DATA 605 HW14

Homework 14

The Taylor series of function \(f(x)\) is:
\(\displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + ...\)

When \(a\) = \(0\), the series is also called a Maclaurin series.
\(\displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = f(0) + \frac{f'(0)}{1!}(x) + \frac{f''(0)}{2!}(x)^2 + \frac{f'''(0)}{3!}(x)^3 + ...\)

Question # 1:

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

Solution:

Function and Derivatives	Function at \(x = 0\)
\(f(x) = \frac{1}{(1-x)}\)	\(f(0) = \frac{1}{(1-0)} = \frac{1}{1} = 1\)
\(f'(x) = \frac{1}{(1-x)^2}\)	\(f'(0) = \frac{1}{(1-0)^2} = \frac{1}{1} = 1\)
\(f''(x) = \frac{-2}{(x-1)^3}\)	\(f''(0) = \frac{-2}{(0-1)^3} = \frac{-2}{-1} = 2\)
\(f'''(x) = \frac{6}{(x-1)^4}\)	\(f'''(0) = \frac{6}{(0-1)^4} = \frac{6}{1} = 6\)
\(f^4(x) = \frac{-24}{(x-1)^5}\)	\(f^4(0) = \frac{-24}{(0-1)^5} = \frac{-24}{-1} = 24\)
\(f^5(x) = \frac{120}{(x-1)^6}\)	\(f^5(0) = \frac{120}{(0-1)^6} = \frac{120}{1} = 120\)
\(f^6(x) = \frac{-720}{(x-1)^7}\)	\(f^6(0) = \frac{-720}{(0-1)^7} = \frac{-720}{-1} = 720\)
\(f^7(x) = \frac{5040}{(x-1)^8}\)	\(f^7(0) = \frac{5040}{(0-1)^8} = \frac{5040}{1} = 5040\)

Use the Maclaurin series (Taylor series centered at point of zero):

\(\frac{1}{(1-x)} = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = f(0) + \frac{f^{'}(0)}{1!}x + \frac{f^{''}(0)}{2!}x^2 + \frac{f^{'''}(0)}{3!}x^3 + ... + \frac{f^{n}(0)}{n!}x^n\)

By substituting terms, we get the following series:

\(\frac{1}{(1-x)} = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = 1 + \frac{1}{1!}x + \frac{2}{2!}x^2 + \frac{6}{3!}x^3 + \frac{24}{4!}x^4 + \frac{120}{5!}x^5 + \frac{720}{6!}x^6 + \frac{5040}{7!}x^7 + ...\)

\(\frac{1}{(1-x)} = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = 1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + ...\)

The series representation of this function is :
\(\frac{1}{(1-x)} = \displaystyle\sum_{n=0}^{\infty} x^n\) for \(|x| < 1\)

Question # 2:

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

Solution:

Function and Derivatives	Function at \(x = 0\)
\(f(x) = e^{x}\)	\(f(0) = e^{0} = 1\)
\(f'(x) = e^{x}\)	\(f'(0) = e^{0} = 1\)
\(f''(x) = e^{x}\)	\(f''(0) = e^{0} = 1\)
\(f'''(x) = e^{x}\)	\(f'''(0) = e^{0} = 1\)
\(f^4(x) = e^{x}\)	\(f^4(0) = e^{0} = 1\)
\(f^5(x) = e^{x}\)	\(f^5(0) = e^{0} = 1\)
\(f^6(x) = e^{x}\)	\(f^6(0) = e^{0} = 1\)
\(f^7(x) = e^{x}\)	\(f^7(0) = e^{0} = 1\)

Use the Maclaurin series (Taylor series centered at point of zero):

\(e^{x} = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = f(0) + \frac{f^{'}(0)}{1!}x + \frac{f^{''}(0)}{2!}x^2 + \frac{f^{'''}(0)}{3!}x^3 + ... + \frac{f^{n}(0)}{n!}x^n\)

By substituting terms, we get the following series:

\(e^{x} = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = 1 + \frac{1}{1!}x + \frac{1}{2!}x^2 + \frac{1}{3!}x^3 + \frac{1}{4!}x^4 + \frac{1}{5!}x^5 + \frac{1}{6!}x^6 + \frac{1}{7!}x^7 + ...\)

\(e^{x} = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} + \frac{x^6}{720} + \frac{x^7}{5040} + ...\)

The series representation of this function is :
\(e^{x} = \displaystyle\sum_{k=0}^{\infty} \frac{x^k}{k!}\)

Question # 3:

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

Solution:

Function and Derivatives	Function at \(x = 0\)
\(f(x) = ln(1+x)\)	\(f(0) = ln(1+0) = ln(1) = 0\)
\(f'(x) = \frac{1}{(1+x)}\)	\(f'(0) = \frac{1}{(1+0)} = \frac{1}{1} = 1\)
\(f''(x) = \frac{-1}{(1+x)^2}\)	\(f''(0) = \frac{-1}{(1+0)^2} = \frac{-1}{1} = -1\)
\(f'''(x) = \frac{2}{(1+x)^3}\)	\(f'''(0) = \frac{2}{(1+0)^3} = \frac{2}{1} = 2\)
\(f^4(x) = \frac{-6}{(1+x)^4}\)	\(f^4(0) = \frac{-6}{(1+0)^2} = \frac{-6}{1} = -6\)
\(f^5(x) = \frac{24}{(1+x)^5}\)	\(f^5(0) = \frac{24}{(1+0)^5} = \frac{24}{1} = 24\)
\(f^6(x) = \frac{-120}{(1+x)^6}\)	\(f^6(0) = \frac{-120}{(1+0)^6} = \frac{-120}{1} = -120\)
\(f^7(x) = \frac{720}{(1+x)^7}\)	\(f^7(0) = \frac{720}{(1+0)^7} = \frac{720}{1} = 720\)

Use the Maclaurin series (Taylor series centered at point of zero):
\(ln(1+x) = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = f(0) + \frac{f^{'}(0)}{1!}x + \frac{f^{''}(0)}{2!}x^2 + \frac{f^{'''}(0)}{3!}x^3 + ... + \frac{f^{n}(0)}{n!}x^n\)

By substituting terms, we get the following series:

\(ln(1+x) = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = 0 + \frac{1}{1!}x + \frac{(-1)}{2!}x^2 + \frac{2}{3!}x^3 + \frac{(-6)}{4!}x^4 + \frac{24}{5!}x^5 + + \frac{(-120)}{6!}x^6 + \frac{720}{7!}x^7 + ...\)

\(ln(1+x) = \displaystyle\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \frac{x^7}{7} + ...\)

The series representation of this function is :
\(ln(1+x) = -\displaystyle\sum_{k=1}^{\infty} \frac{(-1)^k x^k}{k}\) for \(|x| < 1\)