Maple 2020 Questions and Posts

These are Posts and Questions associated with the product, Maple 2020

windows 11 maple 2020 I can load other packages but cannot load the student packages

I need to use convert of complex exponentials to trig, but only to convert exp(I*x) to cos/sin using Euler formula.

The problem is that, since this is done in code without looking at what is inside the exp(), Maple will also convert non complex exponentials as exp(x) to hyperpolic trig which I do not want.  An example will make this clear

For an example, given exp(3*I*x - x)  and applying convert/trig to this it gives 

             (cosh(x) - sinh(x))*(cos(3*x) + sin(3*x)*I)                       --(1)

But I only want to conver the exp(3*I*x) part of the of the above to obtain

          exp(-x) *  (cos(3*x) + sin(3*x)*I)                          ---(2)

I can break  exp(3*I*x - x) first using expand command and obtain  exp(-x) exp(3*I*x) and then parse this and filter out the complex exponentials (may be using select with has I) and then use convert on those terms only leaving the non-complex exponentials alone. But this gets messy for more complex exponentials.

Is there an easy way to tell Maple  to convert expression of the form exp(I*f(x) + g(x)) to trig but only to sin/cos, hence leaving the exp( g(x) ) as a factor? I looked at help but see nothing there so far.

Maple 2021.2

Hi

My integrals are convolutions.and I know I can evaluate this using numerical integration, but I am seeking a numerical solution of this problem using FFT. I have many many integrals of this type to evaluate and I need FFT for speed reasons.

fft.mw

This might inspire you.

https://www.mathworks.com/matlabcentral/answers/228107-how-to-evaluate-a-convolution-integral-by-fast-fourier-transform

I do not understand why what I do in my first example does not work in the problem of integral. Can anyone help me? Thank you very much!

 

restart

assume(x > 0)

solve(x^2+3*x-4 = 0, x)

Warning, solve may be ignoring assumptions on the input variables.

 

1, -4

(1)

solve(x^2+3*x-4 = 0, x, UseAssumptions)

1

(2)
   

Download First_example_Maple.mw

 
 
 

f := proc (x) options operator, arrow; sin(100/(x+13)) end proc

proc (x) options operator, arrow; sin(100/(x+13)) end proc

(2.1.1)

g := proc (x) options operator, arrow; 10/(10*x+3) end proc

proc (x) options operator, arrow; 10/(10*x+3) end proc

(2.1.2)
   
 

A := fsolve(f(x) = g(x), x = 1.1)

1.049742001

(2.2.1)

B := fsolve(f(x) = g(x), x = 1.7)

1.676209051

(2.2.2)

C := fsolve(f(x) = g(x), x = 20)

19.35519742

(2.2.3)

 

 

assume(E > C)

solve(int(f(x)-g(x), x = C .. E) = 3, E, UseAssumptions)

Error, (in depends) too many levels of recursion

 

fsolve(int(f(x)-g(x), x = C .. E) = 3, E)

12.00027508

(2.3.1)

fsolve(int(f(x)-g(x), x = C .. E) = 3, E = 30)

28.05987548

(2.3.2)

La valeur de E cherchée est donc d'environ 28,06 unités.

Download Problem_of_integral.mw

This worksheet has an error when executing EulerLagrange.

How can this error be avoided?

Spring_Pendulum.mw

Hi

I am trying to get a family of curves on one graph. Each curve a different colour and labeled

P=2, Q=0 , P=2,Q=1 , etc

mmcdara 's proposal for inspiration?

https://www.mapleprimes.com/questions/233088-How-Do-I-Plot-A-Function-Of-Two-Variables

Family_of_curves_plot.mw

Hello there, 

When I tried to write an expression with the '1d' subscript, I found that it was automatically translated as '1.'. The following worksheet shows an example. Is there any chance to preserve the expression as I intended to see?

restart;

eq3_130:= Psi[fd] = L[ffd]*i[fd]+L[f1d]*i[1d]-L[ad]*i[d];

Psi[fd] = -L[ad]*i[d]+L[f1d]*i[1.]+L[ffd]*i[fd]

(1)

 

Download Q20220121.mw

I have just come across this curious but really boring problem.
In the code snippet below, expr1 and expr2 are identical.

restart:
data := [a=1, b=2]:
convert(
  piecewise(And(x(t) > a, x(t) < b), 1, 0),
  Heaviside
):
expr1 := eval(eval(%, data),  x(t)=z):
plot(expr1, z=0..3);


convert(
  piecewise(And(z > a, z< b), 1, 0),
  Heaviside
):
expr2 := eval(%, data),:
plot(expr2, z=0..3);


But if I change the parameterization of the problem, expr2 is still correct but expr1 is not

restart:
data := [d=1.5, a=0.5]:
convert(
  piecewise(And(x(t) > d-a, x(t) < d+a), 1, 0),
  Heaviside
):
expr1 := eval(eval(%, data),  x(t)=z):
plot(expr1, z=0..3);


convert(
  piecewise(And(z > d-a, z< d+a), 1, 0),
  Heaviside
):
expr2 := eval(%, data),:
plot(expr2, z=0..3);

Where does this come from?

PS: I'm sorry not to be able loading the mw file

Hello there, 

Would you allow me to ask these two questions?

1) is there any way to collect the expression 'eq14_2' w.r.t. DeltaP? The question is labeled as 'Q1' in the worksheet. 

2) Why does the 'DeltaP' term survive at the RHS of the expression 'eq14_2_3'? In other words, is it possible to cancel out the same term, 'DeltaP', in numerator and denominator?

restart;

eq14_1_3 := w[NET] = Delta*T[Sfg];

w[NET] = Delta*T[Sfg]

(1)

eq14_2 := w[NET] = -((v[2] + v[3])/2)((-DeltaP + P) - P) - ((v[1] + v[4])/2)((P - P) + DeltaP);

w[NET] = -(1/2)*v[2](-DeltaP)-(1/2)*v[3](-DeltaP)-(1/2)*v[1](DeltaP)-(1/2)*v[4](DeltaP)

(2)

simplify(eq14_2);

w[NET] = -(1/2)*v[2](-DeltaP)-(1/2)*v[3](-DeltaP)-(1/2)*v[1](DeltaP)-(1/2)*v[4](DeltaP)

(3)

collect(expand(eq14_2), DeltaP, distributed); # Q1

w[NET] = -(1/2)*v[2](-DeltaP)-(1/2)*v[3](-DeltaP)-(1/2)*v[1](DeltaP)-(1/2)*v[4](DeltaP)

(4)

Desired1 := w[NET] = DeltaP * (-v[2]*(-1)/2 - v[3]*(-1)/2 - v[1]*(1)/2 - v[4]*(1)/2);

w[NET] = DeltaP*((1/2)*v[2]+(1/2)*v[3]-(1/2)*v[1]-(1/2)*v[4])

(5)

eq14_2_2 := rhs(eq14_1_3) = rhs(eq14_2);

Delta*T[Sfg] = -(1/2)*v[2](-DeltaP)-(1/2)*v[3](-DeltaP)-(1/2)*v[1](DeltaP)-(1/2)*v[4](DeltaP)

(6)

eq14_2_3 := eq14_2_2 / DeltaP; # Q2

Delta*T[Sfg]/DeltaP = (-(1/2)*v[2](-DeltaP)-(1/2)*v[3](-DeltaP)-(1/2)*v[1](DeltaP)-(1/2)*v[4](DeltaP))/DeltaP

(7)

Desired2 := w[NET]/DeltaP =  (-v[2]*(-1)/2 - v[3]*(-1)/2 - v[1]*(1)/2 - v[4]*(1)/2);

w[NET]/DeltaP = (1/2)*v[2]+(1/2)*v[3]-(1/2)*v[1]-(1/2)*v[4]

(8)

 

Download Q20220110.mw

Thank you everyone!

I am trying to solve an ODE with nonlinear boundary conditions, it is a BVP. And the maple let me to specify an approximate initial solution. I just don't know how to define the initial solution. What format is the initial solution? I have tried the Help Document told me to, but I still can't figure it out. Please help me, thank you!

Hi. 

Just trying to replicate a calculation, I would expect close to absolute zero, but my units are out

Black_hole_temp.mw

Initial text in the uploaded worksheet describes an attempt to partially tile the Poincare disk and asks to be shown how a known complex transformation can be used to accomplish this.

 Poincare_tiling.mw

The worksheet below contains the desired hyperbolic action and asks for the specific transformation that performs it.

Hyperbolic_Transformation.mw

I have corrected the above worksheet so that the Poincare disk is now centred at [0,0].

I hope this helps you answer my question

Anyone can help me to convert matlab codes to maple, pls, this is my code p

Plotting of a nonlinear differential equation?

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