#1

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

Ans:

f(0) = 1

f’(0) = \(\frac{1}{(1-0)^2} = 1\)

f’’(0) = \(-\frac{2}{(0-1)^3} = 2\)

Using the Maclaurin Series (Taylor series centered at 0), we get

\(\sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}(x^n) = f(0)+\frac{f'(0)}{1!}x+…+\frac{f^{(n)}(0)}{n!}x^n\)

\[ \begin{multline*} \begin{split} &= 1+ \frac{1}{1!}x+ \frac{2}{2!}x^2 + ... \\ &= 1+ x + x^2 + ... \\ \end{split} \end{multline*} \]
The series representation of this function is:

\(\sum_{n=0}^{\infty }x^n\) for |x| < 1

#2

f(x) = e^x

Ans:

f(0) = e⁰ = 1

f’(0) = e⁰ = 1

f’’(0) = e⁰ = 1

Using the Maclaurin series (Taylor series at zero):

\(\sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}(x^n) = f(0)+\frac{f'(0)}{1!}x+…+\frac{f^{(n)}(0)}{n!}x^n\)

\[ \begin{multline*} \begin{split} &= 1+ \frac{1}{1!}x+ \frac{1}{2!}x^2 + \frac{1}{3!}x^3 +... \\ &= 1+ x + \frac{x^2}{2}+ \frac{x^3}{6} + ... \\ \end{split} \end{multline*} \]

The series representation of this function is:

\(\sum_{k=0}^{\infty }\frac{x^k}{k!}\)

#3

f(x) = ln(1+x)

Ans:

f(0) = ln(1+0) = ln(1)= 0

f’(0) = \(\frac{1}{1+0} = 1\)

f’’(0) = \(-\frac{1}{(1+0)^2}= -1\)

f’’’(0) = \(\frac{2}{(1+0)^3}=2\)

f⁽⁴⁾(0) = \(-\frac{6}{(1+0)^4} = -6\)

Using the Maclaurin series (Taylor series at zero):

\(\sum_{n=0}^{\infty }\frac{f^{(n)}(0)}{n!}(x^n) = f(0)+\frac{f'(0)}{1!}x+…+\frac{f^{(n)}(0)}{n!}x^n\)

\[ \begin{multline*} \begin{split} &= 0+ \frac{1}{1!}x+ \frac{(-1)}{2!}x^2 + \frac{2}{3!}x^3 + \frac{(-6)}{4!}x^4 +... \\ &= x - \frac{x^2}{2} + \frac{x^3}{3}- \frac{x^4}{4} + ... \\ \end{split} \end{multline*} \]

The series representation of this function is:

\(\sum_{k=1}^{\infty }\frac{(-1)^{k+1}x^k}{k}\)