Re: Integrate[], Sort[] and Hold[]
- To: mathgroup at smc.vnet.net
- Subject: [mg101192] Re: [mg100994] Integrate[], Sort[] and Hold[]
- From: Neil Stewart <neil.stewart at warwick.ac.uk>
- Date: Fri, 26 Jun 2009 06:51:28 -0400 (EDT)
- Reply-to: Neil Stewart <neil.stewart at warwick.ac.uk>
David and Bob, thanks for your suggestions. I need this to work with lists
of more than two things. I support I could use Min and Max to make a new
Sort function, but wondered if there might be a better way.
So does anyone have any ideas why Min[] and Max[] behave differently to
Sort[]?
I wondered if it ws about attributes.
Attributes[Sort] = {Protected}
Attributes[Min] = {Flat, NumericFunction, OneIdentity, Orderless, Protected}
Attributes[Max] = {Flat, NumericFunction, OneIdentity, Orderless, Protected}
But setting the attributes of Sort[] to match Min[] or Max[] doesn't change
Sort[]'s behaviour in Integrate[].
All comments very welcome.
On Fri, 19 Jun 2009, David Park wrote:
> I was somewhat surprised that the following worked:
>
> Integrate[Min[2^a, 3^a], {a, -1, 1}]
> % // N
> (2 Log[2] + Log[27])/(Log[2] Log[27])
> 2.04952
>
> NIntegrate[Min[2^a, 3^a], {a, -1, 1}]
> 2.04952
>
>
> David Park
> djmpark at comcast.net
> http://home.comcast.net/~djmpark/
>
>
>
> From: Neil Stewart [mailto:neil.stewart at warwick.ac.uk]
>
> I am having trouble integrating a function which contains Sort[]. Here is a
> stripped down (if slightly odd) example:
>
> In = Integrate[Sort[{2, 3}^a], {a, -1, 1}]
> Out = {3/Log[4], 8/Log[27]}
>
> Mathematica is evaluating this as follows. First, Sort[{2, 3}^a] is
> evaluated as {2^a, 3^a}. Then Integrate[2^a, {a, -1, 1}] gives the first
> term 3/Log[4]. Finally Integrate[3^a, {a, -1, 1}] gives the second term
> 8/Log[27]. Note here that Sort[] is sorting 2^a and 3^a without knowing the
> value of a. That is, sort is sorting the raw symbolic expressions.
>
> I would prefer Sort[] to wait until it knows the value of a before sorting.
> For example, when a is -1, then 2^a = 1/2 and 3^a = 1/3, so Sort[{2, 3}^a]
> would be {1/3, 1/2}. However when a is 1, then 2^a = 2 and 3^a = 3, so
> Sort[{2, 3}^a] would be {2, 3}. That is, in the first case the terms are
> swapped, but in the second case they are not. So what I'm after is
>
> In = Integrate[Sort[{2, 3}^a], {a, -1, 1}]
> Out = {2/Log[3] + 1/Log[4], 1/Log[2] + 2/Log[27]}
> (* This does not actually happen *)
>
> [If you prefer to picture this, Plot[{2^a, 3^a}, {a, -1, 1}] draws two
> increasing lines that cross at a = 0. Mathematica is integrating under each
> curve. I'm trying to integrate under the line made from the two lower
> segments, and under the line made from the two upper segments.]
>
> I've tried using Hold[], ReleaseHold[], and Evaluate[] but have got myself
> into a terrible mess. Obviously with this trivial example I could just split
> the integral up myself, but is there a way to achieve delaying Sort[] until
> a is known? I would be very grateful for any comments.
>
>
>
>
>
>