Preben Alsholm

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18 years, 364 days

MaplePrimes Activity


These are replies submitted by Preben Alsholm

The output from sparsematrixplot makes suggests that what you ask for isn't readily available:

restart;
with(plots):
with(LinearAlgebra):

A := Matrix([[2, 1, 0, 0, 3], [0, 2, 1, 0, 0], [0, 0, 2, 1, 0], [0, 0, 0, 2, 1], [0, 0, 0, 0, 2]]);

sparsematrixplot(A, matrixview, color = "DarkGreen"); p:=%:
pd:=plottools:-getdata(p);
pd[1];
plot(pd[1,3],thickness=4);
plot(pd[1,3],style=point,symbol=solidcircle,symbolsize=50);

@vv Maybe I misunderstand you when you say " Unfortunately, Maple cannot compute it."
In Maple 2021.2 I get (using your suggested change of variables):

 

A:=Int(2/(1-x^p)^(1/p), x=0..1);
IntegrationTools:-Change(A,x^p=t) assuming p>1;
value(%) assuming p>1;

The answer given is 2*Beta(1/p, -1/p + 1)/p, i.e. the same as the result by cook.

So what is the correct answer?

@escorpsy I don't see any way of getting a symbolic answer with parameters. But you can certainly find exact answers with concrete values for the parameters. As an example:

M := Int((x*(1 - x)*(-I*epsilon + p^2) + m^2)^((d - 4)/2), x = 0 .. 1);
value(eval(M,[epsilon=1/100,d=3,m=1,p=1])); # Exact integration
evalc(%)=evalf(%);
evalf(eval(M,[epsilon=1/100,d=3,m=1,p=1])); # Numerical integration


 

The use of D is unfortunate since it means the differentiation operator in Maple:

D(sin); # Answer cos

So use another name, e.g. D1.
Secondly, by default the imaginary unit is  'I', not 'i'.

@acer Just an observation: So the following two versions of dividing are not quite the same:

1/4/(x-2);
1/(4*(x-2));

@tomleslie In order to be fair you must divide by 2000 000 in ans2:

ans1:=(654321.987*123456.789)/2;
ans2:=(654321987*123456789)/2000000;

@nm Normally you cannot put a number first followed by a letter; that would produre a syntax error. Example 1w would produce an error.
There are exceptions:  1d and  1e are examples.

@Jaime_mc2 This works with Digits:=50 in Maple 2021.2
 

evalf(Int(abs(u(x) - P__2), x = -1 .. 1,method= _Gquad));

The result is 0.00062767200590347019035308761635491656933673486658951.
Inspired by mmcdara I added simplify like this:
 

r := evalf(allvalues(RootOf(simplify(ChebyshevT(5, x)), x)));

and this time the result was
0.00062767200590347019035308761635491656933673486658622

not a huge difference there.

But the latter version, but not the former, worked in Maple 12 too and with that same result.

@nm Have a look at this simple example:

restart;
f:=x->a*x;
simplify(sqrt(f(a)),assume=[a<0]);
simplify(sqrt(f(a))) assuming a<0;

In the first case sqrt(f(a)) is evaluated to sqrt(a^2) before the assumption kicks in, thus the result -a.

In the second case the expression sqrt(f(a)) is not evaluated until the assumption a<0 is made. But the "version" of the 'a'  that is seen by `assuming` is therefore only the argument to f, not the 'a' appearing in the in the definition of f.
My point here has just been to say that the two versions don't work quite the same.
###
You can see it here from the printout from debug(`assuming`:
 

restart;
f:=x->a*x;
debug(`assuming`);
simplify(sqrt(f(a)),assume=[a<0]);
simplify(sqrt(f(a))) assuming a<0;

 

Obviously the syntax for Int is wrong.
Replace Int with F then both simplifications work.

@oggsait  Yes I was using the recent version of Maple.

But in Maple 18 (which you are using) I get the plot you show.

But as Carl Love points out you should use a finer grid. Besides that it helps to set gridrefine to 2 as in this code:
 

plots:-display(seq(plots:-implicitplot(eval(EQ2,params1 union {C=k/200}),u=-1..1,z=-1..1,grid=[50,50],gridrefine=2),k=-2..2));

If you look at the help for implicitplot you will find other options. Actually it looks to me that this version gives a better result than the one above:
 

plots:-display(seq(plots:-implicitplot(eval(EQ2,params1 union {C=k/200}),u=-1..1,z=-1..1,grid=[50,50],gridrefine=1,crossingrefine=2),k=-2..2));

 

@jennierubyjane Merry Christmas to you too!

@jennierubyjane You can say considerably more.
A few hints at what is going on:
Notice in my code the use of values for initmesh and maxmesh.
So from the interval, in your case 0..10, is chosen an initial mesh (default consisting of 8 points). I used 512 points.
Notice also that an approximate solution can be given as an option (I didn't use that here). If not given the procedure finds an approximate solution itself based only on the boundary conditions (!).
Here is an example of getting some information about the solving process:

restart;
#infolevel[`dsolve/numeric/BVP`]:=2:  # Also try 3 or 4:
infolevel[`numeric/bvp`]:=5: # Another possibility
ode:= diff(y(x),x,x)+sin(y(x)) +y(x) = 1;
bcs:=y(0)=1,D(y)(5)=0;
res:=dsolve({ode,bcs},numeric);
plots:-odeplot(res,[x,y(x)]);

Deliberately I used a much simpler example than yours, but the process is the same.
Notice that `dsolve/numeric/bvp` is different from `dsolve/numeric/BVP`.
The former calls the latter, i.e. `dsolve/numeric/BVP` is used by `dsolve/numeric/bvp`.
You can see that here in line 440.
showstat(`dsolve/numeric/bvp`);

@jennierubyjane When dsolve/numeric is given a boundary value problem it is using the procedure `dsolve/numeric/bvp`

For that procedure see the help page ?dsolve,numeric,bvp.

RKF45 is made for initial value problems.
If one insists on somehow making use of that method (or other intial value procedures) shooting has to be done.
That means starting at one end (e.g. eta = 0) with all the required initial values, which implies "guessing" some.

That can be done. I have no reason to believe that that is going to give a better result.
The "guessing" can be implemented by using the parameters option to dsolve where the parameters will be the unknown initial values. Then fsolve will be involved after that.

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