Below we work out the Taylor series expansiosn for different functions \[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \frac{f''''(a)}{4!}(x-a)^4 + ... \]
\[ f(x) = \frac{1}{1-x} \\ f'(x) = \frac{1}{1-x}^2 \\ f''(x) = \frac{2}{1-x}^3 \\ f'''(x) = \frac{6}{1-x}^4 \\ f''''(x) = \frac{24}{1-x}^5 \\ \implies\\ f(x) = 1 + x + x^2 + x^3 + x^4 + ... \implies\\ \sum_{n=0}^{\infty}x^n \]
\[ f(x) = e^x \\ f'(x) = e^x \\ f''(x) = e^x \\ f'''(x) = e^x \\ f''''(x) = e^x \\ \implies \\ f(x) = \frac{1}{1!}x+ \frac{1}{2!}x^2+ \frac{1}{3!}x^3+ \frac{1}{4!}x^4 + ...\\ \implies \\ \sum_{n=0}^{\infty}\frac{x^n}{n!} \] ## Problem 3
\[ f(x) = ln(1+x) \\ f'(x) = \frac{1}{1+x} \\ f''(x) = -\frac{1}{(1+x)^2} \\ f''' (x) = \frac{2}{(1+x)^3} \\ f''''(x) = -\frac{6}{(1+x)^2} \ \implies \\ f(x) = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4- \frac{1}{5}x^5... \\ \implies \\ \sum_{n=1}^{\infty}(-1)^n\frac{x^n}{n} \]