This week, we’ll work out some Taylor Series expansions of popular functions. The function for Taylor series is:
\[f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n}\]
Here are the functions we’ll be working through: \[f(x) = \frac{1}{(1−x)}\] \[f(x) = e^{x}\] \[f(x) = ln(1 + x)\] For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
\[f(x) = \frac{1}{(1−x)}\] Let’s find the first few derivatives:
\[f'(x) = \frac{1}{(1 - x)^{2}}\] \[f''(x) = \frac{2}{(1 - x)^{3}}\] \[f'''(x) = \frac{6}{(1 + x)^{4}}\] \[f''''(x) = \frac{24}{(1 + x)^{5}}\] So when x = 0, we see that: \[f^{n}(0) = f(0) + f'(0) + f''(x) + f'''(0) + f''''(0) +...\]
\[f^{n}(0) = 1 + 1 + 2 + 6 + 24 +...\]
Substituting back into the Taylor Series function:
\[f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}0}{n!}x^{n}\] \[= 1 + x + \frac{2x^{2}}{2!} + \frac{6x^{3}}{3!} + \frac{24x^{4}}{4!} + ...\] \[= 1 + x + x^{2} + x^{3} + x^{4} + ...\] This clearly shows the Taylor Series expansion summarizes to: \[f(x) = \sum_{n=0}^{\infty}{x^{n}}\]
\[f(x) = e^{x}\] The derivative of e^x is always e^x: \[f(x) = e^{x}\] \[f'(x) = e^{x}\] \[f''(x) = e^{x}\] \[...\] \[f^n(x) = e^{x}\]
So when x = 0, we see that: \[f^n(0) = e^{0}\] \[f^n(0) = 1\] \[f^{n}(0) = f(0) + f'(0) + f''(x) + f'''(0) + f''''(0) +...\]
\[f^{n}(0) = 1 + 1 + 1 + 1 + 1 +...\]
Substituting back into the Taylor Series function:
\[f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}0}{n!}x^{n}\]
\[= x + \frac{x^{2}}{2} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + ...\] \[= x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24} + ...\]
This clearly shows the Taylor Series expansion summarizes to: \[f(x) = \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\]
\[f(x) = ln(1 + x)\]
Let’s find the first few derivatives:
\[f'(x) = \frac{1}{(1 + x)}\] \[f''(x) = \frac{-1}{(1 + x)^{2}}\] \[f'''(x) = \frac{2}{(1 + x)^{3}}\] \[f''''(x) = \frac{-6}{(1 + x)^{4}}\] Evaulating these at f^{n}(0): \[f^{n}(0) = f(0) + f'(0) + f''(x) + f'''(0) + f''''(0) +...\]
\[f^{n}(0) = ln(1+0) + \frac{1}{(1 + 0)} + \frac{-1}{(1 + 0)^{2}} + \frac{2}{(1 + 0)^{3}} + \frac{-6}{(1 + 0)^{4}} + ...\]
\[f^{n}(0) = ln(1) + \frac{1}{(1)} + \frac{-1}{(1)} + \frac{2}{(1)} + \frac{-6}{(1)} + ...\]
\[f^{n}(0) = 0 + 1 -1 + 2 - 6...\]
Plugging this into the taylor series formula we see: \[\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^{n} = f(0) + \frac{f'(0)}{1!}x^{1} + \frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3} + \frac{f''(0)}{4!}x^{4} + ... \] \[ = 0 + \frac{1}{1!}x^{1} + \frac{(-1)}{2!}x^{2} + \frac{2}{3!}x^{3} + \frac{(-6)}{4!}x^{4} + ... \] \[ = 0 + x - \frac{x^{2}}{2} + \frac{2x^{3}}{6} - \frac{6x^{4}}{24} + ... \] \[ = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + ... \]
This clearly summarizes to: \[ = \sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n} \]