Maple Questions and Posts

These are Posts and Questions associated with the product, Maple

hello,

this is my first post here so sorry if it's done wrong...

I'm having difficulties with the following command because maple only returns a solution if the first two boundary conditions of the "if" arguments are true. If the the first or the second condition is false then maple doesn't give a solution solution.

for i to n do if H[i] < 2.7 then if A[i, f] < 12 then if A[i, o] < 1.2 then Q[i, foo] := evalf(610*(A[i, o]*sqrt(H[i])*h[k]*A[i, T])^(1/2)) else Q[i, foo] := evalf(7.8*A[i, t]+378*A[i, o]*sqrt(H[i])) end if end if end if; print(Q[i, foo]); end do;

I've also tried with the elif command but it gives a similar problem

Anyone knows how to solve this?

suppose for example i am working on the function

exp(-I*Pi*(n+2*n*(m-1))/m);

and i wanted to for what ever reason assign a unique symbol to each of the two times the indeterminant 'n' occurs:

exp(-I*Pi*(n[1]+2*n[2]*(m-1))/m)

How could i accomplish this for any function F?

i have tried the method of

map(op, [op(op(exp(-I*Pi*(n+2*n*(m-1))/m)))]);
                [-1, Pi, n, 2 n (m - 1), m, -1]
 

And though i could then use algsubs on the original function for each in the above list, but then ran into problems arising that for any function the number of times i need to map op in iteration is not known, i will actually work this out by the end of the night so i dont know why im bothering asking

 

Edit 2: yep its ok i discovered subsop

I've created a worksheet that outputs a boggle board.  I think it could be more efficient than the method I came up with but the idea is there.  The only way I could figure to rotate the letters was to output them to a bmp format then read them back in and use imagetools for rotation.  I used Times Roman font but the font Boggle uses I think is Tunga, Latha or Mangal.  

Note - remove the colon in the last line to produce the output.  One other thing I believe, in Tools->Options-> (uncheck)Limit Expression Length to 1000000  

Saving the file with the output would have produced a file in the tens of Megabytes and may have caused error loading.


 

restart; gc()

with(plots); with(ImageTools)

a := [seq(k, k = "A" .. "Z")]

["A", "B", "C", "D", "E", "F", "G", "H", "I", "J", "K", "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"]

(1)

for i in a do plotsetup(bmp, plotoutput = cat("c:/test/", i, ".bmp")); img || i := textplot([0, 1, i, font = ["times", "roman", 200]], axes = none, scaling = constrained) end do
NULL

plotsetup = default

(2)

plotsetup(default)

with(combinat)``

``

Setting up the 16 boggle cubes

 

cube1 := ["H", "E", "E", "N", "W", "G"]

cube2 := ["T", "M", "I", "O", "C", "U"]

cube3 := ["D", "E", "X", "L", "R", "I"]

cube4 := ["S", "P", "F", "A", "K", "F"]

cube5 := ["T", "O", "E", "S", "I", "S"]

cube6 := ["H", "N", "L", "N", "Z", "R"]

cube7 := ["R", "L", "T", "Y", "T", "E"]

cube8 := ["D", "E", "Y", "L", "R", "V"]

cube9 := ["C", "A", "O", "S", "P", "H"]

cube10 := ["Qu", "U", "M", "H", "I", "N"]

cube11 := ["D", "Y", "I", "S", "T", "T"]

cube12 := ["S", "N", "I", "E", "E", "U"]

cube13 := ["T", "O", "O", "W", "A", "T"]

cube14 := ["W", "H", "E", "V", "R", "T"]

cube15 := ["J", "B", "O", "O", "A", "B"]

cube16 := ["N", "A", "E", "A", "E", "G"]

cubes := [seq(cat("cube", i), i = 1 .. 16)]

["cube1", "cube2", "cube3", "cube4", "cube5", "cube6", "cube7", "cube8", "cube9", "cube10", "cube11", "cube12", "cube13", "cube14", "cube15", "cube16"]

(3)

c := randperm(cubes)

["cube13", "cube14", "cube11", "cube6", "cube9", "cube1", "cube16", "cube7", "cube3", "cube2", "cube12", "cube5", "cube4", "cube10", "cube8", "cube15"]

(4)

cc := map(parse, c)

[cube13, cube14, cube11, cube6, cube9, cube1, cube16, cube7, cube3, cube2, cube12, cube5, cube4, cube10, cube8, cube15]

(5)

ccf := [seq(op(randcomb(cc[i], 1)), i = 1 .. 16)]

["A", "H", "I", "N", "A", "N", "E", "T", "D", "M", "E", "S", "K", "N", "E", "O"]

(6)

with(ArrayTools)

g := Reshape(Array(ccf), [4, 4])

Array(%id = 18446744074360417206)

(7)

rr := proc () randcomb([0, 90, 180, 270], 1) end proc

Reshape(Array([seq(display(Preview(Rotate(Read(cat("c:/test/", ccf[i], ".bmp")), op(rr()))), axes = none), i = 1 .. 16)]), [4, 4])
 

````

 

 

 

NULL


 

Download Boggle3-6final.mw

Why input fraction to eigenvector and then evalf output are all the same when input different ?

i would like to see more decimal numbers

digits := 36 

command can not show more decimal numbers

 

why can not see the difference?

 

if start from fraction

After set round screen display to 36 digits in options

If start from floating value

the result different from start from fraction

start from which is the most accurate and correct?

if fraction is correct, why all result are the same even if input are different?

I'm absolutely new in maple, but I need to solve PDE and I don't understand why maple do not solve it

restart; with(PDEtools);
U := diff_table(u(x, t));

pde[1] := U[t, t] = U[x, x]+5*sin(3*x);

bc[1] := eval(U[], x = 0) = 0; bc[2] := eval(U[], x = Pi) = 0;
ic[1] := eval(U[], t = 0) = 0; ic[2] := eval(D[2]*U[], t = 0) = 1;

sys[1] := [pde[1], bc[1], bc[2], ic[1], ic[2]];
pdsolve(sys[1]);

 

Error: Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {u(0, t), u(Pi, t), u(x, 0)} in the given boundary conditions {D[2]*u(x, 0) = 1, u(0, t) = 0, u(Pi, t) = 0, u(x, 0) = 0}

 

If I solve it  with only boundary conditions or without any conditions, maple gives me an answer. 

Please, help me to understand, how to solve this pde system with boundary and initials conditions. This is my firs use of maple so I hardly understand how to write code correctly.

Real part + complex part

 

but 

sometimes it display

complex part + real part 

how to consistent display real part + complex part?

Sometimes we have questions that apply to every execution we make, regardless of what the actual nature of the commands included in that execution are.

 

For this reason, I propose that in the same way that we have the feature of "start up" code, which automatically executes upon opening the specified worksheet, we also implement an option of automatically executed code for every execution that is carried out within the specified worksheet. 

for example, code that serves purposes of output such as 

" This execution took xxxx minutes and xx.x seconds to complete, and required an average of x % of your computer's total processor utilization capacity across all cores "

I know how i will be providing this feature in my worksheets, but i just feel that it would be a beneficial feature for those that use maple for purposes that do not involve a knowledge of how maple is programmed, and would not be able to do so.

Any discussion about this would be welcomed, which is why i am posting about it of course.

I am trying to see if Maple can solve Laplace PDE inside the disk in polar coordinates. Standard textbook problem. Radius of disk is `a`. The boundary conditions on the disk is `f(theta)`. One of the conditions needed also is that the solution is finite in the center of the disk.

I do not know how to tell Maple that the solution should be finite in the center of the disk. If I do not give this conditions, Maple gives me strange looking solution, which does not look like anything close to the standard series solution one gets from hand solution. There is not even a series solution.

This is what I tried

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta);
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0

Now, how to tell it that `u(0,theta)` is bounded? So that the `ln(r)` solution do not show up? Adding `u(0,theta)<infinity` to the boundary conditions, gives error

restart;
pde:=diff(u(r,theta),r$2)+1/r*diff(u(r,theta),r)+1/r^2*diff(u(r,theta),theta$2)=0;
bc:=u(a,theta)=cos(theta),u(0,theta)<infinity;
sol:=pdsolve([pde,bc],u(r,theta)) assuming r<=a,r>0

The standard solution to this PDE is

Where `c0` and `cn` and `kn` above are found from boundary conditions at $u(a,\theta)$.

How can one get Maple to give the above solution? How to tell it that $u$ is bounded at $r=0$?

 

 

we know that the exponotial function is represented as a taylor series, that means that the exponontial of any value is approximatvly evaluated ,the same thing of sinus function when i calcute sin(Pi/6) it give me the exacte result,
 

sum(0^i/i!,i=0..infinity);

0

(1)

exp(0);

1

(2)

sin(0.2);

.1986693308

(3)

sin(Pi/6);

1/2

(4)

sin(Pi/2);

1

(5)

 


 

Download trying.mws

,can anyone explain to me these results ?

Came across this issue again.  While working within a worksheet, everything seems fine, using commands like imagetools outputing bmp files etc.  I had to, in the the Options->Precision, uncheck limit expression length to 1000000 during my session.  I saved it multiple times, however and unfortunately did not save as seperate versions, so I was left with one file.

The file size ended up being 64Mb.  The problem is trying to load it into Maple, everytime I try to load the worksheet my computer seems to freeze (laptop, 4Gb RAM, windows 7 64 bit, Maple 2017) I check task manager and my javaw.exe file is consuming 3.7Gb of memory!  So that's why it's locking up. 

I've tried multiple times to open the file and utilmately end up pressing and holding the power button to restart my system. 

 

C: y^2=10161/256*t^2+8829189/25600*t+7266953961/1024000 

how to find the (t, y) points on the curve C ?

THANKS.

Attractor.mw
Hello, anybody could help with this error?

restart

interface(imaginaryunit = j);

I

(1)

with(DEtools):

k := .5; 1; tau := .95; 1; mu := 0.1e-1; 1; pi := 116.1; 1; vartheta := 0.8e-2; 1; phi := 0.25e-2; 1; epsilon := 0.2e-2; 1; rho := 0.5e-1; 1; beta := 0.115e-1; 1; chi := 0.598e-2; 1; q := .5; 1; eta := .2; 1; delta := .1; 1; alpha := 0.57e-1; 1; p := .2; 1; Upsilon := 1.2

.5

 

.95

 

0.1e-1

 

116.1

 

0.8e-2

 

0.25e-2

 

0.2e-2

 

0.5e-1

 

0.115e-1

 

0.598e-2

 

.5

 

.2

 

.1

 

0.57e-1

 

.2

 

1.2

(2)

lambda := k*tau*(C(t)*Upsilon+I(t))/(S(t)+V(t)+C(t)+I(t)+R(t)):

DEplot3d({d*C(t)/dt = rho*lambda*S(t)+rho*`&epsilon;`*lambda*V(t)+(1-q)*eta*I(t)-(mu+beta+chi)*C(t), d*I(t)/dt = (1-rho)*lambda*S(t)+(1-rho)*`&epsilon;`*lambda*V(t)+chi*C(t)-(mu+alpha+eta)*I(t), d*R(t)/dt = beta*C(t)+q*eta*I(t)-(mu+delta)*R(t), d*S(t)/dt = (1-p)*pi+phi*V(t)+delta*R(t)-(mu+lambda+`&vartheta;`)*S(t), d*V(t)/dt = p*pi+`&vartheta;`*S(t)-(lambda*`&epsilon;`+mu+phi)*V(t)}, {C(t), I(t), R(t), S(t), V(t)}, t = 0 .. 300, stepsize = .1, [[S(0) = 8200, V(0) = 2800, C(0) = 200, I(0) = 210, R(0) = 200]], linecolour = t, axes = BOXED, scene = [S, V, C, I, R])

Error, (in DEtools/DEplot/CheckInitial) the 'number' option must be specified before initial conditions

 

``


 

Download Attractor.mw

 

How to convert maple file into the pdf file?For example the attacahed file is maple file.I want to convert this to a pdf file.

conservation_of_wave_eq.mw

If i was a good student i wouldnt be using the is function for these at all, but im quite tired now so i thought that it would be fine this time. But i honest cannot see what additional declarations need to be made here, and why a false is returned for the first query, then FAIL for the negation of the first in the second query, the only thing i can think of is declaring that indeed n and m are elements of N, but if maple was allowing for Gaussian n,m surely both queries would return fail, and so obvious it is indeed acknowledging the arguments are on R, anyway hopefully someone gets the chance to make me feel silly here

 


 

assume(n < m, n > 1)

is((3*n-1)/(n^3+2*n+1)-(3*m-1)/(m^3+2*m+1) < 0)

false

(1)

is((3*n-1)/(n^3+2*n+1)-(3*m-1)/(m^3+2*m+1) >= 0)

FAIL

(2)

``


 

Download MISSING_ASSUMPTION.mw

I'm stuck with applying boundary conditions for a cantilever beam please help with format

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