8 years, 73 days

## Workaround...

 > restart;
 > expr := [ exp(y*LambertW(ln(y))), (ln(y)/LambertW(ln(y)))^y, eval(x^(x^x), x = exp(LambertW(ln(y)))), eval(x^(x^x), x = ln(y)/LambertW(ln(y))) ]:
 > Matrix(4, (i,j)->'is'(expr[i]=expr[j])): simplify(%) assuming y>1;
 (1)

Workaround

 > Y:=solve(LambertW(ln(y))=t, y);
 (2)
 > mat:=simplify(subs(LambertW(ln(y))=t, y=Y, expr)) assuming t>0;
 (3)
 > Matrix(4, (i,j)->is(mat[i]=mat[j]) );
 (4)

## direct...

We don't need Maple for this.

ΔABC is similar to ΔEBC and ΔFDC. So, D is the midpoint of BC. DF=AE=EB=3, DE=AF=FC=4, DB=DC=5 (Pythagoras).

H being the centroid of ΔABC we have AD=BD=DC=5, and finally DH = AD/3 = 5/3.

## Change...

It's possible to use the standard changes of variables (Chebyshev). E.g.:

 > J := Int(x^(1/2)/(x^2+1)^(3/4), x);
 (1)
 > (x^2+1)/x^2 = t^4; X:=solve(%,x)[1]; T:=solve(%%,t)[1];
 (2)
 > simplify( IntegrationTools:-Change(J, x=X) ) assuming t>1;
 (3)
 > simplify(value(%)) assuming t>1;
 (4)
 > J=simplify(eval(%, t=T)) assuming x>0;
 (5)
 > simplify(diff(J - rhs(%), x)); # check
 (6)

 > restart;
 > with(plots): with(plottools): #with(DEtools); N := S(t) + In(t) + C(t); eqn1 := diff(S(t), t) = lambda - (lambda + sigma)*S(t) - (beta + zeta)*S(t)*In(t) - beta__1*S(t)*C(t), S(0) = ic1; eqn2 := diff(In(t), t) = beta*S(t)*In(t) - (lambda + gamma)*In(t), In(0) = ic2; eqn3 := diff(C(t), t) = zeta*In(t) + zeta*In(t)^2 - (rho + lambda)*C(t) - zeta*C(t)*In(t), C(0) = ic3; lambda := 0.117852; mu := 0.035378; beta := 0.11; beta__1 := 0.05; g := 1; rho := 0.1; zeta := 0.02; sigma := 0.066; ic1 := 2390000; ic2 := 753; ic3:= 358500;
 (1)
 > dsol := dsolve([eqn1, eqn2, eqn3], numeric);
 (2)
 > dsol(0.); dsol(0.001);
 (3)
 > odeplot(dsol,[[t,C(t)],[t,In(t)],[t,S(t)]], 0..0.001);

`Remark. gamma in Maple denotes the Euler constant. I hope you used it as such.`

## Not a bug...

 > restart;
 > S:= Sum(1/(4*n^2-4*n+4*100000000^2+1)/10^n,n=1..infinity);
 (1)
 > s:=convert(S, hypergeom);
 (2)
 > v:=value(S);
 (3)
 > #evalf(v);
 > #evalf(s);
 > evalf[100](S);
 (4)
 > evalf[500](S);
 (5)

 > restart

1 < alpha, alpha < 2      are given.

 > f := (a - b + c)^2 + (alpha - 1)*b^2/alpha + (alpha - 1)*c^2/alpha
 (1)

Find M>0 such that  f >= M*a^2,  for any a,b,c>0.

Solution.

Denote  b/a=B, c/a=C. Then we have to determine  min F, where:

 > F := (1 - B + C)^2 + (alpha - 1)*B^2/alpha + (alpha - 1)*C^2/alpha;
 (2)
 > F:=simplify( eval(F, alpha=1/(1-t)) );
 (3)
 > minimize(F, B=0..infinity, C=0..infinity, location) assuming t <= 1/2, t >= 0;
 (4)
 > M__max = simplify(eval(%[1], t=(alpha - 1)/alpha))
 (5)

## explicit, autorange...

 > restart;
 > f:=2.96736996560705*10^(-12)*p^2+1.31319840299485*10^(-13)*t^2- 8.89549693662593*10^(-7)*p+8.53128393394231*10^(-7)*t-3.65558815509970*10^(-30)*p*t-1:
 > tt:=simplify([solve](f,t)):
 > pp:=solve(tt[1]>-infinity,{p}): p1,p2 := lhs(pp[1]), rhs(pp[2]):
 > plot(tt, p=p1 .. p2, scaling=constrained);

## Workaround...

 > #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]);
 (1)
 >

## CartesianProduct...

```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...

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
 > 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...

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);
 (1)
 > f(4)
 (2)
 > f(2.)
 (3)
 > evalf(f(2));
 (4)

## Both are correct...

Both solutions are correct.

 > restart;
 > ode:=x*diff(y(x),x\$2)-3*diff(y(x),x)+x*y(x);
 (1)
 > Order:=16; rhs( dsolve(ode,y(x),'series') ); maple:=convert(%, polynom);
 (2)
 > simplify(eval(ode, y(x)=maple)); # Check OK
 (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);
 (4)
 > simplify( eval(ode, y(x)=mma) ); # Check OK
 (5)

## 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.

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