Entire Complex Functions

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monomocoso

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Jan 25, 2012
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Show there are two entire complex functions f1 and f2 that satisfy
\(\displaystyle \frac {d^2 f }{dz^2}=zf(z)\) where

f1(0)=1
f1'(0)=0
f2(0)=0
f2'(0)=1

Assume the solution has the following form
\(\displaystyle \displaystyle\sum_{n=0}^{\infty} {a_n z^n }\)
 
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I think I would start with the first two derivatives of the summation expression. What was your plan?
 
I'm confused by the question. Is the solution given for f1 or f2?

The soln is a0 +a1z +a2z^2 + ...
If we use initial conditions f1(0)=1 and f1'(0)=0, we can see that a0=1 and a1=0


If we use initial conditions f2(0)=0 and f2'(0)=1, we can see that a0=1 and a1=1
 
Are you going to find the second derivative or aren't you?

The first derivative might look like this, inside the summation \(\displaystyle n\cdot a_{n}\cdot z^{n-1}\)
 
Now what does that equal according to your differential equation? What do you know about the coefficients of a power series?
 
The second derivative = zf(z) so we get

a_0=2*3a_3
a_1=3*4a_4
a_2=4*5a_5 ...
a_n = a_(n-3) / (n-1)n

How do I get a_2?
 
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Hmmm... What are you doing?

\(\displaystyle f(z) = \sum_{n=0}^{\infty} a_{n}\cdot z^{n} = a_{0} + a_{1}\cdot z + a_{2}\cdot z^{2} + ...\)


\(\displaystyle \frac{df(z)}{dz} = \sum_{n=1}^{\infty} n\cdot a_{n}\cdot z^{n-1} = a_{1} + 2a_{2}\cdot z + 3a_{3}\cdot z^{2} + ...\)


\(\displaystyle \frac{d^{2}f(z)}{dz^{2}} = \sum_{n=2}^{\infty} n\cdot (n-1)\cdot a_{n}\cdot z^{n-2} = 2a_{2} + 6a_{3}\cdot z + 12a_{4}\cdot z^{2} + ...\)


\(\displaystyle z\cdot f(z) = \sum_{n=0}^{\infty} a_{n}\cdot z^{n+1} = a_{0}\cdot z + a_{1}\cdot z^{2} + a_{2}\cdot z^{3} + ...\)

Make sure you follow each piece.

Now what?
 
I got that far, then I set f' = zf(z) and started looking at the coefficients of z^i

For z : 6a_3 = a_0
For z^2: 12a_4 = a_1

etc.
 
monomocoso said:
I got that far, then I set f' = zf(z) and started looking at the coefficients of z^i

For z : 6a_3 = a_0
For z^2: 12a_4 = a_1

etc.
Click to expand...

You still have first and second derivatives confused, at least in the notation.

Let's first fix the index.

\(\displaystyle \frac{d^{2}f(z)}{dz^{2}} = \sum_{n=0}^{\infty} (n+2)\cdot (n+1)\cdot a_{n+2}\cdot z^{n} = 2a_{2} + 6a_{3}\cdot z + 12a_{4}\cdot z^{2} + ...\)

Don't forget \(\displaystyle 2a_{2} = 0\)

And the general formula: \(\displaystyle (n+1)(n+2)a_{n+2} = a_{n}\)
 
I still don't see where you got that relation. Can you be more explicit?
 
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I calculated the second derivative and zf(z) and they were the same as his. So I set them equal and compared coefficients. This gave

a_0=2*3a_3
a_1=3*4a_4
a_2=4*5a_5 ...

WHich suggests the relation
a_n /(n+2)(n+3) = a_(n+3)

Which is not what he got, and I don't know where my mistake is
 
tkhunny had a typo. It should have been:


\(\displaystyle (n+2)(n+3)a_{n+3}=a_n\)


Now, since \(\displaystyle 2a_2=0, a_2=0\). Then, \(\displaystyle a_{3k+2}=0\) where \(\displaystyle k\in \mathbb{Z}^+\) (nonnegative integers).

Next, plug in your initial conditions:

\(\displaystyle \displaystyle f_1(z)=\sum_{n=0}^\infty{a_n z^n} = a_0+\sum_{n=1}^\infty{a_n z^n}\).
Therefore, \(\displaystyle f_1(0) = 1 = a_0\)

Then, \(\displaystyle \displaystyle f_1'(z)=\sum_{n=0}^\infty{na_n z^{n-1}} = \sum_{n=0}^\infty{(n+1)a_{n+1}z^n}=a_1 + \sum_{n=1}^\infty{a_{n+1}z^n}\)
So, \(\displaystyle f_1'(0) = 0 = a_1\)

Now, you can use the recurrence relation to have all entries:

\(\displaystyle \displaystyle f_1(z) = \sum_{n=0}^\infty{\left(\prod_{i=1}^n{\frac{1}{3i(3i-1)}}\right)z^{3n}}\)
 
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