[Date Index]
[Thread Index]
[Author Index]
Re: simple question
*To*: mathgroup at smc.vnet.net
*Subject*: [mg73706] Re: simple question
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Sun, 25 Feb 2007 04:34:44 -0500 (EST)
*References*: <eroorg$91o$1@smc.vnet.net>
Quite elementary but very tricky at the same time!
Thanks a lot Andrzej!
Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:
Sorry, I posted not what I had intended. Let me do it all again. We
define:
I1 = Pi/2 - ArcTan[u];
I2 = ArcTan[1/u];
and want to show that I1==I2 for u>0. We check that the derivative of
the difference is 0:
In[42]:=
FullSimplify[D[I1-I2,u],u>0]
Out[42]=
0
and not, to prove that I1==I2 for u>0 we simply need to evaluate the
difference I1-I2 for just one suitable chosen u>0, e.g.
FullSimplify[(I1-I2)/.u->1]
0
(Last time I pasted and copied the wrong line here.)
Andrzej Kozlowski
On 24 Feb 2007, at 10:54, Andrzej Kozlowski wrote:
> *This message was transferred with a trial version of CommuniGate
> (tm) Pro*
> Note that you can prove equality here as follows:
>
>
> I1 = Pi/2 - ArcTan[u];
> I2 = ArcTan[1/u];
>
>
> FullSimplify[D[I1-I2,u],u>0]
>
> 0
>
> And now all you need is one value of u>0 for which I1 and I2 are
> easy to evaluate:
>
> FullSimplify[D[I1 - I2, u], u > 0]
> 0
>
>
> This method works in surprisingly many cases.
>
> Andrzej Kozlowski
>
>
>
> On 24 Feb 2007, at 08:12, dimitris wrote:
>
>> Hello.
>>
>> This post has a connection with a recent post of David Cantrell.
>>
>> (I hope I don't miss anything!)
>>
>> Consider the functions
>>
>> In[37]:=
>> I1 = Pi/2 - ArcTan[u];
>> I2 = ArcTan[1/u];
>>
>> For u>=0 the functions are equal as the following demonstrates
>>
>> In[39]:=
>> Show[Block[{$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
>>
>>
>
Prev by Date:
**Re: Re: is 3?**
Next by Date:
**performance of tail-recursion**
Previous by thread:
**Re: simple question**
Next by thread:
**Re: simple question**
| |