vv

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MaplePrimes Activity


These are replies submitted by vv

@ecterrab 

I think that everybody agrees that a software cannot solve all possible problems.
For a mathematician it is essential to know whether an equation (algebraic, ode, pde, ...) has a solution or not, it is a general one or not.
For example for solve, the message (and the corresponding environment variable)
Warning, solutions may have been lost

is very useful and informative.

No such message exists for PDEs, ODEs.
According to the documentation:
"When pdsolve fails, it returns NULL".
It seems that the converse is not true because you said
"All linear homogeneous equations have the (so-called: trivial) solution u = 0. What would be the purpose of returning that?"

The documentation also says:
"When a solution, but not the most general one, is obtained pdsolve expresses the result using the internal PDESolStruc function..."
It seems that this statement is not always true.

I think that a an analogue of the solve message would be be essential here.
Also, a FAIL result could be more informative that NULL, to separate the situations when the solution does not exist or it cannot be found.

I hope that these thoughts are seen as constructive.

Best regards,
V. Anisiu,
Babeş-Bolyai University, Cluj.

 

@Lali_miani 

Using the change of variables   sin(theta)=sqrt(t)    ==> Beta(p,1-p). This equals GAMMA(p)*GAMMA(1-p).
Finally use Euler's reflection formula.

I doubt very much about this statement.

@Markiyan Hirnyk 

@Rouben Rostamian  

The elliptic problems are indeed studied (existence and uniqueness) mainly for the boundary conditions you have mentioned. But theoretically we are allowed to impose any conditions and look for existence/uniqueness. I wonder whether the Maple solution is reliable as a unique solution for the first 3 conditions only.
Note also that using all 4 conditions Maple needs a much longer time.

@Markiyan Hirnyk 

Probably the best thing is to not bother with an answer in the future.

It is a pity that you don't even try to write the expression S(n) in Maple. It should be easy for a beginner.

@Markiyan Hirnyk 

I don't think that more than this can be said on your question. There are 2 options:
1. Choose the prefered order.
2. Let Maple choose the one which is easier for the program.

Edit. What is unknown is the default order. You don't need it, and the result shows the order which was used as I have explained. If you really want to know the default order you have also two options:
1. Look at the source code
2. Ask the developper.

@Markiyan Hirnyk 

For Maple ю, щ   are just 2 distinct symbols. The default order is not mentioned in the documentation but the user can choose his prefered one if he wants.

@Markiyan Hirnyk 

Here you did not provide an order for the variables, so a the program has chosen [ю, щ] for you
and has decomposed the solution set into 7 disjoint cells.

@Markiyan Hirnyk 

RegularChains is used internally by SemiAlgebraic . You need RegularChains only for special manipulations.
As I said, the order of the variables determines the type of the cells. I think that it is clear.

@Markiyan Hirnyk 

As solid people I can say that the first CPV is 3.80259236620486251646441207770...
and the second one is - infinity.

@Markiyan Hirnyk 

Interesting examples. It is possible to compute them using series; direct computations as in your answer would need a very large value for Digits and probably a prohibitive amount of time.

@Markiyan Hirnyk 

This way for a small variation like

f:=exp(-sec(t))*cos(t)/(-1/4+sin(t)^2) + 1/sin(t-Pi/6)^5;

is complicated.

@Carl Love 

The simplification of floor is very weak. Even
simplify(floor(x)) assuming x<1,x>0;
remains unevaluated.

This is not a big problem because the isolation of the singularity usually works.

Unfortunately there is a bug in series which I did not mention in the answer.

restart;

f:=exp(-sec(t))*cos(t)/(-1/4+sin(t)^2):

series(f, t=Pi/6,1);

Error, division by zero series

 

series(f, t=Pi/6,2);  # ???

series(exp(-(2/3)*3^(1/2))/(t-(1/6)*Pi)+O((t-(1/6)*Pi)^0),t = (1/6)*Pi,0)

(1)

series(f, t=Pi/6,3);  #finally OK

series(exp(-(2/3)*3^(1/2))/(t-(1/6)*Pi)+(2/3)*(-exp(-(2/3)*3^(1/2))-(1/3)*exp(-(2/3)*3^(1/2))*3^(1/2))*3^(1/2)+O((t-(1/6)*Pi)^1),t = (1/6)*Pi,1)

(2)

 

 

series(1/(sin(x)),x=0, 1);
Produces the same error.

 

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