14453 Reputation

11 years, 269 days

Solution...

First, we complete the squares to estimate the ranges for the loop variables.
There are 2 integer solutions:

```restart;
Eq:=7*(x^2+y^2+z^2)+2*x+4*y-8*z-8;
Student:-Precalculus:-CompleteSquare(Eq, [x,y,z]);
Eq1:=%/7; k:=0: T:=table():
for x from -2 to 2 do
for y from -2 to 2 do
for z from -2 to 2 do
if Eq=0 then k:=k+1; T[k]:=[x,y,z] fi;
od: od: od:
T:=convert(T,list);
```

Syntax...

Two syntax errors:
1. N/2  should be a positive integer.

```N:=10:
A:=Array(1..N/2);

sort...

Here is your example automatically. It works for an arbitrary list:

```restart;
L:= [-1+sqrt(7), -1 - sqrt(7)]:

L1:=sort(L,(x,y)->evalf(x)<=evalf(y));
m:=floor(L1[1]);  M:=ceil(L1[-1]);
[\$ m..M, op(L)];
samples:=sort(%, (x,y)->evalf(x)<=evalf(y));
evalf(samples);
```

Edit. If you need to insert something between the list items, you can do so:

`[seq(op([A,s]), s=samples), A];`

parametric=true, real...

`solve(a*x^2 + b*x + c=0, x, parametric=true, real);`

Procedure for plotting graphs in 3D...

Here is a simple procedure that allows you to beautifully plot graphs in 3D. Required parameters: Names - the list of  vertex names, Coords - the list of vertex coordinates, Edges - the set of edges. Opt  is an optional parameter that determines the graphic properties of objects (the color of the vertices, the size of the vertices (as a fraction of the average distance between the vertices), the color of the edges, the thickness of the edges).

 > restart; DrawGraph3D:=proc(Names::list,Coords::list, Edges::set(set), Opt::list:=[gold,0.05,brown,3]) local n, Dist, f, h, V, L, T; uses plottools, plots, ListTools; n:=nops(Edges); Dist:=(A,B)->sqrt((A[1]-B[1])^2+(A[2]-B[2])^2+(A[3]-B[3])^2); f:=x->Coords[Search(x,Names)]; h:=add(map(x->Dist(f(x[1]),f(x[2])), convert(Edges,list)))/n; V:=seq(sphere(c,Opt[2]*h,style=surface,color=Opt[1]), c=Coords); L:=seq(line(f(c[1]),f(c[2]),color=Opt[3],thickness=Opt[4]), c=Edges); T:=textplot3d([seq([f(c)[1],f(c)[2],f(c)[3]+0.05*h,c], c=Names)], font=[times,16],align={left,above}); display(V, L, T, scaling=constrained, axes=none); end proc:

Examples of use

 > DrawGraph3D([a,b,c,d],[seq([cos(2*Pi*k/3),sin(2*Pi*k/3),0],k=0..2),[0,0,1]],{{a,b},{b,c},{c,a},{d,a},{d,b},{d,c}},[red,0.06,blue,2]); DrawGraph3D([\$ 1..7],[seq([cos(2*Pi*k/5),sin(2*Pi*k/5),0],k=0..4),[0,0,1],[0,0,-1]],{seq({i,i+1},i=1..4),{5,1},seq({6,i},i=1..5),seq({7,i},i=1..5)}); # Graphic options by default

Animation of this picture:

plots:-animate(plots:-display,[plottools:-rotate(%,phi,[[0,0,0],[0,0,1]])], phi=0..2*Pi, frames=60);

Edit.

dimension=3...

Use the  dimension=3  option.

Example:

```with(GraphTheory):
H := CompleteGraph(4);
DrawGraph(H, style = spring, dimension = 3);```

plots:-animate...

```plots:-animate(plot3d, [[x, (cos(x)+1)*cos(phi), (cos(x)+1)*sin(phi)], x = 0 .. 4*Pi, phi = (1/2)*Pi .. t, style = surface, color = khaki, labels = [x, y, z], lightmodel = light1], t = (1/2)*Pi .. (-3*Pi)*(1/2), frames = 60, axes = normal, view = [-1 .. 13.7, -2.4 .. 2.4, -2.4 .. 2.4])
```

I guess that OP only meant to animate the process of painting one area (the curves themselves should not move):

```restart;
with(plots):
animate(shadebetween, [sin(x), cos(x), x = (1/4)*Pi .. t, positiveonly],
t = (1/4)*Pi .. 5*Pi*(1/4),background = plot([sin(x), cos(x)], x = 0 .. 2*Pi,
gridlines = true, thickness = 3, labels = ["", ""]), scaling = constrained, size = [900, 300])```

depends...

The answer is obvious - the function  2*sqrt(xyz)  does not depend on  . xyz is just a symbol unrelated to x .
In Maple, this can be tested using the  depends  function:

```depends(x^2,x);
depends(xyz,x);
```

true
false

Explore...

In the usual sense, it is impossible to plot a function of three variables, since it requires 3 dimensions for the arguments and one dimension for the values of the function, that is, we need 4-dimensional space. I propose the following way to solve problem: we fix 1 variable, for example , and then plot the function of two variables  x  and  y . Using  Explore command we can trace how this plot changes as  z  changes.

 > restart;
 > varphi[3/4,2/3]:=(x,y,z)->1/60*GAMMA(7/4)/(sqrt(2*Pi)*GAMMA(5/4))*(((abs(x)^(1/3)+abs(y)^(1/3))^2-abs(z)^(2/3))*(abs(z)^(2/3)-(abs(x)^(1/3)-abs(y)^(1/3))^2))^(1/4)/abs(x*y*z)^(1/2)*(10+66*cos(phi)^3-36*cos(phi));
 (1)
 > A:=eval(varphi[3/4,2/3](x,y,z),phi=Pi);
 (2)
 > Explore(plot3d(A, x=0.01..5, y=0.01..5, grid=[100,100], axes=normal, view=-0.5..0.1), z=0.01..10.);
 > eval(A,[x=1,y=3,z=0.05]);
 (3)
 > evalf(%);
 (4)
 >

The last 2 lines of code show that for some values of arguments your function takes imaginary values. For such values, the plot is not built.

Numerical differentiation...

To solve your problem, I used the simplest formulas for numerical differentiation. Your expression  diff(u(y, t), y)+diff(diff(u(y, t), y), t)  is calculated for the 4 indicated values  b   by replacing the derivatives at the point  (y,t) = (0,1)  with the corresponding difference ratios with the step  h=0.001

 > restart; a := 0.7: L := 8: HAA := [0, 2, 5, 10]: h:=0.001: PDE1 := diff(u(y, t), t) = diff(u(y, t), y, y)+diff(diff(u(y, t), y, y), t)-b*u(y, t)+T(y, t): PDE2 := diff(T(y, t), t) = (1+(1+(a-1)*T(y, t))^3)*(diff(T(y, t), y, y))+(a-1)*(1+(a-1)*T(y, t))^2*(diff(T(y, t), y))^2+T(y, t)*(diff(T(y, t), y, y))+(diff(T(y, t), y))^2: ICandBC := {T(L, t) = 0, T(y, 0) = 0, u(0, t) = t, u(L, t) = 0, u(y, 0) = 0, (D[1](T))(0, t) = -1}: for i from 1 to nops(HAA) do b := HAA[i]; pds[i] := pdsolve({PDE1,PDE2}, ICandBC, numeric); f:=(y0,t0)->rhs((pds[i]:-value(u(y,t),t=t0)(y0))[3]); A[i]:=(f(h,1)-f(0,1))/h+(f(h,1+h)-f(0,1+h)-f(h,1)+f(0,1))/h^2; end do: seq(A[i], i=1..4);
 (1)
 >

combinat:-permute...

Example:

`combinat:-permute([a,b,c,d]);`

[[a, b, c, d], [a, b, d, c], [a, c, b, d], [a, c, d, b], [a, d, b, c], [a, d, c, b], [b, a, c, d], [b, a, d, c], [b, c, a, d], [b, c, d, a], [b, d, a, c], [b, d, c, a], [c, a, b, d], [c, a, d, b], [c, b, a, d], [c, b, d, a], [c, d, a, b], [c, d, b, a], [d, a, b, c], [d, a, c, b], [d, b, a, c], [d, b, c, a], [d, c, a, b], [d, c, b, a]]

Some partial results...

In fact, there are many more solutions than indicated by vv, for example

```restart;
sys := [x^2 + y^2 - x*y - 1 = 0, y^2 + z^2 - y*z - a^2 = 0, z^2 + x^2 - x*z - b^2 = 0];
fsolve(eval(sys,[a=10,b=10.4]), {x=0..infinity,y=0..infinity,z=0..infinity});
;
```

sys := [x^2-x*y+y^2-1 = 0, -a^2+y^2-y*z+z^2 = 0, -b^2+x^2-x*z+z^2 = 0]
{x =0.1989730189, y = 1.084528287, z = 10.49805888}

Further, we strictly show that there are positive solutions  (x, y, z)  for arbitrary a=b>=1. From considerations of continuity, this implies that for any number  a>=1 and a sufficiently small positive number c  for  (a, b=a+c)  there are also a positive solutions  (x, y, z) .

```restart;
sys := [x^2 + y^2 - x*y - 1 = 0, y^2 + z^2 - y*z - a^2 = 0, z^2 + x^2 - x*z - b^2 = 0];
Sol:=simplify(solve(eval(sys,[b=a]), [x,y,z], explicit));
a in solve(a>=1 and `or`(seq((eval(x,k)>0 and eval(y,k)>0 and eval(z,k)>0), k=%)));
```

The result:                                          a in RealRange(1, Open(infinity))

Numerical experiments show that all solutions seem to be in the strip  1<=a<=b<a+1/2 . This does not mean that every point of this strip is a solution. This is only a necessary condition.

Example of use of the module sol...

```restart;
a := 1: b:= 2: l:= 1: alpha[1]:= 1: alpha[2]:= 3:
syspde:= [diff(u(x, t), t)-a+u(x, t)-u(x, t)^2*v(x, t)-alpha[1]*(diff(u(x, t), x\$2)) = 0, diff(v(x, t), t)-b+u(x, t)^2*v(x, t)-alpha[2]*(diff(v(x, t), x\$2)) = 0]:
Bcs:= [u(x,0)=1,v(x,0)=1,D[1](u)(-l, t) = 0,D[1](u)(l, t) = 0,D[1](v)(-l, t) = 0,D[1](v)(l, t) = 0]:
sol:=pdsolve(syspde, Bcs, numeric);

sol:-value(u(x, t)+diff(u(x, t), x)+diff(u(x, t), t), t=0.7)(0.5);
```

sol:=module() ... end module
[x = 0.5, t = 0.7, u(x, t)+diff(u(x, t), x)+diff(u(x, t), t) = 2.34305224942915]

Solutions...

```restart;
s := Sum(a*X[n]+b, n=1..N);
expand(value(s));
op(2,%)%+op(1,%);

s := Sum(X[n]+Y[n], n=1..N);
(expand@value)(s);
```

I use Maple 2018.2

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