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Section 6.4 Working with Taylor Series

References.

The Binomial Series.

The Taylor series for the square root seen in exercise Checkpoint 6.3.10 in the previous section can be viewed as giving a fractional power \(x^{1/2}\text{;}\) in fact, a similar calculation leads to a Taylor series for any power function.

However, most powers \(x^r\) will not have a Maclaurin series (one with center \(a=0\)), due to the function being undefined for \(x < 0\text{,}\) which is why center \(a=1\) was used above.

To get a simpler form in powers of \(x\text{,}\) we shift the function over, and look for the Maclaurin series of \((1+x)^r\text{.}\) For \(r\) a positive integer, this will give a case of the familiar binomial expansion

\begin{equation} (a+b)^r = a^r + ra^{r-1}b + \frac{r(r-1)}{2} a^{r-2}b^2 \cdots + b^r\tag{6.4.1} \end{equation}

Noe that the formula for \(\ds\left(\begin{array}{c} r \\ n \end{array} \right)\) is the same as for the combinatorial quantity denoted \(\mbox{}^rC_n\) or \(\mbox{}_rC_n\text{,}\) giving the number of ways of selecting \(r\) objects from a collection of \(n\) objects, except that now we now allow \(r\) to have any real value.

  1. Verify that the result is as expected for \(r=-1\) and for \(r=2\) (or any positive integer).

  2. Give the Maclaurin series for \(\sqrt[3]{1+x}\) and show that for \(0 < x < 1\) this series is eventually “alternating and decreasing”. (What about for \(x < 0\text{?}\))

Common Functions Expressed as Taylor Series.

Function Maclaurin Series Interval and Radius of Convergence
\(f(x) = \ds\frac{1}{1-x}\) \(\ds\sum_{n=0}^\infty x^n\) \(-1 < x < 1\text{,}\) \(R=1\)
\(f(x) = e^x\) \(\ds\sum_{n=0}^\infty \frac{x^n}{n!}\) \(-\infty < x < \infty\text{,}\) \(R=\infty\)
\(f(x) = \ln(1+x)\) \(\ds\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n\) \(-1 < x \le 1\text{,}\) \(R=1\)
\(f(x) = \sin(x)\) \(\ds\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\) \(-\infty < x < \infty\text{,}\) \(R=\infty\)
\(f(x) = \cos(x)\) \(\ds\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}\) \(-\infty < x < \infty\text{,}\) \(R=\infty\)
\(f(x) = \arctan(x)\) \(\ds\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}\) \(-1 \le x \le 1\text{,}\) \(R=1\)
\(f(x) = (1+x)^r\) \(\ds\sum_{n=1}^\infty \left(\begin{array}{c} r \\ n \end{array}\right) x^n\) \(-1 < x < 1\text{,}\) \(R=1\) (see note)
where \(\ds \left(\begin{array}{c} r \\ n \end{array}\right) = \frac{r(r-1) \cdots (r-n+1)}{n!}\)

Note: the Binomial Series for \((1+x)^r\) also converges for all \(x\) if \(r\) is a natural number (where the series is just a polynomial), and also can converge at the endpoints \(x = \pm 1\) for some other values of \(r\text{.}\)

Evaluating Nonelementary Integrals.

Evaluating integrals for which we do not know an explicit anti-derivative formula is one important use of power series.

The Error Function, very important in statistics is defined as

\begin{equation} \text{Erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt;\tag{6.4.2} \end{equation}

that is, the anti-derivative of \(\ds P(x) = \frac{2}{\sqrt{\pi}} e^{-x^2}\) with the constant of integration chosen to have \(\text{Erf}(0) = 0\text{.}\)

There is no expression for this integral in terms of elementary functions, but a power series can be derived for it, starting from the Maclaurin seried for \(e^x\text{,}\)

\begin{equation*} e^x = 1 + x + \frac{1}{2}x^2 + \frac{1}{3!}x^3 + \cdots + \frac{1}{n!}x^n + \cdots \end{equation*}

First, the substitution \(x \to (-x^2)\) gives

\begin{equation*} \begin{split} e^{-x^2} &= 1 + (-x^2) + \frac{1}{2}(-x^2)^2 + \frac{1}{3!}(-x^2)^3 + \cdots + \frac{1}{n!}(-x^2)^n + \cdots\\ &= 1 -x^2 + \frac{1}{2}x^4 - \frac{1}{3!}x^6 + \cdots + \frac{(-1)^n}{n!} x^{2n} + \cdots \end{split} \end{equation*}

Next, this can be integrated term-by-term to get the indefinite integral

\begin{equation*} \int e^{-x^2} dx = C + x -\frac{1}{3}x^3 + \frac{1}{5 \cdot 2}x^4 - \frac{1}{7 \cdot 3!}x^6 + \cdots + \frac{(-1)^n}{(2n+1)n!} x^{2n} + \cdots \end{equation*}

Finally, multiplying by that factor \(\frac{2}{\sqrt{\pi}}\) and using the fact that \(\text{Erf}(0) = 0\) to show that the constant is \(C=0\text{,}\)

\begin{equation*} \text{Erf}(x) = \frac{2}{\sqrt{\pi}} \left(x -\frac{1}{3}x^3 + \frac{1}{5 \cdot 2}x^4 - \frac{1}{7 \cdot 3!}x^6 + \cdots + \frac{(-1)^n }{(2n+1)n!}x^{2n} + \cdots \right) \end{equation*}

As an example of using this we can approximate the value of the Error Function for \(x=1\) accurate two decimal places. First

\begin{equation*} \text{Erf}(1) = \frac{2}{\sqrt{\pi}} -\frac{2}{3\sqrt{\pi}} + \frac{1}{5\sqrt{\pi}} - \frac{1}{21\sqrt{\pi}} + \frac{1}{108\sqrt{\pi}} + \cdots + \frac{(-1)^n 2}{(2n+1)n!} + \cdots \end{equation*}
This can be verified to meet all conditions the Alternating Series Theorem, and the fifth term is smaller than \(0.01\text{,}\) so the fifth partial sum
\begin{equation*} \frac{2}{\sqrt{\pi}} -\frac{2}{3\sqrt{\pi}} + \frac{1}{5\sqrt{\pi}} - \frac{1}{21\sqrt{\pi}} + \frac{1}{108\sqrt{\pi}} = 0.843449\dots \end{equation*}
is accurate enough, and we also know that thisis an over-estimate. (Note: this is \(p_9(1)\text{;}\) evaluating the degree 9 Maclaurin polynomial.) In fact, that fifth term bounds the error in the fourth partial sum (which is \(p_7(1)\text{;}\) from the degree 7 Maclaurin polynomial), so
\begin{equation*} \frac{2}{\sqrt{\pi}} -\frac{2}{3\sqrt{\pi}} + \frac{1}{5\sqrt{\pi}} - \frac{1}{21\sqrt{\pi}} = 0.838225\dots \end{equation*}
is already accurate enough, and an under-estimate: we have shown that \(0.838 \le \text{Erf}(1) \le 0.844\text{,}\) so to two decimal places, \(\text{Erf}(1) \approx 0.84\text{.}\)

  1. Find a power series for \(\ds \int \frac{\sin x}{x} dx\text{.}\)

  2. Use this series to evaluate \(\ds\int_{-\pi}^\pi \frac{\sin x}{x} dx\) to within \(0.01\text{.}\)

Evaluating “\(0/0\)” Limits Using Power Series.

Sometimes, limits \(\lim_{x \to a} f(x)/g(x)\) that lead to the indeterminate form “\(0/0\)” can be evaluated by expressing the top and bottom in terms of power series with center \(a\text{:}\) there will be a common factor \(x-a\) top and bottom to cancel out.

Evaluate \(\ds \lim_{x \to 0} \frac{e^x -1 - x - x^2/2}{x^2 \sin x}\text{.}\)

(Note: This would involve third derivatives if done with l'Hôpital's Rule.)

Study Guide.

Study Calculus Volume 2, Section 6.4 2 ; in particular

  • all about The Binomial Series

  • Examples 17, 18, 22

  • Checkpoint 16 (Hint: as usual, first rewrite using a negative power), 17, 21

  • and one or several exercises from each of the following ranges: 174–177, 186 and 187, 194 and 195, 210–213.

Note that we omit the topic of Solving Differential Equations with Power Series.

https://openstax.org/books/calculus-volume-2/pages/6-4-working-with-taylor-series
https://openstax.org/books/calculus-volume-2/pages/6-4-working-with-taylor-series