It seems to me that polarform should be an ordered pair of r and theta
representing r*exp(%i*theta).
changing the exp() to an approximation of e to a power is not what I
would expect.
But what we would really need is a whole set of stuff so that
polarform(r1,theta1) <op> polarform(r2,theta2) comes out as
polarform(r3,theta3).
where op = the usual arithmetic. Also functions like sin cos tan...
All of which could, I suppose be implemented as e.g. sin(pf) =
polarform(sin(rectform(pf))) until people
find them inadequate ...
Raymond Toy wrote:
> These were sent to me by Jim FitzSimons, who wrote the bigfloat
> implementation of elliptic integrals that I translated into Lisp.
>
> bfloat(sqrt(%i)) -> 1.0b0*(-1)^(1/4)
>
> Should this be more like .707+.707*%i?
>
> Another issue is
>
> float(polarform(sqrt(1+%i))) -> 1.18*2.718^(%i*%pi/8)
>
> But
>
> bfloat(polarform(sqrt(1+%i))) -> 1.18*(3.82b-1*%i+9.23b-1)
>
> Not sure which is the correct answer. In some ways, the float result is
> better since it preserves the polarform, but doesn't float the phase
> part. The bfloat result is better since everything is a bfloat, but the
> polarform is converted to rectangular.
>
> Ray
>
>
>
>
> _______________________________________________
> Maxima mailing list
> Maxima at math.utexas.edu
> http://www.math.utexas.edu/mailman/listinfo/maxima
>