rsolve(a(n)=2/a(n-1)+1,a(1)=1),

thx in advance.

I want to solve two equations with all roots.

eq1:=-7.506556000*10^(-23)*omega^12+2.243366854*10^(-78)*omega^34+3.651941282*10^(-50)*omega^25+1.145341958*10^(-105)*omega^41-3.837716260*10^(-102)*omega^40+9.952750000*10^(-10)*omega^5+2.520327960*10^(-113)*omega^43-5.961864252*10^(-98)*omega^39+2.934254874*10^(-60)*omega^29+3.877088344*10^(-109)*omega^42-3.482296310*10^(-82)*omega^35+2.846747460*10^(-123)*omega^45-1.089582984*10^(-70)*omega^32-5.927854900*10^(-48...

I have a problem in determine the relation between two variables in a polynomial equation f(l2,v)=0. The results of two methods puzzled me.
The first method I tried is use the fsolve command.
In equation f(l2,v)=0,the range for v is [3405,5054] and l2 is a complex variable.
I plotted the figure for the l2=l2(v) in complex plane.Because the degree of l2 is 8, I have eight curves in the figure.
Obviously, the order in the result of fsolve is indefinite, so it can...

plot3d(sin(sqrt(x^2+y^2))/sqrt(x^2+y^2),x=-3..3,y=-3..3);

the command above makes a figure which has a hole(means discontinuity) at the (0,0).

is it a bug?

athlon 4400+,2GB RAM,win xp 32.

still have problems,
1.
restart:
f := proc (x) options operator, arrow; piecewise(0 <= x and x < 4, 50, 4 <= x and x < 5, 0, 5 <= x, f(x-5)) end proc;

plot(proc (t) options operator, arrow; evalf(Int(f, 0 .. t, epsilon = 0.1e-3, digits = 15)) end proc, 0 .. 20);

in the above figure,there is a discont point in the interval [16..18],but in fact,it's a continuous curve.

2.
how to use fsolve deal with two periodic functions?