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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>
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