2.14 Series Expansion of Functions
General methods for expanding a function in power series in \(x\) and in \((x - a)\) are the Maclaurin’s series and Taylor’s series respectively.
Note that the important requirements is that the function and its derivatives of all orders must exists at \(x = 0\) and \(x = 0\). This functions like \(\dfrac {1}{x}\hspace {0.2cm}, \hspace {0.2cm} \ln x, \) and \(\cot x\) cannot be expanded in power series of \(x\).
- 1.
- MaClaurin’s Series Assuming that a given function can be represented by power series in \(x\). That series is necessarily of the form of MaClaurin’s series. \[f(x) = f(0) + \frac {f'(0)}{1!}x + \frac {f''(0)}{2!}x^2 + \cdots \cdots \cdots + \frac {f^{(n-1)(0)}}{(n-1)!}x^{n-1} + \cdots \cdots \]
- 2.
- Taylor’s Series Assuming that a given function can be represented by a a power series in \((x -a)\), that series is necessarily of the form of Taylor’s series. \[f(x) = f(a) + \frac {f'(a)}{1!}(x-a) + \frac {f''(a)}{2!}(x-a)^2 + \cdots \cdots \cdots + \frac {f^{(n-1)(a)}}{(n-1)!}(x-a)^{n-1} + \cdots \cdots \]
Obtain the MaClaurin’s series of the function \(f(x) = \sin x\) and determine the interval of convergence of the series.
Solution. \[f(x) = f(0) + \frac {f'(0)}{1!}x + \frac {f''(0)}{2!}x^2 + \cdots \cdots \cdots + \frac {f^{(n-1)(0)}}{(n-1)!}x^{n-1} + \cdots \cdots \]
\(f(x) = \sin x \implies f(0) = \sin (0) = 0\)
\(f'(x) = \cos x \implies f'(0) = \cos (0) = 1\)
\(f''(x) = -\sin x \implies f''(0) = -\sin (0) = 0\)
\(f'''(x) = -\cos x \implies f'''(0)= -1\)
\(\vdots \)
\(\vdots \)
Note that the values of the derivatives at \(x = 0\) form cycles of \(0, 1 , 0 , -1, \cdots \cdots \cdots \)
Therefore, \[\sin x = 0 + x + \frac {0}{2!}x^2 + \frac {-1}{3!}x^3 + \frac {0}{4!}x^4 + \frac {1}{5!}x^5+\cdots \cdots \]
\[\implies \sin x = x - \frac {1}{3!}x^3 + \frac {1}{5}x^5 - \frac {1}{7!} + \cdots \cdots \cdots + \frac {(-1)^{n-1}}{(2n -1)}x^{2n-1}\]
By ratio test \[\lim \limits _{n \rightarrow \infty } \left |\dfrac {x^{2n + 1}}{(2n+1)!}\cdot \dfrac {(2n-1)!}{x^{2n-1}}\right |= x^2\lim \limits _{n\rightarrow \infty } \frac {1}{2n(2n+1)} = 0\]
\(\implies \hspace {0.3cm}\) the series converges \(\forall x\).
Find the third Taylor polynomial for \(f(x) = \sin x\) expanded about \(x = \dfrac {\pi }{6}\).
Solution. \[f(x) = f(a) + \frac {f'(a)}{1!}(x-a) + \frac {f''(a)}{2!}(x-a)^2 + \cdots \cdots \cdots + \frac {f^{(n-1)(a)}}{(n-1)!}(x-a)^{n-1} + \cdots \cdots \]
\(f\Big (\dfrac {\pi }{6}\Big ) = \sin \Big (\dfrac {\pi }{6}\Big ) = \dfrac {1}{2}\)
\(f'\Big (\dfrac {\pi }{6}\Big ) = \cos \Big (\dfrac {\pi }{6}\Big ) = \dfrac {\sqrt {3}}{2}\)
\(f''\Big (\dfrac {\pi }{6}\Big ) = -\sin \Big (\dfrac {\pi }{6}\Big ) = \dfrac {-1}{2}\)
\(f'''\Big (\dfrac {\pi }{6}\Big ) = -\cos \Big (\dfrac {\pi }{6}\Big ) = \dfrac {-\sqrt {3}}{2}\)
\[\sin (x) = \dfrac {1}{2} +\dfrac {\dfrac {\sqrt {3}}{2}\Big (x - \dfrac {\pi }{6}\Big )}{1!} + \dfrac {\Big (\dfrac {-1}{2}\Big )\Big (x - \dfrac {\pi }{6}\Big )^2}{2!} + \frac {\Big (\dfrac {-\sqrt {3}}{2}\Big )\Big (x - \dfrac {\pi }{6}\Big )^3}{3!} + \cdots \cdots \cdots \]
Solution.
Let \(\hspace {0.4cm} f(x) = \ln (1 + x)\)
\(f(0) = \ln 1 = 0\)
\(f'(x) = \dfrac {1}{1 + x} \implies f'(0) = 1\)
\(f''(x) = \dfrac {-1}{(1 + x)^2} \implies f''(0) = -1\)
\(f'''(x) = \dfrac {2}{(1 + x)^2}\implies f'''(0) = 2\)
\[f(x) = 0 + 1\cdot - \dfrac {1\cdot x^2}{2!} + \dfrac {2\cdot x^3}{3!} + \cdots \cdots \cdots \]
\[\implies \hspace {0.3cm} f(x) = x - \dfrac {x^2}{2} + \frac {x^3}{3} - \frac {x^4}{4}+ \cdots \cdots \cdots \]
but \(\ln (1.1) = \ln ( 1 + 0.1)\) so that \(x = 0.1\) substituting in power series yields \begin {align*} \ln (1.1) & = 0.1 - \frac {(0.1)^2}{2} + \frac {(0.1)^3}{3} - \frac {(0.1)^4}{4}\\ & = 0.090308\\\\ \end {align*}
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