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MaplePrimes Questions

How do we use on the axes Pi, Pi/2 in Maple 13 plo...

Asked by: mapleatha 160
pi tickmarks axis
October 17 2020
0  6

How do we use on the axes Pi, Pi/2, ...,  in Maple 13 plots without evaluating Pi? When I just put enter Pi, Maple gives me 3.1415... . "piticks" is not available for Maple 13. I do have access to the university ocmputer's Maple 18, but, recently, it has decided that Citrix iwill not serve Windows 7.

Thank you.

mapleatha

Retrieving RGB values from pictures...

Asked by: BennyMopps 45
matrix imagetools
October 17 2020
1  3

Hello, 

im looking to see if there is a way to read in RGB values of the pixels in images and put those values into matrices.

any help or suggestions are appreciated.

How do I input this discrete distribution?...

Asked by: amb043 5
distribution discrete statistics
October 16 2020
1  2

Hi,

I'm new to Maple and was wondering if anyone could help me with how to put a discrete distribution such as Pr(X=1)=0.25, Pr(X=2)=0.65, Pr(X=3)=0.1, into Maple.

Thanks.

Computational Nash equilibrium...

Asked by: omkardpd 50
game equilibrium
October 16 2020
0  0


Dear Colleagues,

Apologies for the generic question below.

I am trying to obtain the Nash equilibrium solutions for a two-person game. I am not sure of any in-built packages that can help me in obtaining the solutions computationally. The algorithms that I created do not seem to give good solutions that are meaningful in my application. Any suggestion would be much appreciated. 

Regards,

Omkar

 

 

solve non-linear differential equation...

Asked by: Dark Energy 100
ode dsolve differential-equations
October 15 2020
1  3

Hi guys

I want to solve the following non-linear differential equation but by using dsolve(), the computer cannot solve it, so please guide me.

Q:=2*diff(a(t), t, t)*a(t)^3 - 3*diff(a(t), t)^4 + diff(a(t), t)^2*a(t)^2

with the best regards

adomian polynomial for nolinear function...

Asked by: usmanjutt 15
differentiation summation duplicate-question
October 15 2020
1  1

i have two nonlinear functions from which one is exicuted properly but 2nd funnction not risponde properly. i can not understan how to overcome it
 

> 

restart
> 

with(LinearAlgebra):
> 

> 

F := proc (x) options operator, arrow; x^2 end proc:
> 

F(u(t));

u(t)^2

 (1)
> 

G := proc (w) options operator, arrow; w*dw/dt end proc:
> 

G(h(t));

h(t)*dw/dt

 (2)
> 

for n from 0 while n <= 6 do V[n] := (diff(F(sum(t^i*u[i], i = 0 .. n)), [`$`(t, n)]))/factorial(n); U[n] := (diff(G(sum(t^i*h[i], i = 0 .. n)), [`$`(t, n)]))/factorial(n) end do:
> 

t := 0;

0

 (3)
> 

for i from 0 while i <= n-1 do A[i] := V[i]; B[i] := U[i] end do;

u[0]^2

  

h[0]*dw/dt

  

2*u[0]*u[1]

  

h[1]*dw/dt

  

2*u[0]*u[2]+u[1]^2

  

h[2]*dw/dt

  

2*u[0]*u[3]+2*u[1]*u[2]

  

h[3]*dw/dt

  

2*u[0]*u[4]+2*u[1]*u[3]+u[2]^2

  

h[4]*dw/dt

  

2*u[0]*u[5]+2*u[1]*u[4]+2*u[2]*u[3]

  

h[5]*dw/dt

  

2*u[0]*u[6]+2*u[1]*u[5]+2*u[2]*u[4]+u[3]^2

  

h[6]*dw/dt

 (4)
> 

for j from 0 while j <= n-1 do u[0] := 1; u[j+1] := int(x*B[j], x)+int(A[j], x) end do;

1

  

(1/2)*x^2*h[0]*dw/dt+x

  

1

  

(1/2)*x^2*h[1]*dw/dt+(1/3)*x^3*h[0]*dw/dt+x^2

  

1

  

(1/2)*x^2*h[2]*dw/dt+(1/3)*x^3*h[1]*dw/dt+(5/12)*x^4*h[0]*dw/dt+x^3+(1/20)*h[0]^2*dw^2*x^5/dt^2

  

1

  

(1/2)*x^2*h[3]*dw/dt+(1/3)*x^3*h[2]*dw/dt+(1/6)*x^4*h[1]*dw/dt+(1/6)*x^5*h[0]*dw/dt+(1/2)*x^4+(13/180)*h[0]^2*dw^2*x^6/dt^2+(2/5)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^5+(1/2)*((1/2)*h[1]*dw/dt+1)*x^4

  

1

  

(1/2)*x^2*h[4]*dw/dt+(1/3)*x^3*h[3]*dw/dt+(5/12)*x^4*h[2]*dw/dt+(1/15)*x^5*h[1]*dw/dt+(1/18)*x^6*h[0]*dw/dt+(1/5)*x^5+(139/1260)*h[0]^2*dw^2*x^7/dt^2+(2/15)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^6+(1/5)*((1/2)*h[1]*dw/dt+1)*x^5+(1/160)*h[0]^3*dw^3*x^8/dt^3+(1/3)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^6+(2/5)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^5+(1/9)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^6/dt+(1/5)*((1/2)*h[1]*dw/dt+1)^2*x^5

  

1

  

(1/2)*x^2*h[5]*dw/dt+(1/3)*x^3*h[4]*dw/dt+(5/12)*x^4*h[3]*dw/dt+(1/6)*x^5*h[2]*dw/dt+(1/45)*x^6*h[1]*dw/dt+(1/63)*x^7*h[0]*dw/dt+(139/5040)*h[0]^2*dw^2*x^8/dt^2+(17/1296)*h[0]^3*dw^3*x^9/dt^3+(1/15)*x^6+(4/105)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^7+(1/15)*((1/2)*h[1]*dw/dt+1)*x^6+(2/21)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(2/15)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^6+(1/15)*((1/2)*h[1]*dw/dt+1)^2*x^6+(1/4)*((13/180)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^8+(2/7)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*((5/12)*h[1]*dw/dt+1+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(2/5)*((1/3)*h[2]*dw/dt+(1/4)*h[0]*dw^2*h[3]/dt^2)*x^5+(1/4)*((1/20)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(5/36)*h[0]^2*dw^2/dt^2)*x^8+(2/7)*((5/12)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/3)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*(((1/2)*h[1]*dw/dt+1)*((1/3)*h[1]*dw/dt+1)+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(1/5)*((1/2)*h[1]*dw/dt+1)*h[2]*dw*x^5/dt+(2/63)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^7/dt

  

1

  

(1/20)*h[2]^2*dw^2*x^5/dt^2+(1/3)*x^3*h[5]*dw/dt+(5/12)*x^4*h[4]*dw/dt+(1/6)*x^5*h[3]*dw/dt+(1/18)*x^6*h[2]*dw/dt+(2/315)*x^7*h[1]*dw/dt+(1/252)*x^8*h[0]*dw/dt+(139/22680)*h[0]^2*dw^2*x^9/dt^2+(2167/90720)*h[0]^3*dw^3*x^10/dt^3+(7/8800)*h[0]^4*dw^4*x^11/dt^4+(1/2)*x^2*h[6]*dw/dt+(2/105)*x^7+(2/9)*((13/180)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(1/3)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^9+(1/4)*(((1/2)*h[1]*dw/dt+1)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)+(1/3)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^8+(2/7)*(((1/2)*h[1]*dw/dt+1)*((5/12)*h[1]*dw/dt+1)+(1/9)*h[0]*dw^2*h[2]/dt^2)*x^7+(1/3)*((1/3)*((1/2)*h[1]*dw/dt+1)*h[2]*dw/dt+(1/6)*h[0]*dw^2*h[3]/dt^2)*x^6+(1/105)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^8+(2/105)*((1/2)*h[1]*dw/dt+1)*x^7+(1/42)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^8+(4/105)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^7+(2/105)*((1/2)*h[1]*dw/dt+1)^2*x^7+(1/18)*((13/180)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^9+(1/14)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^8+(2/21)*((5/12)*h[1]*dw/dt+1+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^7+(2/15)*((1/3)*h[2]*dw/dt+(1/4)*h[0]*dw^2*h[3]/dt^2)*x^6+(1/18)*((1/20)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(5/36)*h[0]^2*dw^2/dt^2)*x^9+(1/14)*((5/12)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/3)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^8+(2/21)*(((1/2)*h[1]*dw/dt+1)*((1/3)*h[1]*dw/dt+1)+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^7+(1/9)*((1/10)*((1/3)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(25/144)*h[0]^2*dw^2/dt^2)*x^9+(1/8)*((1/20)*h[2]*dw^3*h[0]^2/dt^3+(5/6)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^8+(1/7)*((5/12)*h[0]*dw^2*h[2]/dt^2+((1/3)*h[1]*dw/dt+1)^2)*x^7+(2/9)*((139/1260)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((43/180)*h[0]*dw/dt+(8/45)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/6)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)/dt)*x^9+(1/4)*((43/180)*h[0]*dw/dt+(8/45)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/6)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((3/10)*h[1]*dw/dt+4/5+(1/10)*h[0]*dw^2*h[2]/dt^2+(1/5)*((1/2)*h[1]*dw/dt+1)^2)/dt)*x^8+(2/7)*((3/10)*h[1]*dw/dt+4/5+(37/120)*h[0]*dw^2*h[2]/dt^2+(1/5)*((1/2)*h[1]*dw/dt+1)^2)*x^7+(1/3)*((5/12)*h[2]*dw/dt+(1/6)*h[0]*dw^2*h[3]/dt^2)*x^6+(2/5)*((1/3)*h[3]*dw/dt+(1/4)*h[0]*dw^2*h[4]/dt^2)*x^5+(1/5)*((1/2)*h[1]*dw/dt+1)*h[3]*dw*x^5/dt+(1/15)*((1/2)*h[1]*dw/dt+1)*h[2]*dw*x^6/dt+(1/126)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^8/dt+(1/6)*h[2]*dw*((1/3)*h[1]*dw/dt+1)*x^6/dt

 (5)
> 

y := sum(u[l], l = 0 .. n-1);

1+(1/2)*x^4+(1/5)*x^5+(1/2)*x^2*h[0]*dw/dt+(1/2)*x^2*h[1]*dw/dt+(1/3)*x^3*h[0]*dw/dt+(1/2)*x^2*h[2]*dw/dt+(1/3)*x^3*h[1]*dw/dt+(5/12)*x^4*h[0]*dw/dt+(1/20)*h[0]^2*dw^2*x^5/dt^2+(1/2)*x^2*h[3]*dw/dt+(1/3)*x^3*h[2]*dw/dt+(1/6)*x^4*h[1]*dw/dt+(1/6)*x^5*h[0]*dw/dt+(13/180)*h[0]^2*dw^2*x^6/dt^2+(1/2)*x^2*h[4]*dw/dt+(1/3)*x^3*h[3]*dw/dt+(5/12)*x^4*h[2]*dw/dt+(1/15)*x^5*h[1]*dw/dt+(1/18)*x^6*h[0]*dw/dt+(139/1260)*h[0]^2*dw^2*x^7/dt^2+(1/160)*h[0]^3*dw^3*x^8/dt^3+(1/2)*x^2*h[5]*dw/dt+(1/3)*x^3*h[4]*dw/dt+(5/12)*x^4*h[3]*dw/dt+(1/6)*x^5*h[2]*dw/dt+(1/45)*x^6*h[1]*dw/dt+(1/63)*x^7*h[0]*dw/dt+(139/5040)*h[0]^2*dw^2*x^8/dt^2+(17/1296)*h[0]^3*dw^3*x^9/dt^3+x+(1/15)*x^6+x^3+(2/15)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^6+(1/15)*((1/2)*h[1]*dw/dt+1)^2*x^6+(1/4)*((13/180)*h[0]^2*dw^2/dt^2+(1/2)*h[0]*dw*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)/dt)*x^8+(2/7)*((3/10)*h[0]*dw/dt+(1/5)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/2)*h[0]*dw*((5/12)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*((5/12)*h[1]*dw/dt+1+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(2/5)*((1/3)*h[2]*dw/dt+(1/4)*h[0]*dw^2*h[3]/dt^2)*x^5+(1/4)*((1/20)*((1/2)*h[1]*dw/dt+1)*h[0]^2*dw^2/dt^2+(5/36)*h[0]^2*dw^2/dt^2)*x^8+(2/7)*((5/12)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt+(1/3)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(1/3)*(((1/2)*h[1]*dw/dt+1)*((1/3)*h[1]*dw/dt+1)+(1/6)*h[0]*dw^2*h[2]/dt^2)*x^6+(4/105)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^7+(1/15)*((1/2)*h[1]*dw/dt+1)*x^6+(2/21)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^7+(2/15)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^6+(1/5)*((1/2)*h[1]*dw/dt+1)*x^5+(1/3)*((5/12)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/3)*h[1]*dw/dt+1)/dt)*x^6+(2/5)*((1/3)*h[1]*dw/dt+1+(1/4)*h[0]*dw^2*h[2]/dt^2)*x^5+(1/5)*((1/2)*h[1]*dw/dt+1)^2*x^5+(2/5)*((1/3)*h[0]*dw/dt+(1/2)*h[0]*dw*((1/2)*h[1]*dw/dt+1)/dt)*x^5+(1/2)*((1/2)*h[1]*dw/dt+1)*x^4+(1/5)*((1/2)*h[1]*dw/dt+1)*h[2]*dw*x^5/dt+(2/63)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^7/dt+(1/9)*h[0]*dw*((1/2)*h[1]*dw/dt+1)*x^6/dt+x^2

 (6)
> 

``
> 

NULL


 

Download incomplete_example.mw

Help!. Transition probability matrix plot for firs...

Asked by: ogunmiloro 120
statistics probability plotting
October 15 2020
0  15


 

> 

P = (Matrix(5, 5, {(1, 1) = .6199, (1, 2) = .1450, (1, 3) = 0.636e-1, (1, 4) = .1549, (1, 5) = 0.166e-1, (2, 1) = .6550, (2, 2) = 0.2e-3, (2, 3) = 0.870e-1, (2, 4) = .1827, (2, 5) = 0.751e-1, (3, 1) = .2990, (3, 2) = .1294, (3, 3) = 0.10e-2, (3, 4) = .4925, (3, 5) = 0.781e-1, (4, 1) = .2773, (4, 2) = .7184, (4, 3) = 0.43e-2, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.9e-3, (5, 2) = .6229, (5, 3) = .3762, (5, 4) = 0, (5, 5) = 0}))

P = (Matrix(5, 5, {(1, 1) = .6199, (1, 2) = .1450, (1, 3) = 0.636e-1, (1, 4) = .1549, (1, 5) = 0.166e-1, (2, 1) = .6550, (2, 2) = 0.2e-3, (2, 3) = 0.870e-1, (2, 4) = .1827, (2, 5) = 0.751e-1, (3, 1) = .2990, (3, 2) = .1294, (3, 3) = 0.10e-2, (3, 4) = .4925, (3, 5) = 0.781e-1, (4, 1) = .2773, (4, 2) = .7184, (4, 3) = 0.43e-2, (4, 4) = 0, (4, 5) = 0, (5, 1) = 0.9e-3, (5, 2) = .6229, (5, 3) = .3762, (5, 4) = 0, (5, 5) = 0}))

 (1)
> 


 

Download Transition_Probaility_Markov_chain.mw

layouts about DrawGraph...

Asked by: lcz 1029
graphtheory drawgraph
October 15 2020
0  3

I've been studying the  drawing  of graph lately .    One of the themes is  1-planar graph .

A 1-planar graph is a graph that can be drawn in the Euclidean plane in such a way that each edge has at most one crossing point,  where it crosses a single additional edge. If a 1-planar graph, one of the most natural generalizations of planar graphs, is drawn that way, the drawing is called a 1-plane graph or 1-planar embedding of the graph.

 

 

 

 

 

I know it is NP hard to determine whether a graph is a 1-planar . My idea is to take advantage of some mathematical software to provide some roughly and  intuitive understanding before determining .

Now,  the layout of vertices or edges becomes important.  The drawing of a plane graph is a good example.

G1:=AddEdge( CycleGraph([v__1,v__2,v__3,v__4]),{{v__2,v__4},{v__1,v__3}}):
DrawGraph(G1)
DrawGraph(G1,style=planar)

K5 := CompleteGraph(5);
DrawGraph(K5);
vp:=[[-1,0],[1,0],[-0.2,0.5],[0.2,0.5],[0,1]];
SetVertexPositions(K5,vp);  #modified the vertex position
DrawGraph(K5);

My problem is that I see that  Maple2020 has updated a lot of layouts about DrawGraph  graph theory backpack , and I don’t know which ones are working towards the least possible number of crossing of  each edges of graph . 

Some links that may be useful:

https://de.maplesoft.com/products/maple/new_features/Maple2020/graphtheory.aspx

https://de.maplesoft.com/support/help/Maple/view.aspx?path=GraphTheory/SetVertexPositions

I think the software can improve some calculations related to topological graph theory, such as crossing number of graph, etc.

 

Why won't this code display an ellipse?...

Asked by: Earl 955
polar conic plotting
October 15 2020
0  4

The uploaded worksheet plots conics in polar coordinates.

A circle, parabola and hyperbola all plot correctly, but the plot command for the ellipse displays a shifted circle. What's wrong?

Conic_quandry.mw

How to find sgn on maple?...

Asked by: Ahmed111 130
signum
October 15 2020
0  0

How to find sgn on maple?

signum.mw

How to generate all non-isomorphic graphs of a cer...

Asked by: DJJerome1976 540
graphtheory order select
October 14 2020
1  4

I am trying to generate all non-isomorphic graphs of a certain order and size that have the same degree sequence (not necessarily regular). I assume I would have to use the select option in GraphTheory[NonIsomorphicGraphs] command, however, there are no examples (that I could find) of how to use the option. Any idea how this could be done?

How to duplicate rows/columns in a matrix?...

Asked by: DJJerome1976 540
matrix row
October 14 2020
0  2

I have an n x n matrix. I am trying to write a procedure that will randomly choose a row (column) in the matrix, and replace another row (column) with the entries of the chosen row (column). So, the end result will be another n x n matrix that has two identical rows (columns). How could one do this?

RE: Substitution...

Asked by: pik1432 245
simplification
October 14 2020
0  2

Hello there, 

Would you please tell me how to get the 'Desired' expression from the 'Sec_Z2prim' expression?

My 'Attempt' did not work. 


 

> 

restart;
> 

TR_turns_ratio := N = n2 / n1;

N = n2/n1

 (1)
> 

Sec_Z2prim := Z1 = Z2*n1^2/n2^2;

Z1 = Z2*n1^2/n2^2

 (2)
> 

Attempt := subs(TR_turns_ratio, Sec_Z2prim) assuming (n1 > 0, n2 > 0);

Z1 = Z2*n1^2/n2^2

 (3)
> 

Desired := Z1 = Z2*(1/N)^2;

Z1 = Z2/N^2

 (4)
> 

 

 

Thank you, 
 

Download Q20201014.mw

Exporting  a piecewise functions in Maple to Matla...

Asked by: Maple_lover1 125
October 14 2020
1  14

I usually piecewise functions in my maple codes. I want to export  the my maple code to a Matlab code in order to utilize advantages of plots of Matlab.

What methods would you recommend me?

 

For example:

Suppose that I have a piecewise u function.

restart:
u:=1/(1. + exp(x))^2 + 1/(1. + exp(-5.*t))^2 - 0.2500000000 + x*(1/(1. + exp(1 - 5*t))^2 - 1./((1. + exp(-5*t))^2) + 0.1776705118 + 0.0415431679756514*piecewise(0. <= t and t <= 0.5000000000, 1.732050808, 0.) + 0.00922094377856479*piecewise(0. <= t and t <= 0.5000000000, 30.98386677*t - 7.745966692, 0.) + 0.0603742508215732*piecewise(0.5000000000 <= t and t <= 1., 1.732050808, 0.) - 0.00399645630498528*piecewise(0.5000000000 <= t and t <= 1., 30.98386677*t - 23.23790008, 0.)) + (-0.00243051684581302*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000809061198761621*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.0152377552205917*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.00195593427342862*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0. <= t and t <= 0.5000000000, 1.732050808, 0.) + (-0.000433590063316381*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000146112803263678*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.00319022339097685*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.000477063086307787*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0. <= t and t <= 0.5000000000, 30.98386677*t - 7.745966692, 0.) + (-0.00276114805649180*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) - 0.000933166016624500*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) - 0.0207984584912892*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) - 0.00314360556336114*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0.5000000000 <= t and t <= 1., 1.732050808, 0.) + (0.000172746997599710*piecewise(0. <= x and x <= 0.5000000000, 1.732050808, 0.) + 0.0000586775450031145*piecewise(0. <= x and x <= 0.5000000000, 30.98386677*x - 7.745966692, 0.) + 0.00136190009033518*piecewise(0.5000000000 <= x and x <= 1., 1.732050808, 0.) + 0.000211410172315387*piecewise(0.5000000000 <= x and x <= 1., 30.98386677*x - 23.23790008, 0.))*piecewise(0.5000000000 <= t and t <= 1., 30.98386677*t - 23.23790008, 0.);

 

 

I use the following code to transform a Matlab code, but the output is not working in Matlab.

 

with(CodeGeneration):
 	
Matlab(u,resultname="w");

 

how to obtain all solutions with multiplicity from...

Asked by: nm 10229
polynomial solve pdetools
October 14 2020
2  8

Using solve on this example:

restart;
eq:=x^3 - 3*x^2 + 3*x - 1=0;
solve(eq,x);

gives solution x=1, with multiplicty 3

              1, 1, 1

When using PDEtools:-Solve

restart;
eq:=x^3 - 3*x^2 + 3*x - 1=0;
solve(eq,x);
PDEtools:-Solve(eq,x);

it gives

         x = 1

How can one make it show x=1, x=1, x=1 ?

Found that only when using PDEtools:-Solve(eq,x,numeric) it gives 

           x = 1., x = 1., x = 1.

How to make it do the same without using numeric?

Maple 2020.1.1

 

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