Hi,

I'd like to know, if it is possible to define any sort of range for parameters in NonlinearFit. E. g. I know that one of parameters should be somewhere between 0.2 - 0.4. I know there is a possibility of initalvalues, but using it doesn't lead into this range.

Thanks.

Hi. So my question is how can I get Square brackets on the phone app in calculations or is it possible?

In a french magazine written by High Schools teachers I found this problem:

let a, b, p, q four strictly positive integers such that a > b^2 and p > q+1;
find 4-tuples (a, b, p, q) such that 

(a^2 - b^4) = p!/q!

Given the source of this problem I suspect that there is a trick to answering this question.
After some hours spent, I have found no general method to solve it, only a few solutions (first one and second one are almost obvious), for instance

rel := a^2 - b^4 = p!/q!:

eval(rel, [a= 5, b=1, q=1, p=4]);   
eval(rel, [a=11, b=1, q=1, p=5]);
eval(rel, [a=71, b=1, q=1, p=7]);
eval(rel, [a= 2, b=1, q=2, p=3]);
eval(rel, [a=19, b=1, q=2, p=6]);
eval(rel, [a=21, b=3, q=2, p=6]);

Do you have any idea how to solve this problem?
Could it be handled by Maple (without a systematic exploration of a part of N^4)?

Thanks in advance

lets say we have one or two lines 

y=2x-4 and y = -2x+4 is it possible to get Maple to illustrate the angle between these two lines in a plot? Or the angle of inclination in respect to the x-axes for them individually? 

kindly help me to find the inverse Laplace of this function. I tried but maple leaves the integral unevaluated.
 

Download invrslplc.mw

how can revision of error in 

restart;
u0 := proc (x) options operator, arrow; 1+2*x end proc; h := 0;
for k from 0 to 5 do U := proc (k, h) options operator, arrow; eval((diff(u0(x), [`$`(x, k)]))/factorial(k), x = 0) end proc end do;
m := h+1;
for k from 0 to 5 do U := proc (k, h) options operator, arrow; (sum(sum(U(r, h-s)*U(k-r, s), s = 0 .. h), r = 0 .. k)+(k+1)*U(k, h))/m end proc end do;
U(0, 1);
Error, (in U) too many levels of recursion

I have the following expression.

Maple will evaluate this to:

Screenshot:

Plotting these two in 2D on Desmos to demonstrate: https://www.desmos.com/calculator/tvp4rbzxzp

These two are not the same expression. Is Maple broken or am I doing something wrong?

Hello. Please help me solve the ODE system

ODU_v2.mw
 

Download ODU_v2.mw

with(plots);
f := x -> x^3 + 3^x;
complexplot3d(f, -15 - 22*I .. 15 + 22*I);

This gives good color graph however I would like to see it like the way this cat graphed it.

https://www.quora.com/How-do-I-solve-for-x-in-x-3-3-x-17

Ive got a question about the Dutch book again. They sure did their part on finding questions with answers you cannot find in the book (maybe in later chapters or the 2nd book "part 2"). Maybe to make sure you go to the teacher, so he sees you, and he is sure you did your part, and to get some interaction between the student and the teacher.

I guess you guys/girls are the teacher... Hahah :)

It is the 3rd question. It says: "given is the funtion y=f(x) ( see the photo for the function, "voor" means "for")

a. Draw the graph of y=f(x)
b. Draw with the help of the graph of y=f(x), a graph y=f(-x),y=-2f(x),y=f(x-1), and y=(2x)"

This paragraph is about shifting graphs of functions, and how to alter them. I know how it works, ive done that quite some time ago, but now in maple it is a new challange. Is there a way to get the funtion f(x) to adhere to the rules imposed on it by the text in the book? If that can be done, the answering of the questions will be a lot easier!

Thank you!

Greetings,

the function 

Hallo, I would like to ask anyone have encountered this problem. I tried to update a matrix value using recurrence do but end up with "Error, recursive assignment". Here is the code. Any help is appreciated.

 

with(LinearAlgebra);
with(plots);
Nx := 200;
Nt := 200;
dx := 0.5e-2; dt := 0.2e-1;
A := Matrix(Nx, Nx);

for k to Nx do A[k, k] := -2.0 end do;
for j to Nx-1 do A[j+1, j] := 1.0 end do;
for j to Nx-1 do A[j, j+1] := 1.0 end do; A;
zeta := (.1*dt/dx)^2;
phi || 0 := Matrix(Nx, Nx);

fin := proc (t) options operator, arrow; cos(4*Pi*t) end proc;
phi || 0[(1/2)*Nx, (1/2)*Nx] := fin(0);
phi || 1 := (1/2)*zeta^2 . A . phi || 0; phi || 1[(1/2)*Nx, (1/2)*Nx] := fin(dt);

phinow := Matrix(Nx, Nx); phinext := Matrix(Nx, Nx); phibefore := Matrix(Nx, Nx);

phinow := phi || 1; phibefore := phi || 0;

for j from 2 to Nt do phinext := Zeta^2 . (A . phinow+phinow . A)+2.*phinow-phibefore; phi || j := Matrix(Nx, Nx); phinext[(1/2)*Nx, (1/2)*Nx] := fin(j*dt); phi || j := phinext; phibefore := phi || (j-1); phinow := phi || j end do;
Error, recursive assignment

for j to (1/5)*Nt do p || j := matrixplot(phi || j, heights = histogram, gap = 0.5e-2, labels = ["", "", ""]) end do;

plots[display](p || (1 .. 20), insequence = true);

Hi,

I solve numerically an ODE system which depends on 25 parameters.
I want to know the maximum value of the time at which a specific event is triggered, when these parameters vary (independently the one of the other) within a 25 dimensional hyperbox.

To solve this maximization problem, which I assume is local¹, I would like to use NLPSolve

The attached zip (I use Mac OSX) contains:

• a file that gives a full description of the problem (the original ".pages" file and its export as pdf),
• an m file wich contains the solution procedure plus a few other variables needed to solve the problem,
• the mw file which contains:
  • the reading of the m file,
  • the procedure (OBJ) which returns the trigerring time,
  • many attemps to find its maximum value using NLPSolve.

None of my attempts at using NLPSolve gave me the expected answer

Could you help me to fix this?

Thanks in advance

mw_and_m_files.zip

1: Initializing a local method (see the explanation file) with different points gave me different optimizers but all of them led to rather close values of the objective function. Thus the problem is either global instead of local, or convergence might not be achieved for all the initialization points (but keep in mind that I'm not interested in the location of the minimizer(s) but in the maximum value of if the time when the event is triggered).

Hi Every one. I was trying to solve three point boundary value problem but cound't get the solution. any help? 
u1 need to plot  -1..0 and u2 from 0..1 and shown in a single doamin -1..1


 

Hi!

I am so pleased to be here to look for help in using Maple.

I am trying to find solutions to a set of nonlinear systems of equations using fsolve.

This code I am tried to solve is shown as follow:

 nonlinearsystemforallsolution.mw

I can get one solution for three variables, however, it's clear there are some solutions missed.

My questions are listed as:

1) How do I get all solutions for a fixed variable of sigma?

2) How do I create a  loop to obtain the solutions for various values of sigma?

3) How to plot solution of each variables versus sigma? 

3) How do I determine the stability of each solution?

Many thanks in advance.

sincerely

I have some more questions, now i get some more odd situations where an obvious answer cannot be given by solve. And a funtion cannot be plotted somehow. 

Here is the file. I tried to understand why, but it is not happening. They did it on purpose (i guess) to get simple questions, and show you that Maple sometimes does not have the answer right away. So you run into the "limitations" of the program right away so you are aware what could go wrong.

Download Applied_Math_Part_1_questions_part_two.mw

The function