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MathGroup Archive 2007

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Re: simple question

  • To: mathgroup at smc.vnet.net
  • Subject: [mg73723] Re: [mg73661] simple question
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 25 Feb 2007 04:43:52 -0500 (EST)
  • References: <200702240712.CAA09136@smc.vnet.net>

Note that you can prove equality here as follows:


I1 = Pi/2 - ArcTan[u];
I2 = ArcTan[1/u];


FullSimplify[D[I1-I2,u],u>0]

0

And now all you need is one value of u>0 for which I1 and I2 are easy  
to evaluate:

FullSimplify[D[I1 - I2, u], u > 0]
0


This method works in surprisingly many cases.

Andrzej Kozlowski



On 24 Feb 2007, at 08:12, dimitris wrote:

> Hello.
>
> This post has a connection with a recent post of David Cantrell.
>
> (I hope I don't miss anything!)
>
> Consider the functions
>
> In[37]:=
> I1 = Pi/2 - ArcTan[u];
> I2 = ArcTan[1/u];
>
> For u>=0 the functions are equal as the following demonstrates
>
> In[39]:=
> Show[Block[{$DisplayFunction = Identity}, (Plot[{I1, I2}, {u, #1[[1]],
> #1[[2]]}, PlotPoints -> 100, Axes -> False,
>       Frame -> {True, True, False, False}, PlotStyle -> {Red, Blue},
>       FrameTicks -> {Range[-2*Pi, 2*Pi, Pi], Range[-2, 4, 1]}] & ) /@
> Partition[Range[-2*Pi, 2*Pi, Pi], 2, 1]]]
>
> In[43]:=
> I1 /. u -> 0
> (Limit[I2, u -> 0, Direction -> #1] & ) /@ {1, -1}
>
> Out[43]=
> Pi/2
> Out[44]=
> {-(Pi/2), Pi/2}
>
> However
>
> None of these work
>
> In[81]:=
> FullSimplify[I1 - I2, u > 0]
> FullSimplify[I1 == I2, u > 0]
> FullSimplify[I1 - I2 == 0, u > 0]
>
> Out[81]=
> (1/2)*(Pi - 2*ArcCot[u] - 2*ArcTan[u])
> Out[82]=
> 2*(ArcCot[u] + ArcTan[u]) == Pi
> Out[83]=
> 2*(ArcCot[u] + ArcTan[u]) == Pi
>
> Even for specific u we don't have "simplification" to zero
>
> In[111]:=
> FullSimplify[I1 - I2 /. u -> Pi]
> N[%]
>
> Out[111]=
> (1/2)*(Pi - 2*(ArcCot[Pi] + ArcTan[Pi]))
> Out[112]=
> -2.220446049250313*^-16
>
> Any ideas how to show that I1-I2=0 (or I1=I2) symbolically?
>
> I personally tried (based on relevant material on the Help Browser, M.
> Trott's Guidebook for Symbolics and Dana DeLouis' solution to David's
> original post)
>
> In[40]:=
> equ = TrigToExp[I1 - I2 /. u -> Pi/2]
> avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
> _ArcSec | _Log, Infinity] & ;
> FullSimplify[equ, ComplexityFunction -> avoid]
>
> Out[40]=
> Pi/2 - (1/2)*I*Log[1 - (2*I)/Pi] + (1/2)*I*Log[1 + (2*I)/Pi] -
> (1/2)*I*Log[1 - (I*Pi)/2] + (1/2)*I*Log[1 + (I*Pi)/2]
> Out[42]=
> 0
>
> In[43]:=
> equ = TrigToExp[I1 - I2 /. u -> Pi/3]
> avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
> _ArcSec | _Log, Infinity] & ;
> FullSimplify[equ, ComplexityFunction -> avoid]
>
> Out[43]=
> Pi/2 - (1/2)*I*Log[1 - (3*I)/Pi] + (1/2)*I*Log[1 + (3*I)/Pi] -
> (1/2)*I*Log[1 - (I*Pi)/3] + (1/2)*I*Log[1 + (I*Pi)/3]
> Out[45]=
> 0
>
> However this approach is not general; for example
>
> In[84]:=
> equ = TrigToExp[I1 - I2 /. u -> Pi/4]
> avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
> _ArcSec | _Log, Infinity] & ;
> FullSimplify[equ, ComplexityFunction -> avoid]
>
> Out[84]=
> Pi/2 - (1/2)*I*Log[1 - (4*I)/Pi] + (1/2)*I*Log[1 + (4*I)/Pi] -
> (1/2)*I*Log[1 - (I*Pi)/4] + (1/2)*I*Log[1 + (I*Pi)/4]
> Out[86]=
> (-(1/2))*I*(I*Pi + 2*Pi*Floor[(Pi - Arg[(-((-4*I - Pi)/(4*I - Pi)))^(-
> I)] - Arg[((-4*I + Pi)/(4*I + Pi))^(-I)])/(2*Pi)] +
>    I*Log[1/((-((-4*I - Pi)/(4*I - Pi)))^I*((-4*I + Pi)/(4*I +
> Pi))^I)])
>
> Also it fails for u unspecified in advance
>
> In[93]:=
> equ = TrigToExp[I1 - I2];
> avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
> _ArcSec | _Log, Infinity] & ;
> FullSimplify[equ, ComplexityFunction -> avoid]
> FullSimplify[%, u > 0]
>
> Out[95]=
> (1/2)*(Pi - I*Log[1 - I*u] + I*Log[1 + I*u] - I*Log[(-I + u)/u] +
> I*Log[(I + u)/u])
> Out[96]=
> (-(1/2))*I*(Log[I - u] + Log[1 - I*u] - Log[1 + I*u] - Log[I + u])
>
> Thanks in advance for your response!
>
> Dimitris
>
>



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