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Re: Handling branch cuts in trig functions
Posted:
Mar 24, 2013 5:02 PM


"Nasser M. Abbasi" schrieb: > > I tried to simplify sqrt( sec(x)^2 ) but Mathematica will > only do this by assuming x is inside one branch, say > x>Pi/2 && x<Pi/2 but Maple and maxima simplified it > but they gave the answer is terms of sec(x) to take > care of the sign which depends on the branch. > > Here is plot of sec(x) > > http://mathworld.wolfram.com/Secant.html > >  > In[37]:= Assuming[x>Pi/2&&x<Pi/2,Simplify[Sqrt[Sec[x]^2]]] > Out[37]= Sec[x] > > In[39]:= Assuming[x > Pi/2 && x < Pi, Simplify[Sqrt[Sec[x]^2]]] > Out[39]= Sec[x] >  > > If I just tell M that x>0, it will not simplify it. > >  > In[38]:= Assuming[x>0,Simplify[Sqrt[Sec[x]^2]]] > Out[38]= Sqrt[Sec[x]^2] >  > > but Maple did it only with the x>0 assumption: > >  > restart; > simplify(sqrt(sec(x)^2)) assuming x::positive; > > 1 >  > cos(x) > restart; > simplify(sqrt(sec(x)^2)); > / 1 \ > csgn > \cos(x)/ >  > cos(x) >  > > On maxima 12.04.0 > > sqrt(sec(x)^2); > sec(x) > > I think now that answer to sqrt(sec(x)^2) should be > sec(x) without need to give the branch. Since the only > different is the sign. Or is there something else here? >
Mathematica and Maple by default work in the complex plane, and must therefore define ABS(z) = SQRT(RE(z)^2 + IM(z)^2), which agrees with your SQRT(z^2) = SQRT(RE(z)^2 + 2*#i*RE(z)*IM(z)  IM(z)^2) only if IM(z) = 0. So SQRT(SEC(z)^2) can be 'simplified' to ABS(SEC(z)) only where SEC(z) = 1/COS(z) is real, that is for all real z only. Both systems should be able to simplify SQRT(SEC(z)^2)  ABS(SEC(z)) to zero if z is restricted to real.
Martin.



