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April 14

Minimization of three variables

Given that x, y, and z are all real numbers, find the minimum possible value of $x^{4}+y^{4}+z^{4}-4xyz$ . I don't know how to tackle these kinds of problems. Thanks. —Preceding unsigned comment added by 70.111.95.226 (talk) 03:18, 14 April 2008 (UTC)[reply]

Trying to minimise a polynomial expression usually makes use of the fact that any real number raised to an even power must be positive. So, what you need to do is construct an expression that equals x^4+y^4+z^4-4xyz, but written in a form of (some stuff)^2 + a constant. Then you can say that the minimum value occurs when some stuff = 0 (you may have to prove that this is possible), so the minimum of the expression must be the constant. I would suggest looking at maybe expansions that look something like ((x - y)^2 + (y - z)^2 + (z - x)^2)^2, although that's just an educated guess. Confusing Manifestation(Say hi!) 03:32, 14 April 2008 (UTC)[reply]

Although, come to think of it, the fact that you've got a degree 3 term in there suggests that maybe something closer to (x + y + z - 1)^4 would be more helpful. Confusing Manifestation(Say hi!) 03:33, 14 April 2008 (UTC)[reply]

Since the function gets big for big values of x, y and z, the minimum value will be attained at a local minimum. Local minima can only occur at points where the gradient of the function is zero. By my calculations, that only gives you 8 points to check (or 2, if you do it cleverly). Is that any help? 134.173.93.127 (talk) 03:45, 14 April 2008 (UTC)[reply]

The function f(x,y,z)= x⁴+y⁴+z⁴−4xyz is symmetrical. (f(x,y,z)=f(z,y,x)=f(y,x,z)=&c). f(x,x,x)=3x⁴−4x³ has minimum f(1,1,1)=−1. Is that a local minimum for f(x,y,z)? Are there other local minima? Bo Jacoby (talk) 12:34, 14 April 2008 (UTC).[reply]

Yes, in addition to (1,1,1) there are also (-1,-1,1) & (-1,1,-1) & (1,-1,-1). StuRat (talk) 18:41, 14 April 2008 (UTC)[reply]

Check whether the Hessian matrix is positive definite at the singular point or not. The singular point is the one where the gradient vanishes.--Shahab (talk) 17:49, 14 April 2008 (UTC)[reply]

Also, you can use the AM-GM inequality applied to x^4, y^4, z^4 and 1^4.

(x^4 + y^4 + z^4 + 1^4)/4 ≥ xyz

therefore

x^4 + y^4 + z^4 + 1 ≥ 4xyz

x^4 + y^4 + z^4 - 4xyz ≥ -1

so you can get the minimum without finding the actual values of x,y and z. 91.143.188.103 (talk) 21:33, 14 April 2008 (UTC)[reply]

That gives you a bound on the minimum, it doesn't tell you if that bound is attained. It's a good place to start, but you do still need to find the values of x, y and z with give equality (which is trivial, of course). --Tango (talk) 22:41, 14 April 2008 (UTC)[reply]

We were actually only requested to find "find the minimum possible value of

x^{4}+y^{4}+z^{4}-4xyz

" which is −1, not "the values of x, y and z with give equality". Bo Jacoby (talk) 07:09, 15 April 2008 (UTC).[reply]

Yes, but if you've only found a lower bound, then you need to prove that the function attains that bound at some point, otherwise the actual minimum could easily be greater than that lower bound. Confusing Manifestation(Say hi!) 23:02, 15 April 2008 (UTC)[reply]

Yes, but the second part of the AM-GM inequality says that equality is attained if and only if x^4 = y^4 = z^4 = 1 (note that some care is needed over signs and roots to make sure you don't introduce spurious "solutions" - details left to student). Gandalf61 (talk) 11:06, 16 April 2008 (UTC)[reply]

aquestion about function

for quite some time the mathematicians have thought the mean value conditions is quite enough for afunction to be continouse at acertain point until RIEMANN gave his abs(x),where it satisfies the mean value at zero but not continouse at zero.my question is that i recall afunction but i cannot find it,it satisfies the mean value at every pints within it`s range but it is not continouse at all of those points.any one can tell me what is that function?thank you very much.Husseinshimaljasimdini (talk) 09:26, 14 April 2008 (UTC)[reply]

I think you are talking about the mean value theorem. The mean value theorem does indeed not apply to abs(x), because abs(x)=abs(-x) yet there is no point where the derivative of abs(x) is 0. Abs(x) fails to meet the conditions of the mean value theorem because it is not differentiable at 0. Note that this is not because abs(x) is not continous - it is continuous everywhere - differentiability is a stronger condition than continuity.

To construct a function that fails to meet the conditions of the mean value theorem for every interval in its range, you need a function that is nowhere differentiable. Since differentiability requires continuity you could use a function that is nowhere continuous, like the Dirichlet function, or a function that has a dense set of discontinuities, like Thomae's function. There are also functions that are everywhere continuous but nowhere differentiable, like the Koch curve. Gandalf61 (talk) 10:00, 14 April 2008 (UTC)[reply]

Thank you.I got it now.I realize that my informations were mixed up.Husseinshimaljasimdini (talk) 08:36, 15 April 2008 (UTC)[reply]

Scheduling problem

I've just about convinced myself that the following problem has no feasible solution - can someone either confirm this, or find one?

Four people who walk at the same speed are to complete a certain route. There is available one bicycle and one moped, each of which can carry only one person. Each person will ride the bicycle at the same speed and the moped at the same speed. Suppose that walk/bicycle/moped speeds are 5/10/20 mph. It is required that all four people start and end the journey together - this is to be achieved by each walking half the distance, riding the bicycle for one quarter of the distance and riding the moped for one quarter of the distance. These fractions can be made up of any number of smaller components.

Everything I've tried breaks down by the bicycle being required somewhere before it has arrived.—81.132.237.15 (talk) 18:29, 14 April 2008 (UTC)[reply]

Does it have be optimal? I have a solution where everyone waits an hour (call it an 80 mile track) out of 12 hours total -- maybe it's possible to do it in 11 still. 207.148.157.228 (talk) 19:31, 14 April 2008 (UTC)[reply]

I double-checked this, I think it works. Again, assume an 80 mile route.

A: 1hr moped, 8hr walk, 1hr rest, 2hr bike

B: 2hr bike, 1hr moped, 1hr rest, 8hr walk

C: 4hr walk, 2hr bike, 1hr moped, 1hr rest, 4hr walk

D: 8hr walk, 2hr bike, 1hr rest, 1hr moped

207.148.157.228 (talk) 19:59, 14 April 2008 (UTC)[reply]

I'd just break the bike apart, form it into two unicycles, sit a 2nd guy on the back of the moped and make him pull the unicyclists along. Lateral thinking ;) -mattbuck (Talk) 12:28, 15 April 2008 (UTC)[reply]

Better yet, hail a passing cab. --Tango (talk) 23:19, 15 April 2008 (UTC)[reply]

Precisely what would you do after that? Most cabs won't let you to put the bicycle in. – b_jonas 09:26, 18 April 2008 (UTC)[reply]

Leave it behind - who needs a bicycle once the cab gets there? --Tango (talk) 14:30, 19 April 2008 (UTC)[reply]

Euclidean prime numbers do not exist and euclid must should himself know it.

In 1996 Torsten Jensen was awarded the Millennium Leibniz Prize in logic, mathematics, physics, chemistry and medicine by providing a beautifully crafted very short proof that showed that all natural numbers can be divided by 3 and therefore Euclides theorem is false, invalid and worthless.

My question is this: 1. Do you know his ultra simple proof? 2. Why Euclides himself did not think of it? 3. Why did it take about 2300 years before a man saw the error in Euclid's theorem and destroyed about 12,000 rubbish theorems in number theory including Andrew Wiles's attempt at finding a proof for Fermat's last theorem?

Signed: T. Hansen, Lans, German Lutheran Church <email removed>

I do not mind at all if people read and steal my thoughts. —Preceding unsigned comment added by 81.152.51.207 (talk) 20:51, 14 April 2008 (UTC)[reply]

Strangely, our article on the Gottfried Wilhelm Leibniz Prize makes no mention of Herr Jensen and his incredible proof. Gandalf61 (talk) 21:02, 14 April 2008 (UTC)[reply]

Please do not write in all capitals, it's regarded as SHOUTING and is very irritating to the reader - Adrian Pingstone (talk) 21:08, 14 April 2008 (UTC)[reply]

Fixed that. — Kieff | Talk 23:31, 14 April 2008 (UTC)[reply]

Sure, all natural numbers can be divided by 3, but you don't get a natural number at the end of it 67% of the time. What's your point? --Tango (talk) 22:44, 14 April 2008 (UTC)[reply]

There is apparently no mention of +"Torsten Jensen" +Leibniz on the web, which means this whole thing is almost certainly made up and trolling. Not that any of us needed to look it up. — Kieff | Talk 23:31, 14 April 2008 (UTC)[reply]

I didn't know that 2 is divisible by 3... thanks for the post! --123.243.7.17 (talk) 02:03, 15 April 2008 (UTC)[reply]

Sure, you divide 2 by 3 and you get two thirds. --Tango (talk) 23:18, 15 April 2008 (UTC)[reply]