6.2 Properties of Power Series

Big idea

Once you know where a power series converges, it defines a function on that interval, and the natural next question is whether the function inherits the convenience of the expression. If a power series really does behave like a polynomial, then you should be able to add two of them, multiply one by a constant, substitute something into one, and above all differentiate and integrate one term at a time. All of that is true, and the scope of the permission is the point of this section.

The reason the permission is not free is that an infinite sum is a limit, and limits do not always commute with other limits. Differentiation is itself a limit, so differentiating term by term means swapping two limiting processes, and there are settings in analysis where that swap is illegal and produces nonsense. Power series are well behaved enough that the swap is safe everywhere strictly inside the interval of convergence, and that is a genuine theorem rather than a notational convenience.

The practical payoff is that you almost never have to build a series from scratch. One series, the geometric series, is known outright because its partial sums telescope into a closed form. Everything else in this part of the course is reachable from it by four moves: substitute an expression for the variable, multiply by a power, differentiate, integrate. Each move is legal, each move keeps the radius of convergence, and the four of them together generate the standard catalogue.

That catalogue is worth building deliberately, because the alternative is computing derivative after derivative of a complicated function. Getting the series for the inverse tangent from the geometric series takes three lines; getting it from repeated differentiation takes an afternoon and invites arithmetic errors. The skill this section trains is recognizing which known series a new problem is a disguised version of.

Decoder

Inside its interval of convergence a power series may be differentiated and integrated one term at a time, and the resulting series has the same radius of convergence as the original.

The phrase doing the work is inside its interval. The theorem is about the open interval $|x - a| < R$, not the closed one. Endpoint behaviour is not preserved: differentiating can destroy convergence at an endpoint and integrating can create it, because the operations divide and multiply the coefficients by something like $n$. The radius is stable; the two boundary points are not.

The second thing to read carefully is the same radius. This is stronger than it sounds. It says the new series is not merely convergent somewhere, but convergent on exactly the same disc of validity, so you never have to redo the radius calculation after a differentiation or an integration. You do have to redo the endpoint checks if the endpoints matter to you.

The classic mistake in the other direction is substituting without adjusting the interval. If a series in $u$ is valid for $|u| < 1$ and you set $u = -3x^2$, the new series is valid for $|3x^2| < 1$, which is $|x| < 1/\sqrt{3}$, not $|x| < 1$. Substitution preserves the inequality, not the number. Write down the inequality in $u$ first, then translate it.

Definitions and results

The geometric series as a function. For $|x| < 1$,

$$ \sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} $$

Every construction below starts here. The series diverges at both endpoints.

Sums and constant multiples. If $f(x) = \sum a_n x^n$ on $|x| < R_1$ and $g(x) = \sum b_n x^n$ on $|x| < R_2$, then on the smaller of the two intervals

$$ f(x) + g(x) = \sum_{n=0}^{\infty} (a_n + b_n)x^n, \qquad c\,f(x) = \sum_{n=0}^{\infty} c\,a_n x^n $$

Multiplication by a power. Multiplying $f(x)$ by $x^m$ shifts every exponent up by $m$ and changes nothing about the radius. This is how a series that starts at $x^0$ becomes one that starts at $x^m$.

Substitution. If $f(u) = \sum a_n u^n$ for $|u| < R$ and $u = c x^m$, then $f(cx^m) = \sum a_n c^n x^{mn}$, valid exactly when $|c x^m| < R$. Solve that inequality for $x$ to get the new radius.

Products. Two power series multiply the way polynomials do, by collecting like powers. The coefficient of $x^n$ in the product is $\sum_{k=0}^{n} a_k b_{n-k}$. In practice you compute as many of these as you need rather than all of them.

Term-by-term differentiation. If $f(x) = \sum_{n=0}^{\infty} c_n (x-a)^n$ on $|x - a| < R$, then $f$ is differentiable there and

$$ f'(x) = \sum_{n=1}^{\infty} n\,c_n (x-a)^{n-1} $$

with the same radius $R$. The index starts at $1$ because the constant term differentiates away.

Term-by-term integration. On the same interval,

$$ \int f(x)\,dx = C + \sum_{n=0}^{\infty} \frac{c_n}{n+1}(x-a)^{n+1} $$

again with radius $R$. For a definite integral over an interval strictly inside, integrate each term between the same limits and add.

Infinite differentiability. Because the derivative of a power series is again a power series with the same radius, you may differentiate as many times as you like. A function represented by a power series on an open interval has derivatives of every order there.

Worked examples

A substitution into the geometric series

Find a series for $\dfrac{1}{1 + x^2}$.

Write the denominator as $1 - (-x^2)$ so that it matches the geometric pattern with $u = -x^2$:

$$ \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots $$

Validity requires $|-x^2| < 1$, that is $|x| < 1$, so the radius is $1$. Check the arithmetic at $x = 1/2$: the series gives $1 - 0.25 + 0.0625 - 0.015625 + \cdots$, whose partial sums approach $0.8$, and the function gives $1/(1 + 0.25) = 0.8$.

Differentiating to get a new closed form

Start from $\dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ on $|x| < 1$ and differentiate both sides. The left side gives $1/(1-x)^2$, and the right side differentiates term by term:

$$ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} = 1 + 2x + 3x^2 + 4x^3 + \cdots $$

The radius is still $1$. Test at $x = 1/2$: the left side is $1/(0.5)^2 = 4$, and the partial sums $1, 2, 2.75, 3.25, 3.5625, 3.6875, \dots$ do climb toward $4$.

Multiplying through by $x$ gives $\sum_{n=1}^{\infty} n x^{n} = x/(1-x)^2$, which is the form you want when the exponent and the coefficient index should agree.

Integrating to get a logarithm

Replace $x$ by $-x$ in the geometric series to get $\dfrac{1}{1+x} = \sum_{n=0}^{\infty} (-1)^n x^n$ on $|x| < 1$, then integrate from $0$ to $x$. The left side integrates to $\ln(1+x)$, and each term integrates to $(-1)^n x^{n+1}/(n+1)$:

$$ \ln(1+x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots $$

The constant of integration is zero because both sides vanish at $x = 0$. The radius is still $1$, but the endpoints have changed character: at $x = 1$ the integrated series is the alternating harmonic series, which converges, while the series we integrated diverged there. This is the endpoint instability the decoder warned about.

The same move applied to $1/(1+x^2)$ gives the inverse tangent, since integrating $1 - x^2 + x^4 - \cdots$ term by term produces $x - x^3/3 + x^5/5 - \cdots$.

Splitting a product apart first

Find a series for $\dfrac{1}{(1-x)(1-2x)}$.

Multiplying two geometric series directly would work but leaves you collecting convolutions. Partial fractions is cleaner. Write

$$ \frac{1}{(1-x)(1-2x)} = \frac{A}{1-x} + \frac{B}{1-2x} $$

so that $A(1-2x) + B(1-x) = 1$. Setting $x = 1$ gives $-A = 1$, so $A = -1$. Setting $x = 1/2$ gives $B/2 = 1$, so $B = 2$. Now expand each piece:

$$ \frac{-1}{1-x} + \frac{2}{1-2x} = -\sum_{n=0}^{\infty} x^n + 2\sum_{n=0}^{\infty} 2^n x^n = \sum_{n=0}^{\infty} \left( 2^{n+1} - 1 \right) x^n $$

The first piece needs $|x| < 1$ and the second needs $|2x| < 1$, so the combined radius is the smaller one, $1/2$.

Check the first few coefficients against direct multiplication. The product $(1 + x + x^2 + \cdots)(1 + 2x + 4x^2 + \cdots)$ has constant term $1$, coefficient of $x$ equal to $2 + 1 = 3$, and coefficient of $x^2$ equal to $4 + 2 + 1 = 7$. The formula gives $2 - 1 = 1$, $4 - 1 = 3$, and $8 - 1 = 7$.

Practice

First the bookkeeping. After a substitution, a differentiation or an integration, you still have to say where the new series is valid, so these drills ask for radii and intervals of the series that the operations produce.

Then the construction itself. Given a function that is a disguised geometric series, produce its series representation by substituting, multiplying, differentiating or integrating.

Quiz

Five items on combining series, term-by-term calculus, and deriving new representations from the geometric series.