Mixing
Maximizing $x^4+y^4+z^4 + p(xy+xz+yz)$ on a sphere
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What values of $x,y,z$ maximize $f(x,y,z,p) = x^4+y^4+z^4 + p(xy+xz+yz)$ with constant $p geq 0$ with the constraint $x^2+y^2+z^2=1$?
Some preliminary studies in Mathematica showed that the behavior of the solution changes around $p=1$. For $p>1$, the problem seems solved by $x=y=z=frac{1}{sqrt 3}$. For $0 leq p leq 1$ it appears the solution (up to permutations of x,y,z) looks like x > y = z. These solutions appear to favor one or the other of the two terms in $f$, and I find the $p>1$ behavior striking and would appreciate a proof that $x=y=z=frac{1}{sqrt 3}$ indeed maximizes.
UPDATE
For at least $p>4/3$, one can show that $x=y=z=frac{1}{sqrt 3}$ maximizes f.
The proof for $1<p<4/3$ evades me currently (the left bound, as mentioned in comments, may be slightly higher than 1).
I'll assume without loss of generality that $x,y,z$ are all positive. Note that $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3} $, so if we show that for some $p$ we enjoy $f(x,y,z,p) leq p+frac{1}{3}$, we've shown $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
Now consider that via the constraint, $$f(x,y,z,p) = 1-2(x^2y^2+x^2z^2+y^2z^2) + p(xy+xz+yz)$$
The root mean square compared to arithmetic mean inequality tells us that for nonnegative $a,b,c$ we enjoy $sqrt{frac{a^2+b^2+c^2}{3}} geq frac{a+b+c}{3},$ so $-(a^2+b^2+c^2) leq -frac{(a+b+c)^2}{3}$. Then $$f(x,y,z,p) leq 1 + p(xy+xz+yz) -frac{2}{3}(xy+xz+yz)^2.$$
Now, let's call $(xy+xz+yz)=q$. Note that it is easy to prove with our constraints that $0leq q leq 1$.
Then we can bound the maximum of $f(x,y,z,p)$ by finding the maximum of $g(q) = 1+pq-frac{2}{3}q^2$ on $0leq q leq 1$.
The derivative $g'(q) = p-frac{4}{3}q$, so $g(q)$ increases up to $q=frac{3}{4}p$. Then on $0leq q leq 1$, we have $q_{max} = min(frac{3}{4}p, 1)$.
Thus for $p geq frac{4}{3}$ we have $q_{max}=1$, so $$ f(x,y,z,p) leq 1+p*1-frac{2}{3}*1^2 = p + 1/3$$
Thus since $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3}$, we have for $p geq frac{4}{3}$ that $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
I have attempted the method of Lagrange multipliers, which led me to three cubic equations:
I set out to maximize
$4(x^3+y^3+z^3)+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1)$.
This yielded
$$4x^3+2xlambda + p(y+z)=0$$
$$4y^3+2ylambda + p(x+z)=0$$
$$4z^3+2zlambda + p(y+x)=0$$
But now I feel a little at a loss on how to solve this system of equations.
One can verify that $x=y=z=frac{1}{sqrt{3}}$ solves these equations at any p with the right choice of lamda. I'm unsure about pairing Hessian techniques to verify the local nature of the critical point when using lagrange multipliers.
I have also tried incorporating the constraint by writing $z=cos(theta)$, $x=sin(theta)cos(phi)$ and $y=sin(theta)sin(phi)$.
This yields
$$f(theta, phi, p) = cos(theta)^4+(sin(theta)cos(phi))^4+(sin(theta)sin(phi))^4+p(cos(theta)sin(theta)sin(phi)+cos(theta)sin(theta)cos(phi)+(sin(theta))^2sin(phi)cos(phi) ) $$
which I am unsure how to maximize. However, with motivation from Michael Behrend, I've found that studying $phi=frac{pi}{4}$ shows interesting critical point behavior slightly above p=1 where three critical points undergo some sort of bifurcation leaving just a single critical point at the eternal $x=y=z=frac{1}{3}$ .
optimization lagrange-multiplier spheres
asked Jan 20 at 6:13
user196574user196574
1657
$endgroup$
add a comment |
$begingroup$
What values of $x,y,z$ maximize $f(x,y,z,p) = x^4+y^4+z^4 + p(xy+xz+yz)$ with constant $p geq 0$ with the constraint $x^2+y^2+z^2=1$?
Some preliminary studies in Mathematica showed that the behavior of the solution changes around $p=1$. For $p>1$, the problem seems solved by $x=y=z=frac{1}{sqrt 3}$. For $0 leq p leq 1$ it appears the solution (up to permutations of x,y,z) looks like x > y = z. These solutions appear to favor one or the other of the two terms in $f$, and I find the $p>1$ behavior striking and would appreciate a proof that $x=y=z=frac{1}{sqrt 3}$ indeed maximizes.
UPDATE
For at least $p>4/3$, one can show that $x=y=z=frac{1}{sqrt 3}$ maximizes f.
The proof for $1<p<4/3$ evades me currently (the left bound, as mentioned in comments, may be slightly higher than 1).
I'll assume without loss of generality that $x,y,z$ are all positive. Note that $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3} $, so if we show that for some $p$ we enjoy $f(x,y,z,p) leq p+frac{1}{3}$, we've shown $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
Now consider that via the constraint, $$f(x,y,z,p) = 1-2(x^2y^2+x^2z^2+y^2z^2) + p(xy+xz+yz)$$
The root mean square compared to arithmetic mean inequality tells us that for nonnegative $a,b,c$ we enjoy $sqrt{frac{a^2+b^2+c^2}{3}} geq frac{a+b+c}{3},$ so $-(a^2+b^2+c^2) leq -frac{(a+b+c)^2}{3}$. Then $$f(x,y,z,p) leq 1 + p(xy+xz+yz) -frac{2}{3}(xy+xz+yz)^2.$$
Now, let's call $(xy+xz+yz)=q$. Note that it is easy to prove with our constraints that $0leq q leq 1$.
Then we can bound the maximum of $f(x,y,z,p)$ by finding the maximum of $g(q) = 1+pq-frac{2}{3}q^2$ on $0leq q leq 1$.
The derivative $g'(q) = p-frac{4}{3}q$, so $g(q)$ increases up to $q=frac{3}{4}p$. Then on $0leq q leq 1$, we have $q_{max} = min(frac{3}{4}p, 1)$.
Thus for $p geq frac{4}{3}$ we have $q_{max}=1$, so $$ f(x,y,z,p) leq 1+p*1-frac{2}{3}*1^2 = p + 1/3$$
Thus since $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3}$, we have for $p geq frac{4}{3}$ that $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
I have attempted the method of Lagrange multipliers, which led me to three cubic equations:
I set out to maximize
$4(x^3+y^3+z^3)+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1)$.
This yielded
$$4x^3+2xlambda + p(y+z)=0$$
$$4y^3+2ylambda + p(x+z)=0$$
$$4z^3+2zlambda + p(y+x)=0$$
But now I feel a little at a loss on how to solve this system of equations.
One can verify that $x=y=z=frac{1}{sqrt{3}}$ solves these equations at any p with the right choice of lamda. I'm unsure about pairing Hessian techniques to verify the local nature of the critical point when using lagrange multipliers.
I have also tried incorporating the constraint by writing $z=cos(theta)$, $x=sin(theta)cos(phi)$ and $y=sin(theta)sin(phi)$.
This yields
$$f(theta, phi, p) = cos(theta)^4+(sin(theta)cos(phi))^4+(sin(theta)sin(phi))^4+p(cos(theta)sin(theta)sin(phi)+cos(theta)sin(theta)cos(phi)+(sin(theta))^2sin(phi)cos(phi) ) $$
which I am unsure how to maximize. However, with motivation from Michael Behrend, I've found that studying $phi=frac{pi}{4}$ shows interesting critical point behavior slightly above p=1 where three critical points undergo some sort of bifurcation leaving just a single critical point at the eternal $x=y=z=frac{1}{3}$ .
optimization lagrange-multiplier spheres
asked Jan 20 at 6:13
user196574user196574
1657
$endgroup$
add a comment |
$begingroup$
What values of $x,y,z$ maximize $f(x,y,z,p) = x^4+y^4+z^4 + p(xy+xz+yz)$ with constant $p geq 0$ with the constraint $x^2+y^2+z^2=1$?
Some preliminary studies in Mathematica showed that the behavior of the solution changes around $p=1$. For $p>1$, the problem seems solved by $x=y=z=frac{1}{sqrt 3}$. For $0 leq p leq 1$ it appears the solution (up to permutations of x,y,z) looks like x > y = z. These solutions appear to favor one or the other of the two terms in $f$, and I find the $p>1$ behavior striking and would appreciate a proof that $x=y=z=frac{1}{sqrt 3}$ indeed maximizes.
UPDATE
For at least $p>4/3$, one can show that $x=y=z=frac{1}{sqrt 3}$ maximizes f.
The proof for $1<p<4/3$ evades me currently (the left bound, as mentioned in comments, may be slightly higher than 1).
I'll assume without loss of generality that $x,y,z$ are all positive. Note that $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3} $, so if we show that for some $p$ we enjoy $f(x,y,z,p) leq p+frac{1}{3}$, we've shown $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
Now consider that via the constraint, $$f(x,y,z,p) = 1-2(x^2y^2+x^2z^2+y^2z^2) + p(xy+xz+yz)$$
The root mean square compared to arithmetic mean inequality tells us that for nonnegative $a,b,c$ we enjoy $sqrt{frac{a^2+b^2+c^2}{3}} geq frac{a+b+c}{3},$ so $-(a^2+b^2+c^2) leq -frac{(a+b+c)^2}{3}$. Then $$f(x,y,z,p) leq 1 + p(xy+xz+yz) -frac{2}{3}(xy+xz+yz)^2.$$
Now, let's call $(xy+xz+yz)=q$. Note that it is easy to prove with our constraints that $0leq q leq 1$.
Then we can bound the maximum of $f(x,y,z,p)$ by finding the maximum of $g(q) = 1+pq-frac{2}{3}q^2$ on $0leq q leq 1$.
The derivative $g'(q) = p-frac{4}{3}q$, so $g(q)$ increases up to $q=frac{3}{4}p$. Then on $0leq q leq 1$, we have $q_{max} = min(frac{3}{4}p, 1)$.
Thus for $p geq frac{4}{3}$ we have $q_{max}=1$, so $$ f(x,y,z,p) leq 1+p*1-frac{2}{3}*1^2 = p + 1/3$$
Thus since $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3}$, we have for $p geq frac{4}{3}$ that $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
I have attempted the method of Lagrange multipliers, which led me to three cubic equations:
I set out to maximize
$4(x^3+y^3+z^3)+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1)$.
This yielded
$$4x^3+2xlambda + p(y+z)=0$$
$$4y^3+2ylambda + p(x+z)=0$$
$$4z^3+2zlambda + p(y+x)=0$$
But now I feel a little at a loss on how to solve this system of equations.
One can verify that $x=y=z=frac{1}{sqrt{3}}$ solves these equations at any p with the right choice of lamda. I'm unsure about pairing Hessian techniques to verify the local nature of the critical point when using lagrange multipliers.
I have also tried incorporating the constraint by writing $z=cos(theta)$, $x=sin(theta)cos(phi)$ and $y=sin(theta)sin(phi)$.
This yields
$$f(theta, phi, p) = cos(theta)^4+(sin(theta)cos(phi))^4+(sin(theta)sin(phi))^4+p(cos(theta)sin(theta)sin(phi)+cos(theta)sin(theta)cos(phi)+(sin(theta))^2sin(phi)cos(phi) ) $$
which I am unsure how to maximize. However, with motivation from Michael Behrend, I've found that studying $phi=frac{pi}{4}$ shows interesting critical point behavior slightly above p=1 where three critical points undergo some sort of bifurcation leaving just a single critical point at the eternal $x=y=z=frac{1}{3}$ .
optimization lagrange-multiplier spheres
asked Jan 20 at 6:13
user196574user196574
1657
$endgroup$
What values of $x,y,z$ maximize $f(x,y,z,p) = x^4+y^4+z^4 + p(xy+xz+yz)$ with constant $p geq 0$ with the constraint $x^2+y^2+z^2=1$?
Some preliminary studies in Mathematica showed that the behavior of the solution changes around $p=1$. For $p>1$, the problem seems solved by $x=y=z=frac{1}{sqrt 3}$. For $0 leq p leq 1$ it appears the solution (up to permutations of x,y,z) looks like x > y = z. These solutions appear to favor one or the other of the two terms in $f$, and I find the $p>1$ behavior striking and would appreciate a proof that $x=y=z=frac{1}{sqrt 3}$ indeed maximizes.
UPDATE
For at least $p>4/3$, one can show that $x=y=z=frac{1}{sqrt 3}$ maximizes f.
The proof for $1<p<4/3$ evades me currently (the left bound, as mentioned in comments, may be slightly higher than 1).
I'll assume without loss of generality that $x,y,z$ are all positive. Note that $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3} $, so if we show that for some $p$ we enjoy $f(x,y,z,p) leq p+frac{1}{3}$, we've shown $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
Now consider that via the constraint, $$f(x,y,z,p) = 1-2(x^2y^2+x^2z^2+y^2z^2) + p(xy+xz+yz)$$
The root mean square compared to arithmetic mean inequality tells us that for nonnegative $a,b,c$ we enjoy $sqrt{frac{a^2+b^2+c^2}{3}} geq frac{a+b+c}{3},$ so $-(a^2+b^2+c^2) leq -frac{(a+b+c)^2}{3}$. Then $$f(x,y,z,p) leq 1 + p(xy+xz+yz) -frac{2}{3}(xy+xz+yz)^2.$$
Now, let's call $(xy+xz+yz)=q$. Note that it is easy to prove with our constraints that $0leq q leq 1$.
Then we can bound the maximum of $f(x,y,z,p)$ by finding the maximum of $g(q) = 1+pq-frac{2}{3}q^2$ on $0leq q leq 1$.
The derivative $g'(q) = p-frac{4}{3}q$, so $g(q)$ increases up to $q=frac{3}{4}p$. Then on $0leq q leq 1$, we have $q_{max} = min(frac{3}{4}p, 1)$.
Thus for $p geq frac{4}{3}$ we have $q_{max}=1$, so $$ f(x,y,z,p) leq 1+p*1-frac{2}{3}*1^2 = p + 1/3$$
Thus since $f(frac{1}{sqrt 3},frac{1}{sqrt 3},frac{1}{sqrt 3}, p) = p+frac{1}{3}$, we have for $p geq frac{4}{3}$ that $x=y=z=frac{1}{sqrt 3}$ maximizes $f$.
I have attempted the method of Lagrange multipliers, which led me to three cubic equations:
I set out to maximize
$4(x^3+y^3+z^3)+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1)$.
This yielded
$$4x^3+2xlambda + p(y+z)=0$$
$$4y^3+2ylambda + p(x+z)=0$$
$$4z^3+2zlambda + p(y+x)=0$$
But now I feel a little at a loss on how to solve this system of equations.
One can verify that $x=y=z=frac{1}{sqrt{3}}$ solves these equations at any p with the right choice of lamda. I'm unsure about pairing Hessian techniques to verify the local nature of the critical point when using lagrange multipliers.
I have also tried incorporating the constraint by writing $z=cos(theta)$, $x=sin(theta)cos(phi)$ and $y=sin(theta)sin(phi)$.
This yields
$$f(theta, phi, p) = cos(theta)^4+(sin(theta)cos(phi))^4+(sin(theta)sin(phi))^4+p(cos(theta)sin(theta)sin(phi)+cos(theta)sin(theta)cos(phi)+(sin(theta))^2sin(phi)cos(phi) ) $$
which I am unsure how to maximize. However, with motivation from Michael Behrend, I've found that studying $phi=frac{pi}{4}$ shows interesting critical point behavior slightly above p=1 where three critical points undergo some sort of bifurcation leaving just a single critical point at the eternal $x=y=z=frac{1}{3}$ .
optimization lagrange-multiplier spheres
optimization lagrange-multiplier spheres
asked Jan 20 at 6:13
user196574user196574
1657
asked Jan 20 at 6:13
user196574user196574
1657
edited Jan 21 at 21:51
user196574
asked Jan 20 at 6:13
user196574user196574
1657
asked Jan 20 at 6:13
user196574user196574
1657
asked Jan 20 at 6:13
user196574user196574
1657
1657
add a comment |
add a comment |
1 Answer
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Lagrange function is
$$L=x^4+y^4+z^4+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1).$$
Solving system
$$L'_x=0,;L'_y=0,;L'_z=0,;L'_x=0,;L'_lambda=0$$
we get that if $pge1$ global maximum is
$$f_{max}=p+frac13$$
Maximum points is:
if $p>1$ then $[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}]$ or
$[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
if $p=1$ then $[x=frac{sqrt{6}}{3},y=frac{1}{sqrt{6}},z=frac{1}{sqrt{6}}],[x=-frac{sqrt{6}}{3},y=-frac{1}{sqrt{6}},z=-frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{sqrt{6}}{3},z=frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{1}{sqrt{6}},z=frac{sqrt{6}}{3}],[x=-frac{1}{sqrt{6}},y=-frac{sqrt{6}}{3},z=-frac{1}{sqrt{6}}],[x=-frac{1}{sqrt{6}},y=-frac{1}{sqrt{6}},z=-frac{sqrt{6}}{3}],[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}],[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
If $p=0$ then
$$f_{max}=1,$$
$[x=pm1,y=0,z=0],[x=0,y=pm1,z=0],[x=0,y=0,z=pm1]$.
If $0<p<1$ we can't exact solve system.
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
$endgroup$
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
1
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
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1 Answer
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1 Answer
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$begingroup$
Lagrange function is
$$L=x^4+y^4+z^4+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1).$$
Solving system
$$L'_x=0,;L'_y=0,;L'_z=0,;L'_x=0,;L'_lambda=0$$
we get that if $pge1$ global maximum is
$$f_{max}=p+frac13$$
Maximum points is:
if $p>1$ then $[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}]$ or
$[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
if $p=1$ then $[x=frac{sqrt{6}}{3},y=frac{1}{sqrt{6}},z=frac{1}{sqrt{6}}],[x=-frac{sqrt{6}}{3},y=-frac{1}{sqrt{6}},z=-frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{sqrt{6}}{3},z=frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{1}{sqrt{6}},z=frac{sqrt{6}}{3}],[x=-frac{1}{sqrt{6}},y=-frac{sqrt{6}}{3},z=-frac{1}{sqrt{6}}],[x=-frac{1}{sqrt{6}},y=-frac{1}{sqrt{6}},z=-frac{sqrt{6}}{3}],[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}],[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
If $p=0$ then
$$f_{max}=1,$$
$[x=pm1,y=0,z=0],[x=0,y=pm1,z=0],[x=0,y=0,z=pm1]$.
If $0<p<1$ we can't exact solve system.
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
$endgroup$
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
1
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
add a comment |
$begingroup$
Lagrange function is
$$L=x^4+y^4+z^4+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1).$$
Solving system
$$L'_x=0,;L'_y=0,;L'_z=0,;L'_x=0,;L'_lambda=0$$
we get that if $pge1$ global maximum is
$$f_{max}=p+frac13$$
Maximum points is:
if $p>1$ then $[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}]$ or
$[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
if $p=1$ then $[x=frac{sqrt{6}}{3},y=frac{1}{sqrt{6}},z=frac{1}{sqrt{6}}],[x=-frac{sqrt{6}}{3},y=-frac{1}{sqrt{6}},z=-frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{sqrt{6}}{3},z=frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{1}{sqrt{6}},z=frac{sqrt{6}}{3}],[x=-frac{1}{sqrt{6}},y=-frac{sqrt{6}}{3},z=-frac{1}{sqrt{6}}],[x=-frac{1}{sqrt{6}},y=-frac{1}{sqrt{6}},z=-frac{sqrt{6}}{3}],[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}],[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
If $p=0$ then
$$f_{max}=1,$$
$[x=pm1,y=0,z=0],[x=0,y=pm1,z=0],[x=0,y=0,z=pm1]$.
If $0<p<1$ we can't exact solve system.
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
$endgroup$
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
1
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
add a comment |
$begingroup$
Lagrange function is
$$L=x^4+y^4+z^4+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1).$$
Solving system
$$L'_x=0,;L'_y=0,;L'_z=0,;L'_x=0,;L'_lambda=0$$
we get that if $pge1$ global maximum is
$$f_{max}=p+frac13$$
Maximum points is:
if $p>1$ then $[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}]$ or
$[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
if $p=1$ then $[x=frac{sqrt{6}}{3},y=frac{1}{sqrt{6}},z=frac{1}{sqrt{6}}],[x=-frac{sqrt{6}}{3},y=-frac{1}{sqrt{6}},z=-frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{sqrt{6}}{3},z=frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{1}{sqrt{6}},z=frac{sqrt{6}}{3}],[x=-frac{1}{sqrt{6}},y=-frac{sqrt{6}}{3},z=-frac{1}{sqrt{6}}],[x=-frac{1}{sqrt{6}},y=-frac{1}{sqrt{6}},z=-frac{sqrt{6}}{3}],[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}],[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
If $p=0$ then
$$f_{max}=1,$$
$[x=pm1,y=0,z=0],[x=0,y=pm1,z=0],[x=0,y=0,z=pm1]$.
If $0<p<1$ we can't exact solve system.
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
$endgroup$
Lagrange function is
$$L=x^4+y^4+z^4+p(xy+xz+yz)+lambda(x^2+y^2+z^2-1).$$
Solving system
$$L'_x=0,;L'_y=0,;L'_z=0,;L'_x=0,;L'_lambda=0$$
we get that if $pge1$ global maximum is
$$f_{max}=p+frac13$$
Maximum points is:
if $p>1$ then $[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}]$ or
$[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
if $p=1$ then $[x=frac{sqrt{6}}{3},y=frac{1}{sqrt{6}},z=frac{1}{sqrt{6}}],[x=-frac{sqrt{6}}{3},y=-frac{1}{sqrt{6}},z=-frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{sqrt{6}}{3},z=frac{1}{sqrt{6}}],[x=frac{1}{sqrt{6}},y=frac{1}{sqrt{6}},z=frac{sqrt{6}}{3}],[x=-frac{1}{sqrt{6}},y=-frac{sqrt{6}}{3},z=-frac{1}{sqrt{6}}],[x=-frac{1}{sqrt{6}},y=-frac{1}{sqrt{6}},z=-frac{sqrt{6}}{3}],[x=frac{1}{sqrt{3}},y=frac{1}{sqrt{3}},z=frac{1}{sqrt{3}}],[x=-frac{1}{sqrt{3}},y=-frac{1}{sqrt{3}},z=-frac{1}{sqrt{3}}]$
If $p=0$ then
$$f_{max}=1,$$
$[x=pm1,y=0,z=0],[x=0,y=pm1,z=0],[x=0,y=0,z=pm1]$.
If $0<p<1$ we can't exact solve system.
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
answered Jan 20 at 9:08
Aleksas DomarkasAleksas Domarkas
1,37216
1,37216
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
1
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
add a comment |
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
1
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
$begingroup$
I appreciate the answer. Could you explain how you solved the system of equations for p>1, and why you think that we cannot exactly solve the system for p between 0 and 1?
$endgroup$
– user196574
Jan 20 at 9:33
1
1
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
After trying to prove that the crossover is at p = 1, I'm coming to the conclusion that this isn't true. The crossover seems to be somewhere between 1.01 and 1.02. Hope to post more later.
$endgroup$
– Michael Behrend
Jan 21 at 18:32
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
@MichaelBehrend I've been working a bit more with the trigonometric functions under the assumption that phi is pi/4. In that special case, I see interesting behavior with the critical points in theta. No matter the value of p, the equivalent to the x=y=z critical point remains, though whether it's a global maximum or even a local maximum changes with p. In particular, it seems just slightly above 1 that the three critical points become just 1 critical point. If you can find the value of p where that occurs, a 2nd derivative test should show that x=y=z maximizes above that. Thanks for the help!
$endgroup$
– user196574
Jan 21 at 20:55
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
The crossover point is the larger real root of $$9x^4 + 10x^3 + 12x^2 - 24x - 8$$ which according to akiti.ca/PolyRootRe.html is 1.0148293664019847 (to 16 d.p.) This was found by a rather lengthy method and checked empirically by a computer program. If I can find a better proof I'll post it as an answer.
$endgroup$
– Michael Behrend
Jan 22 at 15:00
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
$begingroup$
Sorry, my previous comments were wrong. There is a local maximum at the "pole" $(1/sqrt3, 1/sqrt3, 1/sqrt3)$, and for $p < p_0 = 1.014829ldots$ there are local maxima at three other points (all three with the same function value). For $p < 1$ these three maxima are greater than the maximum at the pole, for p = 1 all four maxima are equal, and for $p > 1$ the maximum at the pole becomes greater. So the crossover is $p = 1$ as originally stated. The change at $p = p_0$ is that the three other local maxima disappear.
$endgroup$
– Michael Behrend
Jan 23 at 9:43
add a comment |
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