Advanced Search

MaplePrimes Questions

How to get revolution surface of polar coords curv...

Asked by: Pepini 255
polar plotting
July 26 2023
0  8

Hi! Do you know how can I plot revolution of this curve:

plot(1 + 1/4*sin(8*(1 + 1/8*sin(16*(1 + 1/16*sin(32*t))))), t = 0 .. 2*Pi, coords = polar)

to get a wrinkled torus like this:

Abelprijs 2015 voor John Nash en Louis Nirenberg - NEMO Kennislink

Do u have any idea? (I've tried to convert this polar curve to cartesian coords but I don't get it)

Mapleflow map won't work with left quotes...

Asked by: Christopher2222 6030
operator prefix
July 25 2023
0  3

map won't work with left quotes

L:=[2,"foo",4]

How to calculate the number of ways a number can ...

Asked by: jasser 95
integer partition
July 25 2023
3  8

Hello

I have a list with n positive integer numbers and a positive integer k.

I would like to write a Malple programm, that calculates, in how many ways k can be written as a sum of numbers in the list.

Easy examles: let the list be [2,3,6,9].

k=9 can be written in 4 ways: 9=9, 9=6+3, 9=3+3+3, 9=3+2+2+2. The result therefore shoud be 4.

k=8 can be written in 3 ways: 8=6+2, 8=3+3+2, 8=2+2+2+2. The result therefore shoud be 3.

Thank you very much your help!

MapleFlow for next loop...

Asked by: Chico 15
syntax do-loop print
July 25 2023
0  6

I cannot get this to work:

Print all the non-string entries in a list:

> 

L := [1, 2, "abc", "a", 7.0, infinity]:

> 

for x in L do
    if type(x, 'string') then
        next;
    end if;
    print(x);
end do;

Generating class tests in Maple?...

Asked by: Caroline Honscha 5
education export quiz
July 25 2023
1  3

Hi everyone,

I am using Maple for quite a while, now I would like to do something new.

I am a teacher at a german highschool and I would like to generate class tests with maple. Say for an example, I need 30 tests,  where the first task is calculating the first derivative of a function. This function should be different for every person, for example f(x)=a*x^3+3x^2, where the value of a is different in every test (probably random).

So what I need is a way to generate nice looking sheets in Maple (including page breaks, titles and so on). Is there a way to generate such a thing in maple or do I have to export to TeX and adjust there quite a bit?

Maybe someone has done a similar thing. Actually I am looking for something like serial letters in Maple.

Any idea, how I could do such a thing?

Thanx for your help,

Axel

GraphTheory['AllPairsDistance']: Which output is c...

Asked by: sursumCorda 1244
graphtheory shortest-distance
July 25 2023
2  5

GraphTheory:-GraphEqual says that G1 and G2 are equal, but GraphTheory:-AllPairsDistance gives different results instead: 

> 

restart;
> 

with(GraphTheory)
> 

M := `<|>`(`<,>`(0, 0, 0), `<,>`(1, 0, 0), `<,>`(1, 1, 0))
> 

G__1 := Graph(convert(-M, Matrix, datatype = integer[8]))
> 

G__2 := Graph(convert(-M, Matrix, datatype = integer))
> 

GraphEqual(G__1, G__2)

true

 (1)
> 

AllPairsDistance(G__1)
> 

AllPairsDistance(G__2)

Matrix(%id = 36893491227039185244)

  

Error, (in GraphTheory:-AllPairsDistanceExt) negative cycle detected

  
> 

 

Download allpairs.mw

So, which one is incorrect? Any reasons?

Is there any bug in the "randpoly" command?!!...

Asked by: MDD 367
duplicate-question randpoly
July 24 2023
0  4
> 

restart
> 

with(PolynomialIdeals):
> 

``
> 

``
> 

randomzero := proc (ListVar) local A, G, i, lm, B, f, g; option trace; A := ListVar; G := NULL; for i to nops(A) do f := A[i]^(i+1)+randpoly([op(`minus`({op(A)}, {A[i]}))], terms = 1, coeffs = rand(-4 .. -1), degree = i); G := G, f end do; print(IsZeroDimensional(`<,>`(G))); RETURN([G]) end proc:
> 

randomzero([x, y])

{--> enter randomzero, args = [x, y]

  

[x, y]

  

"G:="

  

x^2-y-4

  

x^2-y-4

  

y^3-3*x

  

x^2-y-4, y^3-3*x

  

true

  

<-- exit randomzero (now at top level) = [x^2-y-4, y^3-3*x]}

  

[x^2-y-4, y^3-3*x]

 (1)
> 

``

Download bug.mw

Hi

I think there is a bug in the "randpoly" command. please see the attached file line 7 of my procedure "randomzero". Why x^2-y-4 is created while terms=1 is considered and the outputs must contain binomial?

Limitations of `sum`?...

Asked by: sursumCorda 1244
sum hypergeom
July 23 2023
0  4

Some work, while others don't work. 
 

> 

restart;
> 

`assuming`([is(1/(sqrt(n*(n+1))*(sqrt(n)+sqrt(n+1))) = 1/sqrt(n)-1/sqrt(n+1))], [n::posint])

true

 (1)
> 

sum(1/sqrt(n)-1/sqrt(n+1), n = 1 .. infinity)

1

 (2)
> 

sum(1/(sqrt(n*(n+1))*(sqrt(n)+sqrt(n+1))), n = 1 .. infinity)

sum(1/((n*(n+1))^(1/2)*(n^(1/2)+(n+1)^(1/2))), n = 1 .. infinity)

 (3)
> 

`assuming`([sum(binomial(3*n, n)*x^n/(2*n+1), n = 0 .. infinity)], [abs(x) <= 4/27])

(2/3)*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2)))/x^(1/2)

 (4)
> 

`assuming`([eval(binomial(3*n, n)*x^n/(2*n+1), n = 0)+sum(binomial(3*n, n)*x^n/(2*n+1), n = 1 .. infinity, parametric)], [abs(x) <= 4/27])

1+x*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)

 (5)
> 

simplify(convert(1+x*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-hypergeom([1/3, 2/3], [3/2], (27/4)*x), elementary), symbolic)

(1/3)*(3*x^(1/2)+3*x^(3/2)*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-2*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2))))/x^(1/2)

 (6)
> 

plot((1/3)*(3*x^(1/2)+3*x^(3/2)*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-2*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2))))/x^(1/2), x = -4/27 .. 4/27)

 
> 

verify(0, (1/3)*(3*x^(1/2)+3*x^(3/2)*hypergeom([1, 4/3, 5/3], [2, 5/2], (27/4)*x)-2*3^(1/2)*sin((1/3)*arcsin((3/2)*3^(1/2)*x^(1/2))))/x^(1/2), 'equal')

FAIL

 (7)
> 

NULL


 

Download unable_to_sum.mw

How to explain this behavior?
I have read the help page, but I can not get the point.

Integration problem...

Asked by: abiodunowoyemi9 30
July 23 2023
1  4

Please, is there anyway I can solve this problem with a nice output? The solution I got is complicated. Please, helper is need

f := exp(-beta*x)*x^(5-alpha)

int(f, x)

beta^(alpha-6)*(-x^(-alpha)*beta^(-alpha)*(alpha^4-14*alpha^3+71*alpha^2-154*alpha+120)*(alpha-6)*(beta*x)^((1/2)*alpha)*exp(-(1/2)*beta*x)*WhittakerM(-(1/2)*alpha, -(1/2)*alpha+1/2, beta*x)/(-alpha+6)+x^(-alpha)*beta^(-alpha)*(beta^4*x^4-alpha*beta^3*x^3+alpha^2*beta^2*x^2+5*beta^3*x^3-alpha^3*beta*x-9*alpha*beta^2*x^2+alpha^4+12*alpha^2*beta*x+20*beta^2*x^2-14*alpha^3-47*alpha*beta*x+71*alpha^2+60*beta*x-154*alpha+120)*(alpha-6)*(beta*x)^((1/2)*alpha)*exp(-(1/2)*beta*x)*WhittakerM(-(1/2)*alpha+1, -(1/2)*alpha+1/2, beta*x)/(-alpha+6))

Define function by giving different values???...

Asked by: delvin 60
eval substitution
July 21 2023
0  6

How to write a program that obtains the value of the function by giving different values
Like the example below:

 

How to "simplify" a RootOf result?...

Asked by: mmcdara 7896
solve identity erf
July 21 2023
2  2

I rephrased my previous question in a more synthetic form
(there was probably a lot in it that I thought was important for understanding the problem, but I realized afterwards that it only added confusion).

The true question is yellow-highlighted in the code below

> 

restart
> 

# The result below seems natural: we were taught in school that exp 
# being a bijective function we can get rid of it in the equality to
# solve and write simply x=Pi.

x = solve(exp(x)=exp(Pi), x)

x = Pi

 (1)
> 

# But the solution method solve uses is not that natural (and I
# don't really understand it).
# infolevel[solve] := 10:
# x = solve(exp(x)=exp(Pi), x);
> 

# Replacing now exp by some undefined function f produces a
# kind of "no-solution" answer: this seems quite normal because
# not knowing the properties of f one cannot simply get rid of it.

infolevel[solve] := 0:
x = solve(f(x)=f(Pi), x)

x = RootOf(f(_Z)-f(Pi))

 (2)
> 

# Finally replace f by a bijective function with no analytic expression.

s = solve(erf(x)=erf(Pi), x) assuming x::real

s = RootOf(erf(_Z)-erf(Pi))

 (3)
> 

# It would have seem reasonable for Maple to answer x=Pi, or
# at least it is what I would have done given the properties
# of the erf function.
#
# How can I "force" Maple to "simplify" it's RootOf result to get
# x=Pi?


Download solve_erf.mw

For those interested in the motivations of this quastion, see here Where_does_the_question_come_from.mw

The original question is here Original_question.pdf

Expansion of trigonometric unions in Maple???...

Asked by: delvin 60
expand trig
July 21 2023
0  4

How is the expansion of trigonometric functions in Maple like the following function?

Elementwise operations...

Asked by: SiggiN 10
elementwise
July 20 2023
2  6

Hi!

I'm trying to recreate a Mathcad sheet in Maple Flow

I could use some help figuring out how elementwise operations are made:

From mathcad, what i want to reproduce:

What I have in Maple Flow so far:

1) How do i get MF to output numerical valus in sigma_t ?

2) How would H look in MF? 

Thank you!

Ragards
Siggi

 

Did anyone notice...

Asked by: rightClick 115
relativity physics
July 20 2023
3  0

that in Kip Thorne's book Maple is mentioned through out ?

Kip Thorne and Roger Blandford : "Modern Classical Physics: Optics, Fluids, Plasmas, Elasticity, Relativity, and Statistical Physics"

also, courtesy of Caltech in Chapter 24

linearize the complex conjugate function partial d...

Asked by: bashar27 55
linearize differential-equations conjugate
July 20 2023
0  0

restart;
alias(u = u(x, z, t), f = f(x, z, t));
                              u, f
u := (f+sqrt(R))*exp(I*R*x);
                    /     (1/2)\           
                    \f + R     / exp(I R x)
pde1 := I*(diff(u, z))+diff(u, x, x)+diff(u, t, t)+u*abs(u)*abs(u)-(u*abs(u)*abs(u))*abs(u)*abs(u);
    / d   \              / d  / d   \\           
  I |--- f| exp(I R x) + |--- |--- f|| exp(I R x)
    \ dz  /              \ dx \ dx  //           

           / d   \                /     (1/2)\  2           
     + 2 I |--- f| R exp(I R x) - \f + R     / R  exp(I R x)
           \ dx  /                                          

       / d  / d   \\           
     + |--- |--- f|| exp(I R x)
       \ dt \ dt  //           

                                                            2
       /     (1/2)\                           2 |     (1/2)| 
     + \f + R     / exp(I R x) (exp(-Im(R x)))  |f + R     | 

                                                            4
       /     (1/2)\                           4 |     (1/2)| 
     - \f + R     / exp(I R x) (exp(-Im(R x)))  |f + R     | 

simplify(%);
         / d   \              / d  / d   \\           
       I |--- f| exp(I R x) + |--- |--- f|| exp(I R x)
         \ dz  /              \ dx \ dx  //           

                / d   \                 2             
          + 2 I |--- f| R exp(I R x) - R  exp(I R x) f
                \ dx  /                               

             (5/2)              / d  / d   \\           
          - R      exp(I R x) + |--- |--- f|| exp(I R x)
                                \ dt \ dt  //           

                                               2  
                                   |     (1/2)|   
          + exp(I R x - 2 Im(R x)) |f + R     |  f

                                               2       
                                   |     (1/2)|   (1/2)
          + exp(I R x - 2 Im(R x)) |f + R     |  R     

                                               4  
                                   |     (1/2)|   
          - exp(I R x - 4 Im(R x)) |f + R     |  f

                                               4       
                                   |     (1/2)|   (1/2)
          - exp(I R x - 4 Im(R x)) |f + R     |  R     
collect(%, exp(I*R*x));
  /  (5/2)       / d   \      2       / d   \   / d  / d   \\
  |-R      + 2 I |--- f| R - R  f + I |--- f| + |--- |--- f||
  \              \ dx  /              \ dz  /   \ dx \ dx  //

       / d  / d   \\\           
     + |--- |--- f||| exp(I R x)
       \ dt \ dt  ///           

                                          2  
                              |     (1/2)|   
     + exp(I R x - 2 Im(R x)) |f + R     |  f

                                          2       
                              |     (1/2)|   (1/2)
     + exp(I R x - 2 Im(R x)) |f + R     |  R     

                                          4  
                              |     (1/2)|   
     - exp(I R x - 4 Im(R x)) |f + R     |  f

                                          4       
                              |     (1/2)|   (1/2)
     - exp(I R x - 4 Im(R x)) |f + R     |  R     
 

First 181 182 183 184 185 186 187 Last Page 183 of 2428