Work out the Taylor Series expansions of these following functions:
\[f(x)=\frac{1}{1-x}\]
For simplicity, I will begin by calculating the several derivations of the original function for later use:
\(f(x)=\frac{1}{1-x}\) \(f'(x)=\frac{1}{(1-x)^2}\) \(f''(x)=\frac{2}{(1-x)^3}\) \(f'''(x)=\frac{6}{(1-x)^4}\)
Using definition 39 from the textbook, the first few elements of the expansion can be calculated:
\[\frac{\frac{1}{1-0}}{0!}x^0 + \frac{\frac{1}{(1-0)^2}}{1!}x^1 + \frac{\frac{2}{(1-0)^3}}{2!}x^2 + \frac{\frac{6}{(1-0)^4}}{3!}x^3 + ...\]
This can be simplified to:
\[1 + x + x^2 + x^3\]
And finally:
\[\sum_{n=0}^{\infty}x^n\]
\[f(x)=e^x\]
Using the definition 39 from the textbook, the first few elements of the expansion can be calculated:
\[\frac{e^0}{0!}x^0 + \frac{e^0}{1!}x^1 + \frac{e^0}{2!}x^2 + \frac{e^0}{3!}x^3 + ...\]
This can be simplifed as:
\[1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...\] And finally:
\[\sum_{n=1}^{\infty}\frac{x^n}{n!}\]
\[f(x)=ln(1+x)\]
For simplicity, I will begin by calculating the several derivations of the original function for later use:
\(f(x)=ln(1+x)\) \(f'(x)=\frac{1}{1+x}\) \(f''(x)=\frac{-1}{(1+x)^2}\) \(f'''(x)=\frac{2}{(1+x)^3}\)
Using the definition 39 from the textbook, the first few elements of the expansion can be calculated:
\[\frac{ln(1+0)}{0!}x^0 + \frac{\frac{1}{1+0}}{1!}x^1 + \frac{\frac{-1}{(1+0)^2}}{2!}x^2 + \frac{\frac{2}{(1+0)^3}}{3!}x^3 + ...\]
This can be simplifed as:
\[0+x+\frac{-1}{2}x^2 + \frac{1}{3}x^3 + ...\]
And finally:
\[\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n}x^n\]