Markiyan Hirnyk

Markiyan Hirnyk

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11 years, 61 days

MaplePrimes Activity


These are answers submitted by Markiyan Hirnyk

Look here to this end.

This can be done as follows. We see a great potential of Maple.


 

restart; J := Int(exp(-(1/2)*v/k)*v^3*exp((1/2)*v/m)*Ei(1, -(1000*I)*v+(1/2)*v/m), v = 0 .. infinity)``

Int(exp(-(1/2)*v/k)*v^3*exp((1/2)*v/m)*Ei(1, -(1000*I)*v+(1/2)*v/m), v = 0 .. infinity)

(1)

L := `assuming`([IntegrationTools:-Change(J, x = v/k, [x])], [k > 0, m > 0])

Int(exp((1/2)*x*(-m+k)/m)*x^3*k^4*Ei(1, -((1/2)*I)*x*k*(2000*m+I)/m), x = 0 .. infinity)

(2)

N := `assuming`([value(L)], [k > 0, m > 0])

16*(48000000000*ln(m*(I+2000*k)/(k*(2000*m+I)))*k^3*m^3-36000*ln(m*(I+2000*k)/(k*(2000*m+I)))*k*m^3+(72000000*I)*ln(m*(I+2000*k)/(k*(2000*m+I)))*k^2*m^3+6000*k^3*m-36000*k^2*m^2+30000*k*m^3+(24000000*I)*k^3*m^2-(24000000*I)*k^2*m^3-(6*I)*ln(m*(I+2000*k)/(k*(2000*m+I)))*m^3-(2*I)*k^3+(9*I)*k^2*m-(18*I)*k*m^2+(11*I)*m^3)*k^4*m/((I+2000*k)^3*(-m+k)^4)

(3)

limit(eval(N, [k = 1]), m = 1)

383999424000024/256000256000096000016000001-(767999808000/256000256000096000016000001)*I

(4)

evalf(%)

0.1499996250e-11-0.2999996250e-14*I

(5)

Digits := 15; -1; int(exp(-(1/2)*v)*v^3*exp((1/2)*v)*Ei(1, -(1000*I)*v+(1/2)*v), v = 0 .. infinity, numeric)

0.1499991612e-11-0.2992397e-14*I

(6)

``


 

Download parametric_integral.mw

 

It is possible, making use of the Physics package. I start from the cites

  • The Inverse command, when applied to an object, represents the object's (noncommutative) multiplicative inverse; that is, Inverse(Z) * Z = Z * Inverse(Z) = 1, where * herein represents the Physics[*] product, whose commutativity depends on the operands (see also type, commutative).
     
  • When the Physics package is loaded, a more general product operator, represented by the same symbol *, is loaded. Unlike the default Maple * operator (herein called the commutative product operator), the * operator loaded with Physics does not assume commutativity of its operands and instead automatically operates like a commutative, anticommutative, or noncommutative product, depending on the corresponding commutation properties of its operands.
  • As with the general Maple simplifier, simplify, when you call the Physics[Simplify] command with no extra arguments, all of the simplifications are attempted. When you call it with extra arguments specifying different simplifications, any of algebrarules, bracketrules, indices, noncommutativeproducts, sum, and int, only the specified simplifications are attempted.
     

In view of the above the following code does the job.

>with(Physics):Setup(mathematicalnotation = true):Setup(noncommutativeprefix={a}):Simplify(a2*Inverse(a2*Inverse(a1)*a2)*a2);
                               a1

Simplify.mw 

Edit. The description of the Simplify command is added.

Confirm this Maple defect. I use Task Manager to cancel Java(TM)... in Processes to this end.

Here is an example.


 

with(PDEtools):``

PDE := r*cos(theta)+r*cos(phi)-1 = 0:

 

 tr := {phi = arctan(y, x), r = sqrt(x^2+y^2+z^2), theta = arccos(z/sqrt(x^2+y^2+z^2))}:

PDE1 := PDEtools:-dchange(tr, PDE);

z+(x^2+y^2+z^2)^(1/2)*x/(x^2+y^2)^(1/2)-1 = 0

(1)
plots:-implicitplot3d(PDE, r = 0 .. 5, phi = 0 .. 2*Pi, theta = 0 .. Pi, coords = spherical, grid = [20, 20, 20], style = surface, shading = z)

 

plots:-implicitplot3d(PDE, r = 0 .. 5, phi = 0 .. 2*Pi, theta = 0 .. Pi, coords = spherical, grid = [20, 20, 20], style = surface, shading = z);

 
plots:-implicitplot3d(PDE1, x = -5 .. 5, y = -5 .. 5, z = -5 .. 5, style = surface, shading = z);

 

plots:-implicitplot3d(PDE1, x = -5 .. 5, y = -5 .. 5, z = -5 .. 5, style = surface, shading = z)

 

``


 

Download dchange.mw

For example, the output of

restart; printlevel := 10: Student[MultivariateCalculus]:-LagrangeMultipliers(x*y, [x^2+y^2-1], [x, y]);

contains, in particular,

{x^2+y^2-1, -2*y*lambda[1]+x = 0, -2*x*lambda[1]+y = 0}

LM1.mw

because s and N may be  complex numbers so branch cuts should be taken into consideration. If we simplify matters, then Maple does the job

simplify((1/s^2)^N, symbolic);
                             (-2 N)
                            s      

 

if I correctly understand you. The matter is the same plot is demonstrated from different viewing angles. Here is a simple example 


 

plot3d(x^2+2*y^2, orientation = [20, 40, 70]);plot3d(x^2+2*y^2, orientation = [20, 40, 70])

 

plot3d(x^2+2*y^2, orientation = [20, 70, 40]);plot3d(x^2+2*y^2, orientation = [20, 70, 40])

 

``


 

Download orientation.mw

This can be done in Maple 2016 as follows:


 

with(NumberTheory):``

q := NumberTheory:-RepeatingDecimal(-37/31)

_m400140804

(1)

``


 

Download recurring_decimal.mw

-


 

Try


NULL

NULL

NULL

NULL

NULL

Digits := 15; -1; Fct := (1/8)*sqrt(3)*(-4*exp((2*I)*beta)*beta^2*Ei(1, (2*I)*beta)+beta^2*(-(7/4*I)*beta+(5/8)*beta^2-1)*exp(I*beta)*Ei(1, I*beta)+(5/8*I)*beta^3-(5/2*I)*beta+(9/8)*beta^2+9/4); -1; with(plots); -1; semilogplot(Re(Fct), beta = 10^(-3) .. 10^4, 0 .. .6)

 

NULL

NULL

NULL

 

 

NULL


Download digits.mw

The apply rule command was likely introduced in Maple V R5 (see screen05.09.16.docx).

As far as I understand your question, you are interested only in real solutions. The answer is complicated. The parametric plane (m,u) can be splitted into 24 domains in each of those the number of the real solutions of

F(z):=m*z^4-4*m*z^3+(3*m+3)*z^2-6*u in z is constant. This can be done as follows.



with(RootFinding[Parametric]):

sys := [m*z^4-4*m*z^3+(3*m+3)*z^2-6*u = 0]:

s := CellDecomposition(sys, [z]);

[[[Equations,=, [m z^4-4 m z^3+(3 m+3) z^2-6 u]],[Inequalities,=, []],[Filter,=, 0<>1],[Variables,=, [z]],[Parameters,=, [m,u]],[DiscriminantVariety,=, [[m],[u],[3 m^4-48 m^3 u-64 m^2 u^2+192 m^2 u-18 m^2-48 m u-24 m-9]]],[ProjectionPolynomials,=, [[u,2 u-9,16 u-9,2 u-1],[m,3 m^4-48 m^3 u-64 m^2 u^2+192 m^2 u-18 m^2-48 m u-24 m-9]]],[SamplePoints,=, [[m=-21,u=-1],[m=-10,u=-1],[m=2,u=-1],[m=5,u=-1],[m=-1,u=1/4],[m=-117438644421/1099511627776,u=1/4],[m=1,u=1/4],[m=3,u=1/4],[m=-1,u=17/32],[m=-41246484769/549755813888,u=17/32],[m=1,u=17/32],[m=147009845109/68719476736,u=17/32],[m=3,u=17/32],[m=5,u=17/32],[m=-1,u=3],[m=-14774057851/549755813888,u=3],[m=24,u=3],[m=49,u=3],[m=-3,u=5],[m=-1,u=5],[m=-1054275706337/4398046511104,u=5],[m=-5162024001/274877906944,u=5],[m=41,u=5],[m=83,u=5]]]]

(1)

RootFinding:-Parametric:-CellPlot(s, 'samplepoints')

 

NumberOfSolutions(s)

[[1, 4], [2, 2], [3, 0], [4, 2], [5, 2], [6, 4], [7, 2], [8, 4], [9, 2], [10, 4], [11, 2], [12, 4], [13, 2], [14, 4], [15, 2], [16, 4], [17, 2], [18, 4], [19, 2], [20, 0], [21, 2], [22, 4], [23, 2], [24, 4]]

(2)

``


See RootFinding,Parametric for more info.

Download number_of_real_solutions.mw

Similar questions were asked and answered a mountain.

You wrote "Only difference is that I wrote  (a+b)/sqrt(b^2-a^2) in one, and  sqrt(a+b)/sqrt(b-a) in the other. But these two expressions are the same".

This statement is not true. The parameters a and b are treated as complex by default. Let us consider

evalc(eval((a+b)/sqrt(-a^2+b^2) = sqrt(a+b)/sqrt(b-a), [a = -I, b = -3*I]))

-(1/2)*sqrt(8) = sqrt(2)

evalf(%);
-1.414213562 = 1.414213562

substitution.mw

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