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Re: Making a function out of repeated hyperbola integrations?
*To*: mathgroup at smc.vnet.net
*Subject*: [mg122077] Re: Making a function out of repeated hyperbola integrations?
*From*: Patrick Scheibe <pscheibe at trm.uni-leipzig.de>
*Date*: Wed, 12 Oct 2011 03:43:45 -0400 (EDT)
*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com
*References*: <201110110824.EAA00824@smc.vnet.net>
Hi,
with the starting value n-a everything you have to do is to
iterate the the following two steps again and again
1. replacing n->n/x
2. integrating
and that is as simple as it sounds:
NestList[Function[
Integrate[# /. n -> n/x, {x, a, n/a},
GenerateConditions -> False]],
n - a, 4]
If you want the first starting element dropped, use Rest on the resulting
list.
Cheers
Patrick
On Oct 11, 2011, at 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?
>
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