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

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 Robert Israel Posts: 11,902 Registered: 12/6/04
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 UTC-7, 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(
> >     (1-1/(8*r*q)+9/128/r^2/q^2-75/1024/r^3/q^3+3675/32768/r^4/q^4
> >    -59535/262144/r^5/q^5)/
> >     (1-1/(8*a*q)+9/128/a^2/q^2-75/1024/a^3/q^3+3675/32768/a^4/q^4
> >    -59535/262144/a^5/q^5), q)) assuming r>0, a>0;

>
> > 1-1/8*(a-r)/r/a/q+1/128*(9*a^2-7*r^2-2*r*a)/r^2/a^2/q^2
> > -1/1024*(-7*r^2*a-59*r^3+75*a^3-9*r*a^2)/a^3/r^3/q^3
> > +1/32768*(-236*r^3*a-3013*r^4+3675*a^4-126*r^2*a^2-300*r*a^3)/a^4/r^4/q^4
> > -1/262144*(-3013*r^4*a-50735*r^5+59535*a^5-1062*r^3*a^2-1050*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 same--but 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 + (r-a)/(8*a*r*q) + (9*a^2-2*a*r-7*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 need--I 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

Date Subject Author
2/14/11 Paul Skoczylas
2/15/11 Robert Israel
2/15/11 Paul Skoczylas
2/16/11 Robert Israel