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MathGroup Archive 2005

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Re: Thanks! Re: Integrate is driving me crazy, please help!

  • To: mathgroup at smc.vnet.net
  • Subject: [mg56277] Re: Thanks! Re: Integrate is driving me crazy, please help!
  • From: "Peltio" <peltio at trilight.zone>
  • Date: Wed, 20 Apr 2005 05:32:59 -0400 (EDT)
  • References: <d3vn1h$jm9$1@smc.vnet.net> <d42lb6$3ki$1@smc.vnet.net>
  • Reply-to: "Peltio" <peltioNOSPAM at despammed.com.invalid>
  • Sender: owner-wri-mathgroup at wolfram.com

"Jim Martin" wrote

> My end goal is to get that
>solution written into a spreadsheet but I have my doubts about any
>spreadsheet (excel or gnumeric) being able to handle the imaginary numbers.

That's unfortunate, because a solution that seems not to suffer from
some sort of discontinuity for positive values of the parameters is that
involving the logartithm. For example this

intF[a_, ha_, hb_, f_, f0_] := With[
    {g = Sqrt[hb^2 - 4a^2 f^2 Pi^2]},
        (f0 + ha) hb (1/(2 f)
        - hb/(2g f Pi)I  Log[(2a f Pi - I g)/(2 a f Pi + I g)])
        - 2a ha
    ]

This is a real quantity, since the log a ratio of complex conjugate numbers
is imaginary and so -I Log[(x+I y)/(x+I y)] happens to be real.

    -I Log[(x-I y)/(x+I y)] =
     = -I (Log[Abs[x-I y]/Abs[x+I y]] +I Arg[(x-I y)/(x+I y)])
     =  -I Log[1] + Arg[(x-I y)/(x+I y)]
     = 0 + Arg[(x-I y)/(x+I y)]
     = -2 Arg[x+I y]
     = -2 ArcTan[y/x]

The problem is that Arg, ArcTan[] and ArcTan[.,.] are affected by different
forms of discontinuity due to the modular world of angles (or the polidromy
of Log, if you prefer). So some of the different equivalent forms that can
be found by expressing log in some other form, are only valid modulo Pi or 2
Pi. Have a look at this graphics, for example

    funs = {
        Chop[(-I) Log[(x-I y)/(x+I y)]],
        Arg[(x-I y)/(x+I y)],
        ArcTan[(2 x y)/(y^2-x^2)],
        ArcTan[y^2-x^2, 2 x y],
         -2 ArcTan[y/x],
         - 2 ArcTan[x, y]};
    Block[{$DisplayFunction = Identity},
        plots = (Plot[Evaluate[#1 /. x -> 1], {y, -10, 10},
            Frame -> True, PlotRange -> {-Pi, 2*Pi},
            PlotLabel -> #1] & ) /@ funs
    ];
    Show[GraphicsArray[Partition[plots, 3]]];

The following definition for your integral solution seems to do well for
positive valued parameters, and it could be plotted in a spreadsheet with a
well implemented arctanget function:

intF[a_, ha_, hb_, f_, f0_] := With[
    {g = Sqrt[hb^2 - 4a^2 f^2 Pi^2]},
        (f0 + ha) hb (1/(2 f) - hb/(g f Pi) ArcTan[ g/( 2 a f Pi)]) - 2a ha
]

Try to plot it into your spreadsheet to see if you experience some kind of
discontinuity that should not be there. Something like this should come up:

    Plot[intF[a, 3, 50, 1, 8], {a, 0, 30}, Frame -> True]

cheers,
Peltio
Invalid address in reply-to. Crafty demunging required to mail me.




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