Homework 02

5. Power series

Consider a power series of the form

$a = 1 + \sum_{n = 1}^{\infty} a_{n} z^{n}$

Show that there is another power series

$b = \sum_{n = 0}^{\infty} b_{n} z^{n}$

such that, as a formal expression

$a b = 1$

Do this by giving a recurrence relation for the coefficients $b_{n}$ . Solve this recurrence relation for $1 / e^{z}$ and show it gives the power series for $e^{- z}$ .

The product $a b$ can be written as:

$a b = (1 + \sum_{n = 1}^{\infty} a_{n} z^{n}) (\sum_{n = 0}^{\infty} b_{n} z^{n}) = 1$

By expanding, we get:

$a b = b_{0} + (b_{1} + a_{1} b_{0}) z + (b_{2} + a_{1} b_{1} + a_{2} b_{0}) z^{2} + \dots + (\sum_{k = 0}^{n} a_{k} b_{n - k}) z^{n} + \dots$

For $a b = 1$ , all the coefficients of $z^{n}$ (for $n \geq 1$ ) must be zero, and the constant term (the coefficient of $z^{0}$ ) must be 1. This gives us the following conditions:

$b_{0} = 1$ (since the constant term of $a b$ must be 1).
For $n \geq 1$ , we have $\sum_{k = 0}^{n} a_{k} b_{n - k} = 0$ .

For $a_{k}$ , we have $a_{0} = 1$ and $a_{n}$ for $n \geq 1$ . Thus, the recurrence relation for $b_{n}$ becomes:

$b_{n} = - \sum_{k = 1}^{n} a_{k} b_{n - k} for n \geq 1$

For $a = e^{z}$ . The power series for $b = 1 / e^{z}$ is given by:

$b_{0} = 1$ (from the initial condition).
For $n \geq 1$ :

$b_{n} = - \sum_{k = 1}^{n} \frac{1}{k!} b_{n - k}$

And if $b_{n} = \frac{(- 1)^{n}}{n!}$ . Then, it is exactly the power series for $e^{- z}$ :

$e^{- z} = \sum_{n = 0}^{\infty} \frac{(- z)^{n}}{n!}$

To prove $b_{n} = \frac{(- 1)^{n}}{n!}$ , we use induction method. Suppose it is true for $n$ , then using

$(1 + (- 1))^{n + 1} = \sum C_{k}^{n + 1} (- 1)^{n + 1 - k} = (n + 1)! \sum_{k = 0}^{n + 1} \frac{(- 1)^{n + 1 - k}}{k! (n + 1 - k)!}$

By substituting $b_{k} = \frac{(- 1)^{k}}{k!}$ for $k \leq n$ , we have

$(- 1)^{n + 1} / (n + 1)! = - \sum_{k = 1}^{n + 1} \frac{1}{k!} b_{n - k}$

So $b_{n + 1} = \frac{(- 1)^{n + 1}}{(n + 1)!}$ . Q.E.D

6. Expansion near pole

Assuming the result of problem 5 and the bonus problem, show that if $f (z) = g (z) / h (z)$ has a pole at $z = 0$ , that we can expand $f (z)$ in a series

$f (z) = \sum_{n = - m}^{\infty} a_{n} z^{n}$

for some $m$ , and that this converges in a “punctured disc” ${z | | z | < R, z \neq 0}$ . Make such an expansion for $f (z) = z / (1 + z^{2})$ around $z = i$ (hint: for this example we use tricks with geometric series, we don’t need the general recurrence relation).

Since $f (z)$ has a pole at $z = 0$ , $h (z)$ must have a zero of order $m$ at $z = 0$

Given this, $h (z)$ can be written as $z^{m} k (z)$ where $k (z)$ is analytic and non-zero at $z = 0$ . Therefore, $f (z)$ can be written as:

$f (z) = \frac{g (z)}{z^{m} k (z)}$

Since $k (z)$ is analytic and non-zero at $z = 0$ , there exists a power series $\sum_{n = 0}^{\infty} b_{n} z^{n}$ such that $\frac{1}{k (z)} = \sum_{n = 0}^{\infty} b_{n} z^{n}$ . This leads to:

$f (z) = \frac{g (z)}{z^{m}} \sum_{n = 0}^{\infty} b_{n} z^{n} = \sum_{n = 0}^{\infty} b_{n} g (z) z^{n - m}$

for $z \neq 0$ . As $g (z)$ is analytic at $z = 0$ , it can also be expanded into a power series around $z = 0$ . Thus, the product of these two series results in a series of the form:

$f (z) = \sum_{n = - m}^{\infty} a_{n} z^{n}$

This series converges in some punctured disc ${z | | z | < R, z \neq 0}$ because it is the product of two convergent power series in that region.

For the specific function $f (z) = \frac{z}{1 + z^{2}}$ , we want to expand around $z = i$

$f (z) = \frac{z}{(z - i) (z + i)}$

To expand around $z = i$ , let’s set $w = z - i$ , so $z = w + i$ . Then:

$f (w + i) = \frac{w + i}{w (w + 2 i)}$

Now, observe that:

$\frac{1}{w + 2 i} = \frac{1}{2 i} \cdot \frac{1}{1 + \frac{w}{2 i}}$

Since $| \frac{w}{2 i} | < 1$ for $| w | < 2$ , we can use the geometric series expansion:

$\frac{1}{1 + \frac{w}{2 i}} = \sum_{n = 0}^{\infty} {(- \frac{w}{2 i})}^{n}$

Thus, we get:

$f (z) = \frac{w + i}{2 i w} \cdot \sum_{n = 0}^{\infty} {(- \frac{w}{2 i})}^{n}$

which will be valid in a punctured disc around $z = i$ , specifically for $| z - i | < 2$ (or $| w | < 2$ ).

--- title: Homework 02 --- [Mathematica notebook](./homework02.nb) ## 5. Power series > Consider a power series of the form > > $$ > a=1+ \sum_{n=1}^{\infty} a_n z^n > $$ > > Show that there is another power series > > $$ > b = \sum_{n=0}^{\infty} b_n z^n > $$ > > such that, as a formal expression > > $$ > ab = 1 > $$ > > Do this by giving a recurrence relation for the coefficients $b_n$. Solve this recurrence relation for $1/e^z$ and show it gives the power series for $e^{−z}$. The product $ab$ can be written as: $$ ab = \left(1 + \sum_{n=1}^{\infty} a_n z^n \right) \left( \sum_{n=0}^{\infty} b_n z^n \right) = 1 $$ By expanding, we get: $$ ab = b_0 + (b_1 + a_1 b_0) z + (b_2 + a_1 b_1 + a_2 b_0) z^2 + \dots + \left( \sum_{k=0}^{n} a_k b_{n-k} \right) z^n + \dots $$ For $ab = 1$, all the coefficients of $z^n$ (for $n \geq 1$) must be zero, and the constant term (the coefficient of $z^0$) must be 1. This gives us the following conditions: 1. $b_0 = 1$ (since the constant term of $ab$ must be 1). 2. For $n \geq 1$, we have $\sum_{k=0}^{n} a_k b_{n-k} = 0$. For $a_k$, we have $a_0 = 1$ and $a_n$ for $n \geq 1$. Thus, the recurrence relation for $b_n$ becomes: $$ b_n = -\sum_{k=1}^{n} a_k b_{n-k} \quad \text{for } n \geq 1 $$ For $a = e^z$. The power series for $b= 1/e^z$ is given by: 1. $b_0 = 1$ (from the initial condition). 2. For $n \geq 1$: $$ b_n = -\sum_{k=1}^{n} \frac{1}{k!} b_{n-k} $$ And if $b_n = \frac{(-1)^n}{n!}$. Then, it is exactly the power series for $e^{-z}$: $$ e^{-z} = \sum_{n=0}^{\infty} \frac{(-z)^n}{n!} $$ To prove $b_n = \frac{(-1)^n}{n!}$, we use induction method. Suppose it is true for $n$, then using $$ (1 + (-1))^{n+1} = \sum C_{k}^{n+1}(-1)^{n+1-k} = (n+1)! \sum_{k=0}^{n+1} \frac{(-1)^{n+1-k}}{k! (n+1-k)!} $$ By substituting $b_k = \frac{(-1)^k}{k!}$ for $k \le n$, we have $$ (-1)^{n+1} / (n+1)~! = -\sum_{k=1}^{n+1} \frac{1}{k!} b_{n-k} $$ So $b_{n+1} = \frac{(-1)^{n+1}}{(n+1)!}$. Q.E.D ## 6. Expansion near pole > Assuming the result of problem 5 and the bonus problem, show that if $f (z) = g(z)/h(z)$ has a pole at $z = 0$, that we can expand $f(z)$ in a series > > $$ > f (z) = \sum_{n=-m}^{\infty} a_n z^n > $$ > > for some $m$, and that this converges in a “punctured disc” $\{ z | |z| < R, z \ne 0 \}$. Make such an expansion for $f (z) = z/(1 + z^2)$ around $z = i$ (hint: for this example we use tricks with geometric series, we don’t need the general recurrence relation). Since $f(z)$ has a pole at $z = 0$, $h(z)$ must have a zero of order $m$ at $z = 0$ Given this, $h(z)$ can be written as $z^m k(z)$ where $k(z)$ is analytic and non-zero at $z = 0$. Therefore, $f(z)$ can be written as: $$ f(z) = \frac{g(z)}{z^m k(z)} $$ Since $k(z)$ is analytic and non-zero at $z = 0$, there exists a power series $\sum_{n=0}^{\infty} b_n z^n$ such that $\frac{1}{k(z)} = \sum_{n=0}^{\infty} b_n z^n$ . This leads to: $$ f(z) = \frac{g(z)}{z^m} \sum_{n=0}^{\infty} b_n z^n = \sum_{n=0}^{\infty} b_n g(z) z^{n-m} $$ for $z \ne 0$. As $g(z)$ is analytic at $z = 0$, it can also be expanded into a power series around $z = 0$. Thus, the product of these two series results in a series of the form: $$ f(z) = \sum_{n=-m}^{\infty} a_n z^n $$ This series converges in some punctured disc $\{ z \,|\, |z| < R, z \ne 0 \}$ because it is the product of two convergent power series in that region. For the specific function $f(z) = \frac{z}{1 + z^2}$, we want to expand around $z = i$ $$ f(z) = \frac{z}{(z - i)(z + i)} $$ To expand around $z = i$, let's set $w = z - i$, so $z = w + i$. Then: $$ f(w + i) = \frac{w + i}{ w (w + 2i)} $$ Now, observe that: $$ \frac{1}{w + 2i} = \frac{1}{2i} \cdot \frac{1}{1 + \frac{w}{2i}} $$ Since $|\frac{w}{2i}| < 1$ for $|w| < 2$, we can use the geometric series expansion: $$ \frac{1}{1 + \frac{w}{2i}} = \sum_{n=0}^{\infty} \left(-\frac{w}{2i}\right)^n $$ Thus, we get: $$ f(z) = \frac{w + i}{2i w } \cdot \sum_{n=0}^{\infty} \left(-\frac{w}{2i}\right)^n $$ which will be valid in a punctured disc around $z = i$, specifically for $|z - i| < 2$ (or $|w| < 2$).