## What happened to Normal distribution in Maple 2019...

Earlier smoothly working generation of normal distribution in v. 2019 unexpectedly shows the error:

RandV  := Statistics[RandomVariable](Normal(0, 1));
Statistics[Sample](RandV, 10);

Error, (in p) unable to convert Float(undefined) to an integer

## Error, (in ExcelTools:-Export) while processing re...

Help create file Excel in ExcelTools, but error row 564?

thu_file.mw

## How to resolve the problem??...

Hii,

I am using a command -NLPSolve(Ecost1, Q = 10 .. 20, initialpoint = {Q = 10}, assume = nonnegative, maximize = false). I am looking for solution that find the Q value at the minimum value of Ecost1. But Ecost1 should not go below 0.

and also I am getting an error -Warning, initialpoint option ignored by solver.

Kindly tell how to deal with these issues.

Thanks

## Doing some Substitution ...

How I can do ?

Thank you.

Substitution of . 5,6,7) into Eqs. 1–(4), gives the new equation as functions of the generalized coordinates,
u_m,n(t);  v_m,n ( t), and w_m,n ( t). These expressions are then inserted in the Lagrange equations (see Eq. 8)) a set of N second-order coupled ordinary differential equations with both quadratic   and cubic nonlinearities.

In Eq (8) q are generalized coordinate such as uvw  and .

\where the elements of the vector, are the time-dependent generalized coordinates.

 >
 (1)
 >
 (2)
 >
 (3)
 >
 (4)
 >
 (5)
 >
 (6)
 >
 (7)
 >
 (8)
 >

## how do i make a comparison or 3 different schemes?...

I want to made a comparison via plots of RK-4, NSFD and LWM.

## 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.)

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