MaplePrimes Questions

I am setting up a practice set on how to find a common denominator, add, and reduce.

If I do not set "allow arithmatic", there is no way for students to correctly type in the answer, which is CURRENTLY the return value for a frac() function call. If I DO set "allow arithmatic", the student can just literally type the question into the answer box to get full credit / never has to reduce, etc, which defeats the purpose of the lesson. This is also the case with the Maple Graded answer type unless I do something fancier than I'm doing.

I can set up the answer properly without using frac() and reduce the fraction, but then--when the answer should be a whole number--the student has to type in "3/1" as an example.

I've also tried setting up an if statement for the answer, but when i convert the numbers to strings, then "2/3" becomes .2.0/3.0". 

So, at this point, I'm super open to ideas. 

 

 

# Define the denominators. We want them to be between 2 and 8.
$B=rint(2,8);
$D=rint(2,8);

# Make it so that the numerators can be between 1 and 1 less than the denominators.
# We also want to make sure they're not 0.
# Hashtag dividingby0.
$R=rint(-$B+1,$B-1);
$A=if($R,$R,1);
$S=rint(-$D+1,$D-1);
$C=if($S,$S,1);

# Calculate the numerator multiplied by the denominator. We'll do this step expressly
# because we'll want it for our feedback.
$X=($A*$C);
$Y=($B*$D);

#Calculate the greatest common factor and the answer.
$G=if(gcd($X,$Y),gcd($X,$Y),1);
$ANS=frac($X,$Y);

Dear All

Say I have created a Maple module called MYMODULE. This means that I have a .mpl file called

mymodule.mpl, with inside

   MYMODULE := module()

   export myfunction;

   ****

   ****

   end module:

   savelib(MYMODULE, "MYMODULE.mla"):

Hence I use it in such a way:

> with(MYMODULE);

                                                              [myfunction]

I want to associate a help to the function myfunction. I know that with Maple18 things have changed

but I am not able to find enough information online. Say I have a .txt file called help_myfunction.txt.

How can I enable this help for my function, so that at the command

> ?myfunction;

the content of help_myfunction.txt shows up?

 

Many thanks,Simone

 

I have two sets 

f:={1,2,3,4};

h:={1,2,4,5}

L=seq(i,i=1..4):

I want to program , if it is true that f[i] = h[i], then it prints f[i]. So the output should be {1,2}.

for i in L do
if evalb(f[i]=h[i]) then
print(f[i]);
end if;
end do;

I get no output. https://i.imgur.com/qA5hU3i.png

I tried changing the set f to list, f:= [1,2,3,4], still no output.

Hi...

 

I want to solve  this ODEs using Maple...

f’’’ + f f’’ – f’2 + λ θ = 0

(1/Pr) θ’’ + f θ’ – f’ θ = 0


and boundary conditions

f(0) = s,  f’(0) = σ + a f’’(0) + b f’’’(0),  θ(0) = 1,

f’(η) = 0, θ(η) = 0  as  η → ∞.

 

Can anyone help me to solve this problem? Thank you... =)

I want do loop with comand for to generate 20 arrays with radon variable 

I do

for i from 1 to 20 do
with(Statistics):
N:=RandomVariable(Normal(0,1)):
F_i:=Sample(N,20):
for k from 1 to 20 do
X[k] := evalf(F[k])
end do:
end do;

 

But bug all

hi 

how i can apply this differential in maple?

thabks...

 

hi .please help me for solve this equations.

bbb2.mw

restart; d[11] := 1; mu[11] := 1; q[311] := 1; d[33] := 1; mu[33] := 1; a[11] := 1; e[311] := 1; a[33] := 1; A := 1; g[111111] := 1; c[1111] := 1; g[113113] := 1; f[3113] := 1; beta[11] := 1; `ΔT` := 1; II := 1; L := 1

J := d[11]*(diff(Phi(x, z), x, x))+mu[11]*(diff(psi(x, z), x, x))+q[311]*(diff(w(x), x, x))+d[33]*(diff(Phi(x, z), z, z))+mu[33]*(diff(psi(x, z), z, z));

diff(diff(Phi(x, z), x), x)+diff(diff(psi(x, z), x), x)+diff(diff(w(x), x), x)+diff(diff(Phi(x, z), z), z)+diff(diff(psi(x, z), z), z)

(1)

B := a[11]*(diff(Phi(x, z), x, x))+d[11]*(diff(psi(x, z), x, x))+e[311]*(diff(w(x), x, x))+a[33]*(diff(Phi(x, z), z, z))+d[33]*(diff(psi(x, z), z, z));

diff(diff(Phi(x, z), x), x)+diff(diff(psi(x, z), x), x)+diff(diff(w(x), x), x)+diff(diff(Phi(x, z), z), z)+diff(diff(psi(x, z), z), z)

(2)

R := A*(g[111111]*(diff(u[0](x), x, x, x, x))-c[1111]*(diff(u[0](x), x, x)+(1/2)*(diff((diff(w(x), x))^2, x)))+e[311]*(diff(diff(Phi(x, z), z), x))+q[311]*(diff(diff(psi(x, z), z), x)));

diff(diff(diff(diff(u[0](x), x), x), x), x)-(diff(diff(u[0](x), x), x))-(diff(w(x), x))*(diff(diff(w(x), x), x))+diff(diff(Phi(x, z), x), z)+diff(diff(psi(x, z), x), z)

(3)

S := -II*g[111111]*(diff(w(x), x, x, x, x, x, x))-II*c[1111]*(diff(w(x), x, x, x, x))+A*g[113113]*(diff(w(x), x, x, x, x))-A*f[3113]*(diff(diff(Phi(x, z), z), x, x))-A*(c[1111]*(diff(u[0](x), x, x)+(1/2)*(diff((diff(w(x), x))^2, x)))+e[311]*(diff(diff(Phi(x, z), z), x))+q[311]*(diff(diff(psi(x, z), z), x)))*(diff(w(x), x))-A*(diff(w(x), x, x))*(c[1111]*(diff(u[0](x), x)+(1/2)*(diff(w(x), x))^2)+e[311]*(diff(Phi(x, z), z))+q[311]*(diff(psi(x, z), z))-beta[11]*`ΔT`);

-(diff(diff(diff(diff(diff(diff(w(x), x), x), x), x), x), x))-(diff(diff(diff(Phi(x, z), x), x), z))-(diff(diff(u[0](x), x), x)+(diff(w(x), x))*(diff(diff(w(x), x), x))+diff(diff(Phi(x, z), x), z)+diff(diff(psi(x, z), x), z))*(diff(w(x), x))-(diff(diff(w(x), x), x))*(diff(u[0](x), x)+(1/2)*(diff(w(x), x))^2+diff(Phi(x, z), z)+diff(psi(x, z), z)-1)

(4)

dsys := {B, J, R, S}; BCS := {D@@2*w(0) = 0, D@@2*w(L) = 0, Phi(x = 0) = 0, Phi(x = L) = 0, Phi(z = -(1/2)*h) = 0, Phi(z = (1/2)*h) = 0, psi(x = 0) = 0, psi(x = L) = 0, psi(z = -(1/2)*h) = 0, psi(z = (1/2)*h) = 0, w(x = 0) = 0, w(x = L) = 0, u[0](x = 0) = 0, u[0](x = L) = 0, (D(w))(0) = 0, (D(w))(L) = 0, (D(u[0]))(0) = 0, (D(u[0]))(L) = 0}

{D@@2*w(0) = 0, D@@2*w(L) = 0, Phi(x = 0) = 0, Phi(x = L) = 0, Phi(z = -(1/2)*h) = 0, Phi(z = (1/2)*h) = 0, psi(x = 0) = 0, psi(x = L) = 0, psi(z = -(1/2)*h) = 0, psi(z = (1/2)*h) = 0, w(x = 0) = 0, w(x = L) = 0, u[0](x = 0) = 0, u[0](x = L) = 0, (D(w))(0) = 0, (D(w))(L) = 0, (D(u[0]))(0) = 0, (D(u[0]))(L) = 0}

(5)

dsol5 := dsolve(dsys, numeric)

Error, (in dsolve/numeric/process_input) missing differential equations and initial or boundary conditions in the first argument: dsys

 

NULL

NULL

NULL

if former equations are not solvable , please help me for another way, in which at first two equation solve..in this way in equation [J and B] assume that q[311]=e[311]=0 and dsolve perform to find Φ and  ψ

after by finding Φ and  ψ is use for detemine w and u0

please see attached file below[bbb2_2.mw]

bbb2_2.mw

Download bbb2.mw

I have the following PDE system steaming from Flash Photolysis:

pdesys := [diff(J(x, t), x) = varepsilon*J(x, t)*C(x, t), diff(C(x, t), t) = phi*varepsilon*J(x, t)*C(x, t)]

when I use pdsolve(pdesys,[J,C]) I get:

{C(x, t) = 0, J(x, t) = _F1(t)}, {C(x, t) = _F1(x)*_C1, J(x, t) = 0}

The solution appears to be either C(x,t) = 0 or J(x,t) = 0. This are the obvious solutions (0 = 0). I have the analytical solution to this PDE system where neither C(x,t) nor J(x,t) are 0.

How to solve this system in maple? Thanks.

 

Hi

In attached mw file I have a equation that "a" is in unknown parameter. X and Y are matrixes of data. How can I calculate the "a" parameter?

Best regard

question.mw

Example 6 from Maple Help.

 

restart:

with(Optimization):

LPSolve(2*x+5*y, {3*x-y = 1, x-y <= 5}, assume = {integer, nonnegative})

 

Kernel crash the same with Maple 2015.

 

Does anyone can confirm?

Mariusz Iwaniuk

hi.please help me for solve this equation

i encounter with error''

Error, (in StringTools:-IsPrefix) second argument must be a string''

equations which be solved attached as pdf file

thanks

Kernel4.mw

root.pdf


 

restart

with(LinearAlgebra):

Typesetting:-Settings(functionassign=false):

NULL

Constants

 

landa := 0.404e11; -1; mu := 0.27e11; -1; alpha := 0.23e-4; -1; rho := 2707; -1; k := 204; -1; c := 903; -1; nu := .3; -1; E := 0.70e11; -1; T0 := 293; -1; omega := 0.1e-1

0.1e-1

(1.1.1)

beta := alpha*(3*landa+2*mu):

NULL

varpi := 0.1e-1; -1; No := 15

15

(1.1.2)

 

 

Eq[1] := besselj(0, xi*b)*(eval(diff(bessely(0, xi*r), r), r = a))-(eval(diff(besselj(0, xi*r), r), r = a))*bessely(0, xi*b):

 

wf1 := unapply(Eq[1], xi):

1

 

1.794010904

 

1

 

2

 

1.794010904

 

1

 

3

 

4.802060761

 

2

 

4

 

4.802060761

 

2

 

5

 

4.802060761

 

2

 

6

 

7.908961712

 

3

 

7

 

7.908961712

 

3

 

8

 

7.908961712

 

3

 

9

 

11.03509457

 

4

 

10

 

11.03509457

 

4

 

11

 

11.03509457

 

4

 

12

 

11.03509457

 

4

 

13

 

14.16798650

 

5

 

14

 

14.16798650

 

5

 

15

 

14.16798650

 

5

 

16

 

17.30400975

 

6

(1.2.1)

Eq[2] := MTM:-besselj(1, eta*b)*(eval(diff(MTM:-bessely(1, eta*r), r), r = a))-(eval(diff(MTM:-besselj(1, eta*r), r), r = a))*MTM:-bessely(1, eta*b):

wf2 := unapply(Eq[2], eta):

1

 

1.958510605

 

1

 

2

 

1.958510605

 

1

 

3

 

4.857021628

 

2

 

4

 

4.857021628

 

2

 

5

 

4.857021628

 

2

 

6

 

7.941288451

 

3

 

7

 

7.941288451

 

3

 

8

 

7.941288451

 

3

 

9

 

11.05802155

 

4

 

10

 

11.05802155

 

4

 

11

 

11.05802155

 

4

 

12

 

11.05802155

 

4

 

13

 

14.18576207

 

5

 

14

 

14.18576207

 

5

 

15

 

14.18576207

 

5

 

16

 

17.31852918

 

6

(1.2.2)

 

for m to MM do K0[m] := proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc; KK0[m] := proc (r, m) options operator, arrow; diff(K0[m](r, m), r) end proc; K1[n] := proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc; KK1[n] := proc (r, n) options operator, arrow; diff(K1[n](r, n), r) end proc end do

proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc

 

proc (r, m) options operator, arrow; MTM:-diff(K0[m](r, m), r) end proc

 

proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc

 

proc (r, n) options operator, arrow; MTM:-diff(K1[n](r, n), r) end proc

 

proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc

 

proc (r, m) options operator, arrow; MTM:-diff(K0[m](r, m), r) end proc

 

proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc

 

proc (r, n) options operator, arrow; MTM:-diff(K1[n](r, n), r) end proc

 

proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc

 

proc (r, m) options operator, arrow; MTM:-diff(K0[m](r, m), r) end proc

 

proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc

 

proc (r, n) options operator, arrow; MTM:-diff(K1[n](r, n), r) end proc

 

proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc

 

proc (r, m) options operator, arrow; MTM:-diff(K0[m](r, m), r) end proc

 

proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc

 

proc (r, n) options operator, arrow; MTM:-diff(K1[n](r, n), r) end proc

 

proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc

 

proc (r, m) options operator, arrow; MTM:-diff(K0[m](r, m), r) end proc

 

proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc

 

proc (r, n) options operator, arrow; MTM:-diff(K1[n](r, n), r) end proc

 

proc (r, m) options operator, arrow; BesselJ(0, xi[m]*r)*BesselY(0, xi[m]*b)-BesselJ(0, xi[m]*b)*BesselY(0, xi[m]*r) end proc

 

proc (r, m) options operator, arrow; MTM:-diff(K0[m](r, m), r) end proc

 

proc (r, n) options operator, arrow; BesselJ(1, eta__n*r)*BesselY(1, eta__n*b)-BesselJ(1, eta__n*b)*BesselY(1, eta__n*r) end proc

 

proc (r, n) options operator, arrow; MTM:-diff(K1[n](r, n), r) end proc

(1.2.3)

U1 := -(int(r*K0[m]*(diff(K1[n], r)+K1[n]/r), r = a .. b))/(int(r*K0[m]^2, r = a .. b)); -1; U2 := -(int(r*K1[n]*(diff(K0[m], r)), r = a .. b))/(int(r*K1[n]^2, r = a .. b)); -1; U3 := (int(r^2*omega^2*K1[n], r = a .. b))/(int(r*K1[n]^2, r = a .. b))

0.1555555555e-3/K1[0]

(1.2.4)

m := 0; -1; for m to MM do M__m := int(r*K1[m](r, m)^2, r = a .. b); bb__m := 1/M__m end do

int(r*K1[1](r, 1)^2, r = 1 .. 2)

 

1/(int(r*K1[1](r, 1)^2, r = 1 .. 2))

 

int(r*K1[2](r, 2)^2, r = 1 .. 2)

 

1/(int(r*K1[2](r, 2)^2, r = 1 .. 2))

 

int(r*K1[3](r, 3)^2, r = 1 .. 2)

 

1/(int(r*K1[3](r, 3)^2, r = 1 .. 2))

 

int(r*K1[4](r, 4)^2, r = 1 .. 2)

 

1/(int(r*K1[4](r, 4)^2, r = 1 .. 2))

 

int(r*K1[5](r, 5)^2, r = 1 .. 2)

 

1/(int(r*K1[5](r, 5)^2, r = 1 .. 2))

 

int(r*K1[6](r, 6)^2, r = 1 .. 2)

 

1/(int(r*K1[6](r, 6)^2, r = 1 .. 2))

(1.2.5)

MM; 1; n; 1; m; 1; U1; 1; U2; 1; U3; 1; xi

6

 

0

 

7

 

-(2/3)*K1[0]/K0[7]

 

0

 

0.1555555555e-3/K1[0]

 

xi

(1.2.6)

for m to MM do for n to MM do dsys := {diff(S[m][n](t), t, t, t)+xi^2*[m]*(diff(S[m][n](t), t, t))+(-U1*U2+`&eta;__&eta;__n__`^2)*(diff(S[m][n](t), t))+xi[m]^2*`&eta;__&eta;__n__`^2*S[m][n](t) = -(2*U2*bb[m]/(Pi*xi[m])*(-BesselJ(0, xi[m]*b)/BesselJ(1, xi[m]*a)))*q+xi^2*[m]*U3} end do end do; sol := dsolve(dsys)

Error, (in StringTools:-IsPrefix) second argument must be a string

 

 

NULL

for m to MM do for n to MM do dsys2 := {diff(Q__mn(t), t, t, t)+xi[m]^2*(diff(Q__mn(t), t, t))+(-U1*U2+eta__n^2)*(diff(Q__mn(t), t))+xi[m]^2*eta__n^2*Q__mn(t) = -2*BesselJ(0, xi[m]*b)*U1*U2*b__m*(1-exp(-xi[m]^2*t))/(BesselJ(1, xi[m]*a)*Pi*xi[m]^3)} end do end do;

sol2 := dsolve(dsys2)

Error, (in dsolve) invalid input: `PDEtools/sdsolve` expects its 1st argument, SYS, to be of type Or(set({`<>`, `=`, algebraic}), list({`<>`, `=`, algebraic}), `casesplit/ans`(list, list)), but received [{Q__mn(t)*pochhammer(1-n, n)+(1497143767/5000000)*(diff(Q__mn(t), [`$`(t, t)]))+eta__n^2*(diff(Q__mn(t), t))+(1497143767/5000000)*eta__n^2*Q__mn(t) = 0}]

 

``

NULL

NULL

 

Download Kernel4.mw

Hi

First question is, given a list of some positive integers, how can I normalize this list? Normalize here is in the sense that, for example, if 

L := [1,3,4,4,5,7,7,7,8]

then there really are only 6 different integers appear, so I would like to assign each part an integer from 1 to 6 in ascending order. So 1 becomes 1, 3 becomes 2, 4 becomes 3, 5 becomes 4, 7 becomes 5 and so on. Normalized list will be

NormalizedL := [1,2,3,3,4,5,5,5,6]

 

Second question is given a list, let's say [1,3,1,3,2,2,4,4], how can I normalize it in a similar way but now we assign each integer upon occurrence of a part. So [1,3,1,3,2,2,4,4] will be [1,2,1,2,3,3,4,4]. This is necessary for me because lists have repeating parts. 

Another example will be [2,4,4,1,2,2,3,3] will be [1,2,2,3,1,1,4,4]. 

Thank you 

How to know how long the animation last and how to stop gif at the end.

I created a GIF from exporttool but I don't know how long is the animation ? I mean when it stops at the end of value. Or how to make GIF run as real time. because I make a gif for showing how a thing fall down from a attitude. And I need to show some figure like t (time) , h ( high ), v( speed ) thing like that while it falling down.

Thanks

My query is related to Maple packages "Lie algebra" and "DifferentialGeometry", I wonder can these packages help me to check isomophism between to Lie algebras, for example, suppose I have a 4-dim Lie algebra with commutation relations:

[e2, e3] = e3, [e2, e4] = - e4, [e3, e4] = - e1 

and another 4-dim Lie algebra with commutation relations

[e2, e3] = e1, [e4, e2] = e2, [e3, e4] = - e3

Can I check for isomophism between these Lie algebras if there is any using Lie algebra package ?

 

Regards

How to hum program repeat several times and save what data result from each time it runs?

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