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monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

Show there are two entire complex functions f1 and f2 that satisfy
$\frac {d^2 f }{dz^2}=zf(z)$ where
f1(0)=1
f1'(0)=0
f2(0)=0
f2'(0)=1

Assume the solution has the following form
$\displaystyle\sum_{n=0}^{\infty} {a_n z^n }$

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?

Shadow · **Posted:** Sun, 19 Feb 2012 22:42:49 UTC

monomoco wrote:

Show there are two entire complex functions f1 and f2 that satisfy
$\frac {d^2 f }{dz^2}=zf(z)$ where
f1(0)=1
f1'(0)=0
f2(0)=0
f2'(0)=1

Assume the solution has the following form
$\displaystyle\sum_{n=0}^{\infty} {a_n z^n }$

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?

Where are

and

? All I see is one

.

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

That's what I'm trying to find, the two functions f1 and f2.

Shadow · **Posted:** Sun, 19 Feb 2012 23:08:59 UTC

monomoco wrote:

That's what I'm trying to find, the two functions f1 and f2.

OK, then here they are:

f_1(z)=1

and

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

I don't think that's what I was supposed to find... My prompt tells me I'm supposed to find a recurrence relation for the coefficients a_n and then show the resulting series have an infinite radius of convergence using the ratio test.

Shadow · **Posted:** Sun, 19 Feb 2012 23:42:35 UTC

monomoco wrote:

I don't think that's what I was supposed to find... My prompt tells me I'm supposed to find a recurrence relation for the coefficients a_n and then show the resulting series have an infinite radius of convergence using the ratio test.

Not sure what to tell you, from what you posted the f_1,f_2

I have provided are just fine. Again, I think you should reread what you originally posted and check to make sure you don't have any typos.

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

There are no typos. How does f(z)=1 satisfy f''=zf(z)? Or f(z)=z?

Shadow · **Posted:** Sun, 19 Feb 2012 23:52:04 UTC

monomoco wrote:

There are no typos. How does f(z)=1 satisfy f''=zf(z)? Or f(z)=z?

It doesn't, but you didn't say f(z)=f_1(z)

, they're different function names. If you intend them to be the same function, you should write it as f_1(z)

and not

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

Sorry it was not clear. I am looking for f1(z) and f2(z)that satisfy the DE and initial conditions and have that form of the series.

Shadow · **Posted:** Mon, 20 Feb 2012 00:04:33 UTC

monomoco wrote:

Sorry it was not clear. I am looking for f1(z) and f2(z)that satisfy the DE and initial conditions and have that form of the series.

Both of them need to satisfy the same differential equation or one of them goes on the left side and the other on the right side?

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

They both need to satisfy the same DE

Shadow · **Posted:** Mon, 20 Feb 2012 00:18:27 UTC

monomoco wrote:

They both need to satisfy the same DE

OK, now that makes a lot more sense.

So you have f_i''(z)=zf_i(z)

Expanding

, we get:

$$\sum_{n=1}^\infty {f_i^{(n-1)}(0)\over (n-1)!}z^n=\sum_{n=2}^\infty {f_i^{(n)}(0)\over (n-2)!}z^n$

Can you see where you go from here?

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

I don't understand how you expanded that. I began listing the terms in the series

f''= 2a_2 + z*6a_3 + z^2*12a_4 + ... and set it to
zf(z)=z*a_0 + z^2*a_1 + z^3*a_2+...

and started looking at the coefficients of z^i

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

etc.

Shadow · **Posted:** Mon, 20 Feb 2012 00:32:26 UTC

monomoco wrote:

I don't understand how you expanded that. I began listing the terms in the series

f''= 2a_2 + z*6a_3 + z^2*12a_4 + ... and set it to
zf(z)=z*a_0 + z^2*a_1 + z^3*a_2+...

and started looking at the coefficients of z^i

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

etc.

I did the same thing as you did, I just did it all at once using sigma-notation rather than write individual terms. I mean, one already knows what the a_n

are by Taylor's theorem, so may as well use it.

monomoco · **Joined:** Thu, 26 Jan 2012 05:30:38 UTC **Posts:** 25

OK, so this is where I was stuck when I first posted. I don't know where to go from here.

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