Find the power series in x-x0?

[Math10]

Homework Statement

Find the power series in x-x₀ for the general solution of y"-y=0; x₀=3.

Homework Equations

None.

The Attempt at a Solution

Let me post my whole work:

 

[Math10]

What I did was the substitution method using z=x-x₀.
The answer for this problem is

 

[Ray Vickson]

Homework Statement

Find the power series in x-x₀ for the general solution of y"-y=0; x₀=3.

Homework Equations

None.

The Attempt at a Solution

Let me post my whole work:

No, please don't. The PF standard is for you to type out the problem and the solution, reserving images for things like diagrams or geometric constructions, etc.

I know that some helpers answer questions about handwritten solution images, but most will not bother. You should consult the pinned post "guidelines for Students and Helpers", by Vela, for a good explanation about this and similar issues.

 

[Math10]

Can you please take a look at the work that I posted? It's clearly written.

 

[PeroK]

Can you please take a look at the work that I posted? It's clearly written.

It looks like the right answer in post #2, if that's all you are asking.

 

[Math10]

I know that's the right answer, but what should I do to get to the right answer after the last step in my work? That's where I got stucked.

 

[PeroK]

I know that's the right answer, but what should I do to get to the right answer after the last step in my work? That's where I got stucked.

You need to spot a pattern in the coefficients (and verify it by induction if need be). You can see from the answer that you need to separate $n$ even and odd.

 

[Math10]

You mean this:

n=2m (even index)
a_2m+2=a_2m/[(2m+2)(2m+1)]
----------------------------------------------------------------------------
n=2m+1 (odd index)
a_2m+3=a_2m+1/[(2m+3)(2m+2)]

 

[PeroK]

You mean this:

n=2m (even index)
a_2m+2=a_2m/[(2m+2)(2m+1)]
----------------------------------------------------------------------------
n=2m+1 (odd index)
a_2m+3=a_2m+1/[(2m+3)(2m+2)]

Yes, but can you see the pattern? The answer gives you a big clue!

 

[Math10]

So how do I get to the answer? I know where x-3 comes from.

 

[PeroK]

So how do I get to the answer? I know where x-3 comes from.

You get to the answer by noticing that $1 \times 2 \times 3 \times \dots \times n = n!$

The clue was that the answer has $n!$ in it.

 

[Math10]

I still don't really get it.

 

[PeroK]

I still don't really get it.

You have:

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

Hence:

$a_{n+2} = \frac{a_n}{(n+2)(n+1)}$

For $n$ even this gives:

$a_2 = \frac{a_0}{2}, \ a_4 = \frac{a_2}{12} = \frac{a_0}{24}, \ a_6 = \frac{a_4}{30} = \frac{a_0}{720} \dots$

And, now by insight, inspiration (or looking at the answer) you have to notice that $2, 24, 720 \dots$ are the even factorials and hence $a_n = \frac{a_0}{n!}$

Odd $n$ is much the same.

 

[Math10]

I got it now. Thank you so much.

 

[berkeman]

You have:

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

Hence:

$a_{n+2} = \frac{a_n}{(n+2)(n+1)}$

For $n$ even this gives:

$a_2 = \frac{a_0}{2}, \ a_4 = \frac{a_2}{12} = \frac{a_0}{24}, \ a_6 = \frac{a_4}{30} = \frac{a_0}{720} \dots$

And, now by insight, inspiration (or looking at the answer) you have to notice that $2, 24, 720 \dots$ are the even factorials and hence $a_n = \frac{a_0}{n!}$

Odd $n$ is much the same.

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