Question

Power series question

http://f.imgtmp.com/BLhyz.jpg
(Zoom in the quality is good).

Answer 1

Why is only x = -1 being considered and not x = 0, x = 1 and x = -1.

Answer 2

Also, I am looking at my book and it says: lambda^2 + (p_0 - 1)*lambda + q_0 = 0 is indicial equation formula but I don't seem to get that here.

Answer 3

First off, the singular points occur when the function multiplying the second derivative vanishes, and that's only at the point x = -1

Answer 4

I don't think it's accurate to refer to the entire coefficient of y'' like that. Only the whatever can make a 0 on the denominator but I got your point. Ok so I get that now but what is the significance of the 0, 1 and -1 in the question then?

Answer 5

Oh, it's just different values of alpha.

Answer 6

So I just do the problem 3 times?

Answer 7

Assuming so, I'm still confused with the indicial equation formula in this question differing from that of my book's.

Answer 8

But thats what is its, is it not?
\[ a(x) y'' + b(x) y' + c(x) y = 0\]
\[y'' + \frac{b(x)}{a(x)} y' + \frac{c(x)}{a(x)} y = y'' + p(x)y' + q(x)y = 0\]

Answer 9

That's what what is?

Answer 10

What you just typed in beautiful math?

Answer 11

you want the spots where p(x) is singular, which is exactly those points where a(x) is zero, unless you have some gross b(x) or c(x) like a tangent.

Answer 12

I meant if there was already stuff in the denominator. I probably won't see that in a test though (other than a constant).

Answer 13

But yeah, let's move on to the indicial equation itself please.

Answer 14

p_0 = alpha/2, q_0 = 0

my book's formula is: lambda^2 + (p_0 - 1)*lambda + q_0 = 0 = lambda^2 + (alpha/2 - 1)*lambda = 0

Answer 15

Imagine all my lambdas are r values.

Answer 16

and the alphas are a values.

Answer 17

Alright bear with me it's been awhile since my power series days :)

Answer 18

Oh pellet!

Answer 19

Lol it changed the swearing.

Answer 20

I got r(r - 1 + a/2)

so now what's it saying in the next part in the solution, I don't get it.

Answer 21

So you need to find where that expression equals zero. There are two possible values for r, either r = 0 or r = 1- a/2. You will have problems if those two roots differ by an integer value.

Answer 22

Which they would if a were some even integer

Answer 23

Which is the case, isn't it? And what do you mean by problems?

Answer 24

Not problems per se, but if your two roots are the same or differ by an integer value then the two linearly independent solutions are pretty gross to calculate, whereas if they differ by a non-integer value they're much more straightforward. A priori a could be anything, so you must account for all of the possible cases. If a is zero or an even integer not equal to -2, you have distinct roots that differ by an integer value and you must proceed accordingly. If a is equal to -2 you have repeated roots, and if a is not an even integer, then you have distinct roots that do not differ by an integer value.

Answer 25

The forms of the linearly independent solutions are shown at the bottom, the things with the natural logs and all that. They are very tedious to work through if your roots do not differ by some non-integer value.

Answer 26

Where did the y_1 and y_2 come from exactly? Are they some formulas I am supposed to have memorized?

Answer 27

When I learned them yes, we just memorized them. You can arrive at them via a few different ways but the process is long and (also) tedious so it's better just to memorize them. You can look up the derivations somewhere, I'm sure.

Answer 28

2 secs.

Answer 29

Yeah, I don't want to be caught up with tediousness on a test so what is it exactly that I have to memorize though? What's the formula and what was plugged in here? Maybe I'm just tired from working too hard but, I can't see it.

Answer 30

Yeah power series solutions take their toll quickly. The first solution is the same in all of them, and that's for your first root. If your second root differs by a non-integer value, then your second solution takes the form given by that y2. The same goes for if you have repeated roots, or if you have roots that differ by some integer value. There are three distinct cases listed down there. The tediousness comes in when you expand all of those solutions and generate recursion relations from them.

Answer 31

Wait, no I lied... the tediousness comes in when you solve inhomogeneous equations.... haha

Answer 32

This question doesn't ask anything tedious, right? I remember doing tedious stuff for these but the question seemed differently phrased. Sorry if I am missing something obvious again, I'm tired.

Answer 33

Also, I don't think I saw a nonhomogeneous power series question before.

Answer 34

So, I shouldn't be responsible for them, should I?

Answer 35

No no, this one seems nice because you didn't have to actually determine the coefficients, only the form of the solutions.

Answer 36

K, good, my teacher will be merciful on the exam then and we only had the tedious ones for homework. But again, what do I memorize? (I don't mean to sound rude if this comes out that way).

Answer 37

I'm asking because it gives me the result from plugging in whatever into the formulas but I want to know the formulas before they have anything plugged in. Is there a wikipedia page with this information or something?

Answer 38

http://en.wikipedia.org/wiki/Frobenius_method#Double_roots