This week, we’ll work out some Taylor Series expansions of popular functions.
\(f(x) = \frac {1}{(1−x)}\)
\(f(x) = e^x\)
\(f(x) = ln(1 + x)\)
\(f(x) = x^{1/2}\)
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 an R-Markdown document.
To calculate the expansions of the given functions we use the Theorem below
Theorem: Function and Taylor Series Equality
\[f(x) = \sum_{n=0}^{\infty} \frac {f^{(n)}(c)}{n!} (x-c)^{n}\]
Find first derivatives for the given function f(x):
\(f^{(0)}(c) = \frac {1}{(1-c)}\)
\(f^{(1)}(c) = \frac {1}{(1-c)^{2}}\)
\(f^{(2)}(c) = \frac {2}{(1-c)^{3}}\)
\(f^{(3)}(c) = \frac {6}{(1-c)^{4}}\)
\(f^{(4)}(c) = \frac {24}{(1-c)^{5}}\)
Thus, the Taylor series expansion for f(x) converges over the interval \((-1, 1)\) and can be defined as:
\(f(x) \approx P(x) = \frac {1}{(1-c) 0!}(x-c)^{0} + \frac {1}{(1-c)^{2} 1!} (x-c)^{1} + \frac {2}{(1-c)^{3} 2!} (x-c)^{2} + \frac {6}{(1-c)^{3} 3!} (x-c)^{3} + \frac {24}{(1-c)^{4} 4!} (x-c)^{4} + ...\)
The sum notation for the expansion can be written as:
\[f(x) \approx P(x) = \sum_{n=0}^{\infty} \frac {1}{(1-c)^{n+1}} (x-c)^{n}\]
Setting c = 0 gives the Maclaurin Series of f(x):
\[f(x) \approx P(x) = \sum_{n=0}^{\infty} x^{n} = 1 + x + x^{2} + x^{3} + x^{4} + ...\]
Find first derivatives for the given function f(x):
\(f^{(0)}(c) = e^{c}\)
\(f^{(1)}(c) = e^{c}\)
\(f^{(2)}(c) = e^{c}\)
\(f^{(3)}(c) = e^{c}\)
\(f^{(4)}(c) = e^{c}\)
Thus, the Taylor series expansion for f(x) converges over the interval \((-\infty, \infty)\) and can be defined as:
\(f(x) \approx P(x) = \frac {e^{c}}{0!}(x-c)^{0} + \frac {e^{c}}{1!}(x-c)^{1} + \frac {e^{c}}{2!}(x-c)^{2} + \frac {e^{c}}{3!}(x-c)^{3} + ...\)
The sum notation for the expansion can be written as:
\[f(x) \approx P(x) = e^{c} \sum_{n=0}^{\infty} \frac {(x-c)^{n}}{n!} \]
Setting c = 0 gives the Maclaurin Series of f(x):
\[f(x) \approx P(x) = \sum_{n=0}^{\infty} \frac {x^{n}}{n!} = 1 + x + \frac {x^{2}}{2!} + \frac {x^{3}}{3!} + \frac {x^{4}}{4!} + ...\]
Find first derivatives for the given function f(x):
\(f^{(0)}(c) = ln(1 + c)\)
\(f^{(1)}(c) = \frac {1}{c+1}\)
\(f^{(2)}(c) = -\frac {1}{(c+1)^{2}}\)
\(f^{(3)}(c) = \frac {2}{(c+1)^{3}}\)
\(f^{(4)}(c) = -\frac {6}{(c+1)^{4}}\)
Thus, the Taylor series expansion for f(x) converges over the interval \((-1, 1)\) and can be defined as:
\(f(x) \approx P(x) = P(x) = \frac {ln(1+c)}{0!}(x-c)^{0} + \frac {1}{(c+1)1!}(x-c)^{1} - \frac {1}{(c+1)^{2}2!}(x-c)^{2} + \frac {2}{(c+1)^{3}3!}(x-c)^{3} - \frac {6}{(c+1)^{4}4!}(x-c)^{43} + ...\)
The sum notation for the expansion can be written as:
\[f(x) \approx P(x) = ln(1+c) + \sum_{n=1}^{\infty} (-1)^{n+1} \frac {(x-c)^{n}}{n(c+1)^{n}} \]
Setting c = 0 gives the Maclaurin Series of f(x):
\[f(x) \approx P(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac {(x)^{n}}{n} = x - \frac {x^{2}}{2} + \frac {x^{3}}{3} - \frac {x^{4}}{4} + ...\]
Find first derivatives for the given function f(x):
\(f^{(0)}(c) = c^{1/2}\)
\(f^{(1)}(c) = \frac {1}{2 (c^{1/2})}\)
\(f^{(2)}(c) = -\frac {1}{4 (c^{3/2})}\)
\(f^{(3)}(c) = \frac {3}{8 (c^{5/2})}\)
\(f^{(4)}(c) = -\frac {15}{16 (c^{7/2})}\)
Thus, the Taylor series expansion for f(x) converges over the interval \([0, \infty)\) and can be defined as:
\(f(x) \approx P(x) = c^{1/2} + \frac {1}{2 (c^{1/2}) 1!} (x-1)^{1} - \frac {1}{4 (c^{3/2}) 2!} (x-2)^{2} + \frac {3}{8 (c^{5/2}) 3!} (x-3)^{3} - \frac {15}{16 (c^{7/2}) 4!} (x-4)^{4} + ...\)
The sum notation for the expansion can be written as:
\[f(x) \approx P(x) = c^{1/2} + \sum_{n=1}^{\infty} (-1)^{n+1} \binom{1/2}{n}(x-c)^{n} (c^{1/2-n}) \]
Setting c = 0 gives the Maclaurin Series of f(x):
\[f(x) \approx P(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \binom{1/2}{n}x^{n}\]