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1) start cmaple.exe;

2) type "proc(x, x);", and watch Maple die;

3) Request a refund from Maplesoft.

Hello all,

I am presenting some results in a small meeting tomorrow and I have a rather large symbollic matrix that I was hoping to be able to view in a more readable form (once you see my code, you will see what I mean). This should be a simple fix. Furthermore, when I use the Latex command to recieve code to import into latex, its not working properly, which makes me think I made some kind of mistake. I am really just trying to get this matrix in its full for so that it is easy for other people to read. Thanks for any help.Turns_Latex.mw
 

restart

with(LinearAlgebra)``

A := Matrix(5, 5, [0, 0, 0, 0, 0, -AXX*UU-AXY*VV-AXZ*WW, AXX, AXY, AXZ, 0, -AXY*UU-AYY*VV-AYZ*WW, AXY, AYY, AYZ, 0, -AXZ*UU-AYZ*VV-AZZ*WW, AXZ, AYZ, AZZ, 0, -AXX*UU*UU-AYY*VV*VV-AZZ*WW*WW-(AXY*UU*VV+AXZ*UU*WW+AYZ*VV*WW)*2+(-E+2*UVW)*AE, -AE*UU-VL2, -AE*VV-VL3, -AE*WW-VL4, AE])

A := subs(VL2 = -AXX*UU-AXY*VV-AXZ*WW, VL3 = -AXY*UU-AYY*VV-AYZ*WW, VL4 = -AXZ*UU-AYZ*VV-AZZ*WW, A)

A := subs(AXX = mu*(zeta__x^2+zeta__y^2+zeta__z^2+(1/3)*`#msup(mi("\`zeta__x\`"),mn("2"))`), AYY = mu*(`ζ__x`^2+`ζ__y`^2+`ζ__z`^2+(1/3)*`#msup(mi("\`ζ__y\`"),mn("2"))`), AZZ = mu*(`ζ__x`^2+`ζ__y`^2+`ζ__z`^2+(1/3)*`#msup(mi("\`ζ__z\`"),mn("2"))`), A)

A := subs(AXY = (1/3)*mu*zeta__x*zeta__y, AXZ = (1/3)*mu*zeta__x*zeta__z, AYZ = (1/3)*mu*zeta__y*zeta__z, A)

A := subs(UU = u, VV = v, WW = w, A)

A := subs(AE = mu*gamma*(`ζ__x`^2+`ζ__y`^2+`ζ__z`^2)/Pr, A)

Matrix(%id = 18446744078321522678)

(1)

latex(A)

 \left[ \begin {array}{ccccc} 0&0&0&0&0\\ \noalign{\medskip}-\mu\,
 \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt
\#msup(mi("zeta}}_{\mbox {{\tt x"),mn("2"))}}}/3 \right) u-1/3\,\mu\,
\zeta_{x}\,\zeta_{y}\,v-1/3\,\mu\,\zeta_{x}\,\zeta_{z}\,w&\mu\,
 \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt
\#msup(mi("zeta}}_{\mbox {{\tt x"),mn("2"))}}}/3 \right) &1/3\,\mu\,
\zeta_{x}\,\zeta_{y}&1/3\,\mu\,\zeta_{x}\,\zeta_{z}&0
\\ \noalign{\medskip}-1/3\,\mu\,\zeta_{x}\,\zeta_{y}\,u-\mu\, \left( {
\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt
\#msup(mi("zeta}}_{\mbox {{\tt y"),mn("2"))}}}/3 \right) v-1/3\,\mu\,
\zeta_{y}\,\zeta_{z}\,w&1/3\,\mu\,\zeta_{x}\,\zeta_{y}&\mu\, \left( {
\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt
\#msup(mi("zeta}}_{\mbox {{\tt y"),mn("2"))}}}/3 \right) &1/3\,\mu\,
\zeta_{y}\,\zeta_{z}&0\\ \noalign{\medskip}-1/3\,\mu\,\zeta_{x}\,\zeta
_{z}\,u-1/3\,\mu\,\zeta_{y}\,\zeta_{z}\,v-\mu\, \left( {\zeta_{x}}^{2}
+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt \#msup(mi("zeta}}_{\mbox
{{\tt z"),mn("2"))}}}/3 \right) w&1/3\,\mu\,\zeta_{x}\,\zeta_{z}&1/3\,
\mu\,\zeta_{y}\,\zeta_{z}&\mu\, \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}
+{\zeta_{z}}^{2}+\mbox {{\tt \#msup(mi("zeta}}_{\mbox {{\tt
z"),mn("2"))}}}/3 \right) &0\\ \noalign{\medskip}-\mu\, \left( {\zeta_
{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt \#msup(mi("zeta}}
_{\mbox {{\tt x"),mn("2"))}}}/3 \right) {u}^{2}-\mu\, \left( {\zeta_{x
}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt \#msup(mi("zeta}}_{
\mbox {{\tt y"),mn("2"))}}}/3 \right) {v}^{2}-\mu\, \left( {\zeta_{x}}
^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{\tt \#msup(mi("zeta}}_{
\mbox {{\tt z"),mn("2"))}}}/3 \right) {w}^{2}-2/3\,\mu\,\zeta_{x}\,
\zeta_{y}\,uv-2/3\,\mu\,\zeta_{x}\,\zeta_{z}\,uw-2/3\,\mu\,\zeta_{y}\,
\zeta_{z}\,vw+{\frac { \left( -E+2\,{\it UVW} \right) \mu\,\gamma\,
 \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2} \right) }{\Pr}
}&-{\frac {\mu\,\gamma\, \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta
_{z}}^{2} \right) u}{\Pr}}+\mu\, \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2
}+{\zeta_{z}}^{2}+\mbox {{\tt \#msup(mi("zeta}}_{\mbox {{\tt
x"),mn("2"))}}}/3 \right) u+1/3\,\mu\,\zeta_{x}\,\zeta_{y}\,v+1/3\,\mu
\,\zeta_{x}\,\zeta_{z}\,w&-{\frac {\mu\,\gamma\, \left( {\zeta_{x}}^{2
}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2} \right) v}{\Pr}}+1/3\,\mu\,\zeta_{x}
\,\zeta_{y}\,u+\mu\, \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}
}^{2}+\mbox {{\tt \#msup(mi("zeta}}_{\mbox {{\tt y"),mn("2"))}}}/3
 \right) v+1/3\,\mu\,\zeta_{y}\,\zeta_{z}\,w&-{\frac {\mu\,\gamma\,
 \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2} \right) w}{\Pr
}}+1/3\,\mu\,\zeta_{x}\,\zeta_{z}\,u+1/3\,\mu\,\zeta_{y}\,\zeta_{z}\,v
+\mu\, \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}+\mbox {{
\tt \#msup(mi("zeta}}_{\mbox {{\tt z"),mn("2"))}}}/3 \right) w&{\frac
{\mu\,\gamma\, \left( {\zeta_{x}}^{2}+{\zeta_{y}}^{2}+{\zeta_{z}}^{2}
 \right) }{\Pr}}\end {array} \right]

 

``


 

Download Turns_Latex.mwTurns_Latex.mw

 

Dear Sir

There is error appears when we solve system of differential equations contains 13 equations ,

the form of error

Error, (in DEtools/convertsys) invalid terms in sum: -9*Y[8]+2 .. Y[3]+.3*Y[6]+.4*Y[7]+3.5*Y[11]+3.5*Y[13]

 

Hint we use dsolve, numerically

How to overcome on this error 

 

Thanks

after i tryrring to instell another vergion of maple and it failed when i reinstell the older vergion of maple i get this error 
what is missing and i did wronge 

l := (n+m+sum(a[i], i = 1 .. n)+sum(b[j], j = 1 .. m))*ln(alpha)+n*ln(lambda[1])+m*ln(lambda[2])+lambda[1]*(sum(x[i], i = 1 .. n))+lambda[2]*(sum(y[j], j = 1 .. m))-(sum((2+a[i])*ln(exp(lambda[1]*x[i])-1+alpha), i = 1 .. n))-(sum((2+b[j])*ln(exp(lambda[2]*y[j])-1+alpha), j = 1 .. m));

Error, (in property/ConvertProperty) invalid input: PropRange uses a 2nd argument, b, which is missing


 

 

Even and odd complement each other

how to find other sequence which complement each other?

such as 3 sequences divided integers or 5 sequences divided whole integers

is there monomials creation method such that solve result about coefficient and power are integers when right side columns are sequences?

 

i find even multiplication numbers are always solved into integers coefficient and power.

is there any more other sequences?

if I choose six multiplication table sequence, 

what is this complement of six multiplication table sequence?

How do i change keyboard shortcuts, E.g one of my keyboard keys are broken or have stopped working. How do i redirect my shortcut to another keyboard combination or another hotkey simply.

Hello, is it possible to have a document synchronized in the same way as Onenote? We're a group of engieneer students wanting to share documents, however the only way we can figure out a way to do it. Is uploading to maplecloud groups, sharing a "base" document, and everytime someone "updates" the document you save a new one. So it kind of defeats the purpose of making it a smart idea of collaborating on one document, am i missing something or is this really how oldschool it works?

I was wondering how Maple cope with piecewise functions during forward integration and if it's preferable to use dsolve events option in place of defining a piecewise discontinuos function.

As far as I understood dsolve/events halts the integration each time an event is triggered and subsequently restarts the integration using the pre-trigger outputs as new initial conditions. I suppose that by using a piecewise, if a discontinuity is detected, dsolve proceeds exactly in the same way halting and restarting the integration.

Here a toy example of a 2D rolling dice (idea of a rolling dice from the rolling cube by @one man :P ) in which the reaction forces of the floor can be seen as function of the compenetration dice/ground

Both the appraches (events and piecewise) give the same results

falling_dice.mw

Thanks to Mr. Joe Riel, my previous problem is the port's name and the input signal's name was the same name was solved. It worked well in the new model but when I try to do exactly the same thing in my current model, the custom component can't be generated even if the "Check dimension" button shown that "no issues found". 

This time error is:
(in SetComponent) flattening error: Errors: Input length = 1
Error in Component btnGenComponent with the caption "Generate MapleSim Component"

Does anyone know from what this error "Input length = 1" happens? 

I am very appreciated for your help.

Hi

In mathematics, the inverse problem for Lagrangian mechanics (Helmholtz inverse problem) is the problem of determining whether a given system of ordinary differential equations can arise as the Euler–Lagrange equations for some Lagrangian function. 

For more information read section IV.2. page 65 of the following reference:

http://www.unilim.fr/pages_perso/loic.bourdin/Documents/bourdin-thesis2013.pdf

________________________________________________________________________

 

I need some hints or procedures (if it is possible) for similar (but a little more complex) problem:

1- Assume that you have one ordinary differential equation, ode1(r) in polar coordinate system (i.e. (r, theta)). The ODE is taken to be independent from theta (It is not a PDE).

2- Assume that "Euler" is an operator that gives the Euler-Lagrange equation, I need a procedure to calculate ode2(r) such that

1/(2r)*Euler (ode2(r)) -Laplacian (1/(2r)*Euler(ode1(r)))=0

It is obvious that we need inverse of Euler operator (say IE) to calculate ode2(r).

ode2(r) =IE( 2r*Laplacian (1/(2r)*Euler(ode1(r))))

I calculate ode2(r) for some simpler cases via trial and error method.

s := proc (S) 
subs(w = w(r), w1 = diff(w(r), r), w2 = diff(w(r), r$2), S) 
end proc: 
Euler := proc (f) 
s(diff(f, w))-(diff(s(diff(f, w1)), r))+diff(s(diff(f, w2)), r$2) 
end proc:

Example:

ode1(r) = -r*(diff(w(r),r))^2:

ode2(r) = (diff(w(r),r))^2/r+r*(diff(w(r),r$2))^2:

-1/(2*r)*Euler(w1^2*r):

simplify(1/(2*r)*Euler(w1^2/r+r*w2^2)-VectorCalculus:-Laplacian(%,('polar')[r,theta]))

I will be grateful if you can hint me to write an appropriate procedure.

Thanks

A catenoid is the minimal surface between two 3D circles which are co-axial and parallel.

Is there a technique for finding the formula for the minimal surface if the circles are "stretched" into ellipses with proportional major and minor axes?


 

restart

Eq3 := lambda*((1/2)*cos(mu*xi)^(2*beta)*a*lambda+(-beta^2*mu^2-c)*cos(mu*xi)^beta+mu^2*beta*(beta-1)*cos(mu*xi)^(beta-2))

lambda*((1/2)*cos(mu*xi)^(2*beta)*a*lambda+(-beta^2*mu^2-c)*cos(mu*xi)^beta+mu^2*beta*(beta-1)*cos(mu*xi)^(beta-2))

(1)

NULL


 

Download mapleprime2.mw

Please help me with the following worksheet containg a sample Equation. I need to equate the exponent and co-efficents of the cosine function

 

How I can solve these time delay  differential equations?

please see attatched files.

Best

Doc191.pdf

ny.mw

 


 

restart; d[1] := 1; d[2] := 4; d[3] := 1; r[1] := 1; r[2] := 1; r[3] := 1; r[4] := .5; a[11] := .5; a[12] := 3; a[21] := 2; a[22] := .8; a[23] := 1; a[32] := .5; a[33] := .9; tau := .3

diff(u(t, x), t) = d[1]*(diff(u(t, x), x, x))+u(t, x)*{r[1]-a[11]*u(t, x)-a[12]*v(t, x)}

diff(u(t, x), t) = diff(diff(u(t, x), x), x)+u(t, x)*{1-.5*u(t, x)-3*v(t, x)}

(1)

diff(v(t, x), t) = d[2]*(diff(v(t, x), x, x))+v(t, x)*{r[2]+a[21]*u(t-tau, x)-a[22]*v(t, x)-a[23]*w(t-tau, x)}

diff(v(t, x), t) = 4*(diff(diff(v(t, x), x), x))+v(t, x)*{1+2*u(t-.3, x)-.8*v(t, x)-w(t-.3, x)}

(2)

diff(w(t, x), t) = d[3]*(diff(w(t, x), x, x))+w(t, x)*{r[3]+a[32]*v(t, x)-a[33]*w(t, x)}

diff(w(t, x), t) = diff(diff(w(t, x), x), x)+w(t, x)*{1+.5*v(t, x)-.9*w(t, x)}

(3)

0 < x and x < Pi, t > 0

0 < x and x < Pi, 0 < t

(4)

diff(u(t, x), x) = 0, diff(v(t, x), x) = 0, diff(w(t, x), x) = 0, x = 0, x = Pi, t >= 0

diff(u(t, x), x) = 0, diff(v(t, x), x) = 0, diff(w(t, x), x) = 0, x = 0, x = Pi, 0 <= t

(5)

u(t, x) > 0, v(t, x) > 0, w(t, x) > 0, `in`(t, x, `&x`([-tau, 0], [0, Pi]))

0 < u(t, x), 0 < v(t, x), 0 < w(t, x), `in`(t, x, [-.3, 0]*[0, Pi])

(6)

``

``


 

Download ny.mw


I have a complicated expression which includes RootOf( a quadratic ) but holds for all x what i'd like to do is turn it into a polynomial in x[1], x[2], x[3] so i can start looking at the monomial coefficients.

k[a1]*((x[1]+x[3])*k[d1]+C[T]*k[m])*(R[b]-x[1]-2*x[2])/((R[b]+R[m]-x[1]-2*x[2]-x[3])*k[a1]+k[m])-k[d1]*x[1]-k[a2]*x[1]*(R[b]-x[1]-2*x[2])+2*k[d2]*x[2] = (-R[b]*k[a2]+2*k[a2]*x[1]+2*k[a2]*x[2])*(k[a1]*kh[m]*((x[1]+x[3])*k[d1]+C[T]*k[m])*(R[b]+R[m]-Rh[m]-x[1]-2*x[2])/(k[m]*((R[b]+R[m]-x[1]-2*x[2]-x[3])*k[a1]*kh[m]/k[m]+kh[m]))-k[d1]*x[1]-kh[a2]*x[1]*(R[b]+R[m]-Rh[m]-x[1]-2*x[2])+2*kh[d2]*x[2])/(2*kh[a2]*RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])+(-2*kh[a2]*RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])+2*k[a2]*x[1]+2*k[d2]-2*kh[d2])*(kh[a2]*x[1]*(R[b]+R[m]-Rh[m]-x[1]-2*x[2])-2*kh[d2]*x[2])/(2*kh[a2]*RootOf(kh[a2]*_Z^2+(-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])*_Z-2*k[a2]*x[1]*x[2]-k[a2]*x[1]^2+k[a2]*x[1]*R[b]+2*kh[d2]*x[2]-2*k[d2]*x[2])-R[b]*kh[a2]-R[m]*kh[a2]+Rh[m]*kh[a2]+2*kh[a2]*x[2])

If this were something like q(x)=p1(x)/sqrt(p2(x)) where p1 and p2 are polynomials and q is a quotient- this would be as simple as making sqrt(p2(x)) the subject and squaring both sides, and then movinbg everything onto one and multiplying out denominators. However RootOf is something I'm not used to manipulating.

Is there anyway of converting this expression to a polynomial using maple commands?

Hello people in mapleprimes,

I installed maple 2018 Japanese version.
And, with solve(x^2-1,x), its solution is expressed as _EXPSEQ(1, -1)

I know this expression is an internal represantation.
How can I have maple answer as 1, -1?

Thanks in advance.

Addition: my pc is mac osx 10.13.6.
 

 

 

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