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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, I 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 \(\alpha = 0\text{.}\) Its Taylor series was calculated in Example 2.2.5. These are its polynomial approximations.sd 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 Taylor polynomial approximation is calculating values of transcendental functions. I can't directly calculate their values using basic arithmetic; I 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. Now, computer algorithms calculate these values for uses, but these algorithms still need a method. Taylor polynomials are one such method.

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 I want to program a computer or calculator to calculate values of \(e^x\) or \(\sin x\) or \(\ln x\) or some other transcendental function, a Taylor polynomial is a great choice.

The logarithm is a transcendental function which can't be directly calculated The Taylor series for \(-\ln (1-x)\) was calculated in Example 2.2.7.

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

Using some clever arithmetic, I can write \(\ln 2 = - \ln \frac{1}{2} = - \ln \left( 1 - \frac{1}{2} \right)\) If I truncate the series at degree \(6\text{,}\) I produce the following 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.