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MaplePrimes Activity

These are replies submitted by sand15


My mistake, this phrase about "a loop" is indeed an error.

It's likely  I didn't understand correctelyyour question.
I understood you wanted to compite the decrease/increase of the eights of the peaks.

What do you mean by "the decrease or incease in percentage between curves" ?
If you this

A_ref := eval(A, lambda = 1.3015):
data_1 := [beta=0.1, Q=1.3015, lambda=0.9986];
B_1    := eval(B, data_1):
plot((B_1-A_ref)/A_ref*100, x=-4..0)

you get a curve which represents the relative variation of B_1 wrt A_ref, but this is a function x.
So do you want a number or an algebraic expression or a curve


Right, I didn't pay attention to that.

type "SIR  model" and click on Search

@Carl Love 

Using fsolve(..., complex) as you suggested gives a solution close to yours to within  10-11 (or less) which verifies rel to within 10-42.
Given that diff(Re(rel[1]), A) is infinite at this point, the slightest variation in A makes "huge" differences in the values of rel[1].

So you're right, there is indeed a second real solution

@Carl Love 

Thanks Carl... but this is not the result I get with Maple 2015:

eval(rel, {
    A = -2.7553365135418814642586082436429575890825402826031,
    B = -0.70285804987973303586180028708027467941012949957141

[                                                     -43    
[0.00004788232651393033381187767465396229938775 - 2 10    I, 

  1.8649856419668477410903757547200476549949700479584 10   

                                                           -41  ]
   - 1.2309622539151182340327668587211822233603338843467 10    I]

A version issue?


Are there other real roots than the one @Rouben Rostamian got?
I don't think so (the couple (A, B) @Carl Love found doesn't verify the equations with an enough small error to be considered, IMO, as a solution).

Some details are given here

If the target starts from the left focus with velocity VT and the predator from the left vertex with velocity VP then: the predator will catch the prey only if VP >VT ant it will catch it at the center of the ellipse iif VP / VT = 1/e (e=eccentricity, assumed > 0 and < 1).
This is a particular situation where the capture at the center of the ellipse is not the rule.

So I doubt that, without extra conditions, the capture always takes place at the center of the ellipse


Suppose you have to functions f[1](x) and f[2](x).
Heaviside(f(x)) (let's put aside the "x=0 case") has two values: 0 if f[1](x) < 0 ans 1 if f[1](x) > 0.
Then h(x)=Heaviside(f[1](x))+2*Heavside(f[2](x)) takes 4 values:

  • 0 if f[1](x) < 0 and f[2](x) < 0
  • 1 if f[1](x) > 0 and f[2](x) < 0
  • 2 if f[1](x) < 0 and f[2](x) > 0
  • 3 if f[1](x) > 0 and f[2](x) > 0

Then contourplotting h(x)  with contours=[0, 1, 2, 3] (provided you use a grid dense enough) will display the 4 domains corresponding to each of theconditions above.

I replaced the Heaviside function by a smooth tanh approximation) to ease the computations of the contour levels.
(the smoothing depends on a parameter [set to 1e6] in the tanh.

The generalization of the "Heaviside trick" is

h(x) = add(2^(n-1)*f[n](x), n=1..N)

@Rouben Rostamian  

The values

[E__1 = 0.991324553918355, E__2 = 0.972412189223068, h__1 = 0.999999468863441, nu = 0.473159649082875]

verify eqE.
To get them do

J := (lhs - rhs)(eqE)^2:
opt := Optimization:-Minimize(J, {0 <= nu, nu <= 0.5}, assume = nonnegative)

But I agree that there is probably no solutions:

J := add(`~`[lhs - rhs]([eqA, eqC, eqD, eqE]) ^~ 2);
opt := Optimization:-Minimize(J, {0 <= nu, nu <= 0.5}, assume = nonnegative, iterationlimit = 10000);
       [                      [                          7  
opt := [0.206149604449184010, [E__1 = 3.48734878853157 10 , 

                                                     5         ]]
  E__2 = 13328.4876967435, h__1 = 6.85943362585648 10 , nu = 0.]]

eval([eqA, eqC, eqD, eqE], opt[2]);
              [0.4017000000 = 4.03508624851090, 

                0.1745000000 = -1.93898396374958, 

                0.1517000000 = -1.41034232626533, 

                0.1332000000 = -1.93899197963935]

A simple observation: nu is likely the Poisson coefficient and E1 and E2 are likely Young modulii.`

Thus h__eq (I guess a length?) has the same dimension than h__1.
The relations which define R__C, R__D and R__E do not seem consistent from a dimensional point of view: shouldn't contain h__1^2 instead of h__1

Maybe it could help if you give us the units you use?

More of this the ranges in the fsolve command seem (at least to me) quite weird (if I agree for the nu range, ranges for E are strange).


Thank's a lot.


Didn't you miss a derivative?
Logistic (ordinary) differential equation writes

diff(u(t), t) = u(t)*(1-u(t))

and Logistic Fractional Equation writes

diff(u(t), t^alpha) = u(t)*(1-u(t))

where  0 < alpha < 1.

I didn't see any derivative in what you wrote

Here is a reference:

Why don't you ask your question on a Matlab Q&A forum?


The complete construction of the embedding of a flat torus in the 3D Euclidean space is presented here

Source code avaliable here (hevea.tgz.)

Good luck if you are courageous enough to translate it in Maple

"I didnt know about Borrelli..."
The Nash-Kuiper embedding theorem(s) has been published in the mid fifties.
This theromem implies the existence of
an embedding of a flat torus in the 3D Euclidean space.
Nevertheless, it was not until almost 60 years later that Borrelli, Jabrane, Lazarus and Thibert built a 3D representation of this object.


Look my reply to @Carl Love

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