2473 Reputation

1 years, 275 days

1...

 > restart;
 > with(LinearAlgebra):
 > M := <100, 0, 0; 0, 90, 0;  0, 0, 90>:
 > K := < 7, -3, 0; -3, 5, -2;  0, -2, 2>*10.^4:
 > E,Q:=simplify( [Eigenvectors(K,M)], zero)[];
 (1)
 > interface(displayprecision=6):
 > Q[..,3]:=Q[..,3]/Q[1,3]: #manually; but a procedure is easy to write
 > E,Q;
 (2)
 >

BC...

You must change the BC conditions. As stated the problem cannot be solved because w(x)=0 satisfies the ODE and the BC ; so, for psi you only have the condition psi'(K/2) = psi'(-K/2) = 0, which of course is not enough (the infimum of phi is obviously 0).

int...

The second objection remains. You cannot speak about "the" integration constant.
For example, the integral of 1/(x+1) can be written as ln(x+1) + C or  ln(2*x+2) + C  or ...
Also, what is this condition when w(x)=sin(sin(x))?

Now about the new formulation. In this case the problem seems to be without solution: taking y=0 the ODE for w(x) has order 2.

Formulation...

The formulation of the problem is not clear.
1. Which are the independent variable(s)? w = w(x) and psi=psi(y)? G seems to be constant (at least in the worksheet).

2. What do you mean by

(it is senseless, being defined up to an additive constant).

3. Explain your solving method mathematically in order to follow the Maple code.

int...

Why should Maple give a warning? It's a legitimate integral but hard to compute.

Note that the argument of the hypergeometric function is not in the unit disk and it is considered by analytic continuation; also a branch point is present.

Even if we alter a bit the parameters of the hypergeometric function to [1/2, 1/2], [3/2] (in order to make an idea of its nature) we get something like:

Int(Int(arcsin((1/2)*sqrt(2)/(k-(1/5)*r))*(k-(1/5)*r)/(k+r)^4, k = 0 .. 10), r = 1 .. 2);

It's not simple but evalf works.

Did this function appeared from some computations or you simply took a more or less random complicated function with singularities?

solve...

Actually the answer of solve is correct but incomplete: a=b-5,  but for b in 2..8 only!
A similar situation occurs for
solve(arcsin(sin(x))=x,x);

x

Typical...

Yes, it's a typical conversation with you.

I know what I said and it is true, see Fubini/Tonelli. No need for screenshots.

theta...

Ypu forgot theta = Pi/6.

f...

f is OP's integrand of course.

divergence...

A brief inspection confirms divergence:
int(f, k=1..10) assuming cos(p)>-1, cos(p)<-1/2, Z>0, Z<1/10;
infinity

But the justification for extrema is eas...

But the justification for extrema is easy: the feasible region is compact and the function is differentiable there.

code...

You make a mistake in this case (if you want to use modern Maple and efficient code).

examples...

Do you have an example where delayed evaluation does not work?

BTW,

Tools > Options > Display > Typesetting level = Standard (instead of Extended)

implies that your original construct works!

But indeed, the situation is strange:

 > A:=Vector(2,(a) -> Matrix(2,2,(b,c) -> m||a||b||c)):
 > %;
 > A[1];
 (1)
 > A;
 > lprint(A);
 Vector[column](2, {1 = Matrix(2, 2, {(1, 1) = m111, (1, 2) = m112, (2, 1) = m121, (2, 2) = m122}, datatype = anything, storage = rectangular, order = Fortran_order, shape = []), 2 = Matrix(2, 2, {(1, 1) = m211, (1, 2) = m212, (2, 1) = m221, (2, 2) = m222}, datatype = anything, storage = rectangular, order = Fortran_order, shape = [])}, datatype = anything, storage = rectangular, order = Fortran_order, shape = [])
 >
 >
 > ################
 > restart;
 > f:=(a) -> 'Matrix'(2,2,(b,c) -> m||a||b||c):
 > V:=Vector(2,F);
 (2)
 > eval(V, F=f);
 > V:=Vector(2, 'f');
 (3)
 > V:=Vector(2):
 > for i to 2 do V[i]:=f(i) od: V;
 > for i to 2 do V[i]:='f'(i) od: V;
 (4)
 >

or ......

You forgot the case f(a,b)=0, g(a,b)=0, h(a,b)>0.

int...

Of course the integral can be computed. My point was that without a general algorithm implemented, many functions with simple elementary antiderivatives will need consistent help from the user to be integrated.

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