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These are replies submitted by vv


I'm also waiting for a version where

is(Re(RootOf(Zeta(z),z))=1/2) assuming Re(z)>0, Re(z)<1;

will work. Or, even better:

is(Re(RootOf(Zeta(z),z))=1/2, option=proof) assuming Re(z)>0, Re(z)<1;


I think that a simple simplify should produce


To obtain this it seems that we must enter five commands:



@Markiyan Hirnyk 

I also suspect that FindFormula will try to be for functions what Maple's identify is for numbers.
I.e. identify will act as FindFormula applied to a constant input.
I wonder how accurate will/could it be!



@Preben Alsholm 

A related question about searching the help system

Say we are searching for inert function.
The most relevant entry would be ?value
But in the help system, 'value' appears as the 20th entry.
I also find that the (rather new) organisation of the Table of Contents
in help system is less intuitive.

What keywors should we use instead of inert function
such that "value" appears in the first place?
(except for value, of course).

For now, it seems that a google search is more effective!
For example,

maple inert function

sends us directly to

i.e. the best place!


A box is determined by xmin,ymin,zmin,xmax,ymax,zmax.  E.g.

B:=Box(-1..1,-1..1,-1..1, outward);

is the box having the faces tangent to your sphere.

@Carl Love 

The simplest is


After x=t^2 ==> int(2,t=0..1)



@Carl Love 

Mathematically, the flux is a surface integral; it reduces to a double integral. If the integral exists (i.e. the integrand is an integrable function in any sense - Riemann/improper/Lebesgue etc) then there is no math problem. 

In this case it exists. Note that many improper integrals become proper after a change of variables.

The divergence theorem was not used here; and it cannot be used. Actually Div(vv) is not integrable in any ball centered at <0,0,0>.



The  "By hand" method is the easiest for a beginner: simply copy the output of Flux(...,inert) and paste it in a new execution group. Then swap  phi = 0 .. Pi    and  theta = 0 .. 2*Pi.
(with no escaped variables problem).



That is exactly what evalf does if applied to an inert integral (see ?int/details): the integral is computed numerically even if a closed form exists.


It is easy to isolate the integral which Maple cannot compute. It is

J:=Int(f, phi = 0 .. Pi);

It does not have an elementary primitive (Risch' algorithm is implemented for such functions AFAIK - i.e. not containing algebraic parts).
Your change of variable ==>
IntegrationTools:-Change(J, cos(phi) = t, t);

This an integral of an odd function on a symmetric interval and Maple knows that it is 0.
The simpler change of variable also works (same reason):

IntegrationTools:-Change(J, phi = Pi/2 - t, t);

Notice that cos(phi)=t also works e.g. for phi = 0..Pi/2 (i.e. flux through a hemisphere)

Int(f, phi = 0 .. Pi/2):
IntegrationTools:-Change(%, cos(phi) = t, t);


But not for phi = 0..Pi/3 !!

However changing the order of integration works in these cases too, simply because the integral J vanishes (being multiplied by a zero factor).
Of course the method will also fail if theta runs in other intervals e.g. 0..Pi/6.

Best regards,







Thank you again.

But I want to mention that the LambertW solution was produced even if the library is not updated (yet).


Thank you for the competent analysis.

Could you please also answer to the following questions.

1. Why is dsolve giving in this case different answers? For example, after loading your worksheet, dsolve gave the LambertW solution, not the RootOf one.
2. Is the library contained in Physics_DEs_MathFuncs_Maple2015.15.2.mla.zip used by the whole system or only by the Physics package?


coeffs(a*x+b*y,[x,y]) =~ coeffs(d*x+e,[x,y]);

@Preben Alsholm 

Most probably yes. But solve cannot use your strategy I think.
Since dsolve gives the correct result for ODE with no IC, it is simpler to use:

res:=dsolve({ diff(y(t),t) = y(t)^2 - y(t)^3 });


You must define an F if you want to see the derivatives. Otherwise F stands fo a generic function.

If you want F to be the polynomial   a x^2 + a x + b  , you must enter

F := x -> x^2 + a*x + b;

(in Maple it is called procedure).

After that you can simplify again your expresion (click in the for loop and press ENTER) and this polynomial will be derivated.


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