Power Series

Power series are briefly discussed in differential equations as solutions to some of the problems. We now have the tools to study rigorously about them.

Definition. A power series is a series of the form

\[\sum_{n=0}^{\infty} a_n z^n\]

where $z \in \mathbb{C}$ and $a_n \in \mathbb{C}$ for all $n$.

Convergence

Lemma. If a power series converges for a particular value of $w$, the it converges absolutely for all values of $z$ in the circle $\vert z \vert < \vert w \vert$.

Proof.

We have

\[|a_n z^n| = |a_n w^n| \left| {z \over w} \right|^n\]

Since $\sum a_n w^n$ converges, the terms $a_n w^n$ are bounded, say

\[|a_n w^n| < C\]

Then

\[0 \le \sum_{n=0}^{\infty} |a_n z^n| < \sum_{n=0}^{\infty} C \left| {z \over w} \right|^n\]

which converges as it is a geometric series. By comparison test, $\sum a_n z^n$ converges absolutely.

Theorem. A power series either

  • convergess absolutely for all $z$, or

  • convergess absolutely for all $z$ inside a circle $\vert z \vert = R$ and diverges for all $z$ outside it, or

  • converges for $z = 0$ only.

Proof.

Definition. The circle $\vert z \vert = R$ is called the circle of convergence and $R$ is called the radius of convergence.

Proposition. If $\vert a_{n+1} / a_n \vert$ tends to a limit $l$ as $n \to \infty$, then the radius of convergence of $\sum a_n z^n$ is $1 / l$.

Proof.

By d’Alembert’s ratio test, we have absolute convergence iff

\[\left| {a_{n+1} z^{n+1} \over a_n z^n} \right| = \left| {a_{n+1} \over a_n} \right| |z| < 1 \quad \implies \quad |z| < {1 \over l}\]

It is to be noted that the convergence of values on the circle of convergence requires special investigation for any particular series.

Multiplication

When we have two power series, it is useful if we can write the product of them as

\[(a_0 + a_1z + a_2z^2 + \cdots)(b_0 + b_1z + b_2z^2 + \cdots) = a_0b_0 + (a_1b_0 + a_0b_1)z + (a_2b_0 + a_1b_1 + a_0b_2)z^2 + \cdots\]

The following theorem works for any absolutely convergent series which need not be power series

Theorem. If $\sum u_n$ and $\sum v_n$ converges absolutely to sums $s$ and $t$, then the series consisting of their products (in any other) converges absolutely to $st$.

Proof.

The products of pairs of terms form a doubly infinite array

\[\begin{vmatrix} u_0 v_0 & u_0 v_1 & u_0 v_2 & \cdots \\ u_1 v_0 & u_1 v_1 & u_1 v_2 & \cdots \\ u_2 v_0 & u_2 v_1 & u_2 v_2 & \cdots \\ \cdots & \cdots & \cdots & \cdots \end{vmatrix}\]

The sum of all these terms can be arranged in infinitely many ways, but whatever the arrangement, it doesn’t exceed

\[\left( \sum_0^\infty |u_n| \right)\left( \sum_0^\infty |v_n| \right)\]

so the series consisting of their products converges absolutely. Thus the sum is the same for whatever order of the terms. One particular order, namely, summation by squares, gives an evident sum that all terms with suffices not exceeding $n$ is

\[(u_0 + u_1 + \cdots + u_n)(v_0 + v_1 + \cdots + v_n)\]

and hence the limit of the sum is $st$.

Corollary. If $\sum a_n z^n$ and $\sum b_n z^n$ have radii of convergence $R$ and $S$, then their product is

\[\sum_{n=0} (a_n b_0 + a_{n-1} b_1 + \cdots + a_0b_n) z^n\]

for $\vert z \vert < \min(R, S)$.

Taylor’s Series

In the discussion of Taylor’s Theorem, we have shown the way to approximate a function by polynomials, which is indeed a power series.

Theorem. [Taylor’s Series] Suppose $f$ has derivatives of every order for neighbourhood $(a - k, a + k)$. For $\vert h \vert < k$, if the remainder $r_n$ (the term in $h^n$) tends to $0$ as $n$ tends to $\infty$, then

\[f(a + h) = f(a) + hf'(a) + \cdots + {h^n \over n!}f^{(n)}(a) + \cdots\]

For the case of $a = 0$, the expansion cound be called Maclaurin’s series.

We can use this to determine the radius of convergence of the Taylor’s series of $f(x)$, by checking in what values $r_n \to 0$ as $n \to \infty$. We have proved the binomial series before but this time we focus on the remainder term.

Lemma. If $-1 < x < 1$,

\[{m \choose n} x^n = {m(m-1)...(m-n+1) \over n!} x^n\]

tends to $0$ as $n \to \infty$.

Proof.

\[\begin{align*} \lim_{n \to \infty} \left| {a_{n+1} \over a_n} \right| &= \lim_{n \to \infty} \left| {m-n \over n+1} \right| |x| \\ &= \lim_{n \to \infty} \left| {1 - m/n \over 1 + 1/n} \right| |x| \\ &= |x| \end{align*}\]

As $\vert x \vert < 1$, the series $\sum a_n$ converges by d’Alembert’s ratio test. Hence, the underlying sequence is a null sequence.

Proposition. [Binomial Theorem] Given $m \in \mathbb{Q}$. For $-1 < x < 1$,

\[(1 + x)^m = 1 + {m \choose 1} x + \cdots + {m \choose n} x^n + \cdots\]

Proof.

If $m$ is positive integer, $f^{(m+1)}(x) \equiv 0$ and we have a polynomial of degree $m$.

In the general case, we have

\[r_n = {x^n \over n!} f^{(n)}(\theta x) = {m \choose n} { x^n \over (1 + \theta x)^{n-m} }\]

where $\theta$ depends on $n$.

For $0 \le x < 1$, $(1 + \theta x)^{n - m} \ge 1$ for $n > m$ and ${m \choose n} x^n \to 0$ as $n \to \infty$ so $r_n \to 0$.

For $-1 < x < 0$, the above argument doesn’t work as $(1 + \theta x) < 1$. Instead, consider the Cauchy’s remainder with $k = 1$ (which actually applies for the full range $-1 < x < 1$), we have

\[r_n = { m(m-1)...(m-n+1) \over (n - 1)! } {(1 - \theta)^{n-1} x^n \over (1 + \theta x)^{n-m} }\]

We have $(1 - \theta)/(1 + \theta x) < 1$ and so

\[|r_n| < K_m \left| {m-1 \choose n-1} \right| |x|^n\]

which tends to $0$ as $n \to \infty$.

Differentiation

We wish to show that the derivative of $\sum a_n x^n$ is indeed $\sum n a_n x^{n-1}$ inside the circle of convergence. Practically, there can be more straightforward arguments to find the derivatives of certain power series.

Proposition. Suppose $\sum a_n z^n$ has radius of convergence $R$. Then $\sum n a_n z^{n-1}$ converges absolutely for $\vert z \vert < R$.

Proposition. Suppose $f(z) = \sum a_n z^n$ has radius of convergence $R$. Then $f$ is differential and has derivative $f’(z) = \sum n a_n z^{n-1}$.

References