Re: Making a function out of repeated hyperbola integrations?
- To: mathgroup at smc.vnet.net
- Subject: [mg122078] Re: Making a function out of repeated hyperbola integrations?
- From: Jacopo Bertolotti <jacopo.bertolotti at gmail.com>
- Date: Wed, 12 Oct 2011 03:43:56 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201110110824.EAA00824@smc.vnet.net>
Dear Nathan, Try out something like: Nest[Integrate[(# /. {n -> n/x}), {x, a, n/a}, Assumptions -> {a > 0, n > a^2}] &, n - a, 2] Notice that I extrapolated the assumptions from your mail. If they are not correct you might have to change them (but they give the result you cited). Also performing this operation for a large muber of iterations (in the above code is just 2 but any number will do) can be very slow. Maybe you might want to run it once as a NestList and save the output for later use. Jacopo On 10/11/2011 10:24 AM, Nathan McKenzie wrote: > I'm working with the following repeated integrals. Is there any way > to automate what I'm doing here? > > I start with this: > > Integrate[ n/x - a, {x, a, n/a}] > > The result of that (after a bit of text editing) is a^2 - n Log[a] + n > (-1 + Log[n/a]). That's the first result I want to work with. Then, I > currently manually edit that result by swapping out n with (n/x), and > have > > Integrate[ a^2 - (n/x) Log[a] + (n/x) (-1 + Log[(n/x)/a]), {x, a, n/ > a}] > > And that resolves to -a^3 + n Log[a] + n Log[a]^2 + 1/2 n Log[n/a^2]^2 > + n (a - (1 + Log[a]) Log[n/a]). Which is the second result I want > to work with. The, I manually swap out n with (n/x) and integrate > again on {x, a, n/a}, and repeat this process ad nauseum. It doesn't > take long before the number of n's for me to edit becomes really > unwieldy and error prone... and ideally I would like to do this many > times in a row (say, up 30 or 40) and have the results around for > random use in other contexts. > > What I would really like to be able to do is just have some sort of > function where I can type F[n,a,s], where s is the number of times > integration is performed, and then n and a (which will be actual > numbers) will get evaluated. I feel like the step where I swap out n > with (n/x) points at something problematic, though. Is there any way > in Mathematica for me to construct such a function and get my results > automatically? > >
- References:
- Making a function out of repeated hyperbola integrations?
- From: Nathan McKenzie <kenzidelx@gmail.com>
- Making a function out of repeated hyperbola integrations?