Title: Default for 16.4 Power Series
Course: HELM Workbooks 2023-2024

Power series

A power series is simply a sum of terms each of which contains a variable raised to a non-negative integer power. To illustrate: \[ x - x^3 +x^5-x^7 + \cdots\] \[1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots \] are examples of power series. In a pervious workbook we encountered an important example of a power series, the binomial series: \[1 + px + \dfrac{p(p-1)}{2!}x^2 + \dfrac{p(p-1)(p-2)}{3!}x^3 + \cdots \] which, as we have already noted, represents the function \((1 + x)^p\) as long as the variable \(x\) satisfies \(|x| < 1\).

A power series has the general form \[b_0 + b_1 x + b_2 x^2 + \cdots = \sum_{p=0}^\infty b_p x^p \] where \(b_0, b_1, b_2, \dots\) are constants. Note that, in the summation notation, we have chosen to start the series at \(p = 0\). This is to ensure that the power series can include a constant term \(b_0\) since \(x^0 = 1\).

The convergence, or otherwise, of a power series, clearly depends upon the value of \(x\) chosen. For example, the power series \[1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots \] is convergent if \(x = -1\) (for then it is the alternating harmonic series) and divergent if \(x = +1\) (for then it is the harmonic series).

Radius of convergence

The most important statement one can make about a power series is that there exists a number, \(R\), called the radius of convergence, such that if \(|x| < R\) the power series is absolutely convergent and if \(|x| > R\) the power series is divergent. At the two points \(x = −R\) and \(x = R\) the power series may be convergent or divergent.

Key Point

For a power series \(\displaystyle{\sum_{p=0}^\infty b_p x^p}\) with radius of convergence \(R\) then

- the series converges absolutely if \(|x|<R\)
- the series diverges if \(|x|>R\)
- the series may be convergent or divergent at \(x=\pm R\)

[Image: Convergence of power series]

For any particular power series \(\displaystyle{\sum_{p=0}^\infty b_p x^p}\) the value of \(R\) can be obtained using the ratio test.

To do this, we apply the ratio test for a general \(x\), we can think of this as fixing \(x\) and asking if the series is convergent. We know, from the ratio test that \(\displaystyle{\sum_{p=0}^\infty b_p x^p}\) is absolutely convergent if \[ \lim_{p \to \infty} \dfrac{|b_{p+1} x^{p+1}|}{|b_p x^p|} = \lim_{p \to \infty}\left|\dfrac{b_{p+1}}{b_p}\right| |x|<1 \quad \mathrm{implying} \quad |x|<\lim_{p\to\infty}\left|\dfrac{b_p}{b_{p+1}}\right| \qquad \mathrm{and \; so} \quad R = \lim_{p\to\infty}\left|\dfrac{b_p}{b_{p+1}}\right| \]

Warning: Note that the terms appearing in the numerator and denominator for the ratio are reversed in the definition of \(R\). This for a good reason, it happens when we divide in the final step above.

It can help to remember that the ratio test looks at how the terms grow by considering the ratio of a term with the previous term. If this is less than \(1\) then the terms are decreasing which is necessary for convergence. We essentially compare the series with a geometric series. In a power series, it looks a little bit like a geometric series with common ratio \(x\), the series converges if the growth rate \(x\) is sufficiently slowed by the growth (or decay) of the coefficients. Hence the series converges if the growth of \(x\) is less than "one over the growth of the coefficients" as the product needs to be less than \(1\).

If nothing else, take care to return to the definitions when you need.

Example

1. Find the radius of convergence of the series \[1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots\]
2. Investigate what happens at the end-points \(x=-1, x=+1\) of the region of absolute convergence.

Solution

1. Here \[1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots = \sum_{p=0}^\infty \dfrac{x^p}{p+1} \] so \[b_p = \dfrac{1}{p+1} \quad \therefore \quad b_{p+1}=\dfrac{1}{p+2}.\] In this case, \[R= \lim_{p \to \infty}\left| \dfrac{p+2}{p+1}\right| = 1\] so the given series is absolutely convergent if \(|x|<1\) and is divergent if \(|x|>1\).
2. At \(x=+1\) the series is \[1+\dfrac{1}{2} + \dfrac{1}{3} + \cdots\] which is divergent (the harmonic series). However, at \(x=-1\) the series is \[1-\dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots \] which is convergent (the alternative harmonic series).
Finally, therefore, the series \[ 1 + \dfrac{x}{2} + \dfrac{x^2}{3} + \dfrac{x^3}{4} + \cdots \] is convergent if \(-1\leq x < 1\).

Exercise

Find the range of values of \(x\) for which the following power series converges: \[1 + \dfrac{x}{[value]} + \dfrac{x^2}{[value]^2} + \dfrac{x^3}{[value]^3} + \cdots \]

First, find the coefficient of \(x^p\):

\(b_p=\)

Now find \(R\), the radius of convergence:

\(R=\)

Lastly, find what happens at \(x = \pm R\).

At \(x = R\), the series is:

At \(x = -R\), the series is:
Feedback: We are given the series \(\displaystyle \sum_{p=1}^\infty \frac{x^p}{[value]^p}\). We see then that the coefficient of \(x^p\) is \[b_p=\dfrac{1}{[value]^p}.\] To find \(R\), the radius of convergence, calculate \[R= \lim_{p \to \infty} \left| \dfrac{b_p}{b_{p+1}} \right| = \lim_{p \to \infty} \left|\dfrac{[value]^{p+1}}{[value]^p}\right| = \lim_{p \to \infty}[value] = [value].\] When \(x=\pm [value]\) the series is clearly divergent as the terms do not go to zero and we can apply the divergence test. Hence the series is convergent if and only if \(-[value] < x < [value]\).

Differentiation and integration of power series

It is worth keeping the in mind even if you have not seen differentiation and integration before. This is one of the more elegant aspects of power series but feel free to return here when later.

Let \(P_1\) and \(P_2\) represent two power series with radii of convergence \(R_1\) and \(R_2\) respectively. We can combine \(P_1\) and \(P_2\) together by addition. We find the following properties:

Key Point

If \(P_1\) and \(P_2\) are power series with respective distinct radii of convergence\(R_1\) and \(R_2\) then the sum \((P_1 +P_2)\) is a power series with the radius of convergence being the smaller of \(R_1\) and \(R_2\).

More care needs to be taken if the radii agree. In general we can only say that the radius of convergence of the sum of the series is at least the common radius of convergence of the two summands.

Power series can also be differentiated and integrated on a term by term basis:

Key Point

If \(P_1\) is a power series with radius of convergence \(R_1\) then \[ \dfrac{\mathrm{d}}{\mathrm{d}x}(P_1) \quad \mathrm{and} \quad \int(P_1)\mathrm{d}x\] are each power series with radius of convergence \(R_1\)

Example

Using the known result that \[(1+x)^p = 1 + px + \dfrac{p(p-1)}{2!}x^2 + \cdots \qquad |x|<1,\] choose \(p=\dfrac{1}{2}\) and by differentiating obtain the power series expression for \((1+x)^{-\frac{1}{2}}.\)

Solution

\[(1+x)^{\frac{1}{2}} = 1 + \dfrac{x}{2} + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}x^2 + \dfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}x^3 + \cdots\]

Differentiating both sides \[\dfrac{1}{2}(1+x)^{-\frac{1}{2}} = 1 - \dfrac{1}{2}x + \dfrac{(-\frac{1}{2})(-\frac{3}{1})}{2}x^2 + \cdots \] This result can, of course, be obtained directly from the expansion for \((1+x)^p\) with \(p=-\dfrac{1}{2}\).

Exercise

Using the known result that \[\dfrac{1}{1+x}= [value] \qquad |x|<1,\]

Find the first three terms in the approximation of \(\ln(1+x)\).

Note: in the input box below, type "log(x)" to represent \( \ln x \) (the natural logarithm of \(x\)).

Hint: integrate both sides of the expression.

\(\ln(1+x)\approx\)

Use the expression to obtain an approximation to \(\ln([value])\). Note: give your answer to 3 significant figures.

\(\ln([value])\approx\)
Feedback: Integrate both sides of \(\dfrac{1}{1+x}= [value] \, :\) \[ \int \dfrac{\mathrm{d}x}{1+x}= \ln(1+x)+c,\] where \(c\) is a constant of integration, and \[\int (1-x+x^2-\dots)\mathrm{d}x = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dots + k,\] where \(k\) is a constant of integration. So we conclude \[\ln(1+x) +c = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dots + k \qquad \mathrm{if} \; |x|<1. \] Choosing \(x=0\) shows that \(c=k\) so they cancel from this equation, therefore \[\ln(1+x) = x - \dfrac{x^2}{2} + \dfrac{x^3}{3} - \dots \qquad \mathrm{if} \; |x|<1.\] Now choose \(x = [value]\) to approximate \(\ln(1+[value]) = \ln([value])\) using terms up to cubic: \[\ln([value]) = [value] - \dfrac{( [value])^2}{2} + \dfrac{( [value])^3}{3} - \dots \approx [value]\] which is easily checked by a calculator.

Power series centered at \(x_0\)

A general power series has the form \[ b_0 + b_1(x-x_0) + b_2(x-x_0)^2 + \cdots = \sum_{p=0}^\infty b_p(x-x_0)^p\]

Exactly the same considerations apply to this general power series as apply to the ‘special’ series \(\sum_{p=0}^\infty b_p x^p\) except that the variable \(x\) is replaced by \((x-x_0).\) The radius of convergence of the general series is obtained in the same way: \[R = \lim_{p \to \infty} \left|\dfrac{b_p}{b_{p+1}}\right|\] and the interval of convergence is now shifted to have centre at \(x = x_0\) (see Figure below). The series is absolutely convergent if \(|x − x_0| < R\), diverges if \(|x − x_0| > R\) and may or may not converge if \(|x − x_0| = R\).

[Image: Interval of convergence]

Exercise

Determine the convergence of the general power series \[1-(x-1)+(x-1)^2-(x-1)^3+\cdots\]

First find an expression for the general term:

\(\displaystyle{\sum_{p=0}^\infty }\) so \(b_p=\)

Now obtain the radius of convergence:

\(R=\)

Finally, decide on the convergence at \(|x - 1| = R\).

When \(x - 1 = R\), the series is:

When \(x - 1 = -R\), the series is:
Feedback: First, we find an expresion for the general term: \[1-(x-[value])+(x-[value])^2-(x-[value])^3+\dots = \sum_{p=0}^\infty [value],\] so \( b_p=[value]\). Now we obtain the radius of convergence: \[ \lim_{p \to \infty} \left|\dfrac{b_p}{b_{p+1}}\right| = \lim_{p+1} \left|\dfrac{(-1)^p}{(-1)^{p+1}}\right| = 1\] Hence \(R=[value]\), so the series is absolutely convergent, and thus convergent, if \(|x-[value]|<[value]\). Finally, we decide on the convergence of the series when \(|x - 1| = R = [value]\). For the first case, we have \(x = 1 + R = [value]\), and the series becomes \[1 - 1 + 1 - 1 + \cdots,\] which is divergent. For the second case, we have \(x = 1 - R = [value]\), and the series becomes \[1 + 1 + 1 + 1 + \cdots,\] which is clearly divergent.

Exercise

Determine the convergence of the general power series \[1-(x-[value])+(x-[value])^2-(x-[value])^3+\cdots\]

First find an expression for the general term:

\(\displaystyle{\sum_{p=0}^\infty }\) so \(b_p=\)

Now obtain the radius of convergence:

\(R=\)

Finally, decide on the convergence at \(|x - [value]| = R\).

When \(x - [value] = R\), the series is:

When \(x - [value] = -R\), the series is:
Feedback: First, we find an expresion for the general term: \[1-(x-[value])+(x-[value])^2-(x-[value])^3+\dots = \sum_{p=0}^\infty [value],\] so \(\displaystyle b_p=[value]\). Now we obtain the radius of convergence: \[ \lim_{p \to \infty} \left|\dfrac{b_p}{b_{p+1}}\right| = \lim_{p+1} \left|\dfrac{(-1)^p}{(-1)^{p+1}}\right| = 1\] Hence \(R=[value]\), so the series is absolutely convergent, and thus convergent, if \(|x-[value]|<[value]\). Finally, we decide on the convergence of the series when \(|x - [value]| = R = [value]\). For the first case, we have \(x = [value] + R = [value]\), and the series becomes \[1 - 1 + 1 - 1 + \cdots,\] which is divergent. For the second case, we have \(x = [value] - R = [value]\), and the series becomes \[1 + 1 + 1 + 1 + \cdots,\] which is clearly divergent.

Exercise

Using the known result that \[\dfrac{1}{1-x}= [value] \qquad |x|<1,\]

Find the first three terms in the approximation of \(\ln(1-x)\).

Note: in the input box below, type "log(x)" to represent \( \ln x \) (the natural logarithm of \(x\)).

Hint: integrate both sides of the expression.

\(\ln(1-x)\approx\)

Use the expression to obtain an approximation to \(\ln([value])\). Note: give your answer to 3 significant figures.

\(\ln([value])\approx\)
Feedback: Integrate both sides of \(\dfrac{1}{1-x}= [value] \, :\) \[ \int \dfrac{\mathrm{d}x}{1-x}= -\ln(1-x)+c,\] where \(c\) is a constant of integration, and \[\int (1+x+x^2+\dots)\mathrm{d}x = x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dots + k,\] where \(k\) is a constant of integration. So we conclude \[-\ln(1-x) +c = +x + \dfrac{x^2}{2} + \dfrac{x^3}{3} + \dots + k \qquad \mathrm{if} \; |x|<1. \] Choosing \(x=0\) shows that \(c=k\) so they cancel from this equation, therefore \[\ln(1-x) = -x - \dfrac{x^2}{2} - \dfrac{x^3}{3} - \dots \qquad \mathrm{if} \; |x|<1.\] Now choose \(x = [value]\) to approximate \(\ln(1-[value]) = \ln([value])\) using terms up to cubic: \[\ln([value]) =- [value] - \dfrac{( [value])^2}{2} - \dfrac{( [value])^3}{3} - \dots \approx [value]\] which is easily checked by a calculator.

Exercise

This exercise assumes knowledge of differentiation, this can be ignored if you have not seen differentiation before.

By differentiating the series for [value] find the power series for [value] and give a quadratic approximation. State its radius of convergence of the power series for [value].

\([value] \approx\)

\(R = \)
Feedback: Using the known result from the binomial series \[(1+x)^p = 1 + px + \frac{p(p-1)}{2!}x^2 + \dots + \frac{p(p-1)(p-2)\cdots(p-n+1)}{n!}x^n \qquad |x|<1,\] choose \(p=[value]\) and by differentiating obtain the power series expression for \([value].\) \[ [value] = [value] \] Differentiating both sides: \( \quad \displaystyle [value] = [value]\) Multiplying through by [value]: \( \quad \displaystyle [value] = [value] \) Using the sum notation, the binomial series is: \[(1+x)^p = \sum_{n=0}^{\infty} \frac{p(p-1)(p-2)\cdots(p-n-1)}{n!}x^n\] Here we have: \[b_p = \dfrac{p(p+1)\cdots(p-n+1)}{n!} \quad b_{p+1}=\dfrac{p(p-1)\cdots(p-n+1)(p-n)}{(n+1)!}\] In this case, \[R= \lim_{p \to \infty}\left| \dfrac{b_p}{b_{p+1}}\right| =\lim_{p \to \infty} \left| \dfrac{p(p-1)\cdots(p-n+1)(p-n)}{(n+1)!} \dfrac{n!}{p(p+1)\cdots(p-n+1)}\right|= \] \[ \lim_{p \to \infty} \frac{|p-n|}{n+1} = \lim_{p \to \infty} \frac{\left| 1- \frac{p}{n}\right|}{1+\frac{1}{n}} =1\] As we learned in the "Properties of power series" section, if \(P_1\) is a power series with radius of convergence \(R_1\) then \[ \dfrac{\mathrm{d}}{\mathrm{d}x}(P_1) \quad \mathrm{and} \quad \int(P_1)\mathrm{d}x\] are each power series with radius of convergence \(R_1.\)

Exercise

1. Find the radius of convergence of the series
\[1 + \frac{x}{[value]} + \frac{x^2}{[value]} + \frac{x^3}{[value]} + \cdots\]
\(R = \)
2. Investigate what happens at the two points \(x = -R\) and \(x = R\).
At \(x = R\), the series is:
At \(x = -R\), the series is:
Feedback: 1. The sequence of coefficients of this series is given by \[b_0 = 1 \quad \mbox{and} \quad b_p = \frac{1}{p + [value]} \ \mbox{for all}\ p > 0.\] The radius of convergence is then \[R = \lim_{p \to \infty} \left\lvert \frac{b_p}{b_{p+1}} \right\rvert = \lim_{p \to \infty} \frac{p + [value]}{p + [value]} = 1.\] Note that the fact that \(b_0\) does not follow the general pattern is irrelevant, because we are taking the limit as \(p\) tends to infinity. 2. At \(x = R = 1\), the series becomes \[1 + \frac{1}{[value]} + \frac{1}{[value]} + \frac{1}{[value]} + \cdots\] Notice that this is very similar to the harmonic series, except it is missing [value] terms. This however does not affect convergence, so we conclude that at \(x = 1\) the series is divergent. At \(x = -R = -1\), the series becomes \[1 - \frac{1}{[value]} + \frac{1}{[value]} - \frac{1}{[value]} + \cdots\] This time, it is just the alternating harmonic series with [value] terms missing. Again, this does not affect convergence, so \(x = -1\) the series is convergent.