@sand15 

It works for me.

Have you loaded plots?
Or, use plots:-display

@tomleslie 

For Digits:=21 the imaginary part is still present.

@asa12 

You have successfully found what? The zero matrix or something else?

@asa12 

If you are considering row additions E, then a matrix X is invariant under them (i.e. E.X = X, for each E)
iff X = 0. Probably this is not what you need.

@asa12 

You should explain what you are trying to do. Code only (which does not work correctly) does not say much.

Are you looking for these?

https://en.wikipedia.org/wiki/Elementary_matrix

@_Maxim_ 

Actually, if f and f' are continuous except for a finite number of finite jumps then f has bounded variation and so the Fourier series is always convergent at x to the midpoint (f(x+)+f(x-))/2.

tsunamiBTP
for the rectangle in the open interval (-1,1)
f:=piecewise(x < -1/2, 0, x < 1/2, 1, 1/2 < x , 0);
the sum of the series (n=infinity) is
F:=piecewise(x < -1/2, 0, x=-1/2, 1/2, x < 1/2, 1, x=1/2, 1/2, 1/2 < x , 0):

I have just explained that.

@Markiyan Hirnyk 

Actually the infinity of solutions refers to ode=0, where:

ode:=(1/2)*(diff(y(x), x))^2 - (1-ln(y(x)^2))*y(x)^2;

This should be normal because the  solve(...)  was used and it gives 2 solutions.

Ya verifies this ode:

simplify( eval(ode, y(x)=Ya) );
    0

If we restrict to solve()[1]  then we have only two solutions.

I don't see a bug here. Your ode (y' = sqrt(2-2*ln(y^2))*y, y(0)=1)  has infinitely many solutions in the interval [0,1]. Two of them are in the plot.

@acer 

evalf/int/control is using a scaling factor of 10^-280 (!?).
I would expect evalf/Int to make use of a minimal amount of symbolic computations but instead it does lots of symbolic with evala.

@Axel Vogt 

Same for me for Digits:=15; but for Digits:=10 the crash is fast (mserver.exe).

@acer 

Yes, and doing so, Q1 will be orthogonal!

@tomleslie 

But the two systems are incompatible. Compare the second equations of each.
==> r(x)^2 * T1(x) = 0 ==> r(x)=0  (because T1(x) <>0).