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Kitonum

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17 years, 243 days
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?...

Kitonum 21890
June 21 2020
0 0

@spinoza  I did not understand the meaning of your question.

No...

Kitonum 21890
June 17 2020
0 0

@Carl Love  Your plot  does not prove anything, because it is built on a finite set of points, and the break point is somewhere between them. Here is another plot  on a very small interval, inside which there is a break point (the zero of the denominator):

Digits:=20:
fsolve(denom(x(t)));
plot(x(t), t=-0.00077376199177721..-0.00077376199177719, -5000..5000);

      

 

Adjustment...

Kitonum 21890
June 16 2020
0 0

@666 jvbasha 


 

> 

restart;
ContoursWithLabels := proc (Expr, Range1::(range(realcons)), Range2::(range(realcons)), Number::posint := 8, S::(set(realcons)) := {}, GraphicOptions::list := [color = black, axes = box], Coloring::`=` := NULL)
local r1, r2, L, f, L1, h, S1, P, P1, r, M, C, T, p, p1, m, n, A, B, E;
uses plots, plottools;
f := unapply(Expr, x, y);
if S = {} then r1 := rand(convert(Range1, float)); r2 := rand(convert(Range2, float));
L := [seq([r1(), r2()], i = 1 .. 205)];
L1 := convert(sort(select(a->type(a, realcons), [seq(f(op(t)), t = L)]), (a, b) ->is(abs(a) < abs(b))), set);
h := (L1[-6]-L1[1])/Number;
S1 := [seq(L1[1]+(1/2)*h+h*(n-1), n = 1 .. Number)] else
S1 := convert(S, list)  fi;
print(Contours = evalf[2](S1));
r := k->rand(20 .. k-20); M := []; T := [];
for C in S1 do
P := implicitplot(Expr = C, x = Range1, y = Range2, op(GraphicOptions), gridrefine = 3);
P1 := [getdata(P)];
for p in P1 do
p1 := convert(p[3], listlist); n := nops(p1);
if n < 500 then m := `if`(40 < n, (r(n))(), round((1/2)*n)); M := `if`(40 < n, [op(M), p1[1 .. m-11], p1[m+11 .. n]], [op(M), p1]); T := [op(T), [op(p1[m]), evalf[2](C)]] else
if 500 <= n then h := floor((1/2)*n); m := (r(h))(); M := [op(M), p1[1 .. m-11], p1[m+11 .. m+h-11], p1[m+h+11 .. n]]; T := [op(T), [op(p1[m]), evalf[2](C)], [op(p1[m+h]), evalf[2](C)]]
fi; fi; od; od;
A := plot(M, op(GraphicOptions));
B := plots:-textplot(T);
if Coloring = NULL then E := NULL else E := ([plots:-densityplot])(Expr, x = Range1, y = Range2, op(GraphicOptions), op(rhs(Coloring)))  fi;
display(E, A, B);
end proc:
> 

# Your new example

A3 := .25*y*(-6*x^2+6*x-2.477250468*x*(x-1)^2-2.477250468*x^2*(x-1)-1.476663599*x*(x-1)^3-2.214995399*x^2*(x-1)^2+.3837076420*x*(x-1)^4+.7674152840*x^2*(x-1)^3+1.049305257*x*(x-1)^5+2.623263142*x^2*(x-1)^4+1.470504325*x*(x-1)^6+4.411512974*x^2*(x-1)^5+2.062933702*x*(x-1)^7+7.220267957*x^2*(x-1)^6+1.610136961*x*(x-1)^8+6.440547843*x^2*(x-1)^7+.6577852166*x*(x-1)^9+2.960033475*x^2*(x-1)^8);




ContoursWithLabels(A3, 0 .. 1, 0 .. 10, {seq(0.1..5,0.2)}, [color = black, thickness = 2, axes = box, size=[450,450], labels = ["eta", "r"],labeldirections = [horizontal, vertical], labelfont = ['TIMES', 'BOLDOBLIQUE', 16]], Coloring = [colorstyle = HUE,style = surface]);

.25*y*(-6*x^2+6*x-2.477250468*x*(x-1)^2-2.477250468*x^2*(x-1)-1.476663599*x*(x-1)^3-2.214995399*x^2*(x-1)^2+.3837076420*x*(x-1)^4+.7674152840*x^2*(x-1)^3+1.049305257*x*(x-1)^5+2.623263142*x^2*(x-1)^4+1.470504325*x*(x-1)^6+4.411512974*x^2*(x-1)^5+2.062933702*x*(x-1)^7+7.220267957*x^2*(x-1)^6+1.610136961*x*(x-1)^8+6.440547843*x^2*(x-1)^7+.6577852166*x*(x-1)^9+2.960033475*x^2*(x-1)^8)

  

Contours = [.1, .3, .5, .7, .9, 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7, 2.9, 3.1, 3.3, 3.5, 3.7, 3.9, 4.1, 4.3, 4.5, 4.7, 4.9]

  

 
> 

 


 

Download ContoursWithLabels11.mw

Example...

Kitonum 21890
May 27 2020
0 0

@radaar  For example, when you define a procedure using  arrow-notation:

x:=10: t:=20:
f:=x->x^2:
g:=t->t^2:
f(3); g(3);

f  and  g  are  the same function.

Automated...

Kitonum 21890
May 27 2020
0 0

The solution to similar and much more complex examples is automated in this post
https://www.mapleprimes.com/posts/145922-Perimeter-Area-And-Visualization-Of-A-Plane-Figure-

It is right...

Kitonum 21890
May 25 2020
0 0

@SanzharMukatay  This is correct, but a little cumbersome. For example, if you set the law of motion in a list, the components of which represent the corresponding coordinates, then you can immediately differentiate this list and get speed components in the form of a list too. The same goes for acceleration:

xy:=[3*t,4*t^2+1]:
v_xy:=diff(XY,t);
a_xy:=diff(v_xy,t);

                                               v_xy := [3, 8 t]
                                                a_xy := [0, 8]

Swap...

Kitonum 21890
May 22 2020
0 0

@Axel Vogt  Probably, vv deliberately swapped the numerator with the denominator in order to obtain the desired result for OP. I actually did the same in my answer.

The polynom in terms of lamda...

Kitonum 21890
May 20 2020
0 0

@AHSAN  You can use the same data, only each sublist needs to be reversed:

 

Download help_new1.mw

assuming integer...

Kitonum 21890
May 19 2020
0 0

@Rouben Rostamian  Thank you. I dont know. The case  assuming integer is more complicated because we must consider the case separately when one parameter is  0 .

int(cos(m*t)*cos(0*t), t=-Pi..Pi, allsolutions) assuming integer;
convert(%, piecewise, m);

 

Adjustment...

Kitonum 21890
May 15 2020
0 0

@Grigoriy Yashin  I added 1 short line:


 

restart; PDEtools[declare](`&theta;__si`(t), `&beta;__si`(t), psi(t), x(t), z(t)); PDEtools[declare](prime = t); V__1six := diff(x(t), t)-`l__1&scy;`*sin(`&theta;__si`(t))*cos(psi(t))*(diff(psi(t), t))-sin(psi(t))*`l__1&scy;`*cos(`&theta;__si`(t))*(diff(`&theta;__si`(t), t)); V__1siy := -`l__1&scy;`*sin(`&theta;__si`(t))*(diff(`&theta;__si`(t), t)); V__1siz := diff(z(t), t)+`l__1&scy;`*sin(`&theta;__si`(t))*sin(psi(t))*(diff(psi(t), t))-cos(psi(t))*`l__1&scy;`*cos(`&theta;__si`(t))*(diff(`&theta;__si`(t), t)); V__1si := simplify(V__1six^2+V__1siy^2+V__1siz^2, size); V__1si := map(simplify, collect(V__1si, [diff, diff(`&theta;__si`(t), t), `l__1&scy;`, `l__1&scy;`^2])); V__1si := simplify(V__1si, size)

`l__1с`^2*(diff(theta__si(t), t))^2-2*cos(theta__si(t))*`l__1с`*((diff(z(t), t))*cos(psi(t))+sin(psi(t))*(diff(x(t), t)))*(diff(theta__si(t), t))+`l__1с`^2*(diff(psi(t), t))^2*sin(theta__si(t))^2+2*sin(theta__si(t))*`l__1с`*((diff(z(t), t))*sin(psi(t))-cos(psi(t))*(diff(x(t), t)))*(diff(psi(t), t))+(diff(z(t), t))^2+(diff(x(t), t))^2

 (1)
> 

 

NULL


 

Download trigon_new.mw

It's impossible...

Kitonum 21890
May 15 2020
0 0

@Tyttemus  You yourself see that on the right side of the equality  k = ...  the variable  V  is not reduced, that is,  k  depends on  V . Therefore, it is impossible to remove  V .

Here is a very simple example illustrating your problem. Let you have the equation  3*x+2*y=0 . Then it is easy to find  x/y = -2/3. But if we slightly change the equation  3*x+2*y+1=0 , then it is impossible to find  x/y , that it does not depend on either  x  or  y .

Thanks...

Kitonum 21890
May 08 2020
0 0

@Carl Love  Thank you for this improvement.

For recent versions only...

Kitonum 21890
May 04 2020
0 0

@Carl Love  This probably only works in recent versions of Maple.

In Maple 2018.2 :

[for i from 11 to 20 do i^2 od];

     Error, reserved word `for` unexpected

color="DarkSlateGray"...

Kitonum 21890
May 03 2020
0 0

@nm   Try  color="DarkSlateGray"  instead of color=gray .

Need text, not picture...

Kitonum 21890
April 28 2020
0 0

You should present all of this in text form using Maple syntax. Why should someone do this for you?

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