Re: Value of E

*To*: mathgroup at smc.vnet.net*Subject*: [mg77349] Re: [mg77301] Value of E*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Thu, 7 Jun 2007 03:47:13 -0400 (EDT)*References*: <200706061122.HAA00476@smc.vnet.net>

On 6 Jun 2007, at 20:22, Jeff Albert wrote: > Now we all know that: > > In[112]:=N[E,10] >> From In[112]:=2.718281828459045 > > and that > > In[113]:=N[(I*E^(I*ArcSin[3/5])-I*E^-(I*ArcSin[3/5]))/2,10] >> From In[113]:=-0.6 + 0.*I. > > Can someone please explain: > > In[117]:=NSolve[(I*A^(I*ArcSin[3/5])-I*A^-(I*ArcSin[3/5]))/2==3/5,A] >> From In[117]:=Solve::ifun: Inverse functions are being used by >> Solve, so > some solutions may not be found. >> From In[117]:={{A -> 0.020608916569560966 + 0.*I}, {A -> >> 0.36787944117144233 > + 0.*I}} > > Where it turns out that the first solution seems to depend on the > argument > for ArcSin[], but the second does not. Indeed one even finds: > > In[115]:=0.36787944117144233^(I*Pi) >> From In[115]:=-1. - 1.2246063538223773*^-16*I > > Jeff Albert > > I really can't see what there is to explain. NSolve solves polynomial equations and equations that can be converted to polynomial equations only. Your equation I*A^(I*ArcSin[3/5])-I*A^-(I*ArcSin[3/5]))/2==3/5 is equivalent to the followng system: eq1 = (1/2)*(I*x - I/x) == 4/5; and eq2 = x == A^(I*ArcSin[4/5]); The first of these is a rational algebraic equation so NSolve or Solve can solve it with no difficulty: sol1 = Solve[eq1, x] {{x -> -(3/5) - (4*I)/5}, {x -> 3/5 - (4*I)/5}} The second is a transcendental equation for which Solve will find only some solutions by using log: sol2 = (Solve[#1, A] & ) /@ (eq2 /. sol1) Solve::ifun:Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >> Solve::ifun:Inverse functions are being used by Solve, so some solutions may not be found; use Reduce for complete solution information. >> {{{A -> (-(3/5) - (4*I)/5)^(-(I/ArcSin[4/5]))}}, {{A -> (3/5 - (4*I)/5)^(-(I/ArcSin[4/5]))}}} Chop[N[%]] {{{A -> 0.09182139893651517}}, {{A -> 0.36787944117144233}}} The second answer is the same as the one you got. The first one is different, but this is to be expected since NSolve uses a different method from Solve and none of them give you all the answers. To get that you need to use Reduce: sols=Reduce[(I*A^(I*ArcSin[3/5]) - I/A^(I*ArcSin[3/5]))/ 2 == 3/5, A] Element[C[1], Integers] && A != 0 && (A == E^(-((I*(2*I*Pi*C[1] + Log[4/5 - (3*I)/5]))/ ArcSin[3/5])) || A == E^(-((I*(2*I*Pi*C[1] + Log[-(4/5) - (3*I)/5]))/ ArcSin[3/5]))) by letting C[1] take various values you get the various solutions above, for exmple: Reduce[FullSimplify[sols /. C[1] -> 0], A] A == (-(4/5) - (3*I)/5)^(1/Log[4/5 + (3*I)/5]) || A == 1/E Chop[N[%]] A == 0.020608916569560962 || A == 0.36787944117144233 which is the answer you got from NSolve. By substituting other values for C[1] you get other answers. What else is there to explain? Andrzej Kozlowski

**References**:**Value of E***From:*"Jeff Albert" <albertj001@hawaii.rr.com>