Re: Making a function out of repeated hyperbola integrations?

*To*: mathgroup at smc.vnet.net*Subject*: [mg122071] Re: Making a function out of repeated hyperbola integrations?*From*: Nathan McKenzie <kenzidelx at gmail.com>*Date*: Wed, 12 Oct 2011 03:42:40 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <201110110824.EAA00824@smc.vnet.net>

Wow - you folks really are a helpful bunch. I've learned a ton about mathematica from this thread. I think I've got my bases covered now, so thanks. Just to add one final solution that was e-mailed to me but not the list (but is doing exactly what I want), there is f[0] = n - a; f[m_Integer?Positive] := f[m] = Assuming[{Element[{a, n}, Reals], a > 0, n > a^2}, FullSimplify[Integrate[f[m - 1] /. n -> n/x, {x, a, n/a}]]] Table[{m, f[m]}, {m, 0, 3}] // Grid ?? f And it is indeed getting much slower as my iterations are getting larger. Nathan On Tue, Oct 11, 2011 at 11:48 AM, DrMajorBob <btreat1 at austin.rr.com> wrote: > I don't get your first result, and ignoring that, I'm not sure which of the > two following solutions is what you want. The second is slow. > > Leaving out Assumptions makes all of it REALLY slow, and the results would > be ConditionalExpression instances. > > ClearAll[f, n, k] > f[1, n_] = > Integrate[n/x - a, {x, a, n/a}, Assumptions -> {n > a^2, a > 0}] > f[k_, n_] = > Integrate[n/x^k - a, {x, a, n/a}, > Assumptions -> {n > a^2, a > 0, k \[Element] Integers, k > 0}] > > a^2 - n + n Log[n/a^2] > > a^2 - n - ((a^(1 - k) - (a/n)^(-1 + k)) n)/(1 - k) > > ClearAll[f, n, k] > f[1, n_] = > Integrate[n/x - a, {x, a, n/a}, Assumptions -> {n > a^2, a > 0}]; > f[k_, n_] /; k > 1 := > f[k, n] = > FullSimplify[ > Integrate[f[k - 1, n/x], {x, a, n/a}, > Assumptions -> {a > 0, n > a^2}], {a > 0, n > a^2}] > Table[f[k, n], {k, 1, 3}] > > {a^2 - n + n Log[n/a^2], -a^3 + a n + > Log[a^2/n] (n + 2 n Log[a] - 1/2 n Log[n]), (1/( > 2 a^2))(a^2 - n) (2 ((-2 + a) a^3 + 3 n - 2 a n) + 8 n Log[a] - > 2 n Log[n] + a (-a^2 + n) Log[n/a^2])} > > Bobby > > > On Tue, 11 Oct 2011 03:24:30 -0500, Nathan McKenzie <kenzidelx at gmail.com> > 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? >> >> > > -- > DrMajorBob at yahoo.com >

**References**:**Making a function out of repeated hyperbola integrations?***From:*Nathan McKenzie <kenzidelx@gmail.com>

**Re: Making a function out of repeated hyperbola integrations?**

**Re: Interesting problem looking for a solution.**

**Re: Making a function out of repeated hyperbola integrations?**

**Re: Making a function out of repeated hyperbola integrations?**