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So also have this differential equation: $$A''(z) + 4 A(z) = 0$$

With $A(z)$ stand for this classic formal power series $$A(z) = a_0 + a_1 z + ....$$

I need to show that the differential equation have a unique solution. How should i do this question?

Edit: i forgot to add that $a_0 = 0$ and $a_1 = 1$. My bad

Doh
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2 Answers2

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Hints: within the convergence interval (or ball)

$$A(z)=\sum_{n=0}^\infty a_nz^n\implies \begin{cases}A'(z)=&\sum_{n=1}^\infty na_nz^{n-1}\\{}\\A''(z)=&\sum_{n=2}^\infty n(n-1)a_nz^{n-2}\end{cases}\;\;\implies$$

$$0=A''(z)+4A(z)=4a_0+2a_2+\sum_{n=2}^\infty \left(4a_n+(n+1)(n+2)a_{n+2}\right)z^n$$

Well now, so...

DonAntonio
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The usual definition for the "derivative" of a formal power series is $$ \left(\sum_{k=0}^\infty a_k z^k\right)' = \sum_{k=1}^\infty ka_kz^{k-1} = \sum_{k=0}^\infty a'_k z^k \quad\text{where}\quad a'_k = (k+1)a_{k+1} \text{,} $$ i.e. one simply assumes that the series may be differentiated term-wise.

Now you just need to put that into the differential equation, and show that the resulting equation determines the $a_k$ uniquely. For that, you can use that for formal power series $$ \sum_{k=0}^\infty a_k x^k = \sum_{k=0}^\infty b_k x^k \quad\Leftrightarrow\quad a_k = b_k \text{ for all $k$.} $$ In other words, you reason that for the differential equation to hold, the formal power series on the left-hand side of the ODE must have all coefficients equal to zero.

fgp
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  • Sorry i forgot to add the first two terms of the of power series. I edited the question. The power series should have $a_0 = 0$ and $a_1 = 1$ – Doh Apr 30 '14 at 13:01
  • @Doh I was about to edit to say that there is no unique solution in your case, but with these constraints there is... – fgp Apr 30 '14 at 13:03