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Controlling Variable Distribution in Polynomial Fo...

Asked by: salim-barzani 1555
polynomial expand pde
April 28 2025
2  5

In this work, I do not intend to expand all the variables across the monomials. Instead, I want to restrict the distribution to only the variables x,y,z,tx, y, z, tx,y,z,t, possibly raising them to appropriate powers as needed, until I obtain the desired solution and satisfy the conditions of my PDE tests. However, I am uncertain whether "monomial" is the correct term to use here.

S1.mw

trail-1.mw

why such different result from integrate when pull...

Asked by: nm 11353
integration int indefinite-integration
April 27 2025
2  3

These two expressions are the same, just pulled minus sign out

But look what happens when integrating them. the anti derivative of one is much more complicated than the other and contains complex numbers and logs. And no matter what I tried, I could not convert the complicated one to look same as the simpler result. Also could not verify the complicated one by back differentiating.

> 

integrand_1:=x^2*(-arctan(x) + x)*exp(-arctan(x) + x)/(x^2 + 1);

x^2*(-arctan(x)+x)*exp(-arctan(x)+x)/(x^2+1)

> 

integrand_2:=evala(integrand_1);

-x^2*(arctan(x)-x)*exp(-arctan(x)+x)/(x^2+1)

> 

simplify(integrand_1 - integrand_2)

0

> 

anti_1:=int(integrand_1,x);

(-arctan(x)+x)*exp(-arctan(x)+x)-exp(-arctan(x)+x)

> 

anti_2:=int(integrand_2,x);

-(1-x+((1/2)*I)*ln(1-I*x)-((1/2)*I)*ln(1+x*I))*(1-I*x)^(-(1/2)*I)*(1+x*I)^((1/2)*I)*exp(x)

> 

simplify(diff(anti_1,x)-integrand_1);

0

> 

simplify(diff(anti_2,x)-integrand_2);

Error, (in simpl/simpl/ReIm/sum) too many levels of recursion

> 

simplify(anti_1 - anti_2)

Error, (in simpl/simpl/ReIm/sum) too many levels of recursion

> 

simplify(anti_2);

(1/2)*(I*ln(1+x*I)-I*ln(1-I*x)+2*x-2)*(1-I*x)^(-(1/2)*I)*(1+x*I)^((1/2)*I)*exp(x)

> 

simplify(anti_2,ln);

(1/2)*(I*ln(1+x*I)-I*ln(1-I*x)+2*x-2)*(1-I*x)^(-(1/2)*I)*(1+x*I)^((1/2)*I)*exp(x)

 

 

Download int_strange_result_april_27_2025.mw

I would have expected same anti derivative to show.  To check, I used another software, and that one gave same anti-derivative for both integrands.

The questions I have: Why Maple gives such different result for same integrand? And how could one convert the one with the logs and complex numbers to the first one?

Maple 2025

Compute integral exactly ...

Asked by: Zeineb 540
integration
April 27 2025
0  0

Dear all 
I have a double integral, i want to compute this integral and verify if the pproposed solution verify the proposed equation or not. 
I can modify the right hand side of my equation or the exact solution, so that my equation has an exact solution with simple form of right hand side. 

exact_solution.mw

Thank you for your help 

Plotting contour integral diagrams...

Asked by: Paras31 245
integration contour plotting
April 27 2025
1  7

 I am writing notes on complex analysis, I need to use figures of contour paths to integrate on them, i want to create something like this

I tried to plot the contour for 
\oint_{|z|=2} \frac{1}{z^2+1}\,dz
I need to have connecting lines all around because the poles can not be isolated

> 

with(plots); circle1 := plot([2*cos(t), 2*sin(t), t = 0 .. 2*Pi], color = blue, thickness = 2); circle2 := plot([(1/2)*cos(t), 1+(1/2)*sin(t), t = 0 .. 2*Pi], color = "Green", thickness = 2); circle3 := plot([(1/2)*cos(t), -1+(1/2)*sin(t), t = 0 .. 2*Pi], color = "Red", thickness = 2); sing1 := plottools[disk]([0, 1], 0.2e-1, color = white); sing2 := plottools[disk]([0, -1], 0.2e-1, color = white); label1 := textplot([.1, 1.1, "z = i"], font = [Arial, Bold, 12]); label2 := textplot([.1, -1.1, "z = -i"], font = [Arial, Bold, 12])
> 

display(circle1, circle2, circle3, sing1, sing2, label1, label2, scaling = constrained, labels = ["Re", "Im"])

 
 
> 

restart; f := proc (z) options operator, arrow; 1/(z^2+1) end proc; z := 2*exp(I*t); dz := diff(z, t); integrand := f(z)*dz; simplify(integrand); value(Int(integrand, t = 0 .. 2*Pi))

0

 (1)

Download CIF.mw

about updating the Physics Package...

Asked by: Jean-Michel 40
update physics
April 27 2025
0  2

Hello all,

After updating the Physics package I have this error :

Physics:-Version();
The "Physics Updates" version "1862" is installed in the

   directory C:\Users\jm\maple\toolbox\2025\Physics Updates but

   is not active. The active version of Physics is within the

   library C:/Users/jm/maple/toolbox/2025/Physics Updates/lib\Ph\

  ysics Updates.maple.

What am I supposed to do next?

Thanks a lot and kind regards to all,

Jean-Michel

bug report. Maple 2025 server crash on call to ode...

Asked by: nm 11353
ode differential-equations crash
April 27 2025
2  1

FYI;

 

You might have to try the command more than one time to see the above crash. Here is the worksheet

> 

restart;
> 

interface(version);

`Standard Worksheet Interface, Maple 2025.0, Linux, March 24 2025 Build ID 1909157`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1862 and is the same as the version installed in this computer, created 2025, April 25, 10:33 hours Pacific Time.`

> 

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 13 and is the same as the version installed in this computer, created April 22, 2025, 15:14 hours Eastern Time.`

> 

restart;
> 

ode:=x^2-2*x*y(x)+5*y(x)^2 = (x^2+2*x*y(x)+y(x)^2)*diff(y(x),x);

x^2-2*x*y(x)+5*y(x)^2 = (x^2+2*x*y(x)+y(x)^2)*(diff(y(x), x))

> 

sol:=y(x) = (-1/2*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))^2+3*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))-6+2*(exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))^2-6*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))+9)^(1/2))/(1/2*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6))^2-3*exp(RootOf(-exp(_Z)^2*ln(x*(exp(_Z)-2))+2*_C7*exp(_Z)^2+_Z*exp(_Z)^2+4*exp(_Z)*ln(x*(exp(_Z)-2))-8*_C7*exp(_Z)-4*exp(_Z)*_Z-2*exp(_Z)-4*ln(x*(exp(_Z)-2))+8*_C7+4*_Z+6)))*x:
> 

odetest(sol,ode);
> 

 

Download crash_maple_2025_april_27_2025.mw

Hopefully a fix could be found for this.

Initialization or Library Archive...

Asked by: Christopher2222 6030
April 27 2025
0  2

When generating a file to update parts of maple (for example constants) is it best to put it in an initialization file or make a library archive .mla?

ScientificConstants names ...

Asked by: Christopher2222 6030
subscript scientificconstants rtfm
April 27 2025
0  5

I fail to see the logic of using short form symbols for the scientific constants and then not being able to use that short form.  One manually has to equate the two as I show below.  Anyone see a reason not to do that in the internal programming?  Just wondering. 

with(ScientificConstants):

GetConstant(M__Sun)             

One has to use the names associated with those short forms described by the command

GetConstant(mass_of_Sun)

I would much rather like to use MSun , so what one has to do is manually equate them

MSun:=mass_of_Sun:

GetConstant(MSun)

Separating solutions based on PDE satisfaction...

Asked by: salim-barzani 1555
pde differential-equations select
April 27 2025
2  6

I have a list of candidate solutions. Some of them satisfy my PDE test (i.e., they make the PDE equal to zero), while others do not. How can I separate the solutions that satisfy the PDE from those that do not?

Trail-pdetest.mw

how do i solve a vector question...

Asked by: Hidayat 5
April 27 2025
1  2

ABC is an equilateral triangle of side 3 units. The points P, Q lie on BC, CA re-
spectively and are such that AQ = CP = 2units. If the point R lies on AB produced

so that BR = 1unit, prove that P, Q, R are collinear.

Modifying an Algorithm for Multiple Variables...

Asked by: salim-barzani 1555
differentiation duplicate-question
April 27 2025
5  9

How to modify the ND procedure to handle derivatives with respect to more than three independent variables for higher-dimensional PDEs, it is work for [x,t] i want  it work for [x,y,z,t] , 

> 

restart
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

 (1)
> 

alias(F=F(x, t), G=G(x, t))

F, G

 (2)
> 

with(PDEtools):
undeclare(prime):

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

 (3)
> 

ND := proc(F, G, U)
  local v, w, f, g, a:
  v := op(F):
  if v[1] in U then w := -v[1] else w := v[1] end if:
  if v[2] in U then w := w, -v[2] else w := w, v[2] end if:
  f := op(0, F):
  g := op(0, G):
  a := diff(f(w)*g(v), U);
  convert(subs([w]=~[v], a), diff)
end proc:
> 

ND(F, G, [x]);
ND(F, G, [t]);

-(diff(F, x))*G+F*(diff(G, x))

  

-(diff(F, t))*G+F*(diff(G, t))

 (4)
> 

ND(F, F, [x]);
ND(F, F, [x, x]);

0

  

2*F*(diff(diff(F, x), x))-2*(diff(F, x))^2

 (5)
> 

ND(F, G, [x$3]);

-(diff(diff(diff(F, x), x), x))*G+3*(diff(diff(F, x), x))*(diff(G, x))-3*(diff(F, x))*(diff(diff(G, x), x))+F*(diff(diff(diff(G, x), x), x))

 (6)
> 

ND(F, F, [x$3, t]);

2*F*(diff(diff(diff(diff(F, t), x), x), x))-2*(diff(diff(diff(F, x), x), x))*(diff(F, t))-6*(diff(diff(diff(F, t), x), x))*(diff(F, x))+6*(diff(diff(F, x), x))*(diff(diff(F, t), x))

 (7)
> 

NULL
> 

NULL
> 

#if i collect P1+P1+...+P7 it must get equation 26 in paper so i want define the up proc to open but is not for (3+1) dimesnion,
> 

P1 := 9*ND(F, F, [x, t])

18*F*(diff(diff(F, t), x))-18*(diff(F, x))*(diff(F, t))

 (8)
> 

NULL
> 

P2 := -5*ND(F, F, [`$`(x, 3), y])

0

 (9)
> 

P3 := ND(F, F, [`$`(x, 6)])

2*F*(diff(diff(diff(diff(diff(diff(F, x), x), x), x), x), x))-12*(diff(diff(diff(diff(diff(F, x), x), x), x), x))*(diff(F, x))+30*(diff(diff(diff(diff(F, x), x), x), x))*(diff(diff(F, x), x))-20*(diff(diff(diff(F, x), x), x))^2

 (10)
> 

P4 := -5*ND(F, F, [`$`(y, 2)])

0

 (11)
> 

P5 := alpha*ND(F, F, [`$`(x, 2)])

alpha*(2*F*(diff(diff(F, x), x))-2*(diff(F, x))^2)

 (12)
> 

P6 := beta*ND(F, F, [x, y])

0

 (13)
> 

P7 := gamma*ND(F, F, [x, z])

0

 (14)

Download define.mw

Two values associated with their respective (factored...

by: jwaltos 10 Maple
April 27 2025
0  0

L. B. Johnson once said: “I may not know much, but I know chicken sh#t from chicken salad.” And the same goes for mathematical software,  Maple is a good chicken salad.

These two numbers can be used to factor two previously factored RSA Challenge numbers:

4492372899485266683229032112393311539091890452003150017722229708882931615085372733373343061967162688807713966063216561545461119244883848142568154156987418243095913219694108294875951005535802313105656690937568115044857082104972025


252470349467980886727391223577367145704558455488893488785280129051457755334632136343591527439288590916228345021218177497619016135424030834870037054353008183582467637830682000623550252325843511739294850378626625818394419012275747807

I'm curious to see if anyone can identify the numbers and do the factorization. If no one has been able to solve this, I will post the solution context at a future date. There exists some very interesting mathematics behind this question that goes beyond a simple recreational diversion.
 

What is wrong with my logic in interpreting this s...

Asked by: zenterix 405
plot evalf floating-point
April 26 2025
0  1

I am very confused by the y-value of the rightmost point on the plot below.

restart

I'd like to find the values of x for which x^2/(10^(-8)-x) = 5*10^(-3).

So I ask Maple to solve this equation.

evalf(solve(x^2/(10^(-8)-x) = 5*10^(-3)))

-0.5000010000e-2, 0.10000e-7

 (1)

Do these solutions work?

eval(x^2/(10^(-8)-x), x = 1.0000*10^(-8)) = Float(infinity) 

eval(x^2/(10^(-8)-x), x = -0.5000010000e-2) = 0.5000000000e-2 NULL

Suppose I define the function

f := proc (x) options operator, arrow; x^2/(1/100000000-x) end proc = proc (x) options operator, arrow; x^2/(1/100000000-x) end proc 

f(10^(-8))

Error, (in f) numeric exception: division by zero

  

f(.999999*10^(-8)) = 0.9999980000e-2NULL

f(.99999*10^(-8)) = 0.9999800001e-3NULL

Now, the function seems continuous between these two points

plot(f, .99999*10^(-8) .. .999999*10^(-8))

  

It is late, and perhaps I am just tired and not seeing things clearly. I expected the topmost point on the right to have a y-value of 0.00999998, ie almost 0.01.

I expected that the bottom leftmost point to be 0.0009999800001, ie almost 0.001, and it is.

And I thus expected to show that there must be some x for which we have f(x)=0.005, which if I am not mistaken is between the two numbers. After all, 0.999998e-2-0.5e-2 = 0.499998e-2NULL

0.5e-2-0.9999800001e-3 = 0.4000020000e-2NULL

what am i missing here?


Download plotq.mw

How to solve this chemical equilibrium equation wi...

Asked by: zenterix 405
solve floating-point cubic
April 26 2025
0  2

I have a question about a calculation I was just trying to do. For some context, I am trying to calculate the pH of a solution of a weak acid in water. When the concentration of the weak acid is low enough, we need to consider the effect of ionization of the water itself (ie, autoprotolysis of water), since the order of magnitude of this ionization is the same as the order of magnitude of the ions due to the weak acid in this case of very low concentration.

There is a specific formula that can be used for this, which is shown below.

K_a is a constant that depends on the weak acid being considered, and K_w is a constant as well (for the autoprotolysis of water). I am interested in solving for the concentration of hydronium, given an initial concentration of the weak acid [HA]_initial.

I'd like to solve the equation for different values of [HA]_initial.

In the calculations below, I am trying to solve for [H__3*LinearAlgebra:-Transpose(O)] in the expression

K__a = ([H__3*LinearAlgebra:-Transpose(O)])([H__3*LinearAlgebra:-Transpose(O)]-K__w/[H__3*LinearAlgebra:-Transpose(O)])/(HA__initial+[-H__3*LinearAlgebra:-Transpose(O)]+K__w/[H__3*LinearAlgebra:-Transpose(O)])

Below, I use x as the concentration of hydronium [H__3*LinearAlgebra:-Transpose(O)] and K__a = 5*10^(-3), K__w = 10^(-14), HA__initial = 10^(-3)

evalf(solve((x^2-10^(-14))/(10^(-3)-x+10^(-14)/x) = 5*10^(-3), x))

0.854101976e-3-0.4e-11*I, -0.5854101969e-2-0.465063510e-12*I, -0.9e-11+0.3865063510e-11*I

 (1)

If I use a manual simplification (noting that K__w is extremely small compared to reasonable values of [H__3*LinearAlgebra:-Transpose(O)], then I am able to get real solutions.

evalf(solve(x^2/(10^(-3)-x) = 5*10^(-3), x))

-0.5854101966e-2, 0.854101966e-3

 (2)

Now, looking at the complex solutions, the imaginary parts are very small and the real parts of two of the three solutions match the real solutions above, so I guess perhaps I could have just ignored those complex parts.

On the other hand, if I try to do similar calculations but with HA__initial = 10^(-6), I get

evalf(solve((x^2-10^(-14))/(10^(-6)-x+10^(-14)/x) = 5*10^(-3), x))

0.1009702e-5-0.3e-11*I, -0.5000999802e-2-0.332050808e-12*I, -0.9902e-8+0.3132050808e-11*I

 (3)

evalf(solve(x^2/(10^(-6)-x) = 5*10^(-3), x))

-0.5000999800e-2, 0.999800e-6

 (4)

So here, I don't see a complex solution that looks like the positive real solution above. Which means that I am not sure if the equation with the complex solutions (which is not simplified) is useful to me (I am doing various such calculations with different values of "`HA__initial`)".

Download concentrations.mw

new regression in Maple 2025 found. in solve ident...

Asked by: nm 11353
solve recursion
April 26 2025
1  4

Just found new regression in Maple 2025. This internal error can not be cought and was not there in Maple 2024.2. This is new from the ones reprted earlier in Collection-Of-Problems-In-Maple-2025 

Here it is , using latest SupportTools

> 

restart;
> 

interface(version);

`Standard Worksheet Interface, Maple 2025.0, Linux, March 24 2025 Build ID 1909157`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1862 and is the same as the version installed in this computer, created 2025, April 25, 10:33 hours Pacific Time.`

> 

SupportTools:-Version();

`The Customer Support Updates version in the MapleCloud is 13 and is the same as the version installed in this computer, created April 22, 2025, 15:14 hours Eastern Time.`

> 

restart;
> 

#17593
eq:=2*A[6]*exp(-2*t)*cos(t)+2*A[3]*exp(-2*t)*sin(t)-2*A[4]*exp(-2*t)*cos(t)-4*A[7]*exp(-t)*sin(2*t)+4*A[8]*exp(-t)*cos(2*t)-2*A[5]*exp(-2*t)*sin(t)+4*A[2]*exp(-t)+4*A[1]*t*exp(-t)+4*A[3]*exp(-2*t)*cos(t)+4*A[4]*exp(-2*t)*sin(t)+2*A[9]*exp(-t)*cos(2*t)-8*A[9]*t*exp(-t)*sin(2*t)+2*A[10]*exp(-t)*sin(2*t)+8*A[10]*t*exp(-t)*cos(2*t)-2*A[5]*exp(-2*t)*cos(t)+2*A[5]*t*exp(-2*t)*sin(t)-2*A[6]*exp(-2*t)*sin(t)-2*A[6]*t*exp(-2*t)*cos(t)+4*A[5]*t*exp(-2*t)*cos(t)+4*A[6]*t*exp(-2*t)*sin(t) = 3*t*exp(-t)*cos(2*t)-2*t*exp(-2*t)*cos(t):


trial_solution_constants:=[A[1], A[2], A[3], A[4], A[5], A[6], A[7], A[8], A[9], A[10]]:
> 

try
     timelimit(30,[solve(identity(eq,t),trial_solution_constants) ]);
catch:
     print("OK cought error");
end try;

Error, (in type/trig) too many levels of recursion

 

 

Download regression_maple_2025_april_26_2025.mw

Here is the same code in Maple 2024.2 but using windows. No internal error and timeout was cought as expected.

> 

restart;
interface(version);

`Standard Worksheet Interface, Maple 2024.2, Windows 10, October 29 2024 Build ID 1872373`

> 

Physics:-Version();

`The "Physics Updates" version in the MapleCloud is 1862. The version installed in this computer is 1849 created 2025, March 12, 12:37 hours Pacific Time, found in the directory C:\Users\Owner\maple\toolbox\2024\Physics Updates\lib\`

> 

#17593
eq:=2*A[6]*exp(-2*t)*cos(t)+2*A[3]*exp(-2*t)*sin(t)-2*A[4]*exp(-2*t)*cos(t)-4*A[7]*exp(-t)*sin(2*t)+4*A[8]*exp(-t)*cos(2*t)-2*A[5]*exp(-2*t)*sin(t)+4*A[2]*exp(-t)+4*A[1]*t*exp(-t)+4*A[3]*exp(-2*t)*cos(t)+4*A[4]*exp(-2*t)*sin(t)+2*A[9]*exp(-t)*cos(2*t)-8*A[9]*t*exp(-t)*sin(2*t)+2*A[10]*exp(-t)*sin(2*t)+8*A[10]*t*exp(-t)*cos(2*t)-2*A[5]*exp(-2*t)*cos(t)+2*A[5]*t*exp(-2*t)*sin(t)-2*A[6]*exp(-2*t)*sin(t)-2*A[6]*t*exp(-2*t)*cos(t)+4*A[5]*t*exp(-2*t)*cos(t)+4*A[6]*t*exp(-2*t)*sin(t) = 3*t*exp(-t)*cos(2*t)-2*t*exp(-2*t)*cos(t):


trial_solution_constants:=[A[1], A[2], A[3], A[4], A[5], A[6], A[7], A[8], A[9], A[10]]:
 
> 

try
     timelimit(30,[solve(identity(eq,t),trial_solution_constants) ]);
catch:
     print("OK cought error");
end try;


Download no_problem_in_maple_2024_april_26_2025.mw

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