vv

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Testzero:=testeq:
M^(-1);

Error, (in rtable/Power) singular matrix

 

It is an interesting and important problem.
Unfortunately the post is not easy to read because you decided to keep the notations (and the example) from the cited question.
This reminds me of a joke about Cauchy: when dealing with three similar quantities such as a,b,c they were denoted: NO2, ∑t1,2   and ∏7p,q .

 

@digerdiga 

'''sol(x)'''  cannot work because the first eval evaluates sol(X)  giving infinity  ( = min([]) ).
In '''sol'''(x), the outer '...' is for the evaluation of the argument, the second is for eval itself and the third is for the final result which must be unevaluated, to be used by eval(%, X = 1).

 

@acer 

hmm is simply wrong (bug).
Actually, appending assuming alpha>0, beta>0, hmm remains unevaluated.

@Markiyan Hirnyk 
But you have considered alpha=beta (=1) and the integral is divergent in this case.

@_Maxim_ 

But isn't this normal?
The result of 
Digits:=30: evalf[10](expr);
is and should be the same as:
evalf[10](expr): evalf[30](%);

 

@Markiyan Hirnyk 

Why? It can be computed by hand using the convolution formula.

@_Maxim_ 

It works, but by chance, for Digits=30 the coefficient of the 1/s term is 0.0  instead of  x.y*10^(-...).

evalf[10]( evalf[40](series(f(1.*s), s = 0, 1)) );
       
(1.*10^(-40)*I)/s+O(s)

 

@Ian Jones 

You must have typos; it is not validated numerically.

@Carl Love 

I mean this:

restart;

N:=5*10^6:

f:=proc()
local k,S:=0.;
for k to N do S:=S+1/evalf(k) od:
end:

f(); #Digits=10, constant

16.00206064

(1)

evalf[10](evalf[30](f()));

16.00216424

(2)

evalf[30](f());evalf[10](%);

16.0021642352999040059818866772

 

16.00216424

(3)

restart;

N:=5*10^6:

evalf[10](add(1/evalf(k),k=1..N)); #Digits increased by add

16.00216423

(4)

 


 

Download digits.mw

@Carl Love 

But in "most" situations evalf[d1](e)  is enough because Digits is automatically increased by evalf if necessary.

1. The third argument of Cubes must be an m-dimensional list. So, use:
Cubes(3, 3, [a, b, c], h);

2. For an enumeration of N^m  see ?vectoint, ?inttovec

@_Maxim_ 

Nice analysis!

Have you used Maple for the convolution, or you did it by hand?

Cubes:=proc(m::posint,n::posint,a::list,h:=1) # for the cube [a, a+~h] in R^m
local k,u,T,C;
T:=combinat:-cartprod([[seq(k/n*h, k=0..n-1)] $ m ]):
C:=('T[nextvalue]()')$(n^m):
seq( [a+u, a+u+~h/n], u=[C])
end:

Cubes(2,2,[a,b],h);

This is probably because you like the symmetry.
But  [x=3,y=2], [x=2, y=10] etc, are also correct.

@Christian Wolinski 

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