Re: Value of E
- To: mathgroup at smc.vnet.net
- Subject: [mg77386] Re: [mg77301] Value of E
- From: DrMajorBob <drmajorbob at bigfoot.com>
- Date: Thu, 7 Jun 2007 04:06:41 -0400 (EDT)
- Organization: Deep Space Corps of Engineers
- References: <30949593.1181133668382.JavaMail.root@m35>
- Reply-to: drmajorbob at bigfoot.com
Both solutions depend on the ArcSin argument:
Off[Solve::"ifun"]
Solve[(I*a^(I*ArcSin[3/5]) - I*a^-(I*ArcSin[3/5]))/2 == 3/5, a]
N@%
{{a -> (-(4/5) - (3*I)/5)^(-(I/ArcSin[3/5]))}, {a -> (4/5 - =
(3*I)/5)^(-(I/ArcSin[3/5]))}}
{{a -> 0.020608916569560962 + 0.*I}, {a -> 0.36787944117144233 + 0.*I}}
Solve[(I*a^(I*ArcSin[arg]) - I*a^-(I*ArcSin[arg]))/2 == arg, a]
{{a -> (-\[ImaginaryI] arg - Sqrt[1 - arg^2])^-\[ImaginaryI]/
ArcSin[arg]}, {a -> (-\[ImaginaryI] arg + Sqrt[
1 - arg^2])^-\[ImaginaryI]/ArcSin[arg]}}
Bobby
On Wed, 06 Jun 2007 06:22:44 -0500, Jeff Albert <albertj001 at hawaii.rr.co=
m> =
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 Solv=
e, 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
>
>
>
>
-- =
DrMajorBob at bigfoot.com