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Legendre's equation with Frobenius method

  1. May 2, 2006 #1
    I know you don't need Frobies method per se, but he wanted us to practice, well I got the right recursion formula, but I realize I got to it wrong...After I plugged in the assumed series solution and all its derivatives and stuff and got that big long equation, it had two terms out in front and then a third term that was the infinite series(err, the indicial equation?) I took the first two terms and set them equal to 0 seperately(which you can't do) and got r=0 from that, do that for the series term and the recursion formula comes out right. So what was I SPOSED to do at that point? Take the whole thing and solve for r or something?
  2. jcsd
  3. May 3, 2006 #2


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    What??? Please post the actual problem and what you did so we can make some sense out of this!
  4. May 3, 2006 #3
    Heh, ok, sorry. I don't have my work anymore or the actual problem but it's Legendre's equation, so like (1-x^2)y''-2xy'+uy=0 where u is a parameter. He wanted us to do it with the method of Frobenius, which is where you assume a solution of the form y= sum(n->0,infinity) Anx^(n+r) An is a constant depending on n, r is ANY real number(not just integers)now I know the problem can be done with just a power series so there's no need for this method, but this is what he wanted us to do.

    So, if I differentiate it twice and once and plug it back into the equation itself, I ended up ultimately with r=0 as being necessary for the whole equation to equal 0, is that what I'm sposed to expect to happen? Everything ended up as the right answer but in class he mumbled about how you were sposed to get r=0 AND r=1 or something, and I wasn't sure about that, and I fear I did an illegal operation in getting my r=0(I don't have my work though)
  5. Nov 12, 2009 #4
    well if you take the indicial equation should be like (r-1)r=0 for the case of legendre, that mean that the roots are 1 and 0 like you professor said and not only r=0, because is not a double root problem.
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