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:

$$f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots.$$

Taylor series are extremely powerful tools for approximating functions that can be difficult to compute otherwise, as well as evaluating infinite sums and integrals by recognizing Taylor series.

If only concerned about the neighborhood very close to the origin, the $n=2$ approximation represents the sine wave sufficiently, and no higher orders are direly needed.^[1]

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 $\sqrt[3]{8.1}:$

$$\begin{aligned} f(8.1) = \sqrt[3]{8.1} &\approx 2 + \frac{(8.1 - 8)}{12} - \frac{(8.1 - 8)^2}{288} \\ &=\color{#3D99F6}{2.008298}\color{#D61F06}{61111}\ldots \\ \\ \sqrt[3]{8.1} &={ \color{#3D99F6}{2.008298}\color{#D61F06}{85025}\dots}. \end{aligned}$$

With just three terms, the formula above was able to approximate $\sqrt[3]{8.1}$ to six decimal places of accuracy. $_\square$

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:

$$f(x) = \frac{1}{x^2},\quad f'(x) = \frac{-2}{x^3},\quad f''(x) = \frac{6}{x^4}.$$

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 $x = 2.1.$

$$\begin{aligned} P_2(2.1) &= f(2)+\frac {f'(2)}{1!} (2.1-2)+ \frac{f''(2)}{2!} (2.1-2)^2\\ &= \frac14 +\frac {\hspace{3mm} \frac{-2}{8}\hspace{3mm} }{1!} (2.1-2)+ \frac{\hspace{3mm} \frac{6}{16}\hspace{3mm} }{2!} (2.1-2)^2 \\ &= \frac14 + \frac {-1}{4}(0.1) + \frac{3}{16}(0.01)\\ &= 0.25 - 0.025 + 0.001875 \\ &= 0.226875. \end{aligned}$$

The actual value is

$$\frac{1}{4.41} = 0.226757...,$$

so the approximation is only off by about 0.05%. $_\square$

1. IkamusumeFan, . Sine GIF. Retrieved June 1, 2016, from https://commons.wikimedia.org/wiki/File:Sine_GIF.gif