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, 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. 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, forever, is therefore worth today

$$\sum_{n=1}^{\infty}\frac{100}{1.05^{n}}=2000$$

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