Power series

A power series is an infinite sum with the general form

\sum_{n=0}^{\infty}a_{n}(x-c)^{n}=a_{0}+a_{1}(x-c)+a_{2}(x-c)^{2}+\cdots

The numbers $a_0$, $a_1$, $a_2$, … are constants, known as the coefficients, and $c$, also a constant, is known as the centre of the series.

The simplest form: The geometric series

The simplest power series to study is the one in which every coefficient is 1 and the centre is 0:

$$1 + x + x^2 + x^3 + \cdots$$

As seen, each term multiplies the previous one by $x$. This is known as a geometric series.

For a geometric series, we can calculate the partial sum of the first $N + 1$ terms, $S_N = 1 + x + x^2 + \cdots + x^N$:

Multiply both sides of this equation by $(1 - x)$ and eliminate the brackets:

$$(1-x)S_{N}=(1-x)(1+x+x^{2}+\cdots+x^{N})=(1+x+x^{2}+\cdots+x^{N})-(x+x^{2}+\cdots+x^{N+1})=1-x^{N+1}$$

Dividing both sides again by $(1 - x)$, we therefore have

$$S_{N}=\frac{1-x^{N+1}}{1-x}\qquad(x\neq 1)$$ (1)

When $N \to \infty$, we have

$$\sum_{n=0}^{\infty}x^{n}=\lim_{N \to \infty}S_N=\lim_{N \to \infty}\frac{1-x^{N+1}}{1-x}\qquad(x\neq 1) $$

• When $-1 < x < 1$: the series converges. As $N$ grows, the number $x^{N+1}$ shrinks towards 0, so the partial sums approach a definite number:

$$\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}\qquad\text{whenever }-1<x<1$$

With $x = 0.1$, for instance, the partial sums run 1; 1.1; 1.11; 1.111; …, settling on 1.111…, and indeed $1/(1 - 0.1) = 1/0.9 = 10/9 = 1.111\ldots$

• When $x = 1$: the series diverges. Notice (1) breaks down here, because $1 - x = 0$ and we cannot divide by 0. The sum is, however, easy to calculate: every term equals 1, so $\lim_{N \to \infty} S_N = \lim_{N \to \infty} (N + 1) = \infty$

• Case $x = -1$: the series diverges by oscillation. The partial sums run 1, 0, 1, 0, 1, … and never settle on a single number. The formula $1/(1 - x)$ would give $1/2$ at $x = -1$, but notice that is the average of 1 and 0. The formula (1) does not guarantee convergence.

• Case $|x| > 1$: the series diverges, as expected. When $x > 1$, the infinite sum is the sum of an infinite number of increasingly large positive numbers, which goes to infinity. When $x < -1$, the infinite sum oscillates between positive and negative infinity, as it is dominated by the last term, which may be positive or negative.

Real-world application of geometric series

If someone promises to pay you €100 each year until the end of your life on earth, with the first payment made today, how much is that promise worth?

On first thought, this promise is worth quite a lot of money, especially if you count on living for decades more. However, you must take into account inflation, which in Europe averages out to about 5% per year. The €100 paid today is worth its full €100, a payment of €100 one year from now is worth only €$100/1.05$ in today's money, and a payment in two years is worth only €$(100)/(1.05)^2$ in today's money, and so on. A payment of €100 every year, with the first payment made today, forever, is therefore worth today

$$\sum_{0}^{n \to \infty}\frac{100}{1.05^n}= \frac{100}{1-\frac{1}{1.05}}= 2100 $$

It is still a handsome sum, if you can live on forever.

Radius of convergence of a power series

In the previous section, we have seen that even the simplest power series, a geometric series, may or may not converge depending on the value of $x$. In general, for a power series centred at $c$, there is a number $R \geq 0$, called the radius of convergence, such that:

• the series converges for every $x$ with $|x - c| < R$ (closer to the centre than $R$);
• the series diverges for every $x$ with $|x - c| > R$.

The set $|x - c| < R$, an interval of length $2R$ centred at $c$, is called the interval of convergence of the series.

For the two boundary points where $|x - c| = R$, the convergence is uncertain and depends on specific characteristics of a given power series.

In general, consecutive terms of a power series $a_{n+1} (x - c)^{n+1}$ and $a_n (x - c)^n$ have magnitudes in the ratio $|a_{n+1}/a_n| \cdot |x - c|$. If, in the long run, this ratio stays below 1, intuitively the series behaves like a convergent geometric series; if the ratio stays above 1, terms grow larger and larger and the series diverges. Therefore we have

$$R=\lim_{n\to\infty}\left|\frac{a_{n}}{a_{n+1}}\right|\quad\text{when the limit exists}$$

A quick sanity check with geometric series: For the geometric series, every coefficient is 1, so the ratio is 1 and $R = 1$, matching the interval $-1 < x < 1$ found by direct calculation in the section above.

Two special cases exist:

• Infinite radius, $R = \infty$: the series converges for every $x$. The exponential and sine series in the next section are the standard examples.
• Zero radius, $R = 0$: the series converges only at the centre itself. For instance the series $1 + x + 2!\,x^2 + 3!\,x^3 + \cdots$ (whose coefficient of $x^n$ is $n!$) has $|a_n/a_{n+1}| = n!/(n + 1)! = 1/(n + 1) \to 0$, so $R = 0$; for any $x \neq 0$ the terms eventually grow without bound.

The exponential and sine series

The power series can be used to approximate known functions that are difficult to calculate directly, such as $e^x$ or $\sin x$.

Imagine we would like to approximate some function $f(x)$ with a power series. We would therefore write

$$f(x)=\sum_{n=0}^{\infty}a_{n}(x-c)^{n}$$

where $c$ and $a_n$ are unknown.

Taking the $k$-th derivative of $f(x)$, we have

$$f^{(k)}(x)=\sum_{n=k}^{\infty}\frac{n!}{(n-k)!}\,a_{n}\,(x-c)^{n-k}$$

Evaluating at $x=c$, we have

$$a_{k}=\frac{f^{(k)}(c)}{k!}$$

Generalising to every $n$, we obtain

f(x)=\sum_{n=0}^{\infty}a_{n}(x-c)^{n}\qquad\Longrightarrow\qquad a_{n}=\frac{f^{(n)}(c)}{n!} (2)

The exponential series

The exponential function $e^x$ has defining properties $\frac{d e^x}{dx}=e^x$ and $e^0 = 1$. Consequently, choosing $c=0$ we have

$$f^{(n)}(0) = 1$$

for every $n$. From (2) we have $a_n = 1/n!$, so

e^{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots

Intuitively, the factorial $n! = 1\cdot 2\cdot 3\cdot\cdots\cdot n$ in the denominator grows much faster than any power function towards infinity, so the terms shrink and the series should converge for every $x$.

Applying the ratio test confirms $R = \infty$:

$$R=\lim_{n\to\infty}\left|\frac{a_{n}}{a_{n+1}}\right|=\lim_{n\to\infty}\frac{(n+1)!}{n!}=\lim_{n\to\infty}(n+1)=\infty$$

A quick sanity check: setting $x = 1$ in the series gives the number $e$ itself: $e = 1 + 1 + 1/2 + 1/6 + 1/24 + \cdots = 2.718\,281\,828\ldots$

The sine series

Again choosing $c=0$:

\sin x=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}-\cdots

Apply once again the ratio test:

$$\lim_{k \to \infty} \frac{\frac{x^{2k+1}}{(2k+1)!}}{\frac{x^{2k+3}}{(2k+3)!}}=\lim_{k \to \infty}\frac{(2k+2)(2k+3)}{x^{2}}= \infty$$

which means the series converges as expected, to $\sin x$.

The binomial series

One more family of power series is useful enough to know by name. For any fixed exponent $p$,

(1+x)^{p}=1+px+\frac{p(p-1)}{2!}x^{2}+\frac{p(p-1)(p-2)}{3!}x^{3}+\cdots\qquad(-1<x<1)

Power series solutions of differential equations

A further application of power series is to solve differential equations whose coefficients vary with $x$. Such equations rarely have solutions built from a finite combination of familiar functions, but a solution can sometimes be written down as a power series. Substituting the series into the equation and equating the coefficients of like powers of $x$ turns the differential equation into recurrence relations that fix the coefficients one after another; the constants left free by the recurrence are exactly the arbitrary constants of the equation, and the resulting series solves the equation exactly on its interval of convergence.

Case 1: expanding about an ordinary point, the Airy equation

The Airy equation

$$y^{\prime\prime} - xy = 0$$ (3)

is named after George Biddell Airy, who employed it in 1838 while studying the intensity of light near a caustic. Its solutions cannot be written as finite combinations of elementary functions, so it is the standard test case for series methods.

Setting

$$y=C\sum_{n=0}^{\infty}a_{n}x^{n}$$

where $a_n$ are unknown coefficients. We have therefore

$$\quad\Rightarrow\quad y^{\prime\prime}=C\sum_{n=0}^{\infty}(n+2)(n+1)\,a_{n+2}\,x^{n},\qquad xy=C\sum_{n=1}^{\infty}a_{n-1}\,x^{n}$$

Substituting these two expressions back into (3) and collecting terms, noting that the equality must hold for every value of $x$, gives

$$\quad a_{2}=0,\qquad (n+2)(n+1)\,a_{n+2}=a_{n-1}\qquad(n\geq 1)$$

We can thereby express all the coefficients in terms of $a_0$ and $a_1$:

$$a_3 = a_0/(3\cdot 2) = a_0/6, \qquad a_6 = a_3/(6\cdot 5) = a_0/180,$$

$$a_4 = a_1/(4\cdot 3) = a_1/12, \qquad a_7 = a_4/(7\cdot 6) = a_1/504,$$

$$a_5 = a_2/(5\cdot 4) = 0, \qquad a_8 = a_5/(8\cdot 7) = 0,$$

The solution is therefore:

$$y=C (a_{0}\left(1+\frac{x^{3}}{6}+\frac{x^{6}}{180}+\cdots\right)+a_{1}\left(x+\frac{x^{4}}{12}+\frac{x^{7}}{504}+\cdots\right))$$

The ratio test gives $R = \infty$, so each series converges for every $x$. Setting $C_1 = Ca_0$ and $C_2 = Ca_1$, and denoting the two bracketed series by $A(x)$ and $B(x)$, we arrive at the well-known form

y''-xy=0\qquad\Longrightarrow\qquad y=C_{1}\operatorname{Ai}(x)+C_{2}\operatorname{Bi}(x)

Further reading

• Power series, Wikipedia
• Power Series, Wolfram MathWorld
• Frobenius method, Wikipedia
• Power Series, Paul's Online Math Notes