Re: showing an expression equal to 0

*To*: mathgroup at smc.vnet.net*Subject*: [mg73723] Re: showing an expression equal to 0*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Mon, 26 Feb 2007 06:17:04 -0500 (EST)*References*: <errm70$8bp$1@smc.vnet.net>

That's why this forum is so wonderful! Questions about (...Mathematics and...) Mathematica answered so clear and quick! Thanks again Andrzej! Best Regards Dimitris Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote: *This message was transferred with a trial version of CommuniGate(tm) Pro* 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] 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 > > 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 > show that > expr is equal to zero. > > Any ideas? > > Thanks!