       simple question

• To: mathgroup at smc.vnet.net
• Subject: [mg73661] simple question
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Sat, 24 Feb 2007 02:12:50 -0500 (EST)

```Hello.

This post has a connection with a recent post of David Cantrell.

(I hope I don't miss anything!)

Consider the functions

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

For u>=0 the functions are equal as the following demonstrates

In:=
Show[Block[{\$DisplayFunction = Identity}, (Plot[{I1, I2}, {u, #1[],
#1[]}, 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:=
I1 /. u -> 0
(Limit[I2, u -> 0, Direction -> #1] & ) /@ {1, -1}

Out=
Pi/2
Out=
{-(Pi/2), Pi/2}

However

None of these work

In:=
FullSimplify[I1 - I2, u > 0]
FullSimplify[I1 == I2, u > 0]
FullSimplify[I1 - I2 == 0, u > 0]

Out=
(1/2)*(Pi - 2*ArcCot[u] - 2*ArcTan[u])
Out=
2*(ArcCot[u] + ArcTan[u]) == Pi
Out=
2*(ArcCot[u] + ArcTan[u]) == Pi

Even for specific u we don't have "simplification" to zero

In:=
FullSimplify[I1 - I2 /. u -> Pi]
N[%]

Out=
(1/2)*(Pi - 2*(ArcCot[Pi] + ArcTan[Pi]))
Out=
-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:=
equ = TrigToExp[I1 - I2 /. u -> Pi/2]
avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
_ArcSec | _Log, Infinity] & ;
FullSimplify[equ, ComplexityFunction -> avoid]

Out=
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=
0

In:=
equ = TrigToExp[I1 - I2 /. u -> Pi/3]
avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
_ArcSec | _Log, Infinity] & ;
FullSimplify[equ, ComplexityFunction -> avoid]

Out=
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=
0

However this approach is not general; for example

In:=
equ = TrigToExp[I1 - I2 /. u -> Pi/4]
avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
_ArcSec | _Log, Infinity] & ;
FullSimplify[equ, ComplexityFunction -> avoid]

Out=
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=
(-(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:=
equ = TrigToExp[I1 - I2];
avoid = Count[#1, _ArcTan | _ArcCot | _ArcSin | _ArcCos | _ArcCsc |
_ArcSec | _Log, Infinity] & ;
FullSimplify[equ, ComplexityFunction -> avoid]
FullSimplify[%, u > 0]

Out=
(1/2)*(Pi - I*Log[1 - I*u] + I*Log[1 + I*u] - I*Log[(-I + u)/u] +
I*Log[(I + u)/u])
Out=
(-(1/2))*I*(Log[I - u] + Log[1 - I*u] - Log[1 + I*u] - Log[I + u])