Why does

5103*`Ω`*(5*sqrt(7)+16)/((1+2*sqrt(7))^(3/2)*(7+2*sqrt(7))*(-2+sqrt(7))*(14+sqrt(7))*(2*sqrt(7)-1)*(4+sqrt(7)))

not simplify to

sqrt(1+2*sqrt(7))*`Ω`

???

Hey guys,

I have the following occurence:

ii_inf:=x^(2-s)*(x^(-s)*GAMMA(3-s)*GAMMA(2-2*s)/(GAMMA(2-s)*GAMMA(3-2*s))+x^(s-2)*GAMMA(3-s)*GAMMA(2*s-2)/GAMMA(s))/(2-s)+(1/2)*(2*s*x-x+1)*(x+1)/(((x+1)^s)^2*(2*s^2-3*s+1))+x^(1-s)*(x^(-s)*GAMMA(2-s)*GAMMA(-2*s+1)/(GAMMA(1-s)*GAMMA(2-2*s))+x^(-1+s)*GAMMA(2-s)*GAMMA(2*s-1)/GAMMA(s))/(1-s)+(x+1)/((2*s-1)*((x+1)^s)^2)+x/((-1+s)*x^s)-(x+1)/((-1+s)*(x+1)^s);

ii_inf=simplify(ii_inf);

asympt(ii_inf,x,3);

Multiseries:-asympt(ii_inf,x,1);
gives different results...the last one however seems to be the correct one...

What is happening here?

If I calculate the integral:

z:=exp(I*t)

evalf(Int(z^(1/2)*(diff(z, t)), t = 0 .. 2*Pi))

I get -1.33333333*I

If I calculate

int(z^(1/2)*(diff(z, t)), t = 0 .. 2*Pi)

I get -4/3

so where is the I coming from? Am I doing sth  something   wrong?

I might add: if I calculate the same for z:=0.5+exp(I*t) I get 0 and -I*0.4714....

so what is going different here?

I want to plot the argument for a complex function. The input (x,y) represented in polar coordinates (r,phi) by default puts the cut at -I*Pi. Likewise the argument function:

argument(f(x)) plots the range -Pi..Pi.

However the function f(x)=x^2 could typically be plotted with 2 riemann surfaces on top of each other. When phi becomes 2Pi f(x) becomes 4Pi and only then I want to identify the 0 with 4Pi again since the points are equivalent in the preimage.

On the other hand the function f(x)=sqrt(x) never surpasses its own domain. The values always stay within the argument range of (0,2Pi) (in fact it only goes till Pi, or -Pi/2..Pi/2 in maple) when the preimage is taken to be (0,2Pi). Thus when plotting a preimage value of (x,y) with argument phi and 2Pi+phi they will have the same value since phi=2Pi+phi and I see a step in the plot. This step is actually there since the function has a cut at this point.

This step in the plotting image is also shown for f(x)=x^2 (e.g. at phi=+-Pi/2) but it is not of importance since it just comes from the argument function being constrained to -Pi..Pi.

So is it possible to change this behaviour?

Is it possible to animate plot like this parametically:
animate(plot, [[cos(t), sin(t), t = 0 .. A]], A = 0 .. 2*Pi, scaling = constrained, frames = 50)

however given the x and y components as solutions of an implicit equation.

I know I could run RootOf. But it seems that there is a part missing due to a branch cut :-/