## 12600 Reputation

8 years, 245 days

## arctrig(cotrig())...

Probably because f-1(f(x)) is mathematically simpler than g-1(f(x)). This is used for all trig/arctrig pairs.
(Of course, here, f-1 is the inverse only for a subdomain).

 > restart;
 > poly := x^6 - 3*x + 3
 (1)
 > firstroot := [solve](poly, x)[1];
 (2)
 > alias(alpha=firstroot);
 (3)
 > G := GroupTheory:-GaloisGroup(poly, x);
 (4)
 > GroupOrder(G);
 (5)
 > quo(poly,x-alpha,x);
 (6)
 > S:=simplify([solve(%)]):
 > alpha[0]=alpha; for i to 5 do print(alpha[i]=S[i]) od:
 (7)

## Solution...

The answer is x_1 = ... = x_n = 1/n  (n=2014) [actually easy to guess].
The problem is designed for a math competition, not for a CAS, but here is a solution.

The function f : [-1,1] -> R, f(x) = (1+x)^(1/2) is continuous, strictly increasing and strictly concave (f''(x)<0 for |x|<1).
For x_1, ..., x_n in [-1,1], Jensen's inequality implies
f(x_1) + ...+ f(x_n) <= n f((x_1 + ... +x_n)/n).
Denoting x* = (x_1 + ... +x_n)/n, from the 1st equation it results
f(1/n) <= f(x*).

Using the 2nd equation:
f(-1/n) <= f(-x*).

f being strictly increasing, 1/n <= x*, -1/n <= -x*, so, x*=1/n.
It results from the first equation that we have "=" in Jensen's inequqlity.
f being strictly concave, we must have x_1 = ...= x_n = 1/n.
Q.E.D.

## hypergeom...

Hypergeometric0F1Regularized[a, z]  corresponds to  hypergeom([], [a], z) / GAMMA(a)  in Maple.

## Digits...

To work correctly, MatrixPower needs a much higher precision, e.g. take Digits=40 in your example.

BTW, a matrix C has a square root if det(C) <>0.  <0,1; 0,0> has not a square root.

## Logical equivalence...

The equivalence is logical equivalence indeed. For the simple expression q := (x[1] &or x[2]) &and x[3],  the only nonidentical permutation is f defined by f(x1)=x2, f(x2)=x1, f(x3)=x3 [so, f(not x1) = not x2 etc].

As the help file (and  Christian Wolinski) says, the expression must be in CNF form. It is strange that Normalize (recommended in the help) does not work and we must use Canonicalize:

```with(Logic);
q := (x[1] &or x[2]) &and x[3];
qn := Normalize(q, form=CNF);                       # wrong!
qn := Canonicalize(q, {x[1],x[2],x[3]}, form=CNF);  # ok
G, L := SymmetryGroup(qn, output = [group, expressions]);
#g1, g2 := Generators(G)[];
GroupOrder(G);  # 2,  ok```

## O...

Here `O` is simply a local variable lost from some procedure and landed at top level, due to a bug.

## Invariance of domain...

According to a well known theorem, the Invariance of domain, an interval such as (380,780) cannot be homeomorphic to  (0,1)^3, so, a correspondence between WaveLength and (r,g,b) will be always problematic.

## eliminate...

 > Pn := [ (5*X + 12*Y)*(-384 + 48*X + 55*Y)/(225*X^2 + 225*Y^2 - 1800*X - 602*Y),          2*(12*X - 5*Y)*(-384 + 48*X + 55*Y)/(225*X^2 + 225*Y^2 - 1800*X - 602*Y)]:
 > (x,y) =~ Pn[]
 (1)
 > eliminate( {%, Y= 2+3*X}, {X,Y})
 (2)
 > conic:=%[-1][];
 (3)
 > plots:-implicitplot(conic, x=-30..30,y=-30..30);
 >

## By hand...

 > B:=<0,-2,1;-1,0,0;-2,0,0>; A:=B+1;
 (1)
 > # By hand:
 > B^2,B^3
 (2)
 > 1 + n*B + binomial(n,2)*B^2:  'A'^n = expand~(%)
 (3)
 > # By Maple:
 > LinearAlgebra:-MatrixPower(A, n);
 (4)

 > S:=Sum(Sum( a^2*b^2/sinh(Pi*(a+b))*(-1)^(a+b),  a=1..infinity), b=1..infinity); # The double series converges very fast!
 (1)
 > evalf[30](S);
 (2)
 > S=identify(%);
 (3)
 > # The equality holds but the proof is to be found!

## sqrt(Pi)...

 > str:="(1/(2^        j) ((k*Gamma[5 + 2 j] Gamma[           1 + l] HypergeometricPFQ[{1, 5/2 + j, 3 + j}, {3 + j + l/2,            7/2 + j + l/2}, -1])/        Gamma[6 + 2 j +          l] + ((k + m) Gamma[7 + 2 j] Gamma[           1 + l] HypergeometricPFQ[{1, 7/2 + j, 4 + j}, {4 + j + l/2,            9/2 + j + l/2}, -1])/Gamma[8 + 2 j + l]))/(2^(-5 - 3 j -        l) Gamma[5 + 2 j] Gamma[      1 + l] (k HypergeometricPFQRegularized[{1, 5/2 + j,          3 + j}, {3 + j + l/2, 7/2 + j + l/2}, -1] +       1/2 (3 + j) (5 + 2 j) (k + m) HypergeometricPFQRegularized[{1,          7/2 + j, 4 + j}, {4 + j + l/2, 9/2 + j + l/2}, -1]))";
 (1)
 > ex:=convert(str, FromMma);
 (2)
 > simplify(ex);
 (3)

## outline only...

For a surface with color=red, the color of the interior of the polygons is red and the color of their borders is black.

But in your case, the surface is actually a curve (it does not depend on c), so the area of each polygon is zero! Hence, the color reduces to black

## typesetting...

You are using interface(prettyprint=0). To get rid of those Typesetting:-mprintslash, use at the beginning:

interface(typesetting=standard);

## extrema...

Note first that the maxima/minima can be found directly using the command  extrema:

 > F:=x^2-2*x*y+2*y^2-12;
 (1)
 > extrema(x, {F}, {x,y}, 'Sx');  # min x   and  max x
 (2)
 > Sx;
 (3)
 > extrema(y, {F}, {x,y}, 'Sy');  # min y   and  max y
 (4)
 > Sy;
 (5)

Now, if you want the math details and use the implicit function theory,

then dy/dx equals indeed   (x-y)/(x-2*y);  you can use

 > dydx:=implicitdiff(F,y,x)
 (6)

Then you have to solve the system  {F=0,  dydx=0}.

 > solve({F=0,  dydx=0}, explicit);
 (7)

Similarly for y (i.e. dx/dy=0).

You may use (9)  in your worksheet only to check that the solutions correspond to min/max rather than inflexion points.

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