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A few problems with the operator method .

  1. Dec 1, 2004 #1
    A few problems with the operator method.....

    Hey there, I've just started doing the operator method etc in Maths and I was hoping someone could give me a hand here cos I've got a couple of qs (relating to the homework sheet).


    In question 1 part 3, with the Hermite equation, I really don't know what to do with it cos it just kinda gets stuck. I multiplied through by x^2 so I could have X^2 D^2 y so I could use the operator (sorry I can't write it up, but if O were the operator, this would look like O(O - 1)). So that part is fine but then that leaves us with x^2 ( 2n + 1 - x^2 )y, and when you've to bring that over to the other side of the eqt, I'm not sure you split it up so you have x^h G(O) y. The others are fine but I don't know about that one because all the powers of x aren't the same and I don't think you can have an x^2 in our G(0). If someone could explain to me where I'm going wrong or where to proceed from here it would be great.

    Second of all, and my last question for the moment cos I think the rest of the h/work is ok, do you think there's a specific reason why in question 3 when it asks for the recurrence relation it only gives us one of them? There are 2 values of alpa (0 and 1-b) and what they've given is only applicable for the 0 one. Am I missing the point of something here? Oh, and one other teeny thing, just about when it says find the series solution for around x=0. We don't actually have tso sub anything to do with zero in, do we? Does this just come from the way we initially write the Frobenius series? I was looking at the past papers and they seemed to take it this way but I just wanted to check....

    Thank you! :smile:
  2. jcsd
  3. Dec 2, 2004 #2
    Btw, I was at the library this afternoon for about an hour and I checked through all their differential equation books, and none of them had the Hermite eqt as this. They all had the same one and it wasn't this so do you think it's possible there has been a misprint here?
  4. Dec 3, 2004 #3

    matt grime

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    I would help but I'm a little puzzled as to what the [tex]\delta[/tex] symbol means here, would you mind elaborating? It doens't appear as though it can just be d/dx
  5. Dec 3, 2004 #4
    No problem. We have D = d/dx and that symbol that you stated is xD (just x times d/dx). Sorry, I should have stated it.
  6. Dec 3, 2004 #5

    matt grime

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    I would start from the expression involving [tex]\delta[/tex] and subs in xd/dx and see what happens. You may then work "backwards" to see who you should do it. There's no harm in doing this, as long as your final write up all goes in one direction, and all the steps are reversible.

    Have no answer for the rest, atm.
  7. Dec 3, 2004 #6
    I'd do that but I don't know what I'm supposed to get in the end. I mean the others were simple but I just dunno how you're supposed to work with this one. Hopefully there is an error with this one. I can always badger the poor lecturer about iton Monday after class. It's no wonder the poor man has turned to drink having me around!

    Thanks for your help, btw. :smile:
  8. Dec 3, 2004 #7
    y''-2xy'+2ny=0 is what seems to be given evereywhere else for the Hermiet eqt btw, which is fine to solve by this method....
  9. Dec 3, 2004 #8
    I'd do the last eqt I mentioned by multiplying through by x^2 so we'd have 0(0-1)y - 2x^20y + 2nyx^2 where 0=xD. Then rearranging it to get it into the form that we want, we'd have 0(0-1)y= x^2 (2(0) +2n)y. Sorry I can't make it any clearer. Someday I will learn how to type mathemtical formulas on here correctly (it'll be my project for the summer hols, I'm already looking forward to them! :p).
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