Jacques, May I suggest that kronprod in Robert Israel's Maple Advisor Database is more versatile - it handles matrices, Matrices, vectors and Vectors. Moreover matrices need not be square and it seems to be very nearly as fast as your code. J. Tarr

Jacques, May I suggest that kronprod in Robert Israel's Maple Advisor Database is more versatile - it handles matrices, Matrices, vectors and Vectors. Moreover matrices need not be square and it seems to be very nearly as fast as your code. J. Tarr

According to the help page, the display command takes a set, list or array as its argument. But it seems to work in cases like this without creating a set or list. J. Tarr

According to the help page, the display command takes a set, list or array as its argument. But it seems to work in cases like this without creating a set or list. J. Tarr

Can anyone show that the limit of this integral is zero as n->infinity? J. Tarr

Can anyone show that the limit of this integral is zero as n->infinity? J. Tarr

Be sure to check the solutions by evaluating the original equations with each solution - please see ?eval - or by substituting the solutions into the original equations - please see ?subs. At first sight, I don't think all the solutions found above are valid. Hope this helps. J. Tarr

Be sure to check the solutions by evaluating the original equations with each solution - please see ?eval - or by substituting the solutions into the original equations - please see ?subs. At first sight, I don't think all the solutions found above are valid. Hope this helps. J. Tarr

George, Your worksheet defines something whose middle component is a[3]*b[1] - a[3]*b[1] That's identically zero. Perhaps you intended a[3]*b[1] - a[1]*b[3] ? If so, you intended to define a cross product that is identical with that in LinearAlgebra. J. Tarr

George, Your worksheet defines something whose middle component is a[3]*b[1] - a[3]*b[1] That's identically zero. Perhaps you intended a[3]*b[1] - a[1]*b[3] ? If so, you intended to define a cross product that is identical with that in LinearAlgebra. J. Tarr

If you know that the values of the variables fall within a certain range, you can insert this as option in fsolve - please see ?fsolve,details. Something like this: restart: Digits:=15: eq[1]:=add((-1)^(i+1)*cos(omega*t[i]),i=1..5)=0: eq[2]:=add((-1)^(i+1)*sin(omega*t[i]),i=1..5)=0: eq[3]:=add((-1)^(i+1)*cos(r*omega*t[i]),i=1..5)=0: eq[4]:=add((-1)^(i+1)*sin(r*omega*t[i]),i=1..5)=0: eq[5]:=t[1]=0: eq[6]:=t[4]-2*t[3]+t[2]=0: eq[7]:=t[5]-2*t[3]=0: s1:=fsolve({seq(eq[i],i=1..7)},{seq(t[i],i=1..5),r,omega},0..4): Solutions:= [seq(t[i]=eval(t[i],s1),i=1..5),r=eval(r,s1),omega=eval(omega,s1)]; check_Solutions := seq(eval(eq[i],s1),i=1..7); Hope this helps, J. Tarr

Maple 10.06 should not confuse omega with the built-in function Wrightomega, unless it has been given as an alias for the latter. AFAIK there were no omega functions in previous versions of Maple. Hope this helps, J. Tarr

George, Your results were only slightly out in the three drill-holes case. Finding the volume common to intersecting cylinders was a problem known to the ancients – Archimedes produced the solution for two cylinders intersecting perpendicularly! The volume common to two, or three, right circular cylinders of equal radius intersecting at right angles is called the Steinmetz solid. If r is the radius of all the cylinders, the common volumes are: 16*(r^3)/3 for two cylinders, and (16 – sqrt(128))*(r^3) for three cylinders. These snippets of information were obtained here Hope this helps. J. Tarr

Following Jean-Marc's steps, the total material removed by three holes mutually at right angles through the cube is: Pi*b^2/2*(3*a/2-b/3) But that assumes someone drilling the holes much more accurately than I can. (My efforts would result in at least 3*a*b^2*Pi/4 material removed and about 50 ml of blood!) J. Tarr

You could get the result you want with one more step: restart; e1 := 2*sqrt(z*(2*r - z))*r -2*sqrt(z*(2*r - z))*z - 2*r^2*arcsin( (-r + z)/r); e2 := subs((z*(2*r - z))=g,e1); e3 := collect(e2,g^(1/2)); e4 := subs(g=(z*(2*r - z)),e3); Hope this is what you want. J. Tarr