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
\[ ab = 1 \]
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:
- \(b_0 = 1\) (since the constant term of \(ab\) 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:
\[ b_n = -\sum_{k=1}^{n} a_k b_{n-k} \quad \text{for } n \geq 1 \]
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 \(k \le n\), 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 \(|w| < 2\)).