Using power series expansions,

find a function $f$ which is holomorphic on the unit disk $D:=$ {$z\in\mathbb C:|z|<1$} and solves the differential equation


for $z\in D$ along with the initial conditions

$f(0)=0$, and $f'(0)=1$

I have been given the hint "for any $c_0,c_1 \in\mathbb C$, we have

$$\sum_{j=0}^{\infty} (c_0z^{2j} +c_1z^{2j+1}) =\frac{c_0+c_1z}{1-z^2} $$ as long as $|z|<1$, why?"

But i still don't understand how to answer the question or what the hint means!

Any help would be much appreciated!

Many thanks


To check your hint, just develop $\frac{c_0+c_1 z}{1-z^2}$ in power series. Now to see why it helps, you have to find power series expansion of $f(z)$.

To achieve this, write $f(z)=\sum_{n=0}^{\infty} a_n z^n$, then replace $f$ with this series in the differential equation. Develop, and manage to get a recurrence equation between, $a_{n}$, $a_{n-1}$ and $a_{n-2}$, and also conditions on $a_0$ and $a_1$, using your knowledge that $f(0)=0$ and $f'(0)=1$.

It's the usual way to solve differential equations with power series. If you have difficulties with this, I'll develop a bit more.

You will have to play with indices, like $$f'(z)=\sum_{n=1}^\infty n a_n z^{n-1}=\sum_{n=0}^\infty (n+1) a_{n+1} z^n$$ $$f''(z)=\sum_{n=2}^\infty n (n-1) a_n z^{n-2}=\sum_{n=0}^\infty (n+1) (n+2) a_{n+2} z^{n}$$

And you will also have to use the fact that power series expansion is unique (that is, if you have $\sum_{n=0}^\infty [expression \ in \ a_n, a_{n-1}, a_{n-2}] z^n = 0$, then the expression is null for all $n$, thus you have a recurrence equation to solve.

Here is a more detailed approach.

First we just replace $f$, $f'$ and $f''$ with their expressions above, in the differential equation: $$(1-z^2)f''(z)-4zf'(z)-2f(z)=f''(z)-z^2f''(z)-4zf'(z)-2f(z)$$ $$=\left(\sum_{n=0}^\infty (n+1) (n+2) a_{n+2} z^{n}\right) - \left(\sum_{n=2}^\infty n (n-1) a_n z^{n}\right) - 4 \left( \sum_{n=0}^\infty n a_n z^{n} \right) - 2 \left(\sum_{n=0}^\infty a_{n} z^n \right)$$

$$=\left(\sum_{n=0}^\infty (n+1) (n+2) a_{n+2} z^{n}\right) - \left(\sum_{n=0}^\infty n (n-1) a_n z^{n}\right) - 4 \left( \sum_{n=0}^\infty n a_n z^{n} \right) - 2 \left(\sum_{n=0}^\infty a_{n} z^n \right)$$

In the preceding last step, I only replace $n=2$ with $n=0$ in the bound of one of the sums, which does not change it, since coefficient $n(n-1)a_n$ guarantees we add $0$. But it's necessary, in order to write the following recurrence equation for all $n \geq 0$.

Since our differential equation is supposed to hold, this whole series is everywhere null. Hence its coefficients in $z^n$ are all zero.

Now we have to put together terms in $z^n$. This yields $$\left( (n+1) (n+2) a_{n+2} - n (n-1) a_n -4n a_n -2 a_n \right) z^n$$

So, we have the recurrence equation $$ (n+1) (n+2) a_{n+2} - n (n-1) a_n -4n a_n -2 a_n = 0$$

Or, $$(n+1) (n+2) a_{n+2} - (n^3 + 3n +2) a_n = 0$$ $$(n+1) (n+2) a_{n+2} - (n+1)(n+2) a_n = 0$$

Hence $a_{n+2}=a_{n}$ for $n \geq 0$. Now we have to find $a_0$ and $a_1$ to know the whole series. But since $f(0)=0$ we have $a_0=0$, and $f'(0)=1$ gives $a_1=1$.

We have therefore $a_{2n}=0$ and $a_{2n+1}=1$, which means

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

As a final step, you always have to check that your series really exists, e.g. that its radius of convergence is not null. But here it's immediate that this radius is $1$.

Of course, you may want to check directly that $\frac{z}{1-z^2}$ is the solution of your equation, but it's less necessary (except to make sure there is no mistake in the above proof).

  • $\begingroup$ hey, i still dont understand the question or how to answer it! i dont get how to put that into a power series either, sorry i have only just started learning this subject and not too sure how to do much more than find the radius of convergence for a complex power series! could you please explain a bit more? many thanks :) $\endgroup$
    – user67411
    Mar 27 '13 at 12:20
  • $\begingroup$ ok, I'll show you then :-) $\endgroup$ Mar 27 '13 at 12:24
  • $\begingroup$ thank you so much for your help! That is almost making sense now.. i will read through it a few times. :) thank you so much! $\endgroup$
    – user67411
    Mar 27 '13 at 12:36

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