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 > >
- References:
- simple question
- From: "dimitris" <dimmechan@yahoo.com>
- simple question