Maple 2018 Questions and Posts

These are Posts and Questions associated with the product, Maple 2018

Hello,

While working for an assignment I had to use a piecewise function which gives a result based on a randomly generated value. The following is a much simpler version of what I've been working on, but it gives the same error.

A random value is uniformly distributed between 0 and 100. The piecewise returns a 1 if the value is between 0 and 50 and it returns 2 if the value is between 50 and 100. As far as I understand this function should only ever give 1 or 2, never something else. However when I loop this a few times Maple regularly returns a 0, which doesn't make sense to me. I've printed the values that return a 0 but none of these should break the piecewise. Can someone please explain to me what's going wrong here and how to fix it?

My code:

restart; randir := piecewise(0 <= r1() and r1() < 50, 1, 50 <= r1() and r1() < 100, 2);
for i to 1000 do r1 := rand(0. .. 100.0); if randir = 0 then print(fail[i], r1()) end if end do;
 

I've included a failure check to see when and at what values it returns a 0, and as you can see it happens very often.

I'm trying to create a graph using a matrix that has numbers and text values, how can I specify that some values of the matrix are strings and others are numbers? I'd like to create something like this

Hello

I use Maple 20018.2.

When I use "Data Set Search" and press Search I get following Error Message. I check that the network access to the internet is on enable. Does anybody has an Idea?

thank you

Murad

z1 := a1+I*b1; z2 := a2+I*b2; abs(z1) = 1; abs(z2) = 1; argument(z1) = alpha; argument(z2) = beta; On considère dans ℂ les complexes z1 et z2 de module 1 et d'argument α et β Show that (z1+z2)^2/(z1+z2) est un réel positf ou nul. Dans quel cas est-il nul ? is((z1^2+2*z1*z2+z2^2)/(z1+z2) = z1/z2+z2/z1+2);#wrong answer z1/z2 = exp(I*(alpha-beta)); z2/z1 = exp(I*(beta-alpha)); is(z1/z2+z2/z1+2 = 2*(1+cos(alpha-beta)));#wong answer Miscalculations. Thank you for your help.

I'm trying to obtain the dynamical response of a simply-supported beam with a cantilever extension, coupled to a spring-mass system. In mathematical terms, this system is ruled by three PDEs (relative to each bare part of the main structure) and one ODE (relative to the spring-mass system). I think my mathemical model is finely formulated, but Maple keeps telling me this:

Error, (in pdsolve/numeric/process_IBCs) improper op or subscript selector

I believe it is because my PDEs depend on "x" and "t", while the ODE depends solely on "t". I have tried to transform my ODE into a "PDE", making it also dependent of "x", but without imposing any boundary conditions relative to "x". However, after this Maple points a new error message:

Error, (in pdsolve/numeric) initial/boundary conditions must be defined at one or two points for each independent variable

Could someone help me finding a solution? My algorythm in shown in the attached file below.

Worksheet.mw

Here's a very simple example, working as intended:

 

 

Now, if I try to use the exception indexing function:

 

 

We can notice that b[1] isn't using exception anymore (as if the fourth input overwrote the second one) and that b[1][2] isn't linked to the value "two". However, if I define the tables first, I get the expected result:

 

 

Why does the exception indexing function prevent me from getting an existing table entry in the second case?

Let E all triplets as X=(p,q,r) such as p^2+q^2=r^2. We define the application f of E dans C complex as X in E f(X)=(p+Iq)/r=Z. Calculate abs(Z). Show that in E the law noted * defined by
X1*X2=(p1*p2-q1*q2,p2*q1+p1*q2,r1*r2) is an internal law. Calculate f(X1*X2). Then if X0=(3,4,5), find
X0*X0, X0*(X0*X0).Thank you for the help.

help me for solve the error of plot

p1.mw

I'm using Maple to solve academic problems.
At a certain stage, I need a loop to perform numerical calculations, from where I extract a graph. However, each step takes a long time and therefore, I would like to evaluate the performance of the result at each step of time.
To simulate my calculation, I created a test code:

restart:                                                                  # Restart
with(Threads):                                                      # To use Sleep()
a := 2*((ceil(rand()/10^10)-50)*(1/10)):                # Random Value
b := 2*(ceil(rand()/10^10)-50):                             # Random Value
c := 2*ceil(rand()/10^10)-100:                             # Random Value
f := proc (x) options operator, arrow; a*x^2+b*x+c end proc;                  # Generic function that i'm ploting.
np := 10:                                                              # Number of Points i'm ploting
vmax := 10:                                                         # Maximum funcxtion value
VAR1 := Matrix(np+1, 2, proc (m, n) options operator, arrow; if n = 1 then (m-1)*vmax/np else 0 end if end proc):       # Declaring VAR1 as a "null" matrix

for i to np+1 do                                                  # Loop Start
VAR1[i, 2] := f((i-1)*vmax/np):                           # VAR1 value update
print(plot([VAR1], x = 0 .. vmax,  gridlines)):                           # Ploting VAR1
Sleep(.25):                                                         # Simulating longer processing time
od;

 

However, all graphs are displayed simultaneously after the last step of the for loop.
I would like to print the result (output) after each time step, plotting a new graph for each time an iteration ends.
Would anyone know how to do this?


Thank you for your help.

 

Can anyone help me to frame the equations in Fractional Reduced Differential Transform Method 

system of nonlinear ordinary differential equations
ds/ dt = b−γ s(t)− (δ s(t)(i(t) + βa(t)) /N − ε s(t) m(t) 
de/ dt = δ (s(t)(i(t) + βa(t))/ N + ε s(t) m(t) − (1−ϑ) θ e(t) − ϑ α e(t) − γ e(t) 
di/ dt = (1−ϑ) θ e(t) − (ρ + γ) i(t)
da/ dt = ϑ α e(t) − (σ + γ) a(t)
dr /dt = ρ i(t) + σ a(t) − γ r(t)
dm /dt = τ i(t) + κ a(t) − ω m(t) 

Good day to all.

I have a function, z(x,y), that I am interested in exploring. The contour plot returns the z-contours on the x-y grid (see attached).

Can anyone suggest a way in which I can now assign z to the vertical axis (instead of y), x to the horizontal axis, and plot the contours for various y-values?

 

Thanks in advance.

Contour_Plot_MaplePrimes.mw

Dear experts

I am interested to solve the following equation numerically by Maple. I would appreciate it if you let me how I can do and what the boundary conditions and initial values are needed


eq:= diff(2*diff(eta(x,y,t),t)+3*eta(x,y,t)*diff(eta(x,y,t),x)+(1/3-1/epsilon/B)*diff(eta(x,y,t),x,x,x),x)+diff(eta(x,y,t),y,y)-1/sqrt(Pi*R)*int(diff(eta(x+zeta,y,t),x,x)/sqrt(zeta),zeta=0..t/epsilon)=0;
where

1) epsilon, B and r are constant

2) 1/epsilon/B is not equal to 1/3 at all

how to show which region of the plan belongs to the argument points between 0 and Pi/2 and the module points between 0 and 2 ?

Can anyone please help me out with the following integration

 


                      f(x) = tanh(x)/sqrt(x^2+1)

 

limits x_initial =1, x_final =100

 

Thanks

 

I am trying to work through an example in a textbook, but its a few years old and uses maple 2015. I am currently using the 2018 edition of maple. The code is an example of how to generate the points on an elliptic curve given a specific input. 
Here is the example code from the textbook:

epoints := proc(ec, x, ub, p)
    local ecurve, z, pct, k, i;
    pct := 0;
    for k from 0 to p-1 while pct <= ub do
        z := subs(x=k, ec) mod p;
        if z = 0 then
           pct := pct+1;
           ecurve[pct] := [k,z];
        fi:
        if z &^ ((p-1)/2) mod p = 1 then
           z := z &^ ((p+1)/4) mod p;
           ecurve[pct+1] := [k,z];
           ecurve[pct+2] := [k, -z mod p];
           pct := pct+2;
        fi:
    od:
    if pct > ub then
       pct := ub:
    fi:
    seq(ecurve[i], i=1..pct):
end:


Here is my code, written to work with Maple 2018:

ecpoints := proc (ec, x, ub, p) local ecurve, z, pct, k, i;
      pct := 0; for k from 0 to p-1 while pct <= ub
         do z := `mod`(subs(x = k, ec), p);
         if z = 0 then pct := pct+1;
            ecurve[pct] := [k, z] end if;
         if `mod`(z^((1/2)*p-1/2), p) = 1 then
            z := `mod`(z^((1/4)*p+1/4), p) = 1;
           ecurve[pct+1] := [k, z];
           ecurve[pct+2] := [k, `mod`(-z, p)];
           pct := pct+2 end if
   end do;
   if ub < pct then pct := ub end if;


   seq(ecurve[i], i = 1 .. pct)
end proc

The problem is with the output. The output should be [0, 5], [0, 14], [2, 4], [2, 15], [3, 6], [3, 13], [4, 6], [4, 13], [6, 0], [10, 16], [10, 3], [12, 6], [12, 13], [14, 16], [14, 3], [18, 17], [18, 2].
What I get is [0, 5 = 1], [0, 14 = 18], [2, 4 = 1], [2, 15 = 18], [3, 6 = 1], [3, 13 = 18], [4, 6 = 1], [4, 13 = 18], [6, 0], [10, 16 = 1], [10, 3 = 18], [12, 6 = 1], [12, 13 = 18], [14, 16 = 1], [14, 3 = 18], [18, 17 = 1], [18, 2 = 18]. 
Any hints on what I could be doing wrong here or what is going on?

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