Maple Questions and Posts

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i try find some part of solution of this kind of pde but i can't get results my openion is maybe this pde is wronge when i defined 

pde.mw

Let eps be a real number greater than zero. Calculate the limit lim(x-->oo)[x^(1-eps)*Integral(from x to x+1) sin(t^2)dt] .

my solution is a little bit long which when i click on pdetest command i wait at least more than a hour but still is runing and i don't get any result and not give me error , which i don't know my result is true or not so How i can find that my pde by this solution it will be zero or not ?

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)

declare(u(x, y, t))

u(x, y, t)*`will now be displayed as`*u

(2)

declare(f(x, y, t))

f(x, y, t)*`will now be displayed as`*f

(3)

pde := diff(u(x, y, t), t)-(diff(diff(u(x, y, t), `$`(x, 4))+5*u(x, y, t)*(diff(u(x, y, t), `$`(x, 2)))+(5/3)*u(x, y, t)^3+5*(diff(u(x, y, t), x, y)), x))-5*u(x, y, t)*(diff(u(x, y, t), y))+5*(int(diff(u(x, y, t), `$`(y, 2)), x))-5*(diff(u(x, y, t), x))*(int(diff(u(x, y, t), y), x))

diff(u(x, y, t), t)-(diff(diff(diff(diff(diff(u(x, y, t), x), x), x), x), x))-5*(diff(u(x, y, t), x))*(diff(diff(u(x, y, t), x), x))-5*u(x, y, t)*(diff(diff(diff(u(x, y, t), x), x), x))-5*u(x, y, t)^2*(diff(u(x, y, t), x))-5*(diff(diff(diff(u(x, y, t), x), x), y))-5*u(x, y, t)*(diff(u(x, y, t), y))+5*(int(diff(diff(u(x, y, t), y), y), x))-5*(diff(u(x, y, t), x))*(int(diff(u(x, y, t), y), x))

(4)

pde_nonlinear, pde_linear := selectremove(proc (term) options operator, arrow; has((eval(term, u(x, y, t) = a*u(x, y, t)))/a, a) end proc, pde)

-5*(diff(u(x, y, t), x))*(diff(diff(u(x, y, t), x), x))-5*u(x, y, t)*(diff(diff(diff(u(x, y, t), x), x), x))-5*u(x, y, t)^2*(diff(u(x, y, t), x))-5*u(x, y, t)*(diff(u(x, y, t), y))-5*(diff(u(x, y, t), x))*(int(diff(u(x, y, t), y), x)), diff(u(x, y, t), t)-(diff(diff(diff(diff(diff(u(x, y, t), x), x), x), x), x))-5*(diff(diff(diff(u(x, y, t), x), x), y))+5*(int(diff(diff(u(x, y, t), y), y), x))

(5)

thetai := t*w[i]+y*p[i]+x

t*w[i]+y*p[i]+x

(6)

eqw := w[i] = -5*p[i]^2

w[i] = -5*p[i]^2

(7)

Bij := proc (i, j) options operator, arrow; (-6*p[i]-6*p[j])/(p[i]-p[j])^2 end proc

proc (i, j) options operator, arrow; (-6*p[i]-6*p[j])/(p[i]-p[j])^2 end proc

(8)

NULL

theta1 := normal(eval(eval(thetai, eqw), i = 1)); theta2 := normal(eval(eval(thetai, eqw), i = 2))

-5*t*p[1]^2+y*p[1]+x

 

-5*t*p[2]^2+y*p[2]+x

(9)

eqf := f(x, y, t) = (-5*t*p[1]^2+y*p[1]+x)*(-5*t*p[2]^2+y*p[2]+x)-(6*(p[1]+p[2]))/(p[1]-p[2])^2

f(x, y, t) = (-5*t*p[1]^2+y*p[1]+x)*(-5*t*p[2]^2+y*p[2]+x)-6*(p[1]+p[2])/(p[1]-p[2])^2

(10)

eq17 := u(x, y, t) = 6*(diff(diff(f(x, y, t), x), x))/f(x, y, t)-6*(diff(f(x, y, t), x))^2/f(x, y, t)^2

u(x, y, t) = 6*(diff(diff(f(x, y, t), x), x))/f(x, y, t)-6*(diff(f(x, y, t), x))^2/f(x, y, t)^2

(11)

eqt := eval(eq17, eqf)

u(x, y, t) = 12/((-5*t*p[1]^2+y*p[1]+x)*(-5*t*p[2]^2+y*p[2]+x)-6*(p[1]+p[2])/(p[1]-p[2])^2)-6*(-5*t*p[1]^2-5*t*p[2]^2+y*p[1]+y*p[2]+2*x)^2/((-5*t*p[1]^2+y*p[1]+x)*(-5*t*p[2]^2+y*p[2]+x)-6*(p[1]+p[2])/(p[1]-p[2])^2)^2

(12)

``

pdetest(eqt, pde)

NULL

Download test.mw

Hi,

I'm looking to animate the Riemann sum only on the interval [0,2], while keeping a global view of the curve (G). I tried the background option, but without success. Any ideas? Thanks!

S7_Riemann_Animation_Aire.mw

Hi. How to plot the 2d output like this? I don't know how to declare "a" and "b" random constant?
The pair of equations x = a and y = b graphically represents lines which  are, a. parallel , b. intersecting at (b, a), c. coincident, d.  intersecting at (a, b)

a := (rand(3 .. 4))(); b := (rand(0 .. 3))(); plot(x = a); plot(y = b)

3

 

2

 

Error, (in plot) unexpected option: x = 3

 

Error, (in plot) unexpected option: y = 2

 
 

``

Download x=a_y=b_plot.mw

In a procedure what is the way to only allow specific values in a variable?

ie/

change:=proc(a::integer, b={10,20,30})#only values 10, 20 and 30 allowed for variable b
  a*b;
end proc:

Trying to find a formula such that when n is negative the value is +1 and when n is positive the value is 0, without using an if statement. 

The square grid is given in the Euclidean plane. All grid points therefore have integer coordinates. Investigate whether there are three points in this grid plane that form an equilateral triangle.

Consider the following:

MyTableElement1 := proc(L::list(nonnegint))
  ## L will have only two elements
  local M, x, y;
  M:=L;
  x:=convert(M[1],string);
  y:=convert(M[2],string);

  return cat("\\begin{tabular}{c} ",x," \\\\ ",y," \\end{tabular}")

end;

MyTableElement1([2,3]);

will output

"\begin{tabular}{c} 2 \\ 3 \end{tabular}"

as desired. However, if one inserts an \hline into the table,

MyTableElement2 := proc(L::list(nonnegint))
  ## L will have only two elements
  local M, x, y;
  M:=L;
  x:=convert(M[1],string);
  y:=convert(M[2],string);

  return cat("\\begin{tabular}{c} ",x," \\\\ \\hline ",y," \\end{tabular}")

end;

will output

"\begin{tabular}{c} 2 \\ \hline 3 \end{tabular}"

again as desired. However, if I copy and paste this int a document, I get

"\begin{tabular}{c} 2 \\ hline 3 \end{tabular}".

Note that \hline is now just hline. Putting "\\\\" in front of the hline outputs

"\begin{tabular}{c} 2 \\ \\hline 3 \end{tabular}",

but this does not compile properly. How can I get a proper \hline command to appear in the table? Thank uou for your consideration.

After a long chat with ChatGPT I finally received a fully working code for a proc for a generalized Woodbury Identity for the inversion of the sum of two or more positive definite matrices.
I was inspired by a family member who is a trained professional programmer, who told me that in his professional work he uses ChatGTP for an initial draft of his program.  
I did find out that ChatGTP makes errors: from simple ones like writing 'Simplify' instead of 'simplify' to serious conceptual errors, for example in recursive loops. However, ChatGTP seems to 'understand' the error after given specific feedback. Although, this does not mean that the next proposal does not contain the same logical error. But after a long chat I received a nice proc that seems to work. 
My second surprise was that Gemini suggested a formula for the generalized Woodbury lemma that was unknown to me, and I was unable to find on Scholar Google or https://math.stackexchange.com. Based on a special case of that formula, I was able to write the second proc myself. 
In conclusion, to start working it can be helpful to collaborate with AI friend, a little patience may help, AI may not be astute as someone on Mapleprimes wrote, but neither am I. I am now retired, and it is fun to play with Maple and AI. 
By the way, the search term Woodbury did not give a single hit on Mapleprimes.With_a_little_help_from_my_friends.mw
kind regards,Harry 

 

 Can I solve the Tolman-Oppenheimer-Volkoff equation with Maple ?  I'm having trouble with Einstein's equation with the energy tensor as the second member


Let's suppose I define a procedure P which uses a Maple built-in procedure BP.
BP is a procedure which uses optional parameters, let's say a and b.

I would like to call P with the same optional parameter names than BP, that is to write something like this

P := proc(....., {a::a_type:=a:-default, b::b-type:=b_default)
       ....
       BP(...., 'a'=a, 'b'=b)
     end proc.

The execution of P(..., a=1) leads to the error unexpected option: 1=1
 

A workaround is to define P this way

P := proc(....., {P_a::a_type:=a_default, P_b::b-type:=b_default)
       ....
       BP(...., a = P_a, b = P_b)
     end proc.

# example

P(..., P_a=1, P_b="yes")

This works but it would be easier for the users to use the same option names in the definition of P than in the definition of BP.

Question: Is that possible?

Here is an illustrative example where I would prefer to write

ecdf(S, color="red", thickness=3)

instead of

ecdf(S, ecdf_color="red", ecdf_thickness=3)


 

restart

ecdf := proc(S, {ecdf_color::string:="blue", ecdf_thickness::posint:=1})
  local N   := numelems(S):
  local M   := sort(S):
  local opt := [color=ecdf_color, thickness=ecdf_thickness]:
  plots:-display(
    plot([ seq(op([[M[i], (i-1)/N], [M[i], i/N], [M[i+1], i/N]]), i=1..N-1) ], opt[])
    , plot([ [M[N], (N-1)/N], [M[N], 1] ], opt[])
    , plot([ [M[N], 1], [(max+(max-min)*0.2)(S), 1] ], opt[], linestyle=3)
    , plot([ [M[1], 0], [(min-(max-min)*0.2)(S), 0] ], opt[], linestyle=3)
  )
end proc:

S := Statistics:-Sample(Uniform(0, 1), 10):
ecdf(S)

 

ecdf(S, ecdf_color="red", ecdf_thickness=3)

 

 


 

Download ecdf.mw


Thanks in advance

 

 

Hello,

I tried to apply Berlekamp's theory to the polynomial X^9+X^6-X+1 in F3.
The matrix of the linear application whose kernel dimension is sought is:

 The dimension should be equal to 2 according to the theory since this polynomial decomposes into 2 irreducible polynomials.
However, the nullspace instruction of maple does not give a result consistent with the theory. Please help me find what is happening.

 

I was trying to find the best way to determine what powers of 10 a number is. 

trunc(log10(a)) #a being any number

It works great for integers, however it fails when the number has is some integer with 7 nine's in the decimal.
trunc(log10(99.999999))
                                                  1

trunc(log10(99.9999998))
                                                 1

trunc(log10(99.9999999))
                                                 2

To get around that fail I found length to be another way. 
     length(trunc(a))-1  #a is any number

Any other ways or mayber there's a better way?

restart

with(LinearAlgebra)

A := `<,>`(`<|>`(-2*`&omega;__0`^2, `&omega;__0`^2), `<|>`(`&omega;__0`^2, -`&omega;__0`^2))

Matrix(%id = 36893488151921005020)

(1)

AA := subs(`&omega;__0` = 3, A)

Matrix(%id = 36893488151921002244)

(2)

evalf([Eigenvectors(AA)])

[Vector[column](%id = 36893488151920990196), Matrix(%id = 36893488151920990316)]

(3)

NULL

Below is an eigenvector equation for AA for the eigenvalue -3.4376...

 

S := AA+3.43769410*IdentityMatrix(2)

Matrix(%id = 36893488151920987180)

(4)

When I solve the system below, I expect to get the vector `<,>`(.6180, 1)

Vector[column](%id = 36893488151920981284)

(5)

, just like the result of the Eigenvectors command used above.

 

LinearSolve(S, `<,>`(0, 0))

Vector[column](%id = 36893488151920975988)

(6)

Instead I get the zero vector.

 

In addition, when I compute the reduced row echelon form, I expect to have one of the rows be zero since matrix S has rank 1.

 

Rank(S)

1

(7)

ReducedRowEchelonForm(S)

Matrix(%id = 36893488151959135044)

(8)

NULL

Therefore, I guess my question is maybe about calculations and not about Maple. That being said, I did the calculations manually, then I asked chatgpt, which agreed with me.

 

Why are the calculations not coinciding with Maple?

 

I can do the calculations "manually" with Maple:

 

solve(-14.5623059000000*x+9*y = 0)

{x = x, y = 1.618033989*x}

(9)

solve(9*x-5.56230590000000*y = 0)

{x = x, y = 1.618033988*x}

(10)

 

Now, if I try to solve them simultaneously, it seems I have the same issue as with LinearSolve

NULL

solve({9*x-5.56230590000000*y = 0, -14.5623059000000*x+9*y = 0})

{x = 0., y = 0.}

(11)

I tried something else that worked, but the questions above still remain.

 

We start with A again

 

A = Matrix(%id = 36893488151921005020)NULL

In my problem I have the variable

 

`&omega;__a` := (1/2)*`&omega;__0`*(sqrt(5)-1) = (1/2)*omega__0*(5^(1/2)-1)NULL

NULL

and it turns out that -`#msub(mi("omega",fontstyle = "normal"),mi("a"))`^2 is an eigenvalue of A.

In particular, for `&omega;__0` = 3 we get the values I showed previously.

 

For example, here is the eigenvalue we were looking at beforeevalf(subs(`&omega;__0` = 3, -`&omega;__a`^2))

-3.437694099

(12)

Okay so the eigenvector equation without subbing in a value for `&omega;__0` is (A+I*`&omega;__a`^2)*v = 0.

 

Here is what the matrix on the left looks like

NULL

A+`&omega;__a`^2*IdentityMatrix(2)

Matrix(%id = 36893488151964015780)

(13)

 

And here we solve for the eigenvectors

sol := LinearSolve(A+`&omega;__a`^2*IdentityMatrix(2), `<,>`(0, 0))

Vector[column](%id = 36893488151964001572)

(14)

evalf(subs(_t2[1] = 1, sol))

Vector[column](%id = 36893488151964010244)

(15)

Which is the expected result.

 

So, Maple can solve the eigenvector equation in this more general case but not if I specify `&omega;__0` before asking it to solve.NULL

Download evcalc.mw

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