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12153 Reputation

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8 years, 3 days
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These are answers submitted by vv

Workaround...

vv 12153
April 05 2023
2 3
> 

#Workaround (inequations work in general only with polynomials -- SemiAlgebraic!)
> 

solve({(4*t1+4*t3)>0,  t1^2-4*t2 = t3^2, t3>=0},[t2,t3]);

piecewise(t1 <= 0, [[t2 = (1/4)*t1^2-(1/4)*t3^2, -t1 < t3]], 0 < t1, [[t2 = (1/4)*t1^2-(1/4)*t3^2, 0 <= t3]])

 (1)
> 

 

 

CartesianProduct...

vv 12153
March 25 2023
4 1

 

C:=Iterator:-CartesianProduct([p,q]$3):
[seq]([seq(v[])], v = C);

    [[p, p, p], [q, p, p], [q, q, p], [p, q, p], [p, q, q], [q, q, q], [q, p, q], [p, p, q]]

 

plot extra elements...

vv 12153
March 12 2023
2 1

You could plot some extra elements.

 

> 

restart;
> 

f := (x,y) -> piecewise(x=0, 0, x <> 0, x*y^2/(x^2+y^4)):
> 

Digits:=15:
> 

p1:=plot3d(f,-1..1,-1..1, numpoints=10000, color=green):
> 

f(y^2,y): # 1/2
> 

p2:=plots:-tubeplot([y^2,y,1/2], y=-1..1,radius=0.01, color=red, style=surface):
> 

r:=0.02:
p3:=plots:-tubeplot([r*cos(t), r*sin(t), f(r*cos(t), r*sin(t))],
    t=0..2*Pi, radius=0.01, color=blue,lightmodel=none, numpoints=4000,style=surface):
> 

plots:-display(p1, p2, p3);

 
> 

 

 

Contagion...

vv 12153
March 08 2023
0 0

In general, when you need approximations, use evalf  because the floating-point "contagion" does not apply to non-numeric constants.

> 

f:=x -> sqrt(2) + 1.5 + ln(x);

proc (x) options operator, arrow; sqrt(2)+1.5+ln(x) end proc

 (1)
> 

f(4)

2^(1/2)+1.5+2*ln(2)

 (2)
> 

f(2.)

2^(1/2)+2.193147181

 (3)
> 

evalf(f(2));

3.607360743

 (4)

 

Both are correct...

vv 12153
March 07 2023
2 9

 Both solutions are correct.

> 

restart;
> 

ode:=x*diff(y(x),x$2)-3*diff(y(x),x)+x*y(x);

x*(diff(diff(y(x), x), x))-3*(diff(y(x), x))+x*y(x)

 (1)
> 

Order:=16;
rhs( dsolve(ode,y(x),'series') );
maple:=convert(%, polynom);

16

  

_C1*x^4*(series(1-(1/12)*x^2+(1/384)*x^4-(1/23040)*x^6+(1/2211840)*x^8-(1/309657600)*x^10+(1/59454259200)*x^12-(1/14982473318400)*x^14+O(x^16),x,16))+_C2*(ln(x)*(series(9*x^4-(3/4)*x^6+(3/128)*x^8-(1/2560)*x^10+(1/245760)*x^12-(1/34406400)*x^14+O(x^16),x,16))+(series(-144-36*x^2+(1/2)*x^6-(25/1024)*x^8+(157/307200)*x^10-(91/14745600)*x^12+(709/14450688000)*x^14+O(x^16),x,16)))

  

_C1*x^4*(1-(1/12)*x^2+(1/384)*x^4-(1/23040)*x^6+(1/2211840)*x^8-(1/309657600)*x^10+(1/59454259200)*x^12-(1/14982473318400)*x^14)+_C2*(ln(x)*(9*x^4-(3/4)*x^6+(3/128)*x^8-(1/2560)*x^10+(1/245760)*x^12-(1/34406400)*x^14)-144-36*x^2+(1/2)*x^6-(25/1024)*x^8+(157/307200)*x^10-(91/14745600)*x^12+(709/14450688000)*x^14)

 (2)
> 

simplify(eval(ode, y(x)=maple)); # Check OK

-(1/14982473318400)*(_C1*x^4+435456*_C2*ln(x)-(3675456/5)*_C2)*x^15

 (3)
> 


# Now the solution obtained with MMA, also for Order=16:
 
> 

mma:=(x^4 - x^6/12 + x^8/384 - x^10/23040 + x^12/2211840 - x^14/309657600 + x^16/59454259200 - x^18/14982473318400)* C2
+ C1*((1040449536000 + 260112384000*x^2 + 16257024000*x^4 - 4967424000*x^6 + 218736000*x^8 - 4398240*x^10 + 51940*x^12 - 407*x^14)/
    1040449536000 + (x^4*(-309657600 + 25804800*x^2 - 806400*x^4 + 13440*x^6 - 140*x^8 + x^10)*ln(x))/4954521600);

(x^4-(1/12)*x^6+(1/384)*x^8-(1/23040)*x^10+(1/2211840)*x^12-(1/309657600)*x^14+(1/59454259200)*x^16-(1/14982473318400)*x^18)*C2+C1*(-(407/1040449536000)*x^14+(53/1061683200)*x^12-(187/44236800)*x^10+(31/147456)*x^8-(11/2304)*x^6+(1/64)*x^4+(1/4)*x^2+1+(1/4954521600)*x^4*(x^10-140*x^8+13440*x^6-806400*x^4+25804800*x^2-309657600)*ln(x))

 (4)
> 

simplify( eval(ode, y(x)=mma) ); # Check OK

-(1/14982473318400)*(C2*x^4-3024*C1*ln(x)+(29304/5)*C1)*x^15

 (5)

 

arctrig(cotrig())...

vv 12153
March 04 2023
2 0

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). 

Answer...

vv 12153
March 02 2023
3 3
> 

restart;
> 

poly := x^6 - 3*x + 3

x^6-3*x+3

 (1)
> 

firstroot := [solve](poly, x)[1];

RootOf(_Z^6-3*_Z+3, index = 1)

 (2)
> 

alias(alpha=firstroot);

alpha

 (3)
> 

G := GroupTheory:-GaloisGroup(poly, x);

_m2134450194848

 (4)
> 

GroupOrder(G);

72

 (5)
> 

quo(poly,x-alpha,x);

x^5+alpha*x^4+alpha^2*x^3+alpha^3*x^2+alpha^4*x+alpha^5-3

 (6)
> 

S:=simplify([solve(%)]):
> 

alpha[0]=alpha;
for i to 5 do print(alpha[i]=S[i]) od:

alpha[0] = alpha

  

alpha[1] = (1/102)*((12*alpha^5+28*alpha^4+20*alpha^3+24*alpha^2-12*alpha-30)*(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(1/3)-612*alpha^5-1428*alpha^4-1020*alpha^3+(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(2/3)-1224*alpha^2+612*alpha+3264)/(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(1/3)

  

alpha[2] = 3*(((2/51)*alpha^5+(14/153)*alpha^4+(10/153)*alpha^3+(4/51)*alpha^2-(2/51)*alpha-5/51)*(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(1/3)+(((1/612)*I)*3^(1/2)-1/612)*(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(2/3)+(alpha^5+(7/3)*alpha^4+(5/3)*alpha^3+2*alpha^2-alpha-16/3)*(I*3^(1/2)+1))/(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(1/3)

  

alpha[3] = -3*((-(2/51)*alpha^5-(14/153)*alpha^4-(10/153)*alpha^3-(4/51)*alpha^2+(2/51)*alpha+5/51)*(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(1/3)+(((1/612)*I)*3^(1/2)+1/612)*(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(2/3)+(alpha^5+(7/3)*alpha^4+(5/3)*alpha^3+2*alpha^2-alpha-16/3)*(I*3^(1/2)-1))/(-426564-41616*alpha^5-97104*alpha^4-69360*alpha^3-83232*alpha^2+10404*(408*alpha^5+952*alpha^4+680*alpha^3+816*alpha^2-408*alpha+1581)^(1/2)+41616*alpha)^(1/3)

  

alpha[4] = -(3/17)*alpha^5-(7/17)*alpha^4-(5/17)*alpha^3-(6/17)*alpha^2-(11/34)*alpha+15/34+(1/34)*(-272*alpha^5+136*alpha^4-68*alpha^3-255*alpha^2-306*alpha+969)^(1/2)

  

alpha[5] = -(3/17)*alpha^5-(7/17)*alpha^4-(5/17)*alpha^3-(6/17)*alpha^2-(11/34)*alpha+15/34-(1/34)*(-272*alpha^5+136*alpha^4-68*alpha^3-255*alpha^2-306*alpha+969)^(1/2)

 (7)

 

Download galois6.mw

Solution...

vv 12153
March 02 2023
3 2

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...

vv 12153
March 01 2023
2 2

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

But your ^(1,0)  is nonsense.

Digits...

vv 12153
February 28 2023
1 0

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...

vv 12153
February 25 2023
1 3

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...

vv 12153
February 18 2023
2 0

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

Invariance of domain...

vv 12153
February 16 2023
0 0

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...

vv 12153
February 04 2023
2 1
> 

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[]

x = (5*X+12*Y)*(-384+48*X+55*Y)/(225*X^2+225*Y^2-1800*X-602*Y), y = 2*(12*X-5*Y)*(-384+48*X+55*Y)/(225*X^2+225*Y^2-1800*X-602*Y)

 (1)
> 

eliminate( {%, Y= 2+3*X}, {X,Y})

[{X = (4/3)*(38274*x+38914*y-126451)/(30102*x-25053*y-131066), Y = -(213300*x+105550*y-767936)/(-30102*x+25053*y+131066)}, {738*x^2+2133*x*y+1240*y^2-2952*x-8173*y}]

 (2)
> 

conic:=%[-1][];

738*x^2+2133*x*y+1240*y^2-2952*x-8173*y

 (3)
> 

plots:-implicitplot(conic, x=-30..30,y=-30..30);

 
> 

 


Download conic.mw

By hand...

vv 12153
January 28 2023
3 5
> 

B:=<0,-2,1;-1,0,0;-2,0,0>; A:=B+1;

Matrix(3, 3, {(1, 1) = 0, (1, 2) = -2, (1, 3) = 1, (2, 1) = -1, (2, 2) = 0, (2, 3) = 0, (3, 1) = -2, (3, 2) = 0, (3, 3) = 0})

  

Matrix(%id = 18446745955451478974)

 (1)
> 

# By hand:
> 

B^2,B^3

Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 2, (2, 3) = -1, (3, 1) = 0, (3, 2) = 4, (3, 3) = -2}), Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0})

 (2)
> 

1 + n*B + binomial(n,2)*B^2:  'A'^n = expand~(%)

A^n = (Matrix(3, 3, {(1, 1) = 1, (1, 2) = -2*n, (1, 3) = n, (2, 1) = -n, (2, 2) = n^2-n+1, (2, 3) = -(1/2)*n^2+(1/2)*n, (3, 1) = -2*n, (3, 2) = 2*n^2-2*n, (3, 3) = -n^2+n+1}))

 (3)
   
> 

# By Maple:
> 

LinearAlgebra:-MatrixPower(A, n);

Matrix(3, 3, {(1, 1) = 1, (1, 2) = -2*n, (1, 3) = n, (2, 1) = -n, (2, 2) = n^2-n+1, (2, 3) = -(1/2)*n^2+(1/2)*n, (3, 1) = -2*n, (3, 2) = 2*n^2-2*n, (3, 3) = -n^2+n+1})

 (4)

 

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