9572 Reputation

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9 years, 134 days

MaplePrimes Activity

These are answers submitted by Kitonum

Do := proc( F::list(procedure) , Q::list(list))
local n:=nops(F), m:=nops(Q);
if n<>m then error "Should be nops(F)=nops(Q)" fi;
seq(F[i]~(Q[i]), i=1..n);
end proc:

Example of use:

Do([x -> x , y -> y^2], [[0,1,2,3], [4,5,6,7]]);

                                            [0, 1, 2, 3], [16, 25, 36, 49]


I took  Gamma/(2*Pi)=1  and plotted  Re(w)  (in red) and  Im(w)  (in blue):

X, Y:=(Re,Im)(w) assuming theta::real, r>0;
plot3d([[r*cos(theta),r*sin(theta),X], [r*cos(theta),r*sin(theta),Y]], theta=0..2*Pi, r=0..3, color=[red,blue], axes=normal, view=[-4.3..4.3, -4.3..4.3, -1..6.7]); 


See help on  ?plot,tickmarks

plot(x^2, x=0..2, 1..1.00000000001); 
# Long values
plot(x^2, x=0..2, 1..1.00000000001, tickmarks=[default, [seq(1+2*10^(-12)*k=1+2*k*10^(`-12`), k=1..5)]]);  # Short values

Use nested  seq  command.


seq(seq(A[i,j] >= 0, j=1..4), i=1..3);

You can not find a limit with your procedure, because it for any particular n simply returns a number, but an exact dependence of the sum on n is needed.
The limit can be found as follows:

S:=unapply(sum(2*i/n*2/n, i=1..n), n);
limit(%, n=infinity);


By  p  procedure the limit can be found only numerically with  a specific accuracy:






gl_inv:=(v,alp)->fsolve(gl_inveq(t,v,alp)=0, t):
# The calculation of a value of the function
evalf(Int(gl_inv(2,t), t=0..1));  # The calculation of the value of the integral


You have an equation of the 9th degree with respect to lambda with two parameters  x  and  y . It is well known that in the general case the roots of such an equation can not be expressed in principle in terms of its coefficients. You can solve this equation numerically, but only by first setting the parameters.


x:=1: y:=2:
fsolve(P, _Z);

   0.1348905839, 0.8646312171, 1.143198734, 1.706737179, 3.176390681, 4.304915397, 4.893289594, 5.142364314, 6.633582298

f:= x-> diff(g(x),x)/(1+g(x)):
applyop(eval, 2, f(x), g(x)=h);
subs(h=g(x), series(%, h=0, 4));







I do not have Maple 18 and so I can not confirm it, but try these options:

plot(x^2, x=-2.5..2.5, tickmarks=[spacing(0.5), default]);
# Or   
plot(x^2, x=-2.5..2.5, tickmarks=[[seq(-2.5+0.5*i=-2.5+0.5*i, i=0..10)], default]);

Usually I do it this way: in Maple I press  "prt sc"  key, then I paste it into the Word, then I crop it, removing the extra one.

Suppose we have a product of some members and we want members  with a lesser degree to stand in front of of the members with a greater degree. Here is the procedure that does this:

local L, L1;
if type(Expr,`^`) then return Expr else
L1:=sort(L, (x,y)->degree(x)<=degree(y));
`*`(map(t->`if`(degree(t)<=1,t,`if`(type(t,`^`),(``(op(1,t)))^op(2,t),t)), L1)[]) fi;
end proc:

Example of use:





In  plots:-matrixplot  command, the matrix must consist of numbers, not functions.

See help on this command.

Use  plots:-spacecurve  command.


plots:-spacecurve([1, t, 1], t=0..10, color=red, thickness=3, axes=normal);

In brackets - parametric equations of this line.

I called your red subexpression as Expr :

Expr:=(A^2/nu)^(1/3)*g*beta*q2[w]*sqrt(nu^2/A)*T(y, t)/((nu*A)^(1/3)*k)+(A^2/nu)^(1/3)*g*beta*T[infinity]/(nu*A)^(1/3)-(A^2/nu)^(1/3)*g*beta*V[infinity]/(nu*A)^(1/3)+(A^2/nu)^(1/3)*g*beta1*C(y, t)/(nu*A)^(1/3)-(A^2/nu)^(1/3)*g*beta1*C[infinity]/(nu*A)^(1/3);

simplify(Expr)  assuming A>0, nu>0;



Using procedure  P and  Explore  command, we can investigate how the solution changes when the variable  A  is changed:

local omega, mu, B, eq1, eq2, eq3, dsys3;
omega := -2.667; mu := 10; B := 1;
eq1 := diff(x(t), t) = omega*x(t)-y(t)^2;
eq2 := diff(y(t), t) = mu*(z(t)-y(t));
eq3 := diff(z(t), t) = A*y(t)-B*z(t)+x(t)*y(t);
dsys3 := {eq1, eq2, eq3, x(0) = 10, y(0) = 10, z(0) = 10};
dsolve(dsys3, numeric);
end proc:

Explore(plots:-odeplot(P(A), [t,y(t)], t=0..10), A=1...10.);

Addition. The same we can do by  plots:-animate  command:

plots:-animate(plots:-odeplot,['P(A)', [t,y(t)], t=0..10], A=1..10, frames=91);


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