Hi!

I want to implement to attached fortran program in Maple 2015 (the procudure starts at the end of the first page). 

localmin.pdf

The code does not seem dificult, but I don't know how to interpret the instructions "go to" of fortran. Reading Maple's doc about the "goto" instruction, I don't understand how to implement it.

Can somebody help with this code, please?

Many thanks in advance for your comments.

Hi, 
I met a an unexpected behaviour of a procedure when the parameter sequence contains the type ':-RandomVariable':  

restart:
with(Statistics):
f1 := proc(A::':-RandomVariable')
  Mean(A)
end proc:

Z := RandomVariable(Normal(mu, sigma)):
hastype(Z, ':-RandomVariable');
f1(Z)
                              true
Error, invalid input: f1 expects its 1st argument, A, to be of type 'RandomVariable', but received _R
# Another attempt
f2 := proc(A)
  if hastype(A, ':-RandomVariable') then Mean(A) end if;
end proc:

f2(Z)
                               mu

Why does f1 generate this error?

Type_RandomVariable.mw

Version used MAPLE 2015.2

I use dualaxisplot and want a logarithmic horizontal axis.
I can get the expected graphic using either of these two methods

restart:
with(plots):
dualaxisplot(
  plot(x, x=0..1, color=blue, axis[1]=[mode="log"]),
  plot(1-x, x=0..1, color=red, axis[1]=[mode="log"]),
  gridlines=true
):

dualaxisplot(
  semilogplot(x, x=0..1, color=blue),
  semilogplot(1-x, x=0..1, color=red),
  gridlines=true
):

This worksheet contains my true code
(pink lines correspond to < plot + axis[1]=[mode="log"] > and blue lines to < semilogplot >).
You will see that the vertical axis for the plot declared in second position is logarithmic, whatever the way the logarithmic x-axis is defined. 
DecisionProb.mw

I am unable to find out if I made a mistake (which is likely) or if it is a bug?
Could you please have a look at this code and give me an answer?
Thanks in advance

I get a problem while selecting elements from sets.

I have a list of two sets apparently ordered in the same order.
When I pick this, for instance, first set from each of these two samples I get the first one from the first set, and the second one from the second set.

Download Selection_from_a_set.mw

Running this command

map(sort, LDF)

displays LDF, which suggests that LDF is already sorted according to some order.
What is this order which makes diff(Theta(x), x$2) the successor of diff(Theta(x), x)  but diff(g(x), x) the successor of diff(g(x), x$2)  ?

Dear Users,

I hope everyone here is fine. I want to extract dat file from the attached contour plot file. Please help me to resolve this issue. Many thanks in advance

Dat_file.mw

Hi everyone, I'm studying physics at college and we have to do lab reports. Of course we have to fit data sets. They only taught us how to calculate errors of the parameters of lineaer distributions.

Unforutnately I have to fit points that follow these functions:

First of all, which command do I use to fit for the second function? (Its the formula of  in function of the frequence in RC circuits)

I know for the first one there's the PowerFit command.

Also, is there a command that returns the errors for the variaous fit parameters? (i.e. a,b,c,d)

Or at least, someone knows the methods and formulas to find them?

Dear Users,
I hope you are doing well. The following is the code to solve a nonlinear PD equation numerically and I plotted the graphs for T(y,t) sucessfully.

restart; with(plots); PDE1 := Pr*(diff(T(y, t), t)-Ree*(diff(T(y, t), y))) = (1+Nr*(T(y, t)+1)^3)*(diff(T(y, t), y, y))+3*Nr*(T(y, t)+1)^2*(diff(T(y, t), y))^2; ICandBC := {T(1, t) = 1, T(y, 0) = 1, (D[1](T))(0, t) = T(0, t)}; Ree := .1; Pr := 6.2; HA1 := [0, 1, 10]; AA := [red, green, blue, cyan, purple, black];
printlevel := 2; for i to nops(HA1) do Nr := op(i, HA1); print("Nr = ", %); PDE[i] := {PDE1}; pds[i] := pdsolve(PDE[i], ICandBC, numeric, spacestep = 1/200, timestep = 1/100); PlotsT[i] := pds[i]:-plot[display](T(y, t), t = 1, linestyle = "solid", labels = ["y", "u"], color = op(i, AA), numpoints = 800) end do;
display([`$`(PlotsT[j], j = 1 .. nops(HA1))], size = [1000, 600], axes = boxed, labels = [x, (convert("T", symbol))(x, T)], labelfont = ["Times", 14, Bold], labeldirections = [horizontal, vertical], axesfont = ["Arial", 14, Bold], thickness = 3)

I want to plot the graphs for (1+Nr*(T(y, t)+1)^3)*(diff(T(y, t), y)), at t = 1. Also want to plot diff(T(y, t), y) at y = 0 and y = 1 against Nr. Kindly help me in this matter.

Dear Users!
I hope are fine here. I got the following expression after a lot of computations

((1/2)*r*(r-1)+(1/6)*r*(r-1)*(r-2))*`&Delta;y`[-1]^3+(1/2)*r*(r-1)*`&Delta;y`[-1]^2+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4)*`&Delta;y`[-2]^7+((1/6)*r*(r-1)*(r-2)+(1/12)*r*(r-1)*(r-2)*(r-3)+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4))*`&Delta;y`[-2]^5+((1/6)*r*(r-1)*(r-2)+(1/24)*r*(r-1)*(r-2)*(r-3))*`&Delta;y`[-2]^4+((1/24)*r*(r-1)*(r-2)*(r-3)+(1/60)*r*(r-1)*(r-2)*(r-3)*(r-4))*`&Delta;y`[-3]^7+((1/24)*r*(r-1)*(r-2)*(r-3)+(1/60)*r*(r-1)*(r-2)*(r-3)*(r-4))*`&Delta;y`[-3]^6+r*`&Delta;y`[0]+y[0]

Actually, for the above, I want the factorization of each coefficient of `&Delta;y`[0], `&Delta;y`[-1], `&Delta;y`[-2] etc and the above expression shoud be in descending order given as:

y[0]+r*`&Delta;y`[0]+(1/2)*r*(r-1)*`&Delta;y`[-1]^2+(1/6)*r*(r-1)*(1+r)*`&Delta;y`[-1]^3+(1/24)*r*(r-1)*(r-2)*(1+r)*`&Delta;y`[-2]^4+(1/120)*r*(r-1)*(r-2)*(r+2)*(1+r)*`&Delta;y`[-2]^5+(1/120)*r*(r-1)*(r-2)*(r-3)*(r-4)*`&Delta;y`[-2]^7+(1/120)*r*(r-1)*(r-2)*(r-3)*(-3+2*r)*`&Delta;y`[-3]^6+(1/120)*r*(r-1)*(r-2)*(r-3)*(-3+2*r)*`&Delta;y`[-3]^7

I am waiting for your positive response. Thanks

Does anyone have any idea why Maple can obtain a closed form of 

Int(f, t=1..3);

but doesn't for 

Int(g, t=1..3);

?

The Int forms are quite close and I don't understand what makes Maple's task that difficult in the second case (the issue seems to come from the conditions in the piecewise function).
Can we force Maple to perform the second integration ?

TIA

Download MyIntegral.mw

I'm working on finding the analytic expression of the PDF of a sum of abstract Uniform Random Variables URV).
Here "abstract" means that the supports are not numeric but litteral.

Maple is capable to find such a PDF for numeric supports but unable to determine the PDF of U1+U2 where 

U1 := RandomVariable(Uniform(a1, b1)):
U2 := RandomVariable(Uniform(a2, b2)):

A way to deal with abstract URV is to complute explicitely the convolution product of th PDFs.
Ir seems that this fails (MAPLE 2015.2) for these PDF are piecewise functions.
A workaround is to convert them first into Heaviside(s).

One done the explicit expression of the convolution product can be obtained for a sum of 2 abstract URVs, but not for a sum of a larger number of abstract URVs.

Thus the second workaround which consists in using direct and inverse Fourier transform.

The question is :
obviously, the PDF f(t ; a1...aN, b1...bN) of  U1 + ... + UN is a continuous function of t: why does discont(f(t ; ...), t) returns the non empty set of the values where the Heaviside functions are undefined ?

Here is a very simple result

u1 := RandomVariable(Uniform(-1, 1)): 
u2 := RandomVariable(Uniform(-1, 1)):
p := PDF(u1+u2, t):
discont(p, t);

print("-------------------------------------");

f1 := convert(PDF(u1, t), Heaviside):
f2 := convert(PDF(u2, t), Heaviside):
g1 := fourier(f1, t, xi):
g2 := fourier(f2, t, xi):
g  := g1*g2:
f  := invfourier(g, xi, t):
discont(f, t);

                               {}
            "-------------------------------------"
                           {-2, 0, 2}

The problem is (IMO) that discont(f, t) should return { }, but that some function (does it exists) should say that d is undefined at points t=-2, t=0, t=2.
The output of discont(f, t) doesn't seem consistent with the definition of the continuity

limit(f, t=-2, left);
limit(f, t=-2, right);
eval(f, t=-2)
                               0
                               0
                           undefined

Download Sum_of_Uniform_RVs.mw

Consider this piece of program

restart:
interface(version)
Standard Worksheet Interface, Maple 2015.2, Mac OS X, December 21 2015 Build ID 1097895
with(Units):
a := 3*Unit('m');
                       3 Units:-Unit('m')

I would like to define a new quantity b which is the dimensionless variant of a.
I browsed the Units package to look for a function that would "remove" the dimension of a dimensional quantity like a above (that is to get '3' alone).

As I couldn't find such a function I use to use this workaround.

b := remove(has, a, Unit);
                               3

Is this a robust strategy?
What would you propose to "remove" the dimension of a?

Thanks in advance

Using plot3d(..., style=surfacecontour, ...) or contourplot3d(...) displays wrong level curves when some axis are switched to a log mode.

Example:

Download WrongLevelCurves.mw

The problem is not dramatic because there is a workaround.
 

Download WrongLevelCurves_Workaround.mw

Even though this question is related to this one 
https://www.mapleprimes.com/questions/234781-How-Can-I-Get-The-Desired-Answer-From-solve
feel it is about a different issue. If any of you feel otherwise feel free to move it to the original one.

In this notional example  the name _Z1~ is created by RootOf: and here is an ad hoc way to catch it.

Download Example_1.mw

In this more complex example an assumption must be made on M to obtain ths desired solution g and the previous method no longer works.

Download Example_2.mw

I have tried using select to "capture" the name _Z2~ but I can't know how to distinguish M~ from _Z2~ (is there a type which could be used?).

Can you helpm fix this?
TIA

I don't understand why the solution of sys_2 isn't those of sys_1 when M__p=1 and M__a=0 ?

Traces of the computation seem to indicate that dsolve proceeds exactly the same for sys_2 and sys_1 .

Please note that sol_1 contains a term of the form t*cos(t) that sol_2 doesn't, thus the question: "Is sol_2 correct?"

Could you help me to fix this?
TIA

Download SomethingWrong.mw

PS: Already, in the following case, dsolve doesn't return the solution of sys_1.

sys_3 := {(A+B)*diff(x(t), t$2)=(A+B)*sin(t)-x(t), x(0)=0, D(x)(0)=0};
sol_3 := dsolve(sys_3)

If I do this

sys_4 := {(A+B)*diff(v(t), t)=(A+B)*sin(t)-x(t), diff(x(t), t)=v(t), x(0)=0, v(0)=0}:
sol_4 := dsolve(sys_4)

I get a very complex solution wich contains a piecewise function which separates the cases A+B=1 and A+B<>1.
Evaluating sol_4 for A+B=1 gives the same expression than sys_1:

simplify(eval(sol_4, A=1-B), trig)
       /       1                  1          1         \ 
      { v(t) = - sin(t) t, x(t) = - sin(t) - - cos(t) t }
       \       2                  2          2         / 

Here is a workaround to get the correct solution of sys_2:

sys_5 := {(M__P+M__A)*diff(v(t), t)=(M__P+C)*sin(t)-x(t), diff(x(t), t)=v(t), x(0)=0, v(0)=0}:
sol_5 := dsolve(sys_5):
simplify(eval(sol_5, [M__P=1, M__A=0, C=0]), trig)
       /       1                  1          1         \ 
      { v(t) = - sin(t) t, x(t) = - sin(t) - - cos(t) t }
       \       2                  2          2         / 

e

I compute the solution of this differential system

shock := piecewise(t <0, 0, t < 1, 10, 0):
sys   := {(M__p+M__a)*diff(x(t), t$2)=M__p*shock-x(t), x(0)=0, D(x)(0)=0}
sol   := unapply(rhs(dsolve(sys)), (M__p,M__a))

I'm interested in 3 quantities:

• the first time tend > 0 such that sol(tend) = 0,
• the time tmax in (0..tend) where sol(tmax) reaches its maximum value,
• the value xmax = sol(tmax).

Since sol has a relatively simple expression, I first attempted to use solve for calculating tend, but that didn't work.
The conclusion is still the same for tmax and xmax.

The values of these 3 quantities that I expect solve to provide, are those obtained using fsolve.

Can you explain me the failures I faced and show me how to force solve to get these values?
TIA

ToyProblem.mw