Section 6.4 Working with Taylor Series
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}
Theorem 6.4.1. The Binomial Series.
For any real number \(r\) and \(|x| < 1\text{,}\)
\begin{equation*}
(1+x)^r = \sum_{n=0}^\infty \left(\begin{array}{c} r \\ n \end{array} \right)x^n
= 1 + rx + \frac{r(r-1)}{2!}x^2 + \cdots + \frac{r(r-1)\cdots (r-n+1)}{n!}x^n + \cdots
\end{equation*}
where
\begin{equation*}
\left(\begin{array}{c} r \\ n \end{array} \right) := \frac{r}{1}\frac{r-1}{2} \cdots \frac{r-n+1}{n}
= \frac{r (r-1) \cdots (r-n+1)}{n!}
\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.
Checkpoint 6.4.2.
Verify that the result is as expected for \(r=-1\) and for \(r=2\) (or any positive integer).
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.
| \(f(x) = \ds\frac{1}{1-x}\) |
\(\ds\sum_{n=0}^\infty x^n\) |
\(-1 < x < 1\text{,}\) \(R=1\)
|
| \(f(x) = e^x\) |
\(\ds\sum_{n=0}^\infty \frac{x^n}{n!}\) |
\(-\infty < x < \infty\text{,}\) \(R=\infty\)
|
| \(f(x) = \ln(1+x)\) |
\(\ds\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}x^n\) |
\(-1 < x \le 1\text{,}\) \(R=1\)
|
| \(f(x) = \sin(x)\) |
\(\ds\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}\) |
\(-\infty < x < \infty\text{,}\) \(R=\infty\)
|
| \(f(x) = \cos(x)\) |
\(\ds\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}\) |
\(-\infty < x < \infty\text{,}\) \(R=\infty\)
|
| \(f(x) = \arctan(x)\) |
\(\ds\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}\) |
\(-1 \le x \le 1\text{,}\) \(R=1\)
|
| \(f(x) = (1+x)^r\) |
\(\ds\sum_{n=1}^\infty \left(\begin{array}{c} r \\ n \end{array}\right) x^n\) |
\(-1 < x < 1\text{,}\) \(R=1\) (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.
Example 6.4.3.
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, \(\text{Erf}(1) \approx 0.84\text{.}\)
Checkpoint 6.4.4.
Find a power series for \(\ds \int \frac{\sin x}{x} dx\text{.}\)
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.
Checkpoint 6.4.5.
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.
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.
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