Maple 2015 Questions and Posts

These are Posts and Questions associated with the product, Maple 2015

Hi, 

Here is the minimal situation that reveals which that could be a (little) bug in the MAPLE 2015 version of Explore.

In the attached file y is a list of numbers and val(r) a procedure that just print the value the rth element of y.
Changing the value of r is done with Explore (of course of no interest at all).
If I define the parameter r as a list ( r=[1..numelems(y)] ), only the value of y[1] is displayed: changing the value of r generates an error.
This doesn(t happen if r is defined as a slider ( r=1..numelems(r) ).

I discovered than the initial instance of Explore defines r as an integer while all the others (due to a change of r) define r as a string.
In the last command of the attached file you will see hjow I have circumvent this problem.

Is it a bug in Explore or does it exist some way to force the values of r to the implicit type they have in r=[$1..numelems(y)] ?

TIA

Download explore.mw

 

 

 

Hello,

I am looking help for solving this integral equation using the collocation method with 1.5<= x<=3.5 ? 

I have used the successive approximation method and the solution seems to be increasing.

Thanks

Hi, 

I have a list of integers (>1) and for all of them I define an alias (in the attached file I've tried two different names for them : a[n] or a||n) wich represents the nth roots of the unity.
When I apply the procedure allvalues to a specific alias it returns the algebraic values of the corresponding roots of the unity.
But when aplied to the list of aliases it gives me back only the name of the alias, not the algebraic values.

How can I fix this ?

TIA

restart:

interface(version)

`Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895`

(1)

for n from 2 to 3 do
  alias(a[n]=RootOf(z^n-1)):
end do:

alias();

a[2], a[3]

(2)

allvalues(a[2]);
allvalues(a[3]);

seq(allvalues(a[n]), n=2..3)

1, -1

 

1, -1/2+((1/2)*I)*3^(1/2), -1/2-((1/2)*I)*3^(1/2)

 

a[2], a[3]

(3)

for n from 2 to 3 do
  alias(a||n=RootOf(z^n-1)):
end do:

alias();

allvalues(a2);
allvalues(a3);

seq(allvalues(a||n), n=2..3)

a[2], a[3], a2, a3

 

1, -1

 

1, -1/2+((1/2)*I)*3^(1/2), -1/2-((1/2)*I)*3^(1/2)

 

a2, a3

(4)

A := [alias()]:
map(allvalues, A);

[a[2], a[3], a2, a3]

(5)

 


 

Download allvalues.mw

 

 

Hello,

I have the general function F(x,g(y)), and I don't understand the following notation of the derivative of  F w.r.t  y in Maple ?  What is D2(F) ?  Can we provide a concrete example ?

Thank you!

Dear Users!

Hope you would be fine with everything. I want to evaluate an expression (diff(u(y, t), y)+diff(diff(u(y, t), y), t)) for various values of b at y = 0, t=1. Please help me to evaluate it. Thanks in advance,

restart; with(plots); a := .7; L := 8; HAA := [0, 2, 5, 10];

for i to nops(HAA) do

b := op(i, HAA);

PDE1[i] := diff(u(y, t), t) = diff(u(y, t), y, y)+diff(diff(u(y, t), y, y), t)-b*u(y, t)+T(y, t);

PDE2[i] := diff(T(y, t), t) = (1+(1+(a-1)*T(y, t))^3)*(diff(T(y, t), y, y))+(a-1)*(1+(a-1)*T(y, t))^2*(diff(T(y, t), y))^2+T(y, t)*(diff(T(y, t), y, y))+(diff(T(y, t), y))^2;

ICandBC[i] := {T(L, t) = 0, T(y, 0) = 0, u(0, t) = t, u(L, t) = 0, u(y, 0) = 0, (D[1](T))(0, t) = -1};

PDE[i] := {PDE1[i], PDE2[i]}; pds[i] := pdsolve(PDE[i], ICandBC[i], numeric)

end do;
 

with(RegularChains):
with(ChainTools):
with(MatrixTools):
with(ConstructibleSetTools):
with(ParametricSystemTools):
with(SemiAlgebraicSetTools):
with(FastArithmeticTools):
R := PolynomialRing([x,y,z,a,b]):
sys := [x^2 + y^2 - x*y - 1 = 0, y^2 + z^2 - y*z - a^2 = 0, z^2 + x^2 - x*z - b^2 = 0,x > 0, y > 0, z > 0, a - 1 >= 0, b-a >= 0, a+1-b > 0]:

dec := RealTriangularize(sys,R): # very slow
Display(dec, R);

dec := LazyRealTriangularize(sys,R): # it is faster
dec2 := value(dec): # very slow
value(dec2); 

find a , b to satisfy sys have real solution

expect  one of solution is below, but above function are very slow, load a very time still no result, where is wrong?

R1 = a^2+a+1-b^2;
R1 = a^2-1+b-b^2;

[R1 > 0, R2 > 0]


 

Hi,

Is it possible to force Maple to simplify these Sum(s) ?
SimplifySum.mw
 

s := Sum(a*X[n]+b, n=1..N);
simplify(s);
value(s);  # part of the job done  but...


IWouldLikeToHave = a*Sum(X[n], n=1..N) + b*N; # or +N*b, it doesn't matter

Sum(a*X[n]+b, n = 1 .. N)

 

Sum(a*X[n]+b, n = 1 .. N)

 

N*b+sum(a*X[n], n = 1 .. N)

 

IWouldLikeToHave = a*(Sum(X[n], n = 1 .. N))+N*b

(1)

s := Sum(X[n]+Y[n], n=1..N);
(expand@value)(s);


IWouldLikeToHave = Sum(X[n], n=1..N) + Sum(Y[n], n=1..N);

Sum(X[n]+Y[n], n = 1 .. N)

 

sum(X[n]+Y[n], n = 1 .. N)

 

IWouldLikeToHave = Sum(X[n], n = 1 .. N)+Sum(Y[n], n = 1 .. N)

(2)

 

 

 

Thanks in advance

 

Hi, 

I can't find a unique way to define a mixture of two random variables that enables at the same time to compute its PDF and generate a sample.
In the attached file you will find two methods :

  1. the first one is the most formal and only CAS are supposed to be able to allow its implementation:
     it enables computing the PDF and the CDF (trivial) but fails to generate a sample.
    (PS : in a first attempt I had defined only the PDF, which should have been enough for method=envelope does work... ... at my opinion)
     
  2. the second method is the one one would use in non CAS languages (for instance Matlab, R, ...). It enables generating a sample (of course) but fails to compute the PDF (which is not very surprising).

So my question: does anyone here would have some suggestions to make one these two methods capable to compute both the PDF (eventually the CDF) and generate samples ?

PS: still have this kind of problems (tomleslie suggested months ago that using interface(rtablesize=10) could help but it's not the case here)
Maple Worksheet - Error
Failed to load the worksheet /maplenet/convert/Mixture_of_random_variables.mw .

 

Download Mixture_of_random_variables.mw

 

Hi 

I got, a bit by mistake, a weird result that seems to come from the way "Maple" manipulates infinity (please have a look at the results below)
I couldn't find a correct explanation to these results in the help pages dedicated to infinity or evaluation.

Could anyone explain me why, in some circumstances, 'infinity' seems to be a name (for instance infinity*Pi; does not return  infinity but infiniy/Pi).
Thanks in advance

 

restart:

with(Statistics):

f := PDF(Cauchy(0, 1), x)

1/(Pi*(x^2+1))

(1)

m := int(x*f, x=-infinity..+infinity);

undefined

(2)

p := int(x*f, x);

(1/2)*ln(x^2+1)/Pi

(3)

# Naive "proof" : eval(..., infinity) seems to consider infinity as a name (consistent with
# the first line in help(infinity) without any particular property.
# This seems to be confirmed by the value of p1 and p2 : I expected to get infinity instead
# of infinity/Pi.

p1 := eval(p, x=-infinity);
p2 := eval(p, x=+infinity);
p2 - p1;

infinity/Pi

 

infinity/Pi

 

0

(4)

# Correct "proof" :

a := limit(p, x=-infinity);
b := limit(p, x=+infinity);
b-a

infinity

 

infinity

 

undefined

(5)

# Does Maple treat 'infinity' as a number ?

infinity         - infinity;
infinity*2       - infinity*2;
infinity*(1/3)   - infinity*(1/3);
infinity*0.333   - infinity*0.333;
infinity*sqrt(2) - infinity*sqrt(2);
infinity/Pi      - infinity/Pi;

undefined

 

undefined

 

undefined

 

Float(undefined)

 

0

 

0

(6)

limit(sqrt(2)/x, x=0) - limit(sqrt(2)/y, y=0);
limit(Pi/x, x=0)      - limit(Pi/y, y=0);

undefined

 

undefined

(7)

 


 

Download Infinity.mw

Dear Users!

Hope you would be fine with everything. I have following code to generate marix A of order M by M

restart; with(LinearAlgebra); with(linalg); Digits := 30; M := 10; nu := 1;

for k1 while k1 <= M do

C[k1] := simplify(sum((-1)^(k1-1-i1)*GAMMA(k1-1+i1+2*nu)*GAMMA(nu+1/2)*x^i1/(GAMMA(i1+nu+1/2)*factorial(k1-1-i1)*factorial(i1)*GAMMA(2*nu)), i1 = 0 .. k1-1))

end do;

A := evalm(Matrix(M, M, proc (i, j) options operator, arrow; eval(C[j], x = (i-1)/(M-1)) end proc))

I want to split (or decompose) A into two parts Ad and Ab 

A = Ab + Ad

where Ad is M by M matrix of all entries of A but first and last rows of Ad shoud be zero

and Ab is M by M matrix with zero entries expect first and last rows.

For exmaple for M = 5, A, Ab and Ad are given as,

Ab := Matrix(5, 5, {(1, 1) = 1, (1, 2) = -2, (1, 3) = 3, (1, 4) = -4, (1, 5) = 5, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (2, 4) = 0, (2, 5) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0, (3, 4) = 0, (3, 5) = 0, (4, 1) = 0, (4, 2) = 0, (4, 3) = 0, (4, 4) = 0, (4, 5) = 0, (5, 1) = 1, (5, 2) = 2, (5, 3) = 3, (5, 4) = 4, (5, 5) = 5});

Ad := Matrix(5, 5, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (1, 4) = 0, (1, 5) = 0, (2, 1) = 1, (2, 2) = -1, (2, 3) = 0, (2, 4) = 1, (2, 5) = -1, (3, 1) = 1, (3, 2) = 0, (3, 3) = -1, (3, 4) = 0, (3, 5) = 1, (4, 1) = 1, (4, 2) = 1, (4, 3) = 0, (4, 4) = -1, (4, 5) = -1, (5, 1) = 0, (5, 2) = 0, (5, 3) = 0, (5, 4) = 0, (5, 5) = 0});

Please help me to fix this problem.
Special request @acer @Carl Love @Kitonum @Preben Alsholm

Hi,

How can I force the command InsertContent(Worksheet(Group(Input( T )))) to display the variable eq as it appears in label (2) ?

(a screen capture of the output of InsertContent(Worksheet(Group(Input( T )))) is given after the Maple code)

restart:

interface(version)

`Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895`

(1)

with(DocumentTools):

with(DocumentTools[Layout]):

eq := piecewise(t < 1, sin(t), cos(t));

C := Cell( Textfield(style=TwoDimOutput,Equation(eq)) ):
T := Table(Column(), widthmode=percentage, width=40, Row(C)):
InsertContent(Worksheet(Group(Input( T )))):

eq := piecewise(t < 1, sin(t), cos(t))

(2)

 



Download Layout.mw

Hi User!

Hope you would be fine with everything. I have a vector "POL" of M dimension obatined for the following expression

restart; with(LinearAlgebra); nu := 1; M := 3;
for k while k <= M do
Poly[k] := simplify(sum(x^i*GAMMA(nu+1)/(factorial(i)*GAMMA(2*nu)), i = 0 .. k-1))
end do;
POL := `<,>`(seq(Poly[k], k = 1 .. M))

and I want to construct a matrix of M by M by collocating it on the points x=i/(M-1) for i=0,1,2,...,M-1 like the following way,

For M=3 I need

Matrix(3, 3, {(1, 1) = Poly[1](0), (1, 2) = Poly[1](1/2), (1, 3) = Poly[1](1), (2, 1) = Poly[2](0), (2, 2) = Poly[2](1/2), (2, 3) = Poly[2](1), (3, 1) = Poly[3](0), (3, 2) = Poly[3](1/2), (3, 3) = Poly[3](1)});

For M=4 I need

Matrix(4, 4, {(1, 1) = Poly[1](0), (1, 2) = Poly[1](1/3), (1, 3) = Poly[1](2/3), (1, 4) = Poly[1](1), (2, 1) = Poly[2](0), (2, 2) = Poly[2](1/3), (2, 3) = Poly[2](2/3), (2, 4) = Poly[2](1), (3, 1) = Poly[3](0), (3, 2) = Poly[3](1/3), (3, 3) = Poly[3](2/3), (3, 4) = Poly[3](1), (4, 1) = Poly[4](0), (4, 2) = Poly[4](1/3), (4, 3) = Poly[4](2/3), (4, 4) = Poly[4](1)})

 

and general form is like this

[[[Poly[1](0/(M-1)),Poly[1](1/(M-1)),Poly[1]((2)/(M-2)),...,Poly[1]((M-1)/(M-1))],[Poly[2](0/(M-1)),Poly[2]((1)/(M-1)),Poly[2]((2)/(M-1)),...,Poly[2]((M-1)/(M-1))],[Poly[3]((0)/(M-1)),Poly[3]((1)/(M-1)),Poly[3]((2)/(M-1)),...,Poly[3]((M-1)/(M-1))],[...,...,...,...,...],[Poly[M]((0)/(M-1)),Poly[M]((1)/(M-1)),Poly[M]((2)/(M-1)),...,Poly[M]((M-1)/(M-1))]]];

Another problem is I want to define a vector of M dimension using a function f(x)=sin(x) and two points a=1, b=2 like the following way,

Vec:=[[[a],[f((1)/(M-1))],[f((2)/(M-1))],[f((3)/(M-1))],[...],[f((M-1)/(M-1))],[b]]]
Please fix my problem. I'm waiting for your kind response.
Special request @acer @acer @Carl Love @Kitonum @Preben Alsholm

Dear Users!

Hope you would be fine with everything. I want the simpliest for of the following expression in two step:

diff(U(X, Y, Z, tau), tau)+U(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), X))+V(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), Y))+W(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), Z))+u[delta]*lambda[1]*(diff(U(X, Y, Z, tau), tau, tau))/L[delta]+u[delta]*lambda[1]*(diff(U(X, Y, Z, tau), tau))*(diff(U(X, Y, Z, tau), X))/L[delta]+u[delta]*lambda[1]*U(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), tau, X))/L[delta]+u[delta]*lambda[1]*(diff(V(X, Y, Z, tau), tau))*(diff(U(X, Y, Z, tau), Y))/L[delta]+u[delta]*lambda[1]*V(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), tau, Y))/L[delta]+u[delta]*lambda[1]*(diff(W(X, Y, Z, tau), tau))*(diff(U(X, Y, Z, tau), Z))/L[delta]+u[delta]*lambda[1]*W(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), tau, Z))/L[delta];
Step 1:
diff(U(X, Y, Z, tau), tau)+U(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), X))+V(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), Y))+W(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), Z))+u[delta]*lambda[1]*(diff(diff(U(X, Y, Z, tau), tau)+U(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), X))+V(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), Y))+W(X, Y, Z, tau)*(diff(U(X, Y, Z, tau), Z)), tau))/L[delta];
Step 2: (final form I need)
(1+(u[delta] lambda[1])/(L[delta]) (&PartialD;)/(&PartialD;tau)) ((&PartialD;)/(&PartialD;tau) U(X,Y,Z,tau)+U(X,Y,Z,tau) ((&PartialD;)/(&PartialD;X) U(X,Y,Z,tau))+V(X,Y,Z,tau) ((&PartialD;)/(&PartialD;Y) U(X,Y,Z,tau))+W(X,Y,Z,tau) ((&PartialD;)/(&PartialD;Z) U(X,Y,Z,tau)));
I'm waiting for your response.
Special request:
@acer @Carl Love @Kitonum @Preben Alsholm

Hi

I would like to use  the Liebniz notation that someone from the technical support posted here
Writing Derivatives at a Point Using Leibniz Notation
to display a formula that is not just a partial derivative but a more complex expression invoking partial derivatives. 
Typically an expression like this one:

2*(Diff(f(mu__1, mu__2), mu__1))^2*lambda__1^2-(Diff(f(mu__1, mu__2), mu__1))^2*mu__1^2+2*(Diff(f(mu__1, mu__2), mu__2))^2*lambda__2^2-(Diff(f(mu__1, mu__2), mu__2))^2*mu__2^2+2*(Diff(f(mu__1, mu__2), mu__1))*(Diff(f(mu__1, mu__2), mu__2))*lambda__1*lambda__2-2*(Diff(f(mu__1, mu__2), mu__1))*mu__1*(Diff(f(mu__1, mu__2), mu__2))*mu__2

Could anyone help me to do this?
Thanks in advance

(PS: I'm still using Maple 2015.2)

Dear Users!

Hoped everything going fine with you. I want to make animation of ten solutions as given bellow but fail to do that. Please see it fix the problem. I shall be very thankful to u.
SOLNSuy[1, 1] := 2.5872902469406659197*10^(-20)-.65694549571241255901*y+1.9708364871372376767*y^2-1.3138909914248251176*y^3-1.6010739356637904911*10^(-19)*y^4;
SOLNSuy[2, 1] := -4.002204462000*10^(-20)-1.7879176897079605225*y+5.3637530691192141414*y^2-3.5758353794044226250*y^3-6.8309939211286845440*10^(-12)*y^4;
SOLNSuy[3, 1] := -1.1953264450000*10^(-19)-3.2481690589079594122*y+9.7445071767154794599*y^2-6.4963381177952273213*y^3-1.2292726248071398400*10^(-11)*y^4;
SOLNSuy[4, 1] := -2.6720465500000*10^(-19)-4.9239979672954025921*y+14.771993901873204315*y^2-9.8479959345587718955*y^3-1.9029826928878336000*10^(-11)*y^4;
SOLNSuy[5, 1] := 3.416928541000*10^(-20)-6.7268498492441931137*y+20.180549547714413714*y^2-13.453699698443639810*y^3-2.6580790570532587008*10^(-11)*y^4;
SOLNSuy[6, 1] := -2.554122292000*10^(-20)-8.5884528335125514887*y+25.765358500514014457*y^2-17.176905666966875698*y^3-3.4587270427710613504*10^(-11)*y^4;
SOLNSuy[7, 1] := -9.206107680000*10^(-20)-10.456823708331499352*y+31.370471124965259849*y^2-20.913647416590986491*y^3-4.2774005353527132160*10^(-11)*y^4;
SOLNSuy[8, 1] := 1.9644186790000*10^(-19)-12.293003938471349390*y+36.879011815379230436*y^2-24.586007876856948223*y^3-5.0932823222176363520*10^(-11)*y^4;
SOLNSuy[9, 1] := -3.775112769000*10^(-19)-14.068404975282556550*y+42.205214925807397100*y^2-28.136809950465931724*y^3-5.8908824448577377280*10^(-11)*y^4;
SOLNSuy[10, 1] := 1.146281780000*10^(-19)-15.762658869974768890*y+47.287976609878780960*y^2-31.525317739837422477*y^3-6.6589592851037286400*10^(-11)*y^4;
plots[animate](plot, [SOLNSuy[A, 1], y = 0 .. 1], A = 1 .. 10);

Special request:
@acer @Carl Love @Kitonum @Preben Alsholm

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