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Topic:
Bessel Function Asymptotic Expansions
Replies:
3
Last Post:
Feb 16, 2011 12:34 AM




Re: Bessel Function Asymptotic Expansions
Posted:
Feb 16, 2011 12:34 AM


On Feb 15, 12:29 pm, Paul Skoczylas <pa...@cfertech.com> wrote: > On Monday, February 14, 2011 10:09:27 PM UTC7, Robert Israel wrote: > > >The stuff at the front of the > > > equation is the same, but the stuff in the parentheses (1 + ...) > > > is different > > > It is the same, according to Maple. What do you think it should be? > > > > map(normal,asympt( > > (11/(8*r*q)+9/128/r^2/q^275/1024/r^3/q^3+3675/32768/r^4/q^4 > > 59535/262144/r^5/q^5)/ > > (11/(8*a*q)+9/128/a^2/q^275/1024/a^3/q^3+3675/32768/a^4/q^4 > > 59535/262144/a^5/q^5), q)) assuming r>0, a>0; > > > 11/8*(ar)/r/a/q+1/128*(9*a^27*r^22*r*a)/r^2/a^2/q^2 > > 1/1024*(7*r^2*a59*r^3+75*a^39*r*a^2)/a^3/r^3/q^3 > > +1/32768*(236*r^3*a3013*r^4+3675*a^4126*r^2*a^2300*r*a^3)/a^4/r^4/q^4 > > 1/262144*(3013*r^4*a50735*r^5+59535*a^51062*r^3*a^21050*r^2*a^3 > > 3675*r*a^4)/r^5/a^5/q^5+O(1/(q^6)) > > If I'm reading that right, Maple is saying it's asymptotically the samebut that's not the same as "equal". That means that both Maple and the author of the text I have are doing something to the ratio of the two series beyond simple algebra. > > Just as an example, if we look only at the terms in parentheses, that part of Eq2 is: > 1 + (ra)/(8*a*r*q) + (9*a^22*a*r7*r^2)/(128*a^2*r^2*q^2) > > And the ratio of the Eq3 3 with q*r and q*a as inputs is: > (1  1/8/(q*a) + 9/128/(q*a)^2)/(1  1/8/(q*r) + 9/128/(q*r)^2) > > With r=0.1, a=0.05, and q=10, the first evaluates to 0.9453 and the second evaluates to 1.0909. As q increases, both asymptotically approach the same result (1.0000), as you said.
Well yes, it's algebra of series, not algebra of polynomials.
(1  1/8/(q*a) + 9/128/(q*a)^2 + ...)/(1  1/8/(q*r) + 9/128/(q*r)^2 + ...) = (1  1/8/(q*a) + 9/128/(q*a)^2 + ...) * (1 + (1/8/(q*r)  9/128/ (q*r)^2 + ...) + (1/8/(q*r)+...)^2 + ...) = (1  1/8/(q*a) + 9/128/(q*a)^2 + ...) * (1 + 1/8/(q*r)  7/128/ (q*r)^2 + ...) = 1 + 1/8*(1/r  1/a)/q + (9/128/a^2  1/64/a/r  7/128/r^2)/q^2 + ...
> I'm curious about what is being done to get the simplification in Eq 3. > > However, that being said, your larger series from Maple give sme exactly what I needI don't really *need* to know how it was obtained, and I certainly don't need more terms than that. (Although I'm still curious about the "how".) I am very grateful to you for providing me with that. > > Thank you very much, > > Paul



