{(1-x)}\)

Interval of Convergence: (-1,1)

Taylor series of \(f(x)\) = \(\frac{1}{(1-x)}\):

\(f(x)\) = \(\frac{1}{(1-c)\space0!}(x-c)^0\) + \(\frac{1}{(1-c)^2\space0!}(x-c)^1\) + \(\frac{2}{(1-c)^3\space2!}(x-c)^2\) + \(\frac{6}{(1-c)^4\space3!}(x-c)^3\) + ……

= \(\frac{0!}{(1-c)\space0!}(x-c)^0\) + \(\frac{1!}{(1-c)^2\space0!}(x-c)^1\) + \(\frac{2!}{(1-c)^3\space2!}(x-c)^2\) + \(\frac{3!}{(1-c)^4\space3!}(x-c)^3\) + ……

= \(\frac{1}{(1-c)}\) + \(\frac{1}{(1-c)^2}(x-c)\) + \(\frac{1}{(1-c)^3}(x-c)^2\) + \(\frac{1}{(1-c)^4}(x-c)^3\) + ……

= \(\sum_{n=0}^{\infty}{\frac{1}{(1-c)^{n+1}}(x-c)^n}\)

centered at 0:

\(\frac{1}{1-x}\) = \(\sum_{n=0}^{\infty}x^n\) = 1 + \(x\) + \(x^2\) + \(x^3\) + ……

\(f(x)\) = \(e^x\)

Interval of Convergence: (-\(\infty\),\(\infty\))

Taylor series of \(f(x)\) = \(e^x\)

\(f(x)\) = \(\frac{e^c}{0!}(x-c)^0\) + \(\frac{e^c}{1!}(x-c)^1\) + \(\frac{e^c}{2!}(x-c)^2\) + \(\frac{e^c}{3!}(x-c)^3\) + ……

= \(e^c\) + \(e^c(x-c)\) + \(\frac{e^c}{2!}(x-c)^2\) + \(\frac{e^c}{3!}(x-c)^3\) + ……

= \(\sum_{n=0}^{\infty}{\frac{e^c}{n!}(x-c)^n}\)

centered at 0:

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

\(f(x)\) = \(ln(1 + x)\)

Interval of Convergence: (0,2]

Taylor series of \(f(x)\) = \(ln(1 + x)\):

\(f(x)\) = \(\frac{ln(1+c)}{0!}(x-c)^0\) + \(\frac{1}{(c+1)\space1!}(x-c)^1\) - \(\frac{1}{(c+1)^2\space2!}(x-c)^2\) + \(\frac{2}{(c+1)^3\space3!}(x-c)^3\) - ……
= \(ln(1+c)\) + \(\frac{0!}{(c+1)\space1!}(x-c)^1\) - \(\frac{1!}{(c+1)^2\space2!}(x-c)^2\) + \(\frac{2!}{(c+1)^3\space3!}(x-c)^3\) - ……
= \(ln(1+c)\) + \(\frac{1}{(c+1)}(x-c)\) - \(\frac{1}{2(c+1)^2}(x-c)^2\) + \(\frac{1}{3(c+1)^3}(x-c)^3\) - ……
= \(ln(1+c)\) + \(\sum_{n=1}^{\infty}{(-1)^{n+1}\frac{1}{n(c+1)^n}(x-c)^n}\)

centered at 0:

ln\((1+x)\) = \(\sum_{n=1}^{\infty}{(-1)^{n+1}\frac{x^n}{n}}\) = \(x\)-\(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) - ……