MaplePrimes Questions

 

ode := diff(x(t), t, t) = - sin(x(t)^3)/x(t)^2

ics := x(0) = 0.7, D(x)(0) = 0.8

sol := dsolve({ics, ode}, numeric)

How can I  plot the derivative of the function x(t)?

I tried something like that:

odeplot(diff(sol, t), t = 0 .. 20) 

but not working

 

How to create procedure "mySumOfMatrixDiagonal(M)" which calculates the sum of the square of diagonal components for any given square matrix

Good day.

I have a simple function that I would like to plot, but am finding this awkward to do. My objective is to plot the function for varying k-values and I tried doing this using implicitplot and sequencing values of k.

The function in question is:
y := k*x*sqrt(z*(1-z));

The constant, k, takes the values: 5, 10, 15, ..., 100
The variables lie in the range: 0 ≤ x < 1 and 0 ≤ z < 1

So, if someone can tell me how to construct a plot of multiple solution curves for varying k, I would be most grateful.

Thanks for reading!

MaplePrimes_Plot.mw

 

Hello all, 

When I tried to extract the RHS of the answer from 'solve()' command, the attempt failed. 


 

restart;

sys := U__s^2 = ((1+sigma)*U__T2N - u__rN * U__T2N/(1+sigma))^2+(u__xN * U__T2N/(1+sigma))^2;

U__s^2 = ((1+sigma)*U__T2N-u__rN*U__T2N/(1+sigma))^2+u__xN^2*U__T2N^2/(1+sigma)^2

(1)

answer := solve( sys, {U__T2N} );

{U__T2N = (1+sigma)*U__s/(sigma^4+4*sigma^3-2*sigma^2*u__rN+6*sigma^2-4*sigma*u__rN+u__rN^2+u__xN^2+4*sigma-2*u__rN+1)^(1/2)}, {U__T2N = -(1+sigma)*U__s/(sigma^4+4*sigma^3-2*sigma^2*u__rN+6*sigma^2-4*sigma*u__rN+u__rN^2+u__xN^2+4*sigma-2*u__rN+1)^(1/2)}

(2)

answer[1];

{U__T2N = (1+sigma)*U__s/(sigma^4+4*sigma^3-2*sigma^2*u__rN+6*sigma^2-4*sigma*u__rN+u__rN^2+u__xN^2+4*sigma-2*u__rN+1)^(1/2)}

(3)

rhs(answer[1]);

Error, invalid input: rhs received {U__T2N = (1+sigma)*U__s/(sigma^4+4*sigma^3-2*sigma^2*u__rN+6*sigma^2-4*sigma*u__rN+u__rN^2+u__xN^2+4*sigma-2*u__rN+1)^(1/2)}, which is not valid for its 1st argument, expr

 

 


Perhaps, the failure might be related to the fact that the equation is contained in curly brackets, but I could not come up with a solution. 

Please let me have a chance to learn the way to do the 'rhs()' operation in the correct way. 

Thank you, 

In Kwon Park 

Download Qprime_20200710.mw

I want to define a vector function, then plot it, differentiate it, etc.

The best I've been able to do so far is to define it two different ways: one for evaluating, one for symbolic manipulation

with(plots): with(VectorCalculus):
r:= < cos(t), sin(t), t >;  # symbolic version of function, for example to take derivative
# returns derivative,
diff(r,t);
rf := (t)->evalf( < cos(t), sin(t), t > );   # returns numerical value of function
# we use numerical version to plot
rCurve := spacecurve( rf(t), t = -1 .. 1 );
# But how to return value of derivative?? Next line doesn't work.
df := (t)->evalf( diff(r,t) );

But how do I plot the derivative? The last line doesn't work. I've tried various variations but no luck.

I don't think this approach is write (why define two different ways?) but haven't been able to figure out how to do it.

Various online docs haven't helped.

 

I was trying to compare my solution with Maple on this ode, when I noticed Maple gives solution with no y(x)= in it. 

restart;
ode:=(y(x)-x*diff(y(x),x))/(y(x)^2+diff(y(x),x))=(y(x)-x*diff(y(x),x))/(1+x^2*diff(y(x),x));
sol:=dsolve(ode)

When there is more than one solution, Maple write y(x) on each. For an example

dsolve(diff(y(x),x)^2=x)

How to explain this output?

Maple 2020.1

I think I've seen such a problem before but thought it was fixed.

Here is an example, where depending on what one calls the constant of integration, maple can or not, verify the solution of the ode.

This should not happen.

When calling the constant of integration C[1] or _C Maple does not verify the solution. Only when using _C1 it does.

This is a problem, since I do not use _C1 or _C2  in my solution (so not to conflict with Maple's), I use C[1] and C[2].

Is there a workaround for this? 

Example 1

ode:=diff(z(x),x)+z(x)*cos(x)=z(x)^n*sin(2*x);
mysol:=z(x) = 1/((_C1*exp(sin(x)*(n - 1))*n - _C1*exp(sin(x)*(n - 1)) + 2*sin(x)*n - 2*sin(x) + 2)/(n - 1))^(1/(n - 1));
odetest(mysol,ode);

    0

Now I just changed _C1 to C[1] 

restart;
ode:=diff(z(x),x)+z(x)*cos(x)=z(x)^n*sin(2*x);
mysol:=z(x) = 1/(( C[1]*exp(sin(x)*(n - 1))*n - C[1]*exp(sin(x)*(n - 1)) + 2*sin(x)*n - 2*sin(x) + 2)/(n - 1))^(1/(n - 1));
odetest(mysol,ode);

 

and I get 2 pages full of stuff thrown at the screen. Same when using C as constant of integration.

I could change my C[1] and C[2] to _C1 and _C2 each time just before calling odetest as workaround.

But would like to ask if this is supposed to happen and why it happens. I depend on odetest alot, as I use it to tell me if my solution is correct or not and I do not want false negatives.

I have found additional ones. May be these will help locate the problem

Example 2

restart;
ode:=diff(y(x),x) = (1+x+y(x))^(1/2);
mysol1:=y(x) = LambertW(-exp(-x/2 - 1 + C[1]/2))^2 + 2*LambertW(-exp(-x/2 - 1 + C[1]/2)) - x:
odetest(mysol1,ode);

gives

But when using _C1 instead of C[1], this is the answer

restart;
ode:=diff(y(x),x) = (1+x+y(x))^(1/2):
mysol2:=y(x) = LambertW(-exp(-x/2 - 1 + _C1/2))^2 + 2*LambertW(-exp(-x/2 - 1 + _C1/2)) - x:
odetest(mysol2,ode);

Example 3

restart;
ode:=diff(y(x),x) = f(x)*y(x)+g(x)*y(x)^k:
mysol1:=y(x) = (-exp(-(k - 1)*Int(f(x), x))*(int(g(x)*exp((k - 1)*Int(f(x), x)), x)*k - int(g(x)*exp((k - 1)*Int(f(x), x)), x) - C[1]))^(-1/(k - 1)):
odetest(mysol1,ode);

   #long output is given

Changing C[1] to _C1 gives 0 right away

restart;
ode:=diff(y(x),x) = f(x)*y(x)+g(x)*y(x)^k:
mysol2:=y(x) = (-exp(-(k - 1)*Int(f(x), x))*(int(g(x)*exp((k - 1)*Int(f(x), x)), x)*k - int(g(x)*exp((k - 1)*Int(f(x), x)), x) - _C1))^(-1/(k - 1)):
odetest(mysol2,ode);

     0

 

Example 4

restart;
ode:=3*y(x)*diff(y(x),x)+5*cot(x)*cot(y(x))*cos(y(x))^2 = 0:
mysol1:=y(x) = RootOf(10*ln(sin(x))*cos(_Z)^2 - 3*cos(_Z)*sin(_Z) + 30*C[1]*cos(_Z)^2 + 3*_Z):
odetest(mysol1,ode);

Gives pages long output. Changing C[1] to _C1 gives zero

restart;
ode:=3*y(x)*diff(y(x),x)+5*cot(x)*cot(y(x))*cos(y(x))^2 = 0:
mysol2:=y(x) = RootOf(10*ln(sin(x))*cos(_Z)^2 - 3*cos(_Z)*sin(_Z) + 30*_C1*cos(_Z)^2 + 3*_Z):
odetest(mysol2,ode);

    0

 

Example 5

restart;
ode:=x*(x-a*y(x))*diff(y(x),x) = y(x)*(y(x)-a*x);
mysol1:=y(x) = exp((-a + 1)*RootOf(x*exp(ln(x)*a + _Z*a + C[1]*a + C[1]) + exp(ln(x)*a + _Z*a + C[1]*a - _Z + C[1])*x - 1) - ln(x)*a - C[1]*(a + 1)):
odetest(mysol1,ode);

gives

Changing C[1] to _C1 gives zero

restart;
ode:=x*(x-a*y(x))*diff(y(x),x) = y(x)*(y(x)-a*x);
mysol2:=y(x) = exp((-a + 1)*RootOf(x*exp(ln(x)*a + _Z*a + _C1*a + _C1) + exp(ln(x)*a + _Z*a + _C1*a - _Z + _C1)*x - 1) - ln(x)*a - _C1*(a + 1)):
odetest(mysol2,ode);

              0

 

Example 6

restart;
ode:=x*diff(y(x),x)^2-a*y(x)*diff(y(x),x)+b = 0:
mysol1:=y(x) = x*RootOf(2*x*_Z^2*a - 2*C[1]*_Z^((2*a - 1)/(a - 1))*a - x*_Z^2 + C[1]*_Z^((2*a - 1)/(a - 1)) - b)/a + b/(RootOf(2*x*_Z^2*a - 2*C[1]*_Z^((2*a - 1)/(a - 1))*a - x*_Z^2 + C[1]*_Z^((2*a - 1)/(a - 1)) - b)*a):
odetest(mysol1,ode);

gives many pages of output. Changing C[1] to _C1 gives zero

restart;
ode:=x*diff(y(x),x)^2-a*y(x)*diff(y(x),x)+b = 0:
mysol2:=y(x) = x*RootOf(2*x*_Z^2*a - 2*_C1*_Z^((2*a - 1)/(a - 1))*a - x*_Z^2 + _C1*_Z^((2*a - 1)/(a - 1)) - b)/a + b/(RootOf(2*x*_Z^2*a - 2*_C1*_Z^((2*a - 1)/(a - 1))*a - x*_Z^2 + _C1*_Z^((2*a - 1)/(a - 1)) - b)*a):
odetest(mysol2,ode);

             0

and more. I think the above should hopefully help locate the issue.

Maple 2020.1 with Physics 708

 

The general solution of x2y'' + 3xy' + λy = 0 is (from Example 4 here)

Why does dsolve(x^2*diff(y(x), x, x) + 3*x*diff(y(x), x) + lambda*y(x) = 0) give only the third case instead of the whole solution?

How to find all solutions [xi,yi] to the following system of equations  

x3y=1

y-sin(12x)=0

which satisfies condition 1,2<x<2  

I want to show the animation of the polar equation r=cos(2theta) be plotted from theta=0 to 2pi

Hello,The system of equations is as follows:

I'd like to find all integer solutions in [1,20], when I use isolve ,the results are not good.  Since at least one variable is equal to 0 in evey solution.

{isolve}({a+b+c=a1+b1+c1, a^2+b^2+c^2=(a1)^2+(b1)^2+(c1)^2,a*b*c=2*a1*b1*c1})

 In maple I did not  want to use less efficient for-loop like C programing as following.

#include <math.h>
#include <stdio.h>
void main()
{
    long int a,b,c,d,e,f;
    for(a=1;a<20;a++)
    {
        for(b=1;b<20;b++)
        {
            for(c=1;c<20;c++)
            {
                for(d=1;d<20;d++)
                {
                    for(e=1;e<20;e++)
                    {
                        for(f=1;f<20;f++)
                        {
                            if(a+b+c==d+e+f&&a*a+b*b+c*c==d*d+e*e+f*f&&a*b*c==2*d*e*f)                                             printf("a=%d,b=%d,c=%d,d=%d,e=%d,f=%d\n",a,b,c,d,e,f);
                        }
                    }
                }
            }
        }
    }
}
a=3,b=5,c=16,d=1,e=8,f=15
a=3,b=5,c=16,d=1,e=15,f=8
a=3,b=5,c=16,d=8,e=1,f=15
a=3,b=5,c=16,d=8,e=15,f=1
a=3,b=5,c=16,d=15,e=1,f=8
a=3,b=5,c=16,d=15,e=8,f=1
a=3,b=16,c=5,d=1,e=8,f=15
a=3,b=16,c=5,d=1,e=15,f=8
a=3,b=16,c=5,d=8,e=1,f=15
a=3,b=16,c=5,d=8,e=15,f=1
a=3,b=16,c=5,d=15,e=1,f=8
a=3,b=16,c=5,d=15,e=8,f=1
a=5,b=3,c=16,d=1,e=8,f=15
a=5,b=3,c=16,d=1,e=15,f=8
a=5,b=3,c=16,d=8,e=1,f=15
a=5,b=3,c=16,d=8,e=15,f=1
a=5,b=3,c=16,d=15,e=1,f=8
a=5,b=3,c=16,d=15,e=8,f=1
a=5,b=16,c=3,d=1,e=8,f=15
a=5,b=16,c=3,d=1,e=15,f=8
a=5,b=16,c=3,d=8,e=1,f=15
a=5,b=16,c=3,d=8,e=15,f=1
a=5,b=16,c=3,d=15,e=1,f=8
a=5,b=16,c=3,d=15,e=8,f=1
a=16,b=3,c=5,d=1,e=8,f=15
a=16,b=3,c=5,d=1,e=15,f=8
a=16,b=3,c=5,d=8,e=1,f=15
a=16,b=3,c=5,d=8,e=15,f=1
a=16,b=3,c=5,d=15,e=1,f=8
a=16,b=3,c=5,d=15,e=8,f=1
a=16,b=5,c=3,d=1,e=8,f=15
a=16,b=5,c=3,d=1,e=15,f=8
a=16,b=5,c=3,d=8,e=1,f=15
a=16,b=5,c=3,d=8,e=15,f=1
a=16,b=5,c=3,d=15,e=1,f=8
a=16,b=5,c=3,d=15,e=8,f=1

Does Maple have more good ways to solve that?

Hello

I had to save previous results of a calculation to files as the number of elements is too big for my computer to handle in one go.   Here it is an idea of what I am doing to read the files and perform the calculations.

mainproc:=proc(arg1,arg2,arg3,arg4) ... end proc:

Grid:-Set(mainproc):
Grid:-Set(arg2,arg3,arg4):   #  They don't change ever.

for i from 1 to number_of_files do 
   read(...):  # it reads arg1 from a file
   Grid:-Set(arg1):
   ans:=Grid:-Seq(mainproc(arg1[i],arg2,arg3,arg4),i=1..numelems(arg1))):
   Grid:-Wait():
   save ans, ....:
   unassign('arg1'):
   unassign('ans'):
   gc():  # An attempt
end do:

The actual code works but, for every step in the loop, the memory used by Maple increases by a certain amount that seems to be mostly related to arg1 (as if arg1 is piling up from iteration to iteration). 

I read some of the earlier posts on a similar subject dated 5 to 10 years old.  I wonder if there is something new that can be done to minimize the usage of memory.   

Many thanks

Ed

PS. I am aware of tasksize, numcpus and Threads.   

I have a simple matrix, 11 rows x 3 columns, with header row.  <<

I multiply column 1 and 2 to get <<0.,0.031,0.218,0.657,1.084,1.095,0.654,0.217>>

Now, I just want to add up these elements, but 'sum' doesn't work 'Sum' doesn't work either. Is there a simpel way to do this?
 

m1 := `<|>`(`<,>`("# girls", 0, 1, 2, 3, 4, 5, 6, 7, 8), `<,>`("P(x)", 0.4e-2, 0.31e-1, .109, .219, .271, .219, .109, 0.31e-1, 0.4e-2), `<,>`("x*P(x)", 0, 0, 0, 0, 0, 0, 0, 0, 0))

Matrix(%id = 18446745635438952446)

(1)

m1[2 .. 9, 3] := `~`[`*`](m1[2 .. 9, 1], m1[2 .. 9, 2])

Vector[column](%id = 18446745635417541318)

(2)

"Sum?"

Vector[column](%id = 18446745635493441398)

(3)

"sum?"

Error, (in LinearAlgebra:-Multiply) invalid arguments

 

``


 

Download Untitled_(3).mw

I must do some formula manipulation
 

Classification of conic sections

restart; with(student):

A*x^2+B*y^2+C*x+D*y+E=0;

A*x^2+B*y^2+C*x+D*y+E = 0

(1)

f:= A*x^2+B*y^2+C*x+D*y+E;

A*x^2+B*y^2+C*x+D*y+E

(2)

completesquare(f, x);

A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y+E

(3)

f:= A*x^2+B*y^2+C*x+D*y+E;

A*x^2+B*y^2+C*x+D*y+E

(4)

ans:=Student[Precalculus][CompleteSquare]( (4), [y] );

B*(y+(1/2)*D/B)^2+A*x^2+C*x+E-(1/4)*D^2/B

(5)

ans1:=Student[Precalculus][CompleteSquare]( (4), [x] );

A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y+E

(6)

ans+ans1;

B*(y+(1/2)*D/B)^2+A*x^2+C*x+2*E-(1/4)*D^2/B+A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y

(7)

B*(y + D/(2*B))^2 + A*x^2 + C*x + 2*E - D^2/(4*B) + A*(x + C/(2*A))^2 - C^2/(4*A) + B*y^2 + D*y = 0;

B*(y+(1/2)*D/B)^2+A*x^2+C*x+2*E-(1/4)*D^2/B+A*(x+(1/2)*C/A)^2-(1/4)*C^2/A+B*y^2+D*y = 0

(8)

Now i must  get this form A( )^2 +B( )^2 = M  

 


 

Download vraag_herleiding_conic_sections_formule.mw

 

 

Does this condition happen when the character is too long?
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