Download !strange-bug-int.mw

Correct computatiton for

for reasonable expressions f(x,y), g(x,y) would be very useful in double integrals.

For the moment this is not possible. Too many bugs:

int(Heaviside(1-x^2-y^2), x=-infinity..infinity, y=-infinity..infinity); #should be Pi
                           undefined
int(Heaviside(1-x^2-y^2), x=-1..1, y=-1..1); #should be Pi
                               0
int(Heaviside(y-x^2), x=-1..1, y=-1..1); #should be 4/3
                               -2

int(Heaviside(y-x^2), y=-1..1, x=-1..1); #This one is OK!
                              4/3

A string is wound symmetrically around a circular rod. The string goes exactly
4 times around the rod. The circumference of the rod is 4 cm and its length is 12 cm.
Find the length of the string.
Show all your work.

(It was presented at a meeting of the European Mathematical Society in 2001,
"Reference levels in mathematics in Europe at age16").

Can you solve it? You may want to try before seing the solution.
[I sometimes train olympiad students at my university, so I like such problems].

restart;
eq:= 2/Pi*cos(t), 2/Pi*sin(t), 3/2/Pi*t; # The equations of the helix, t in 0 .. 8*Pi:
               
p:=plots[spacecurve]([eq, t=0..8*Pi],scaling=constrained,color=red, thickness=5, axes=none):
plots:-display(plottools:-cylinder([0,0,0], 2/Pi, 12, style=surface, color=yellow),
                         p, scaling=constrained,axes=none);
 

VectorCalculus:-ArcLength(<eq>, t=0..8*Pi);

                           20

Let's look at the first loop around the rod.
If we develop the corresponding 1/4 of the cylinder, it results a rectangle  whose sides are 4 and 12/4 = 3.
The diagonal is 5 (ask Pythagora why), so the length of the string is 4*5 = 20.

In a recent post, the following inequality was proved with Maple:

(a,b,c,d >= 0).

Here is another direct proof attempt.

f:=(a+b+c+d)^2*(a*b+a*c+a*d+b*c+b*d+c*d)^2-144*(a^2+b^2+c^2+d^2)*a*b*c*d:
g:=expand(eval(f,d=1)):
s:=minimize(g, [a=0..infinity,b=0..infinity,c=0..infinity]):
length(s);   # huge
        304856
map(evalf@evalf[500],s);

map(evalf@evalf[1000],s);

So, Maple returns the expression

min(0,r1,r2,r3)

where r1,r2,r3 are huge expressions containing RootOfs. In order to evaluate them, several hundreds of digits are needed.
The solution seems to be correct, but the question is: may we (mathematically) accept it? What do you think?

eulermac(1/(n*ln(n)^2),n=2..N,1);  #Error
Error, (in SumTools:-DefiniteSum:-ClosedForm) summand is singular in the interval of summation


eulermac(1/(n*ln(n)^2+1),n=2..N,1);  #nonsense