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Hello Guys, I hope you are all fine. I have been struggling with creating an animation of the points (x,y) in maple. I have tried this example 
L := [[1, 1], [3, 2], [3.4, 6], [5, 3, 7], [3, 7, 9, 1], [2, 6, 8, 4, 5]];
animate(PointPlot, [L[trunc(t)]], t = 1 .. 6, frames = 150)
but in my case it shows two points at different location means it takes x and y seperate value and showed it on 1 and 2 on x axis but i want to animate it as the location of point. Please help me. 
Thank you in anticipation.

I have several functional equations in equally many unknown functions of at least two variables, plus parameters.  ("collect" works just for single equations, right?)

I know that for certain parameter ranges, all the functions involved will be quadratic, and I know some coefficients are zero.  That gives me some  coefficients to determine.  I want to

  1. specify the functional equations [done in a very primitive low-tech way in the attachment, using atomic variables rather than indices ... have I done correctly?!?] 
  2. get Maple to collect coefficients (the K's and the L's in the attachment; the variables are (y,z))
  3. get Maple to state an equation system these coefficients have to satisfy (these will unfortunately be coupled quadratics)
  4. get Maple to solve that equation system if possible, and if not: to tell me when (= for what parameter values, parameters being the "remaining letters" in the attachment) I have specified enough coefficients
  5. in case of a solution, get Maple to tell me which coefficients are real and positive (for those that are solution of quadratic eq's: whether a positive solution exists)

Phew. I am still a complete newbie. Edit: Attachment link: STcoeff2match.mw where the equations themselves are EQ0, EQ1 and EQ2 at the bottom. Copying and pasting them, they look like this (download STcoeff2pastedEQs.mw)

0 = -r__0*(K__011*y^2+K__022*z^2-K__012*(y-L__1)*(z-L__2)-K__01*(y-L__1)+K__02*(z-L__2))+(-2*K__011*y+m__1+K__012*(z-L__2)+K__01)*((2/3)*c__1*y-(4/3)*K__11*y+(2/3)*`K__12 `*(z-L__2)+(20/9)*K__011*y-(10/9)*K__012*(z-L__2)-(10/9)*K__01-(10/9)*m__1-(1/3)*c__2*z+(2/3)*K__22*z-(1/3)*`K__21 `*(y-L__1)-(16/9)*K__022*z+(8/9)*K__012*(y-L__1)-(8/9)*K__02+(8/9)*m__2)+(-2*K__022*z+m__2+K__012*(y-L__1)-K__02)*((2/3)*c__2*z-(4/3)*K__22*z+(2/3)*`K__21 `*(y-L__1)+(20/9)*K__022*z-(10/9)*K__012*(y-L__1)+(10/9)*K__02-(10/9)*m__2-(1/3)*c__1*y+(2/3)*K__11*y-(1/3)*`K__12 `*(z-L__2)-(16/9)*K__011*y+(8/9)*K__012*(z-L__2)+(8/9)*K__01+(8/9)*m__1)+(-(4/3)*K__011*y+(2/3)*K__022*z+(2/3)*K__012*(z-L__2)-(1/3)*K__012*(y-L__1)-(1/3)*m__2+(2/3)*m__1+(1/3)*K__02+(2/3)*K__01)^2+((2/3)*K__011*y-(4/3)*K__022*z-(1/3)*K__012*(z-L__2)+(2/3)*K__012*(y-L__1)+(2/3)*m__2-(1/3)*m__1-(2/3)*K__02-(1/3)*K__01)^2:

``

0 = -r__1*(K__11*y^2-`K__12 `*y*(z-L__2))+`K__12 `*y*((2/3)*c__2*z-(4/3)*K__22*z+(2/3)*`K__21 `*(y-L__1)+(20/9)*K__022*z-(10/9)*K__012*(y-L__1)+(10/9)*K__02-(10/9)*m__2-(1/3)*c__1*y+(2/3)*K__11*y-(1/3)*`K__12 `*(z-L__2)-(16/9)*K__011*y+(8/9)*K__012*(z-L__2)+(8/9)*K__01+(8/9)*m__1)+((2/3)*c__1*y-(4/3)*K__11*y+(2/3)*`K__12 `*(z-L__2)-(1/3)*c__2*z+(2/3)*K__22*z-(1/3)*`K__21 `*(y-L__1)-(10/9)*K__022*z+(5/9)*K__012*(y-L__1)-(5/9)*K__02+(5/9)*m__2+(8/9)*K__011*y-(4/9)*K__012*(z-L__2)-(4/9)*K__01-(4/9)*m__1)^2:

``

0 = -r__2*(K__22*z^2-`K__21 `*(y-L__1)*z)+`K__21 `*z*((2/3)*c__1*y-(4/3)*K__11*y+(2/3)*`K__12 `*(z-L__2)+(20/9)*K__011*y-(10/9)*K__012*(z-L__2)-(10/9)*K__01-(10/9)*m__1-(1/3)*c__2*z+(2/3)*K__22*z-(1/3)*`K__21 `*(y-L__1)-(16/9)*K__022*z+(8/9)*K__012*(y-L__1)-(8/9)*K__02+(8/9)*m__2)+((2/3)*c__2*z-(4/3)*K__22*z+(2/3)*`K__21 `*(y-L__1)-(1/3)*c__1*y+(2/3)*K__11*y-(1/3)*`K__12 `*(z-L__2)-(10/9)*K__011*y+(5/9)*K__012*(z-L__2)+(5/9)*K__01+(5/9)*m__1+(8/9)*K__022*z-(4/9)*K__012*(y-L__1)+(4/9)*K__02-(4/9)*m__2)^2:

``

 

 

Hi guys, I am trying to solve a system of differential equations, I have done the hand written calculations and I know the answer however I need to put it in a maple code for a generic system which I will work on over time. Here is what I have so far, 

restart;

eqn[1]:=-1/8*D[4](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[2]:=-1/8*D[5](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[3]:=-1/8*D[6](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[4]:=-1/8*D[7](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[5]:=-1/8*D[8](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[6]:=-1/8*D[9](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[7]:=-1/8*D[10](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

eqn[8]:=-1/8*D[11](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))-1/2=0;

eqn[9]:=-1/8*D[12](a)(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t))=0;

dsolve({seq(eqn[i],i=1..9)},a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(u(x, y, t), x, x), diff(u(x, y, t), y, x), diff(u(x, y, t), x, t), diff(u(x, y, t), y, t), diff(u(x, y, t), t, t)));

Then I get an error return which says:

Error, (in dsolve) too many arguments; some or all of the following are wrong: [{u(x, y, t)}, a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(diff(u(x, y, t), x), x), diff(diff(u(x, y, t), x), y), diff(diff(u(x, y, t), t), x), diff(diff(u(x, y, t), t), y), diff(diff(u(x, y, t), t), t))].

 

I know that if I replace u(x,y,t) with a dummy variable U, and its derivative with Ux,Uy,... and so on then it will work, but I need the function u(x,y,t) to be part of the solution.

I know the solution should give me:

a(x, y, t, u(x, y, t), diff(u(x, y, t), x), diff(u(x, y, t), y), diff(u(x, y, t), t), diff(diff(u(x, y, t), x), x), diff(diff(u(x, y, t), x), y), diff(diff(u(x, y, t), t), x), diff(diff(u(x, y, t), t), y), diff(diff(u(x, y, t), t), t)) = -4*diff(u(x,y,t),x,x) + F(x,y,t),

where F(x,y,t) is the constant function.

Please any help would be great!!
 

The summation takes too long time. Please help me
 

 

 

 

 

Hello people in mapleprimes,

This time question is a sequel of the previous one:
https://www.mapleprimes.com/questions/224346-Batch-File-And-Directory
, where I could obtain the file with the output.
Maple has been terminated after exhausting the output file.

But, I want  to do the next calculation while using the result in the output file.
For that, I think it might be a good way to have maple calculate a mpl file again, after having added 
new expressions to the original mpl file. But, copying and pasting expressions seen in the output file to the original mpl file with my hand, seems a little messy.

So, what I want to know is whether there isn't a good way for continuing calculations one after another,
while having maple termination after each calculation.

I will be very glad if you will give me an answer to this question.

 Thanks in advance.

I wish to calculate connection, curvature, Ricci curvature etc. for a

Riemannian metric given as follows: there is an orthogonal frame of vector

fields with stipulated Lie bracket relations between them. The frame is

orthogonal but not orthonormal, and the lengths of its vector fields are functions

of a single function on the manifold. Given these metric values on the frame and the

Lie bracket relations, the covariant derivatives are in principle computable from the

Koszul formula, hence connection and curvature are all determined.

When I try to define the metric using a dual coframe in ATLAS's Metric

routine, it allows me to define it but claims there is not actual curvature.

From the help it seems the coframes used in this routine are always given

as differentials of coordinates. Is there a way to get the metric via the data

given above without putting in by hand all the different Koszul formulas etc.?

I am considering a Fourier series

$cos (\alpha x) = \frac{1}{2}a_0 + \sum_{k=1}^{\infty}a_k cos(kx)$ for x between -pi and pi.

I have also shown using a different Fourier series that cos (\alpha x) has an alternative representation:

\frac{cos(\alpha x)}{\sin \alpha \pi} = \frac{1}{\pi \alpha} (1 + \frac{(\alpha \ pi)^2}{6} - \frac{\alpha x^2}{2 \pi} + \frac{2*\alpha^3}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k^2(k^2 - \alpha^2}*cos(kx)$.

To show that the second representation is a better approximation, I need to find the number of terms for this series and the original Fourier series needed for there to be a difference of 10^{-3} from the exact value of cos(\alpha \pi), assuming that \alpha = 0.75.  Could someone advise how I might do this?


 

Hi!

 

Is it possible to get step by step solution for:

- Laplace

- taylor

- fourie

Thanks for answers!

Hi,

I would like to do some computations with Maple using elliptic functions. The implementation in Maple wants me to provide the g2 and g3 invariants. However, what I have is the half-periods. Does Maple have a function, that calculates the invariants from the half-periods? I know I can do that myself, but I'd like to write something concise and probably the output of a built-in function will be precisely, what the other functions are looking for.

Thanks in advance

For my differential equations class we have to input the following problem into Maple, I've never used maple before and I was wondering if someone who is experienced with maple could help me out with the code to put into Maple. Thanks I appreciate it!

 

Using the Maple program, Write the procedure RungeKutta(f, a, b, aplpha, n) which use the improved Euler's method to approximate the solution of the initial-value problem y'=f(t,y), a\leqt\leqb, y(a)= alpha at (n+1) equally spaced numbers in the interval [a,b]

The input parameters are as follow:

f is the name of the function f(t,y);

a and b are the end points of the interval of integration;

alpha is the initial condition.

The output: array w is the approximation of y at the (n+1) values of t.

1.

with(Groebner):
K := {r-x^4,u-(x^3)*y,v-x*y^3,w-y^4};
G := Basis(K, 'tord', degrevlex(r,u,v,w));
R1 := eliminate(G, {r,u,v,w}); # eliminate is the reverse of Basis
Ga := Basis({a*G[1],a*G[2],a*G[3],a*G[4],a*G[5],a*G[6],a*G[7],a*G[8],a*G[9],a*G[10],a*G[11],a*G[12],a*G[13],a*G[14], (1-a)*K[1], (1-a)*K[2], (1-a)*K[3], (1-a)*K[4]}, 'tord', deglex(a,r,u,v,w));
Ga := remove(has, Ga, [x,y,a]);
eliminate(Ga, [r,u,v,w]);

how to eliminate Ga to find back K ?

2.

A1A2 and A3A4 parallel
A(0,0), B(u1,0), D(u2,u3), C(x1,x2), E(x3,x4)
#BC = A1A2
xx1 := u1:
xx2 := x1:
yy1 := 0:
yy2 := x2:
#AD = A3A4
xx3 := 0:
xx4 := u2:
yy3 := 0:
yy4 := u3:
eq1 := (xx2 - xx1)*(yy4 - yy3) - (yy2 - yy1)*(xx4 - xx3);
(x1-u1)*u3-x2*u2;
#CD = A1A2
xx1 := x1:
xx2 := u2:
yy1 := x2:
yy2 := u3:
#AB = A3A4
xx3 := 0:
xx4 := u1:
yy3 := 0:
yy4 := 0:
eq2 := (xx2 - xx1)*(yy4 - yy3) - (yy2 - yy1)*(xx4 - xx3);
-(u3-x2)*u1;
with(LinearAlgebra):
#E is on the same line of AC
xx1 := x3:
yy1 := x4:
xx2 := 0:
yy2 := 0:
xx3 := x1:
yy3 := x2:
eq3 := Determinant(Matrix([[xx1,yy1,1],[xx2,yy2,1],[xx3,yy3,1]]));
-x3*x2+x1*x4;
#E is on the same line of BD
xx1 := x3:
yy1 := x4:
xx2 := u1:
yy2 := 0:
xx3 := u2:
yy3 := u3:
eq4 := Determinant(Matrix([[xx1,yy1,1],[xx2,yy2,1],[xx3,yy3,1]]));
-x3*u3+u1*u3-u1*x4+u2*x4;
 

sol := eliminate({eq1,eq2,eq3,eq4},[x1,x2,x3,x4]);

with(Groebner):
K := {(rhs(sol[1][1])-lhs(sol[1][1])),(rhs(sol[1][2])-lhs(sol[1][2])),(rhs(sol[1][3])-lhs(sol[1][3])),(rhs(sol[1][4])-lhs(sol[1][4]))};
G := Basis(K, 'tord', degrevlex(x1,x2,x3,x4));
R1 := eliminate(G, {x1,x2,x3,x4}); # eliminate is the reverse of Basis
Ga := Basis({a*G[1],a*G[2],a*G[3],a*G[4], (1-a)*K[1], (1-a)*K[2], (1-a)*K[3], (1-a)*K[4]}, 'tord', deglex(a,x1,x2,x3,x4));
Ga := remove(has, Ga, [u1,u2,u3,u4,a]);

From Question1, is it possible to find from sol to eq1, eq2, eq3 and eq4 ?

restart;
with(PDEtools);
assume(k::real, x::real, omega::real, t::real, theta::real, c::real);
u := phi(c*(t*upsilon+x))*exp(I*(k*x+omega*t+theta));
PDE := proc (u) options operator, arrow; I*(diff(u, t))+diff(u, x, x)-I*sigma*u*(conjugate(u)*(diff(u, x))-u*conjugate(diff(u, x))) end proc;
Eq1 := PDE(u)

I have resolved the roots of a series both numerically & analytically.  Let me qualify numerical  versus analytical.  Analytically I evaluate the series without substituting values for the various parameters of the series.  I then differentiate the series, then substitute in the appropriate parametric values, & then solve.  By this method I obtained 5 complex roots.

The numerical approach has values already assigned to the parameters of the series.  I then differentiate & solve.  I obtain only REAL roots in this instance.  I then restricted these results to obtain the solution I believe to be correct given by result (7).

I cannot seem to steer the solver in the analytic case to obtain the correct REAL result that I am expecting.  Can anyone help on this?

Before any website moderator thinks this is the same question as Error-in-Isinternal-Too-Many-Levels, it is not.  I have resolved that question.  This is a different question, but on the same problem!

reconcile_solns.mw

This guy thinks the Cayley-Dickson Construction cycles back at 1024D? I don't actually use Maple, is he just encountering a precision error?

https://www.mapleprimes.com/posts/124913-Visualization-Of-The-CayleyDickson

quote in question

"

I found no new mathematics after 1024 because higher hypercomplex numbers greater than 1024 are cyclic (they repeat all over again).  I do not offer dimensions higher than 256D to the publc Maple Application Center because the mathematics is very slow and time cosuming past 64D.  However, I did keep the code up to 256D in the public Maple Application Center.

"

I received a program for constructing Cayley-Dickson tables for my own use, but it's actually written in bc (arbitrary precision). I'm assuming Maple supports this to some extent but maybe he's encountering a precision error that looks like it's repeating? I can't verify yet because he's right about the one thing, even a 256 table took DAYS to compute. The snapshots are intriguing too but I'm assuming even a second-long video at low resolution would take forever as well.

But I still have the bc code and constructing Cayley-Dickson tables is only a few lines of codes/conditions, it's one of those 'easy for a computer, impossible for a human' kind of deals. I don't see enough complexity in the code where it would suddenly start cycling. It's kind of important to me because the implications of the Cayley-Dickson Construction going on forever are more exciting in my opinion.

By the way I'm actually personally using the bc generated tables for making music sequences/MIDI. It's not the usual multiplication and such though because that even when normalized would "stick" so to speak? It's a bit of a secret, sorry

I know how to use numapprox to fit polynomial and rational polynomial approximations to trig functions.  I would like to fit a rational expression using numerators and denominators like this:

a*x^(1/3) + b*x^(1/2) + c*x^(2/3) + d*x + e*x^(3/2) + f*x^(2) + g*x^(3)

 

 

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