Items tagged with riemann


how to use Riemann matrix to output Riemann surface?

and plot this surface?


evalf(pm, 5);
M := rm;
A := proc (x, y) options operator, arrow; RiemannTheta([x, y], M, [], 0.1e-1, output = list)[2] end proc;
plot3d(Re(A(x+I*y, 0)), x = 0 .. 1, y = 0 .. 4, grid = [40, 40]);

is this graph Riemann Surface?

if so, how to convert A into polynomials?

Hi everyone,


I am using the following code (by dr. Corless) to animate a Riemann surface:


B := 1.5;
u2 := r*cos(th); v2 := r*sin(th);
w1 := u1+I*v1; w2 := u2+I*v2;
z1 := evalc(w1^2); z2 := evalc(w2^2);
x1 := evalc(Re(z1)); x2 := evalc(Re(z2));
y1 := evalc(Im(z1)); y2 := evalc(Im(z2));
f1 := proc (theta) options operator, arrow; plot3d([-x1, -y1, v1], u1 = -6 .. 1, v1 = -B .. B, grid = [50, 50], orientation = [theta, 80], color = u1) end proc;
f2 := proc (theta) options operator, arrow; plot3d([-x2, -y2, v2], r = 1 .. 1, th = -Pi .. Pi, grid = [50, 50], orientation = [theta, 80], color = black) end proc;
display(seq([f1(10*k), f2(10*k)], k = -17 .. 18), insequence = true, axes = box, view = [-1 .. 1, -1 .. 1, -B .. B]);

f1 is the Riemann surface and f2 is the winding curve.

The animation works fine on the individual plots, i.e. when I do:

display(seq(f1(10*k), k = -17 .. 18), insequence = true, axes = box, view = [-1 .. 1, -1 .. 1, -B .. B]);


display(seq(f2(10*k), k = -17 .. 18), insequence = true, axes = box, view = [-1 .. 1, -1 .. 1, -B .. B]);

but with both f's (as in my modification)  the animation does not combine correctly the two plots and shows them separately. Is there any way I can combine the plots so that display produces a smooth animation with each rotated frame containing its winding curve?


Many thanks,


the question is as follow:

The partition does not always have to be equal intervals. Consider evaluating f(x)=x3 between 3 and 5, but splitting up the interval into a partition in which the end points of the subintervals are in a geometric progression. The common ratio r has to be chosen so that 3 is the first term and 5 is the last. Also the subintervals must be capable of getting smaller as n the number of subintervals increases. Check that the geometric series

a, ar, ar2, ar3,.....ari, .....arn =b

with r=  and suitable choices for a and b satisfies these criteria. Treating the difference between ari and ar(i-1) as the width of the subinterval and using the right hand endpoint of the subinterval, evaluate the Riemann sum to n terms for f(x)=x3. Find the limit as n tends to infinity to show that the partition does not affect the result.

here is what i have got so far, can anyone check if im doing it right? thanks




>for i from 0 to 5 do a*r^i end do; -> a list of number appear in sequence ie:3, 3.157...,3.323...3.497...etc








>evalf(sum(f(xj^*)dxj,i=1..100)) -> my value is sth like 162.4788870...

I tried to find the limit, but maple 16 freezed so i think i must have done sth seriously wrong?

<math xmlns=''><mrow><mi>b</mi><mo>&coloneq;</mo><mi>a</mi><mo>&sdot;</mo><msup><mi>r</mi><mrow><mn>10</mn></mrow></msup><mo>&#x3b;</mo><mo>&nbsp;</mo></mrow></math>

Hey all,
 I have created a nested while loop to investigate the Riemann zeta function: 

for q to 10 do
 m := 10^(-q);
  for s from 2 to 10 do
   n := 1;
   b := 1;
    while b >= m do
     b := evalf(abs(Zeta(s)-euclid(s, n)));
     n := n+1
    end do;
  print(m, s, n-1)
 end do
end do 



The attached worksheet is a wonderful introduction to the concept of obtaining the area under a curve.

You'll see how easy it is to learn how to find the limit of the sum of a series using Maple.

An interactive video tutorial that shows you how to do Riemann sums really fast is linked below:

(Ctrl+Click on the link to view the video)

Riemann Sums...

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