Taylor Series Approximation
A Taylor series approximation uses a Taylor series to represent a number as a polynomial that has a very similar value to the number in a neighborhood around a specified \(x\) value.
\[f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots.\]
Suggested steps for approximating values:
1) Identify a function to resemble the operation on the number in question.
2) Choose \(a\) to be a number that makes \(f(a)\) easy to compute.
3) Select \(x\) to make \(f(x)\) the number being approximated.
Using the first three terms of the Taylor series expansion of \(f(x) = \sqrt[3]{x}\) centered at \(x = 8\), approximate \(\sqrt[3]{8.1}:\)
\[f(x) = \sqrt[3]{x} \approx 2 + \frac{(x - 8)}{12} - \frac{(x - 8)^2}{288} .\]
The first three terms shown will be sufficient to provide a good approximation for \(\sqrt[3]{x}\). Evaluating this sum at \(x = 8.1\) gives an approximation for \(\sqrt[3]{8.1}:\)
\[\begin{align} f(8.1) = \sqrt[3]{8.1} &\approx 2 + \frac{(8.1 - 8)}{12} - \frac{(8.1 - 8)^2}{288} \\ &=\color{blue}{2.008298}\color{red}{61111}\ldots \\ \\ \sqrt[3]{8.1} &= \color{blue}{2.008298}\color{red}{85025}\dots. \end{align}\]
With just three terms, the formula above was able to approximate \(\sqrt[3]{8.1}\) to six decimal places of accuracy.
Using the quadratic Taylor polynomial for \(f(x) = \frac{1}{x^2},\) approximate the value of \(\frac{1}{4.41}.\)
The quadratic Taylor polynomial is
\[P_2(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2.\]
First, write down the derivatives needed for the Taylor expansion.
\[f(x) = \frac{1}{x^2}\]
\[f'(x) = \frac{-2}{x^3}\]
\[f''(x) = \frac{6}{x^4}\]
But what about \(a\) and \(x?\) Choose \(a\) so that the values of the derivatives are easy to calculate. Rewriting the approximated value as
\[4.41 = (2+0.1)^2\]
implies \(a = 2\) and \(x = 2.1.\)
\[P_2(2.1) = f(2)+\frac {f'(2)}{1!} (2.1-2)+ \frac{f''(2)}{2!} (2.1-2)^2\]
\[P_2(2.1) = \frac14 +\frac {\frac{-2}{8}}{1!} (2.1-2)+ \frac{\frac{6}{16}}{2!} (2.1-2)^2 \]
\[P_2(2.1) = \frac14 + \frac {-1}{4}(0.1) + \frac{3}{16}(0.01)\]
\[P_2(2.1) = 0.25 - 0.025 + 0.001875 = 0.226875\]
The actual value is \[\frac{1}{4.41} = 0.226757...\]
so the approximation is only off by about 0.05%.
References
- IkamusumeFan, . Sine GIF. Retrieved June 1, 2016, from https://commons.wikimedia.org/wiki/File:Sine_GIF.gif