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I have a question that feels rather simple, but I seem to be stumped!

Given that $f$ is entire, use a power series representation of $f$ about $0$ to solve the differential equation $f''(z)-3f'(z)+2f(z)=0$ when $a_0 = 1$ and $a_1 = 2$.

I got so far as the following:

$$\begin{align} \sum_{n=2}^{\infty} n(n+1)a_nz^{n-2}-3\sum_{n=1}^{\infty}na_nz^{n-1}+2\sum_{n=0}^{\infty}a_nz^n & = 0 \end{align}$$

So from this, we can get:

$$a_{n+2} = \frac{3(n+1)a_{n-1} + 2a_{n-2}}{(n+2)(n+3)} : n\ge 2$$

But I'm stuck here. How do I (or is it even intended that I) recover the exponentials that we know solve this differential equation?

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  • $\begingroup$ Hint: let $a_n = b_n/n!$. $\endgroup$ – Ron Gordon Feb 26 '14 at 1:02
  • $\begingroup$ I'm very much unfamiliar with this type of problem. Doing such a substitution seems like it will create another recursive function, correct? It doesn't seem (immediately) easy to solve, unless I'm making a glaring error and the problem is far simpler than I'm treating it. $\endgroup$ – druckermanly Feb 26 '14 at 1:06
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    $\begingroup$ I didn't have enough comments/points/etc. to upvote until recently. I would've earlier if I could've! $\endgroup$ – druckermanly Feb 28 '14 at 20:31
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You diff eq'n produces as a series solution

$$\sum_{n=0}^{\infty} (n+2) (n+1) a_{n+2} z^n - 3 \sum_{n=0}^{\infty} (n+1) a_{n+1} z^n + 2 \sum_{n=0}^{\infty} a_n z^n = 0$$

where the coefficient satisfies

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

Let $a_n = b_n/n!$. Then

$$b_{n+2} - 3 b_{n+1} + 2 b_n = 0$$

which means that

$$b_n = A 2^n + B \implies a_n = A\frac{2^n}{n!} + B \frac1{n!}$$

$$a_0 = 1 \implies A+B=1$$ $$a_1=2 \implies 2 A+B=2$$

which means that $A=1$ and $B=0$. Therefore $a_n = 2^n/n!$ and

$$f(z) = \sum_{n=0}^{\infty} \frac{(2 z)^n}{n!} = e^{2 z}$$

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  • $\begingroup$ Why did I get $(n+2)(n+3)$ where you got $(n+2)(n+1)$? I feel like an algebra mistake is only magnifying my confusion. EDIT: Fixed that. I had the derivative of $z^{n-1}$ was $(n+1)z^{n-2}$ $\endgroup$ – druckermanly Feb 26 '14 at 1:30
  • $\begingroup$ @user2899162: for the second derivative, you add 2 to the index in order that the first term is the constant term. Thus you go from $n (n-1)$ to $(n+2)(n+1)$. $\endgroup$ – Ron Gordon Feb 26 '14 at 1:33
  • $\begingroup$ Yeah, thank you for that. Now I just don't see the logic behind your line that transitions from $b_{n+2} - 3 b_{n+1} + 2 b_n = 0$ to $b_n = A 2^n + B$. Thank you so much for your assistance, for some reason these things just aren't "clicking" today. $\endgroup$ – druckermanly Feb 26 '14 at 1:37
  • $\begingroup$ Sub $b_n = r^n$ so that $r^2-3 r+2=0$ which means that $r=2$ or $r=1$. $\endgroup$ – Ron Gordon Feb 26 '14 at 1:43
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    $\begingroup$ @user2899162: yes. At some point, you just need to know what to do. $\endgroup$ – Ron Gordon Feb 26 '14 at 1:47

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