       • To: mathgroup at smc.vnet.net
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Tue, 19 Apr 2005 04:54:50 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```The symbolic solution can be further simplified for hillb^2>4a^2*f^2*Pi^2

soln2=Simplify[soln,hillb^2>4a^2*f^2*Pi^2]

(1/(2*f*Pi*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]))*
(2*(f0 + hilla)*ArcTan[(2*a*f*Pi)/Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]]*hillb^2 +
Pi*(f0*hillb*(Sqrt[hillb^2 - 4*a^2*f^2*Pi^2] - hillb) +
hilla*(-hillb^2 + Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]*hillb -
4*a*f*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2])))

Bob Hanlon

>
> From: Bob Hanlon <hanlonr at cox.net>
To: mathgroup at smc.vnet.net
> Date: 2005/04/18 Mon AM 06:34:33 EDT
> To: Jim Martin <jim.martin at utah.edu>, <mathgroup at smc.vnet.net>
>
> Use square brackets for all functions
>
> asmp={Element[{a,f,f0,hilla,hillb}, Reals],
>       a>0,hillb>0};
>
> soln = Simplify[Integrate[
>       (hillb*((f0+hilla)/
>                 (2*Pi*f*a*Cos[ArcSin[x/a]]+hillb)))-
>         hilla,{x,-a,a}],asmp]
>
> (1/(2*f*Pi*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]))*
>   (2*(f0 + hilla)*ArcTan[(2*a*f*Pi)/Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]]*hillb^2 -
>    I*(f0 + hilla)*Log[-(I/(2*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]))]*hillb^2 +
>    I*f0*Log[I/(2*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2])]*hillb^2 +
>    I*hilla*Log[I/(2*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2])]*hillb^2 +
>    f0*Pi*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2]*hillb + hilla*Pi*Sqrt[hillb^2 -
> 4*a^2*f^2*Pi^2]*
>     hillb - 4*a*f*hilla*Pi*Sqrt[hillb^2 - 4*a^2*f^2*Pi^2])
>
> Simplify[soln/.{hilla->3,hillb->50,f0->8,a->1,f->1}]
>
> 269 - 6875/Sqrt[625 - Pi^2] + (13750*ArcTan[Pi/Sqrt[625 - Pi^2]])/
> (Pi*Sqrt[625 - Pi^2])
>
> %//N
>
> 14.0372
>
>
> Bob Hanlon
>
> >
> > From: Jim Martin <jim.martin at utah.edu>
To: mathgroup at smc.vnet.net
> > Date: 2005/04/18 Mon AM 03:08:44 EDT
> >
> > Hello Mathematica Experts:
> >
> > I am a biomechanist and work mostly in the area of muscle contraction. I
> > do a lot of numerical computations using excel, but right now I need an
> > analytical solution that represents force as a function of position
> > integrated over a shortening amplitude. I downloaded a trial version of
> > Mathematica and have tried to obtain a solution for this:
> >
> > Integrate[(hillb*((f0 + hilla)/(2*pi*f*a*Cos(ArcSin(x/a)) + hillb))) -
> > hilla, {x, -a, a}]
> >
> > Mathematica returns this:
> > (-4 a ArcSin Cos f hilla pi + (f0 + hilla) hillb (-Log[hillb - 2 a
> >      ArcSin Cos f pi] + Log[hillb + 2 a ArcSin Cos f pi]))/(4 a ArcSin
> > Cos f pi)
> >
> > I know the line wrap makes this hard to read so please feel free to
> > email me and I can send you the output as a picture.
> >
> > In a sample data set, hilla=3, hillb=50, f0=8, a=1, f=1
> >
> > I can numerically integrate this function and obtain a value for that
> > sample data set of 14.04. When I put those sample values into the
> > solution that Mathematica produces, I get 10.01.
> >
> > Can any of you please give a hand here? I must be making some simple
> > Mathematica-beginner error but I just can't see it.
> >
> > In Mathematica, Log is Log to base e, right (LN in excel)? Did I use
> > variables that have intrinsic functions in Mathematica? Maybe I am
> > misunderstanding the output with regard to implicit parentheses etc.
> > Any help appreciated!
> >
> > Thanks,
> >
> > Jim
> >
> >
> >
>

```

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