MathGroup Archive 2011

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Calculus and InterpolatingFunction

  • To: mathgroup at smc.vnet.net
  • Subject: [mg121664] Re: Calculus and InterpolatingFunction
  • From: Brentt <brenttnewman at gmail.com>
  • Date: Sat, 24 Sep 2011 22:34:24 -0400 (EDT)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <201109221125.HAA26698@smc.vnet.net>

Thank you, I forgot about the Numerical functions, which is a more natural
choice. But yea, it is still puzzling why it works for one of the
expressions, but not the other. Problems solved, but the mystery
continues...?

On Fri, Sep 23, 2011 at 12:45 AM, Murray Eisenberg <murray at math.umass.edu>wrote:

> Two remaining problems:
>
> (1) The Documentation Center page for Interpolation says, "Interpolation
> returns an InterpolatingFunction object, which can be used like any
> other pure function."
>
>    Manifestly that is not the case. Thus the following, for a pure
> function, _does_ work:
>
>   f = #^2 &
>    Integrate[f[x] + 1, {x, 1, 10}]
>
> (2) While the solutions you proposed both work, the latter using Map
> would be problematic for integrands involving the InterpolatingFunction
> in more complicated ways, e.g.:
>
>    f = Interpolation[data];
>    Integrate[#, {x, 1, 10}] & /@ (Sin[f[x]])
> 0.576208
>    NIntegrate[Sin[f[x]], {x, 1, 10}]
> 0.607007
>
> Is there some way to obtain an explicit function from an
> InterpolatingFunction object?
>
>
> On 9/22/11 7:25 AM, Bob Hanlon wrote:
> > data = RandomReal[#]*2&  /@ Range[1, 10];
> >
> > f = Interpolation[data];
> >
> > Integrate[f[x], {x, 1, 10}]
> >
> > 52.9041
> >
> > Use NIntegrate
> >
> > NIntegrate[f[x] + 1, {x, 1, 10}]
> >
> > 61.9041
> >
> > Or Map over the expression
> >
> > Integrate[#, {x, 1, 10}]&  /@ (f[x] + 1)
> >
> > 61.9041
> >
> >
> > Bob Hanlon
> >
> > ---- Just A Stranger<forpeopleidontknow at gmail.com>  wrote:
> >
> > =============
> > I'm trying to get a definite integral for an InterpolatingFunction. It
> works
> > if it is the function by itself, but not for some reason arithmetically
> > combining the InterpolatingFunction with another function makes it not
> > return a value. e.g.
> >
> >
> > In[1]:=
> > data = RandomReal[#]*2&  /@ Range[1, 10];
> > f = Interpolation[data];
> >
> >>   Integrate[f[x], {x, 1, 10}]
> > Out[1]:=40.098
> >
> > So far so good. But just a little bit of arithmetic in the integral and
> it
> > doesn't work anymore:
> > In[2]:=
> > Integrate[f[x]+1, {x, 1, 10}]
> > Out[2]:=
> > Integrate[Plus[1, InterpolatingFunction[][x]], List[x, 1, 10]]
> >
> > (That last answer was actually the output with  //FullForm applied)
> > Why won't it give me a numerical evaluation? Is there anyway to make a
> > continuous function from data that will seemlessly work with Integrate?
> I'm
> > thinking of constructing a piecwise function using Fit, Piecwise, and a
> > Table for the arguments to Piecewise. But I would think  Interpolation
> might
> > have worked and been easier. I want to figure out if I am I doing
> something
> > wrong with Interpolation before I start trying to tackle a slightly more
> > complicated piecewise defined function ?
> >
> >
>
> --
> Murray Eisenberg                     murray at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower      phone 413 549-1020 (H)
> University of Massachusetts                413 545-2859 (W)
> 710 North Pleasant Street            fax   413 545-1801
> Amherst, MA 01003-9305
>
>



  • Prev by Date: Re: Calculus and InterpolatingFunction
  • Next by Date: Re: Compilation: Avoiding inlining
  • Previous by thread: Re: Calculus and InterpolatingFunction
  • Next by thread: Re: Calculus and InterpolatingFunction