Kitonum

dsolve(... , numeric), plots:-odeplot...

Answered: Kitonum 21525

June 23 2020

3 2

>	restart; c1:=3.2: c2:=3.3: c3:=3.4: R:=-10: A:=1.6:

>	sol:=dsolve({diff(f(x),x$4) - c1diff(g(x),x$2) + R(diff(f(x),x)* diff(f(x),x$2) - f(x)diff(f(x),x$3))=0,diff(g(x),x$2)+c2(diff(f(x),x$2)-2g(x))-c3(f(x)diff(g(x),x)-diff(f(x),x)g(x))=0,D(f)(-1)=0, D(f)(1)=0,f(-1)=1-A,f(1) =1,g(-1)=0,g(1)=0}, numeric);

(1)

>	plots:-odeplot(sol,[[x,f(x)],[x,g(x)]], x=-1..1, color=[red,blue]);

>

Download diffeq1.mw

simplify, is...

Answered: Kitonum 21525

June 21 2020

0 3

If your goal is to prove the equality of 2 expressions (say A and B) then do simplify(A - B) or is(A = B) :

Sometimes this requires additional conditions:

is(sqrt(x^2)=x) assuming x>=0;
true

Download Qprime_20200621_new.mw

Addition. I took the expansion not in powers of x , but in powers of x-x0 where x0 = 7.5 (this is the middle of the range 5 ..10). If we take the expansion in powers of x , then to obtain a satisfactory result, it is necessary to take a very large n .

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rsolve...

Answered: Kitonum 21525

June 21 2020

1 3

It is obvious that for any x> 0 a(n) = x for any n , so if x=2 then a(100)=2 .

Download rsolve1.mw

collect...

Answered: Kitonum 21525

June 20 2020

1 0

>

restart;

>	sol:=W(x)=_C1(cosh(alphax)-sinh(alphax))+_C2(cosh(alphax)+sinh(alphax))+_C3sin(alphax)+_C4cos(alphax); W:=eval(W(x),sol);

(1)

>	F:=[sinh(alphax),cosh(alphax),sin(alphax),cos(alphax)]: W1:=collect(W, F); assign(([D1,D2,D3,D4]=~[coeffs(W1, F)])[]);

(2)

>	D1, D2, D3, D4;

(3)

>

Download collect_new.mw

combinat:-nextperm, combinat:-nextcomb...

Answered: Kitonum 21525

June 19 2020

0 1

The Iterator package was introduced in Maple 2016. If you have an older version of Maple, then you can use the commands combinat:-nextperm and combinat:-nextcomb .

primpart, content...

Answered: Kitonum 21525

June 19 2020

2 0

primpart(98-28*sqrt(7), sqrt(7));
content(98-28*sqrt(7), sqrt(7));
%*``(%%);
expand(%);

Or manually:

Expr:=98-28*sqrt(7);
d:=igcd(98,28);
d*``(Expr/d);

infinity...

Answered: Kitonum 21525

June 16 2020

3 0

The integral of a periodic function is the same on any period-long interval. The contradiction below is due to limited accuracy in approximate calculation. With increasing accuracy, both methods yield consistent results. Infinity is explained by the fact that the function is discontinuous, its denominator is 0 at some points.

>

restart;
x(t) := -3.703703704*10^(-7)*(0.000111668023*cos(1000/33*sqrt(1122)*t) - 0.0001214712007*sin(1000/33*sqrt(1122)*t) - 0.0002325581396*sqrt(561)*sqrt(2)*(-0.0004467462845*sqrt(1122)*sin(1000/33*sqrt(1122)*t) + 0.0004467462845*sqrt(1122)*cos(1000/33*sqrt(1122)*t)))/((2.074226433*10^14*cos(1000/33*sqrt(1122)*t) + 2.074226433*10^14*sin(1000/33*sqrt(1122)*t))*(4.895037587*10^(-11) + 0.01685634229*(0.00001474262739*cos(1000/33*sqrt(1122)*t) + 0.00001474262739*sin(1000/33*sqrt(1122)*t))^2)^2);
P:= 33*Pi/1000/sqrt(1122):

# Digits=10
RMS:= evalf(sqrt(1/2/P*Int(x(t)^2, t= -P..P)));
RMS:= evalf(sqrt(1/2/P*Int(x(t)^2, t= 0..2*P)));

Digits:=20:
RMS:= evalf(sqrt(1/2/P*Int(x(t)^2, t= -P..P)));
RMS:= evalf(sqrt(1/2/P*Int(x(t)^2, t= 0..2*P)));

(1)

>

Download Infinity.mw

Plotting surfaces of revolution...

Answered: Kitonum 21525

June 15 2020

1 0

To plot surfaces of revolution it is convenient to use their parametric equations in which to use polar coordinates on the plane. The 1 ex. :

restart;
x:=r*cos(phi): y:=r*sin(phi):
plot3d([[x,y,r],[x,y,4]], r=0..4, phi=0..2*Pi, style=surface, color=grey, scaling=constrained, labels=["x","y",z]);

InertForm:-Parse command automates this...

Answered: Kitonum 21525

June 13 2020

1 0

with(InertForm):
Parse("2+3^4");
Parse("(2+3)^4");

With variable and number substitution instead of variable:

Expr:="2*x+3*x-(4*x+5)/(3*x+6)":
Parse(Expr);
Typeset(%, 'inert'=false);
Parse(StringTools:-Subs("x"="8",Expr));
value(%);

plots:-display...

Answered: Kitonum 21525

June 12 2020

0 1

Download 2_graphs.mw

A way...

Answered: Kitonum 21525

June 12 2020

1 4

You can use the ContoursWithLabels procedure from the post https://www.mapleprimes.com/posts/202222-Contour-Curves-With-Labels for labelling. For convenience, I have included the text of this procedure in the worksheet below:

>

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(rhs(Coloring))) fi;
display(E, A, B);
end proc:

>

A3 := 0.1098129220e-1*x-0.1864590943e-1*x^4-0.3780537764e-1+0.5300456762e-1*x^2-0.2252843255e-4*x^7*sin(6.280000000*y-.2083809520)-0.9011373022e-5*x^7*sin(6.280000000*y-1.256000000)-0.1569010873e-4*x^6*sin(6.280000000*y-.2083809520)-0.6276043495e-5*x^6*sin(6.280000000*y-1.256000000)-0.1462906959e-2*x^5*sin(6.280000000*y-1.256000000)-0.3657267398e-2*x^5*sin(6.280000000*y-.2083809520)-0.7271311325e-3*x^4*sin(6.280000000*y-1.256000000)-0.1817827831e-2*x^4*sin(6.280000000*y-.2083809520)-0.8684412118e-1*x^3*sin(6.280000000*y-1.256000000)-.2171103030*x^3*sin(6.280000000*y-.2083809520)-0.2589262297e-1*x^2*sin(6.280000000*y-1.256000000)-0.6473155744e-1*x^2*sin(6.280000000*y-.2083809520)+.3752380081*x*sin(6.280000000*y-1.256000000)+.9380950203*x*sin(6.280000000*y-.2083809520)+0.3750477138e-1*sin(6.280000000*y-1.256000000)+0.9376192845e-1*sin(6.280000000*y-.2083809520)-0.3550918239e-3*x^6-0.760666870e-4*x^5+0.1713654921e-5*x^7-0.7811069699e-2*x^3-3.822344860*10^(-8)*x^8;

# Your example

ContoursWithLabels(A3, -2 .. 2, 0 .. 2, {seq(-2..2,0.2)}, [color = black, thickness = 2, axes = box, size=[1000,500], scaling=constrained], Coloring = [colorstyle = HUE, colorscheme = ["Blue", "Green", "Yellow", "Red"], style = surface]);

>

Of course, you can change the color scheme and other parameters as you wish.

Download ContoursWithLabels1.mw

A way...

Answered: Kitonum 21525

June 11 2020

0 0

Here is an example of how to do this:

Download pick.mw

A generalization of the coeff command....

Answered: Kitonum 21525

June 11 2020

2 0

Here is another approach that directly generalizes the coeff command to the case of polynomials in several variables:

>

restart;

>	coefff:=proc(P,T::{list,set}(name),t) local L,H,i,k: L:=[coeffs(P,T,'h')]: H:=[h]: k:=0: for i from 1 to nops(H) do if H[i]=t then k:=L[i] fi: od: k; end proc:

Examples of use

>	F:= 2x+6y+4x^2+12xy-5y^2;

(1)

>	coefff(F,{x,y},x); coefff(F,{x,y},x^2); coefff(F,{x,y},x*y);

(2)

>	Q:=x^3-3x^2yz+4z^5; coefff(Q,[x,y,z],z); coefff(Q,[x,y,z],x^2yz); coefff(Q,[x,y,z],x^3); coefff(Q,[x,y,z],z^5);

(3)

>

Download coefff.mws

Student:-MultivariateCalculus:-SecondDer...

Answered: Kitonum 21525

June 08 2020

0 1

The plot shows that your function (Ackley function for 2 variables) has a large number of critical points (at which both partial derivatives are equal to 0), among which there are points of a local minimum, maximum, and also saddle points. You can investigate each specific critical point using Student:-MultivariateCalculus:-SecondDerivativeTest command. Thus, the problem boils down to finding a sufficiently large number of critical points. Of course, the best option is to use DirectSearch package (vv"s suggestion). But if you do not have this package installed, you can use thу above command, having previously built the curves to find the necessary ranges for using fsolve command. See the worksheet below for critical points in the ranges x=0..1, y=0..1.

>	restart; f:=(x,y)->-20exp(-0.2sqrt(0.5(x^2+y^2)))-exp(0.5(cos(2Pix)+cos(2Piy)))+exp(1)+20;

(1)

>	plot3d(f(x,y), x=-3..3, y=-3..3, style=surface, grid=[100,100]);

>	plots:-implicitplot([diff(f(x,y),x),diff(f(x,y),y)], x=0..1, y=0..1, color=[red,blue], gridrefine=3);

>	P1:=fsolve({diff(f(x,y),x),diff(f(x,y),y)},{x=0.6..0.8, y=0.6..0.8}); P2:=fsolve({diff(f(x,y),x),diff(f(x,y),y)},{x=0.9..1, y=0.9..1}); P3:=fsolve({diff(f(x,y),x),diff(f(x,y),y)},{x=0.5..0.7, y=0.8..1}); P4:=fsolve({diff(f(x,y),x),diff(f(x,y),y)},{x=0.8..1, y=0.5..0.7});

(2)

>	with(Student[MultivariateCalculus]): SecondDerivativeTest(f(x,y), [x, y] = eval([x,y],P1)); SecondDerivativeTest(f(x,y), [x, y] = eval([x,y],P2)); SecondDerivativeTest(f(x,y), [x, y] = eval([x,y],P3)); SecondDerivativeTest(f(x,y), [x, y] = eval([x,y],P4));