This week, we’ll work out some Taylor Series expansions of popular functions.
\[ f(x) = \frac {1}{(1 - x)} \] \[ f(x) = e^{x} \] \[ f(x) = ln(1 + x) \]
For each function, only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion. Please submit your assignment as a R-Markdown document.
General formula for Taylor series expansion:
\[ f(x) = \Sigma_{n=0}^{\infty} \frac {f^{n}(a)}{n!}(x - a)^{n}\] 1. \[ f(x) = \frac {1}{(1 - x)} \]
First 3 derivatives \(f^{(n)}(a):\)
\[ f(a) = \frac{1}{1-a}\] \[ f'(a) = \frac{1}{(1-a)^{2}}\] \[ f''(a) = \frac{2}{(1-a)^{3}}\] \[ f'''(a) = \frac{6}{(1-a)^{4}}\]
\[ \frac{1}{1-x} = \frac{1}{(1-a)} + \frac {\frac{1}{(1-a)^2}}{1} (x-a) + \frac {\frac{2}{(1-a)^3}}{2!} (x-a)^{2} + \frac {\frac{6}{(1-a)^4}}{3!} (x-a)^{3} + ... \] \[ = \frac{1}{(1-a)} + \frac {(x-a)}{(1-a)^{2}} + \frac {(x-a)^{2}}{(1-a)^{3}} + \frac {(x-a)^{3}}{(1-a)^{4}} + ... \] \[ \Sigma_{n=0}^{\infty} \frac {(x-a)^{n}}{(1-a)^{n+1}} \] Using the ratio test \(lim_{n -> \infty}|\frac{a_{n+1}}{a_{n}}|\), we know that that the series converges when the quotient is less than 1.
For this series the ratio test produces = \(lim_{n -> \infty}|\frac{x-a}{1-a}|\). The series will therefore converge if \(|\frac{x-a}{1-a}| \lt 1\)
\[ e^{x} = e^{a} + e^{a}(x-a) + e^{a}\frac{(x-a)^{2}}{2!} + e^{a}\frac{(x-a)^{3}}{3!} + ...\] \[ e^{a} \Sigma_{n=0}^{\infty} \frac {(x-a)^{n}}{n!} \] Ratio test produces: \[lim_{n -> \infty}|\frac{a_{n+1}}{a_{n}}| = lim_{n -> \infty}\frac{|x-a|}{n+1} = 0\]
This series converges for all values of x and a
\[ f(x) = ln(1 + x) \]
First 3 derivatives:
\[ f(a) = ln(1+a)\] \[ f'(a) = \frac {1}{(1+a)}\] \[ f''(a) = \frac {-1}{(1+a)^{2}}\] \[ f'''(a) = \frac {2}{(1+a)^{3}}\] \[ ln(1+x) = ln(1+a) + \frac{1}{(1+a)} (x-a) + \frac {\frac{-1}{(1+a)^2}}{2!} (x-a)^{2} + \frac {\frac{2}{(1+a)^3}}{3!} (x-a)^{3} + ... \] \[ = ln(1+a) + \frac{x-a}{1+a} -\frac {1}{2} (\frac {x-a}{1+a})^{2} + \frac {1}{3} (\frac {x-a}{1+a})^{3} + ... \] \[ = ln(1+a) + \Sigma_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} (\frac {x-a}{1+a})^{n}\]
The second expression in the equation above is an alternating series. We know that alternating series converge whenever the absolute value of each term in the series is smaller than absolute value of the preceding term. Therefore, this particular series will converge if \(|\frac{x-a}{1+a}| \leq 1\). We also note that a must be greater than -1 for this series to be properly defined.