Skip to main content

Section 2.3 Approximation and Taylor Polynomials

Figure 2.3.1. Polynomials Approximations to \(e^x\text{.}\)

The previous section defined Taylor series for analytic functions. Instead of taking the terms and coefficients all the way to infinity, we could instead truncate the process at some degree. The result is a polynomial which serves as a polynomial approximation to the function.

Definition 2.3.2.

If \(f(x)\) is analytic, its \(d\)th Taylor polynomials centered at \(\alpha\) is the truncation of its Taylor series, stopping at \((x-\alpha)^d\text{.}\)

\begin{equation*} f(x) \cong \sum_{n=0}^d \frac{f^{(n)}(\alpha)}{n!} (x-\alpha)^n \end{equation*}

Taylor polynomials give the best possible polynomial approximations to analytic functions.

Look at the exponential function \(e^x\) centered at \(\alpha = 0\text{.}\) We have its Taylor series from the previous section. These are its polynomial approximations. Their graphs are shown in Figure 2.3.1.

\begin{align*} e^x \amp \cong \sum_{n=0}^1 \frac{1}{n!} x^n = 1 + x = p_1\\ e^x \amp \cong \sum_{n=0}^2 \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2} = p_2\\ e^x \amp \cong \sum_{n=0}^3 \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} = p_3\\ e^x \amp \cong \sum_{n=0}^4 \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} = p_4\\ e^x \amp \cong \sum_{n=0}^5 \frac{1}{n!} x^n = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120} = p_5 \end{align*}
Figure 2.3.4. Polynomials Approximations to \(\sin x\text{.}\)

The approximations for sine only have odd exponents, since there are only odd monomials in the Taylor series for sine. These are the first few approximations. Their graphs are shown in Figure 2.3.4

\begin{align*} \sin x \amp \cong \sum_{k=0}^0 \frac{(-1)^k}{(2k+1)!} x^{2k+1} = x = p_1\\ \sin x \amp \cong \sum_{k=0}^1 \frac{(-1)^k}{(2k+1)!} x^{2k+1} = x - \frac{x^3}{3!} = p_3\\ \sin x \amp \cong \sum_{k=0}^2 \frac{(-1)^k}{(2k+1)!} x^{2k+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} = p_5\\ \sin x \amp \cong \sum_{k=0}^3 \frac{(-1)^k}{(2k+1)!} x^{2k+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} = p_7\\ \sin x \amp \cong \sum_{k=0}^4 \frac{(-1)^k}{(2k+1)!} x^{2k+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} = p_9 \end{align*}

The main application of approximation is calculating values of transcendental functions. We can't directly calculate their values using basic arithmetic; we need a method. Before the convenience of calculator and computer reference, mathematicians, scientists and engineers carried around large books of tables of values of trig, exponential and logarithmic function.

Polynomials are particularly useful as approximation tools since they involve only the basic operations of arithmetic. Computers can calculate with the basic operations of arithmetic, so computers can understand polynomials If we want to program a computer or calculator to calculate values of \(e^x\) or \(\sin x\) or \(\ln x\) or some other transcendental function, Taylor series are one of the best techniques.

The logarithm is a transcendental function which can't be directly calculated We had a Taylor series for the logarithm in the previous section.

\begin{equation*} -\ln (1-x) = \sum_{n=0}^\infty \frac{x^{n+1}}{n+1} \end{equation*}

Using some clever arithmetic, we can write \(\ln 2 = - \ln \frac{1}{2} = - \ln \left( 1 - \frac{1}{2} \right)\) If we truncate the series at degree \(6\text{,}\) we have this approximation for \(\ln 2\text{.}\)

\begin{align*} \ln 2 \amp \cong 1 \cdot \frac{1}{2} + \frac{1}{2} \cdot \left( \frac{1}{2} \right)^2 + \frac{1}{3} \left( \frac{1}{2} \right)^3 + \frac{1}{4} \left( \frac{1}{2} \right)^4 + \frac{1}{5} \left( \frac{1}{2} \right)^5 + \frac{1}{6} \left( \frac{1}{2} \right)^6\\ \ln 2 \amp \cong \frac{1}{2} + \frac{1}{8} + \frac{1}{24} + \frac{1}{64} + \frac{1}{160} + \frac{1}{384}\\ \ln 2 \amp \cong \frac{1327}{1920} = 0.6911458333333 \ldots = 0.6911458\bar{3} \end{align*}

This is not to far off from the value of \(\ln 2 = 0.69314\ldots\text{,}\) accurate to the thousandths place.

There are many ways in mathematics to find approximations to numbers. Recall that the alternating harmonic series also summed to \(\ln 2\text{.}\) If we truncate that series after ten steps, we get this approximation.

\begin{equation*} \ln 2 \cong 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \frac{1}{9} - \frac{1}{10} = \frac{1627}{2520} = 0.645\overline{634920} \end{equation*}

This expression is a poorer approximation for \(\ln 2\text{;}\) we would need to go much farther down the alternating harmonic series to match the precision of the Taylor series.