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Ronno
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Find minimum over all reals
« on: Apr 8th, 2010, 9:00pm » |
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Find the minimum value of the expression
f(a,b,c)=(a+b)^4+(b+c)^4+(c+a)^4- (4/7)*(a^4+b^4+c^4)
as a, b, c vary over the real numbers.
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towr
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Re: Find minimum over all reals
« Reply #1 on: Apr 9th, 2010, 12:51am » |
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Is this medium?
It seems obvious that if there is a minimum, then a=b=c, because of the symmetry of the formula. And then it's easily seen they must all be 0.
Or am I making some mathematical faux pas here?
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Ronno
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Re: Find minimum over all reals
« Reply #2 on: Apr 11th, 2010, 1:30am » |
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If it's an unique minimum then a=b=c has to hold. Otherwise it is not necessary. For example, if the factor 4/7 was removed, your argument would still produce 0 as the minimum but the expression would not be always non-negative (e.g. at (3,-1,-1) ).
If you still think it's not worthy of medium, you're welcome to move it to easy.
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| « Last Edit: Apr 11th, 2010, 1:34am by Ronno » |
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rmsgrey
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Re: Find minimum over all reals
« Reply #3 on: Apr 11th, 2010, 9:51am » |
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The symmetric formula just means that if x,y,z is a solution, so are x,z,y; y,x,z; y,z,x; z,x,y; and z,y,x.
An example of a simpler symmetric expression which manages to have multiple minima, most of which are not symmetric would be:
(x2 + y2 - 1)2
Obviously the minima are at x2 + y2=1 - the circle of unit radius centered on the origin.
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pex
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Re: Find minimum over all reals
« Reply #4 on: Apr 12th, 2010, 12:31am » |
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on Apr 11th, 2010, 9:51am, rmsgrey wrote:The symmetric formula just means that if x,y,z is a solution, so are x,z,y; y,x,z; y,z,x; z,x,y; and z,y,x.
An example of a simpler symmetric expression which manages to have multiple minima, most of which are not symmetric would be:
(x2 + y2 - 1)2
Obviously the minima are at x2 + y2=1 - the circle of unit radius centered on the origin. |
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And then (x2 + y2 - 1)2 + (x + y)2 has only asymmetric minima...
Perhaps (I haven't really thought about it) an interesting question is; what is the maximum value of k for which
(a + b)4 + (b + c)4 + (c + a)4 - k(a4 + b4 + c4)
does not take negative values? We know from Ronno's post that k < 1. [edit] Actually, it shows that k < 48/83, which is only about .007 more than 4/7... [/edit]
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| « Last Edit: Apr 12th, 2010, 3:14am by pex » |
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Obob
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Re: Find minimum over all reals
« Reply #5 on: Apr 12th, 2010, 4:13am » |
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A proof that the minimum of the function occurs at the origin would probably give insight into that question. I'd suspect that if k > 4/7 then the function is unbounded in the negative direction.
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| « Last Edit: Apr 12th, 2010, 4:14am by Obob » |
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pex
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Re: Find minimum over all reals
« Reply #6 on: Apr 13th, 2010, 8:52am » |
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I have found a slightly larger value numerically: the function is unbounded in the negative direction if and only if k > 0.57784613576525140468749613672912.
| hidden: | Note that the objective function is homogeneous of degree four; thus, it is sufficient to find the minimum of
(a+b)4 + (b+c)4 + (c+a)4
over the region
a4 + b4 + c4 = 1.
Using the Lagrange multiplier m, we need to solve
(a+b)3 + (c+a)3 = m a3
(a+b)3 + (b+c)3 = m b3
(b+c)3 + (c+a)3 = m c3
a4 + b4 + c4 = 1
Introducing x = b/a and y = c/a, and eliminating m from the first three conditions:
(1+x)3 + (x+y)3 = x3 [ (1+x)3 + (1+y)3 ]
(1+y)3 + (x+y)3 = y3 [ (1+x)3 + (1+y)3 ]
One trivial solution is (x,y) = (1,1), leading to a=b=c=3-1/4 with function value 16. Another trivial solution is (x,y) = (-1,0) (or the other way around), leading to a=2-1/4, b=-2-1/4, c=0, with value 1.
I had to resort to numerical methods to find the other real solutions. There are three of them:
x=1, y=-2.9506378683979899400582321588892
x=-2.9506378683979899400582321588892, y=1
x=y=-0.33890976954855421150997769094074
All of these lead to the same solution (but with a,b,c ordered differently), with
a=0.99351028288114569673074327517161 and
b=c=-0.33671034101536799238728202302250, with function value 0.57784613576525140468749613672912. It is clear that this is the minimum.
Thus, if k = 0.57784613576525140468749613672912, then (a+b)4 + (b+c)4 + (c+a)4 - k (a4 + b4 + c4) is everywhere nonnegative, and it is zero on all multiples of (a,b,c) = ( 0.99351028288114569673074327517161 ; -0.33671034101536799238728202302250 ; -0.33671034101536799238728202302250 ) and their permutations. For reference, we can scale this vector to (1,x,x), where x is the unique real root of
2x5 + 8x4 + 14x3 + 7x2 + 4x + 1 = 0.
(Obtained by setting x=y and dividing out a factor x-1.) |
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Ronno
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Re: Find minimum over all reals
« Reply #7 on: Apr 13th, 2010, 9:25am » |
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That was straightforward enough. 
But I would expect that there is a simpler, and of course weaker, solution to the original problem using the factor 4/7.
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SWF
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Re: Find minimum over all reals
« Reply #8 on: Jun 2nd, 2010, 9:06pm » |
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It is always possible to rename a, b, c, such that a*b>=0, because there are 3 values and at least two of them have the same sign (or at least one number is zero). f(a,b,c) is the sum of 3 terms, all >=0:
f(a,b,c) = (10/7)*( a^2 + b^2 + c^2 -11/5*a*b + 14/10(a*c + b*c) )^2
+ 72/7*a*b*(a - b)^2
+ ( sqrt(12/35)*(b*c + a*c) +sqrt(588/35)*a*b )^2
f(a,b,c) must therefore be >=0, and the minimum is zero when a=b=c=0.
Coming up with the above was not easy, but expanding to verify the truth of it is.
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