Manuel Kauers Hagenberg, October 14, 2003 RISC-Linz

Analytic Combinatorics

Homework 2

Exercise 3 a) Find a solution of the differential equation

(1 − z)^{3}f′(z) − 2z − 2(1 − z)^{2}f(z) = 0, f(0) = 0

by using your favorite Computer Algebra System.

b) Show that the solution of Mathematica,

f[z] → \frac{−2(−I · Pi + z + Log[−1 + z])}{(−1 + z)^{2}}

“simplifies to”

g(z) = −\frac{2z}{(1 − z)^{2}} − \frac{2}{(1 − z)^{2}} log(1 − z)

Ad a)

(1-z)^3*diff(f(z),z) - 2*z - 2*(1-z)^2*f(z) = 0;

(1 − z)^{3} (\frac{∂}{∂z} f(z)) − 2z − 2(1 − z)^{2}f(z) = 0

dsolve([%, f(0)=0], f(z));

f(z) = \frac{−2z − 2 ln(−1 + z) + 2Iπ}{(−1 + z)^{2}}

simplify(%, ln, assume=real);

f(z) = \frac{−2z − 2 ln(−1 + z) + 2Iπ}{(−1 + z)^{2}}

Ad b) First we change to readable notation and clean up a bit.

f(z) − g(z) = (\frac{2iπ}{(1 − z)^{2}} − \frac{2z}{(1 − z)^{2}} − \frac{2}{(1 − z)^{2}} log(−1 + z))

− (−\frac{2z}{(1 − z)^{2}} − \frac{2}{(1 − z)^{2}} log(1 − z))

= \frac{2iπ}{(1 − z)^{2}} + \frac{2}{(1 − z)^{2}} log(1 − z) − \frac{2}{(1 − z)^{2}} log(−1 + z)

We are done if we show that the last expression is identically zero. This follows, if we manage to prove the more general result

log(x) − log(−x) = iπ (x ∈ ℂ \ {0}).

We write x = |x|e^{iϕ} and use Euler’s formula e^{πi} = −1. It follows

log(x) − log(−x) = log(x) − log(−1 · x) = log(|x|e^{iϕ}) − log(e^{πi}|x|e^{iϕ})
= log |x| + iϕ − πi − log |x| − iϕ = −πi.

As exp is periodic in the imaginary direction with period 2π, we can freely replace log x by 2kπi + log x for k ∈ ℤ. This means we change the branch of log. Adding 2πi to the last equation gives the desired result.