Re: simple question

*To*: mathgroup at smc.vnet.net*Subject*: [mg73704] Re: [mg73661] simple question*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Sun, 25 Feb 2007 04:33:39 -0500 (EST)*References*: <200702240712.CAA09136@smc.vnet.net> <4E41D3AD-48C8-446F-BFCF-7C4A731A363C@mimuw.edu.pl>

Sorry, I posted not what I had intended. Let me do it all again. We define: I1 = Pi/2 - ArcTan[u]; I2 = ArcTan[1/u]; and want to show that I1==I2 for u>0. We check that the derivative of the difference is 0: In[42]:= FullSimplify[D[I1-I2,u],u>0] Out[42]= 0 and not, to prove that I1==I2 for u>0 we simply need to evaluate the difference I1-I2 for just one suitable chosen u>0, e.g. FullSimplify[(I1-I2)/.u->1] 0 (Last time I pasted and copied the wrong line here.) Andrzej Kozlowski On 24 Feb 2007, at 10:54, Andrzej Kozlowski wrote: > *This message was transferred with a trial version of CommuniGate > (tm) Pro* > 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>