Unanswered Questions

This page lists MaplePrimes questions that have not yet received an answer

I have noticed a few times now with Maple 2019. It looses kernel connection when it is sitting there idly. This time I observed it. Had saved a document after an intensive calculation. The memory used was about 30Gig. shortly after saving the cpu fan was running hard. I checked task manager and cpu was cycling to 100%, it was mserever. Then the memory usage droped to about 6gig and message as shown. During this time Maple screen down in the LH corner displayed "Ready", so it didn't think it was doing anything.
 

I am encountering problems solving a system of differential equations.

In the attached file, on the first try the boundary conditions are defined in H: = He, resulting in "Error, (in DEtools / convertsys) unable to compute coeff". On the second try, the boundary conditions are defined in H: = He + 0.00000000000000001 and this works.

What is the possible cause of the problem?

Thank you

problem.mw

Dear Community,

I run a MapleSim model from Maple. The simulation runs fine giving me the correct graphical plots, but I wonder how could I obtain also the numerical values of the probes vs. time? Sorry I could not figure it out. A matrix format would be perfect: 1st column time, the other columns probe 1 .. n values vs. time. (Files attached)

Tx for the kind help in advance,

best regards

Andras

RunMapleSim.mw , RCNetwork.msim

My question has two steps:

STEP 1:  The multiplication  of is defined as follows

 

if m<>l, then

.

if m=l and m<=s,

Question 1: I wrote a code for calculating the multiplication  of. Is it right?

The code for Step 1  

restart;

multiply:=proc(n,m,l,s) local K,g,a: 
a:=unapply(doublefactorial(2*j-1)/factorial(j),j):
g:=unapply((a(m-j)*a(j)*a(s-j)/a(m+s-j))*(2*m+2*s-4*j+1)/(2*m+2*s-2*j+1),j):

if n<>l then 0 else
sqrt((2*m+1)*(2*s+1))*2^(K/2-1).add((g(j)/sqrt(m+s-2*j+1/2))*phii(n,m+s-2*j),j=0..m) 
end if
end proc:
 
n:=2:
l:=2:
m:=1:
s:=1:
multiply(n,m,l,s);

when I compared the results which I got and the results which is given in the book as follows, I think it is right.

Step 2:

We know that the outer product matrix is calculated as follows 

  

We found the elements of the outer product matrix in Step 1. 

Question 2 : I want to write the elements which are derived in step 1 to the outer product matrix in step 2. In here, the outer product matrix is NxN matrix. N=(M+1).2^(K-1) where K, M are any integers.

Hello,

How I can take variation from left-hand side of  5, and reach to right-hand side of  5. After by using integral by part obtained  7?

Thank you

Maple pdsolve supports periodic boundary conditions. So I was hoping it will be able to solve the heat PDE inside disk with periodic boundary conditions. But I am not able to make it work. 

Is there a trick to make Maple solve this, is there something I need to add or adjust something else? or it is just the functionality is not currently implemented?

This is what I tried

restart;

pde := diff(u(r,theta,t),t)=diff(u(r,theta,t),r$2) + 1/r*diff(u(r,theta,t),r)+1/r^2*diff(u(r,theta,t),theta$2);
bc1 := u(a,theta,t)=0;
bc2 := eval(diff(u(r,theta,t),theta),theta=-Pi)=eval(diff(u(r,theta,t),theta),theta=Pi);
bc3 := u(r,-Pi,t)=u(r,Pi,t);
ic  := u(r,theta,0)=f(r,theta);
sol := pdsolve([pde, bc1,bc2,bc3, ic], u(r, theta, t), HINT = boundedseries(r = 0)) assuming a>0,r>0

I solved this analytically by hand using standard separation of variables method. The issue of telling Maple the solution is bounded at center of disk, I assume is being handled automatically by the HINT=boundedseries(r = 0).

If I remove the hint, it also does not solve it. 

Maple 2019, Physics package 338

Hellow, help required to remove the errors

 How to  obtain the pressure drop i am unable to get the output. I am uploading the file  and the equations

help_dp.mw

 

 

 

 


 

For this problem

I'd like to see if Maple can give, or simplify the solution it now gives to look like this solution 

The one it currently gives is

restart;

pde:=diff(w(x,t),t)+c*diff(w(x,t),x)=0; 
ic:=w(x,0)=f(x);
bc:=w(0,t)=h(t);
sol:=pdsolve([pde,ic,bc],w(x,t))  assuming t>0,x>0,c>0

 

And I did not know how to simplify it or obtain the simpler one. I tried strip and TWS hints.  I also do not understand why Maple gives an integral with 0 as upper limit there (the second integral).

Using Physics package cloud version 338 and Maple 2019. On windows 10.

Thank you

Hi,

I have to build a simple model in MapleSim that allows to simulate a mass that goes down along an inclined plane with a certain friction.

Is there a component that simulates the inclined plane? I tried to use a prismatic joint but it has some predifined translational directions so I can't impose the movement along the plane.

Is there a way to take the laplacian of 1/r and get the "physics" answer of -4*pi*delta(\vec{r})?

For a statistical course I will give next year, I would like to demonstrate the calculation of tetrachoric and polychoric correlations. Furthermore, it would be nice to show the students how the expectation-maximization algorithm works.

Has anyone already programmed this in Maple?

best regards,

Harry

 

with(plots):R := 5; alpha := (1/9)*Pi;
C1 := plot([R*cos(t), R*sin(t), t = 0 .. 2*Pi], color = blue);
A := [R*cos(alpha), R*sin(alpha)]; B := [R*cos(alpha+Pi), R*sin(alpha+Pi)]; AB := plot([A, B], scaling = constrained);
display({AB, C1}, scaling = constrained);# bad drawing

 

https://aws.amazon.com/getting-started/projects/deploy-elastic-hpc-cluster/

is it possible to use maple on high performance clusters?

i can only think to use c program to call cmaple with MPI in linux to use high performance clusters.

is there any other official method to do this?

if i upload my maple 2015 version to amazon for this computing, will it used up all license in this first chance of installation leading to that i can not install maple 2015 linux version to other machine?

Hi all

We denote the collecction of sets determined by the first k coin tosses $F_k$

Suppose the imitial stock price is $S_0$ ,with up and down facter being $u$ and $d$.

Up : S1(H)=u S0 and S1(T)=d S0

S_{N+1}= alpha S_N

where alpha =u or d

Let the probability of each $H$ and $T$ be $p$ and $q=1-p$ and   $F_t$ the sigma-lgebra generated by the coin tosses up to (and inchudling) time t:

After three coin tosses.

Can we propose a code computing the element of the filtration F1 and F3 and sigma(S3) (the sigma algebra generated by S3).

For example by hand we have F1={ emptyset, Omega, AH, AT}

Where AH={ w: w1=H}

AT={w: w1=T}

Can we compute

 

$E[ S_2|F_3] \text { and } E[ S_2|\sigma(S_3) ] $

 

$$E[ \frac{S_2}{S_1} | F_1] \text { and } E[ \frac{S_2}{S_1} | \sigma(S_1) ] $$

 

 

restart;
with(Finance);
S := [7.9, 7.5, 7.1, 6.5, 5., 3.7, 3.3, 2.95, 2.8];
         [7.9, 7.5, 7.1, 6.5, 5., 3.7, 3.3, 2.95, 2.8]
T := BinomialTree(3, S, .3);
TreePlot(T, thickness = 2, axes = BOXED, gridlines = true);

 

 

many thanks

Hello all,

I'm trying to do kinetic modeling of sequential dissociations with DE. I'm hitting a snag when modeling the third dissociation. The population should start at zero at t=0, but some of my model functions are non-zero at t=0. Is there anyway to fix this to force the funtions to go through zero?

Scheme:
PPPP -> intermediates -> PPP -> intermediates -> PP -> intermediates -> P  
(where P is a subunit and intermediates are confirmational changes before dissociation of a subunit)

a'..d' is the first dissociation
e' is the second dissociation
f'..l' is the third dissociation
Fits are evaluated by the residual sum of squares.

sol := dsolve([a' = -k1*a(x), b' = k1*a(x)-k1*b(x), c' = k1*b(x)-k1*c(x), d' = k1*c(x)-k1*d(x),
e' = k1*d(x)-k2*e(x), 
f' = k2*e(x)-k3*f(x), g' = k3*f(x)-k3*g(x), h' = k3*g(x)-k3*h(x), i' = k3*h(x)-k3*i(x), j' = k3*i(x)-k3*j(x), k' = k3*j(x)-k3*k(x), l' = k3*k(x)-k3*l(x), 
a(0) = 1, b(0) = 0, c(0) = 0, d(0) = 0, e(0) = 0, f(0) = 0, g(0) = 0, h(0) = 0, i(0) = 0, j(0) = 0, k(0) = 0, l(0) = 0],
{a(x), b(x), c(x), d(x), e(x), f(x), g(x), h(x), i(x), j(x), k(x), l(x)}, method = laplace);

f1 := sol[6];
f1 := rhs(f1);
g1 := sol[7];
g1 := rhs(g1);
h1 := sol[8];
h1 := rhs(h1);
i1 := sol[9];
i1 := rhs(i1);
j1 := sol[10];
j1 := rhs(j1);
kk := sol[11];
kk := rhs(kk);
l1 := sol[12];
l1 := rhs(l1);

xdata := Vector([0,10,20,30,40,50,60,70,80,90,100,110,120,130,140,150,160,170,180,200,210,220,230,240,250,260,270,280,290,300,310,320,330,340,350,360,370,380,390,400], datatype = float);
ydata := Vector([0.0034,0.00392,0.00184,0.00782,0.01873,0.03683,0.11016,0.09838,0.18402,0.24727,0.20901,0.2972,0.37635,0.49235,0.57845,0.4457,0.50285,0.5672,0.62783,0.57264,0.54918,0.44792,0.49795,0.55218,0.47512,0.46473,0.37989,0.32236,0.3323,0.20894,0.28473,0.21273,0.19855,0.13548,0.12725,0.13277,0.0784,0.07969,0.06162,0.03855], datatype = float);

k1 := 0.391491454107626e-1; 
k2 := 0.222503562261129e-1; 


z1:=f1;
z2:=f1+g1;
z3:=f1+g1+h1;
z4:=f1+g1+h1+i1;
z5:=f1+g1+h1+i1+j1;
z6:=f1+g1+h1+i1+j1+kk;
z7:=f1+g1+h1+i1+j1+kk+l1;

Statistics[NonlinearFit](z1,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z1,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z2,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z2,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z3,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z3,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z4,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z4,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z5,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z5,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z6,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z6,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

Statistics[NonlinearFit](z7,xdata, ydata, x, initialvalues = [k3=0.1], output = [parametervalues, residualsumofsquares]); 
A:=plot(xdata, ydata, style=point, symbol=solidcircle, color=blue, symbolsize=12,labels = ["time (minutes)", "Relative Abundance"], labeldirections = [horizontal, vertical]):
F:=Statistics[NonlinearFit](z7,xdata, ydata, x,initialvalues = [k3=0.1]):
B:=plot(F, x=xdata[1]..xdata[-1], color=red):
plots[display](A, B);

3rd_diss.mw

1 2 3 4 5 6 7 Last Page 1 of 240