Re: Function Equivalence of ArcTan and Log
- To: mathgroup at smc.vnet.net
- Subject: [mg114331] Re: Function Equivalence of ArcTan and Log
- From: Murray Eisenberg <murray at math.umass.edu>
- Date: Wed, 1 Dec 2010 02:12:06 -0500 (EST)
The equation s1==s2 need not be true! For example:
s1 /. {x->0, y->2} //InputForm
-Pi - I*Log[3]
s2 /. {x->0, y->2} //FullSimplify//InputForm
(-I)*Log[3]
On 11/22/2010 7:38 AM, Dr. Robert Kragler wrote:
> Hi,
>
> I have a problem to show with Mathematica (using already TrigToExp and
> ComplexExpand) that
>
> In[1]:= s1=(ArcTan[x-I y]-ArcTan[x+I y])//TrigToExp
> Out[1]:= 1/2 I Log[1-I (x-I y)]-1/2 I Log[1+I (x-I y)]-1/2 I Log[1-I
> (x+I y)]+1/2 I Log[1+I (x+I y)]
> and
>
> In[2]:= s2=I/2 Log[(x^2+(1-y)^2)/(x^2+(1+y)^2)]//ComplexExpand
> Out[2]:= I (1/2 Log[x^2+(1-y)^2]-1/2 Log[x^2+(1+y)^2])
>
> are identical, i.e. s1===s2 should give "True". But FullSimplify applied to lhs
> or rhs does not help.
> Of course, the identity wanted can be derived by hand making use of the inverse
> function Tan[s1]==I z and Sin[s1], Cos[s1] etc.
> which gives rise to the following replacement rule :
>
> ArcTanToLog = {(ArcTan[x_-I y_]-ArcTan[x_+I y_])-> I/2
> Log[(x^2+(1-y)^2)/(x^2+(1+y)^2)]}
>
> It is, however, non-trivial to prove this identity purely with term rewriting by
> means of Mathematica.
>
> Any suggestion is appreciated. Thanks in advance,
>
> Robert Kragler
>
>
--
Murray Eisenberg murray at math.umass.edu
Mathematics & Statistics Dept.
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University of Massachusetts 413 545-2859 (W)
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