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

Referring to the definition in page 485 of the textbook

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

Interval of Convergence is (-1,1)

In order to find the pattern, we find the first few derivatives and evaluate them at 0.

\[ f(x) = \frac{1}{(1-x)} = (1-x)^{-1} \] \[ f(0) = \frac{1}{(1-0)} = (1-0)^{-1} = 1\] \[ f'(x) = (1-x)^{-2} \] \[ f'(0) = (1-0)^{-2} = 1 \]

\[ f''(x) = 2(1-x)^{-3} \] \[ f''(0) = 2(1-0)^{-3} = 2\] \[ f'''(x) = 6(1-x)^{-4} \] \[ f'''(0) = 6(1-0)^{-4}=6 \] \[ f''''(x) = 24(1-x)^{-5} \] \[ f''''(0) = 24(1-0)^{-5} =24\] \[ f'''''(x) = 120(1-x)^{-6} \] \[ f'''''(0) = 120(1-0)^{-6} =120\]

Based on the factorial pattern (1,2,6,24,120…) from the first few derivatives, we can conclude the following:

\[ f^{n}(x) = n!(1-x)^{(-n-1)}=\frac{n!}{(1-x)^{n+1}} \] Thus \[f^{n}(0) = n!\]

Plug \(n!\) into Maclaurin Series definition

\[\sum_{n = 0}^{\infty}\frac{{f^{n}(0)}}{n!} (x)^{n} \ \]

After plugging in, we are left with

\[\sum_{n = 0}^{\infty}(x)^{n} \ \] \[\sum_{n = 0}^{\infty}(x)^{n} = 1 + x + x^2 + x^3 + etc.\ \]

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

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

The function \(f(x) = e^x\) is equivalent to its derivative functions.

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

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

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

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

We can conclude the following:

\[f^{n}(x) = e^x\] Thus \[f^{n}(0) = 1\]

Plug 1 into Maclaurin Series definition

\[\sum_{n = 0}^{\infty}\frac{{f^{n}(0)}}{n!} (x)^{n} \ \]

After plugging in, we are left with

\[\sum_{n = 0}^{\infty}\frac{1}{n!}(x)^{n} \ \]

\[\sum_{n = 0}^{\infty}\frac{(x)^{n}}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}+etc.\ \]

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

\[ f(x) = ln(1+x) \] \[ f(0) = ln(1+0) =0\] \[ f'(x) = \frac{1}{1+x} = (1+x)^{-1} \] \[ f'(0) = \frac{1}{1+0} = (1+0)^{-1} = 1 \]

\[ f''(x) = -(1+x)^{-2} \] \[ f''(0) = -(1+0)^{-2} = -1\] \[ f'''(x) = 2(1+x)^{-3} \] \[ f'''(0) = 2(1+0)^{-3}=2 \] \[ f''''(x) = -6(1+x)^{-4} \] \[ f''''(0) = -6(1+0)^{-4} =-6\] \[ f'''''(x) = 24(1+x)^{-5}\]

\[ f'''''(0) = 24(1+0)^{-5} =24\]

Based on the pattern from the first few derivatives, we can conclude the following:

\[ f^{n}(x) = (-1)^{n+1}(n-1)!(1+x)^{-n}=(-1)^{n+1}\frac{(n-1)!}{(1+x)^n} \] Thus \[f^{n}(0) = (-1)^{n+1}(n-1)!\] Plug \((-1)^{n+1}(n-1)!\) into Maclaurin Series definition

\[\sum_{n = 0}^{\infty}\frac{{f^{n}(0)}}{n!} (x)^{n} \ \]

After plugging in, we are left with

\[\sum_{n = 0}^{\infty}\frac{((-1)^{n+1}(n-1)!)}{n!}(x)^{n} \ \] It is important to note that after simplifying, the denominator is n. Since n cannot be equal to zero, I will start at n = 1 \[\sum_{n = 0}^{\infty}\frac{(-1)^{n+1}(x)^{n}}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ etc.\ \]