## Odditity in Maple 2019...

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.

## simplify equation ...

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

## can Maple 2019 solve the heat PDE inside disk when...

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

## how to obtain this solution for wave PDE in 1D in ...

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

## Laplacian or 1/r...

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

## problem with plot or display...

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

## is it possible to use maple on high performance cl...

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?

https://docs.aws.amazon.com/AWSEC2/latest/WindowsGuide/ConfigWindowsHPC.html#ComputeNode

which virtual machine should i install the maple 12? on one virtual machine or all compute nodes?

how many compute nodes are need to compute dsolve 100,000 systems which may or may not have solution in maple 12?

## Conditional expectation and filtration...

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

## Unable to get DE functions to start at zero...

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

## Solution of the boundary value problem for ODE usi...

Example of Duffing equation with boundary conditions.
y'' + 0.2y' + y^3 - 0.3cos(s) = 0;
y(0) = y (2Pi);
y'(0) = y'(2Pi);
For convenience, we replace the original equation with a system of two first order equations:
--------------------------------------------------------------------------
x1'(t) = 2*Pi*x2(t);
x2'(t) = - 0.4*Pi*x2(t) - 2*Pi*x1(t)^3 +0.6*Pi*cos(2*Pi*t);
x1(0) = x1(1);
x2(0) = x2(1);
--------------------------------------------------------------------------
I have long wanted to apply an optimization package to solve a boundary value problem for ODE. The decision helped procedure for solving ODE, written by forum member vv.
It seems to me that two solutions have been found and that the solutions are weakly sensitive to the initial approximations. These are two closed trajectories. For example, these are points that belong to these solutions:
(0.5966963,  1.0482816) , ( - 0.3132584, 0.0664941).
I am wondering: are the solutions right, and how justified is the use of optimization methods for such tasks?
At the end of the program, the solution is checked on the original Duffing equation using standard Maple functions.   Duffing_equation_BC.mw

(In the figures, the trajectory bypass occurs three times.)

## Maple hangs when computer is locked during workshe...

If worksheet execution was started and computer is locked before execution ends, Maple will stop responding and all work is lost. Execution of whole worksheet takes very long time, sometimes over 3 hours, so "do not lock computer during execution" is not solving the problem.

## efficient coding for procedure...

What are the most efficient way to write and evaluate a procedure in maple? Thank you.

## Symbolic x and y in an equation - How do I retain...

I have made a document to give the results for a problem in table form and graph form.

The tables work out well. The graph results also come out well when I have that section alone in the document. But it does not give the updated curve here. Please correct me where I made mistake.

 Cost in $: Present Value: Interest in % per year: Period in years: Additional Months  Principal$ Interest % per year Year Present value at the year end

UpdateExpr procedure does not update equation, instead turns out a value on LHS

 Cost: Interest % per year (float)

Startup Code Region contains all the codes.

Thanks.

## How can we prevent Maple/MaplePlayer to auto-exit...

Hello every one,

My slideshow now contains a lot of animation sequences and take
a very long time to load and a large amount of physical memory (>20GB).

To reduce this size I have decided to convert these animations to GIF format.
Unfortunatly Maple/MaplePayer do not play GIF animations.

I decide to put URLs (image icons pointing to GIF files) in the document.
Now the GIF are play by a WEB browser (program to operate in fullscreen
mode to minimize any disturbance in the audience with IE11).

Now the issue:
Using F11 to start the Slideshow, cliking the URL start the browser with the requested
animated GIF, that is fine, but now Maple/MaplePayer both auto-exit fullscreen
mode just after that click. The speaker (me) now need to manage the return to
fullscreen (F11) at the current slide while
thinking about what i have to
say next.

Is there any trick to prevent auto-exit from fulls-screen mode in this situation?

LL

## Babylonian square root calculation...- proc binomi...

Des suggestions de solutions? Merci.

A) a) Écrire une procédure qui produit une itération du calcul babylonien de la racine carrée d'un nombre positif k à partir     d'une première approximation x0 .

Entrée: k , x0 .

Sortie:  x1 = (x0+ k/x0)/2 .

b) En utilisant la procédure trouvée en a), en écrire une autre, qui prend en entrée un entier positif n en plus des entiers k   et x0 vus en a) et qui retourne en sortie n itérations du calcul babylonien de la racine carée de k.

B)

Écrire une procédure récursive Maple qui prend en entrée deux nombres n et k et qui utilise l'identité (n k) = (n -1k)+(n-1 k-1) pour retourner en sortie le coefficient binomial (n k) , cette procédure ne doit pas utiliser la commande Maple binomial.

Suggestions for solutions? Thank you.

A) a) Write a procedure that produces an iteration of the Babylonian calculus of the square root of a positive number k from a     first approximation x0.

Input: k, x0.

Output: x1 = (x0 + k / x0) / 2.

b) Using the procedure found in a), write another one, which takes as input a positive integer n in addition to the integers k and   x0 seen in a) and which returns in exit n iterations of the Babylonian calculation of the square root of k.

B) Write a Maple recursive procedure that takes as input two numbers n and k and uses the identity (n k) = (n-1 k)+ (n-1 k-1) to   return the binomial coefficient (n k), this procedure should not use the binomial Maple command.

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