Re: showing an expression equal to 0

*To*: mathgroup at smc.vnet.net*Subject*: [mg73703] Re: [mg73719] showing an expression equal to 0*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Mon, 26 Feb 2007 06:06:10 -0500 (EST)*References*: <200702250941.EAA08417@smc.vnet.net>

Actually, in this case you only need the observation that the function is complex analytic in y for R>0 in the interior of the disk Abs[y]<R. Now: expr = ArcTan[y/Sqrt[R^2 - y^2]] - ArcSin[y/R] = Assuming[R>0,FullSimplify[D[expr,y]]] 0 and Assuming[R>0,FullSimplify[expr/.y->0]] 0 This tells us not only that the derivative vanishes but also that all the higher derivatives at all the points in the interior of the disk vanish in the disk (and in particular at 0). In other words, the Taylor series at 0 is 0, which means that the function is 0 in the interior of the disk. So in fact this case is also easy to deal with by using the method "differentiate and FullSimplify". Andrzej Kozlowski On 25 Feb 2007, at 10:41, dimitris wrote: > Consider > > In[87]:= > expr = ArcTan[y/Sqrt[R^2 - y^2]] - ArcSin[y/R]; > > The expression for R>0 and -R<y<R is equal to zero > > In[72]:= > (Plot[Chop[Evaluate[expr /. R -> #1]], {y, -#1, #1}, Axes -> False, > Frame -> True] & ) /@ Range[5] > > Based on Andrzej's resposnse on a recenet message I am able to show > this equality for > given R. > > E.g. for R=4 the following demonstrates that expr is equal to zero > > In[99]:= > Simplify[expr /. R -> 4 /. y -> 2] > > Out[99]= > 0 > > In[97]:= > D[expr /. R -> 4, y] > Factor[%] > > Out[97]= > -(1/(4*Sqrt[1 - y^2/16])) + (y^2/(16 - y^2)^(3/2) + 1/Sqrt[16 - y^2])/ > (1 + y^2/(16 - y^2)) > Out[98]= > 0 > > However not specifying R (but assuming R>0&&-R<y<R) I am not able to > show that > expr is equal to zero. > > Any ideas? > > Thanks! > >

**References**:**showing an expression equal to 0***From:*"dimitris" <dimmechan@yahoo.com>