Talk:Maxima and minima

From Wikipedia, the free encyclopedia

Contents

[edit] Problem with Counterexample

It is not true that the function mentioned in the counterexample only has one critical point. It clearly has several saddle points at which the gradient vanishes. As far as I know, these are also classified as critical points. Thus, it's really not a counterexample at all. christianjb —Preceding unsigned comment added by 98.198.43.82 (talk) 03:14, 12 April 2008 (UTC)

I get
\frac{\partial f}{\partial y}(x,y)=2y(1-x)^3 =0\iff x=1\text{ or } y=0
and
\frac{\partial f}{\partial x}(x,y)=2x-3y^2(1-x)^2,
in particular
\frac{\partial f}{\partial x}(1,y)=2\not=0
and
\frac{\partial f}{\partial x}(x,0)=2x=0\iff x=0.
Therefore, (x,y) = (0,0) is the only point where both partial derivatives of f are zero, as required for a critical point of f. Schmock (talk) 14:52, 12 April 2008 (UTC)

Thanks for the detailed answer. I have to confess I messed up the graphing of the function last night, which is why I thought there were 'obvious' critical points. However, I'm still not convinced that there doesn't exist a very unobvious pair of stationary points!


We have:

\frac{\partial{f}}{\partial{y}}=2y(1-x)^3
\frac{\partial{f}}{\partial{x}}= 2x-3y^2(1-x)^2

Let x = 1 + ε and y=(1/\epsilon)\sqrt{2/3}, then

\frac{\partial{f}}{\partial{y}}=2\epsilon^2 \sqrt{2/3}
\frac{\partial{f}}{\partial{x}}= 2\epsilon

In the limit \epsilon\rightarrow 0, both derivatives go to zero, and so x = 1 + ε , y=(1/\epsilon)\sqrt{2/3} is a stationary point in this limit.

OK, if my algebra is correct, is it OK to take this limit? The stationary point is located at y\rightarrow \infty , which may not be in the domain of the function. I'd appreciate others' opinions on this. christianjb —Preceding unsigned comment added by 98.198.43.82 (talk) 22:46, 12 April 2008 (UTC)

OK, I asked someone who knows a bit more about this stuff than me. The above is correct, but you can't really define the stationary point in the limit y goes to infinity- it's really an asymptote. So- the way I see it, the function doesn't magically go from a minimum to the infinite drop off without an intervening barrier. Along any finite path, the function must first overcome a barrier, but that barrier is never an exact stationary point, it's always got some small curvature along the transverse direction. However, that transverse curvature goes to zero in the limit y goes to infinity. (I think the barrier region also gets squished to become infinitely thin in this limit).

Another way of looking at it is simply that the stationary point has been pushed out to infinity, beyond the range of the function, which is defined only for finite values of x. (Maybe that's a bit naive, but it sort of makes sense to me.) Christianjb —Preceding unsigned comment added by 98.198.43.82 (talk) 22:56, 23 April 2008 (UTC)

[edit] Request for a figure with a 3-dimensional local maximum in this article

Such a figure would be of great heuristic value in generalizing visually the notion of maxima and minima. Perhaps it could be borrowed or adapted from the Saddle point article. Thanks. Thomasmeeks 13:12, 4 January 2007 (UTC)

I think you should post this request at Wikipedia:Requested images in the math section. There it will be more likely to be found. Oleg Alexandrov (talk) 16:40, 4 January 2007 (UTC)
OK, will do. Thx. BW,Thomasmeeks
May I help? I've generated a simple minimum point of a paraboloid with a tangent plane. Try: this image. I'm a new user, so if there's any problems, tell me on my user page. --Freiddy 11:38, 8 January 2007 (UTC)
That's actually a local maximum, not minimum. I think the picture looks good. Feel free to add it in. Oleg Alexandrov (talk) 15:57, 8 January 2007 (UTC)
Sorry, it's "Maximum". Typo. —The preceding unsigned comment was added by Freiddy (talk • contribs) 16:13, 8 January 2007 (UTC).
I made another version according to your suggestions. The image is image. Note: the extension is no longer in caps. The image is not quantitatively accurate since I didn't use a 3D program to draw the box, but it should explain the basic qualitative concept.--Freiddy 17:13, 8 January 2007 (UTC)
If you haven't noticed on my user talk page, I've already uploaded the two pictures you requested. They are Image:Maximum_tangentplane_boxed.png (just for now, please use the Image:Maximum_tangentplane_boxedN.png as a temporary substitute before this image is moved and replaces the other image) and Image:Maximum_boxed.png. The issues are dealt in my user page. --Freiddy 09:51, 12 January 2007 (UTC)
Alright, everything is okay now. Just use these two images (both 100% ready): Image:Maximum_boxed.png and Image:Maximum_tangentplane_boxed.png (problem fixed now). --Freiddy 17:08, 12 January 2007 (UTC)

Looks good like a thumb (see right). Nice. Feel free to add it (them) in. Oleg Alexandrov (talk) 17:27, 12 January 2007 (UTC)

Thank you so much, Freiddy. -- Thomasmeeks 18:33, 12 January 2007 (UTC)

If you want me to make the dot a little bigger, I can do it now. --Freiddy 12:43, 13 January 2007 (UTC)

I also corrected a few spelling errors and improved the structure of the section on Maxima_and_minima#Functions_of_more_variables. Is it necessary to include a few examples in this section? --Freiddy 12:55, 13 January 2007 (UTC)

Probably helpful. Yes, a bigger max bull's eye (and possibly fading the colors to bring out the max). My thx. Thomasmeeks 21:24, 13 January 2007 (UTC)
I'll do that, but what do you mean "fading the colors to bring out the max[imum]"? --Freiddy 09:32, 19 January 2007 (UTC)
Well, my thought was that if the blue-to-green faded more from the x-y axes approaching the max, it would be easier to pick up the red at the max. Alternately, instead of the reddish glow above the z values, possibly it could on the z values as it approached the solid red of the max with a black dot in the center.
On another matter, I had thought that lengthening the z axis so that the z label would not be in the web of the figure would be a good idea, but I see that to avoid obstructing the max the entire figure wouuld have to be rotated. And rotation would be a lot of work. So, that's probably out.
I also have 2nd thoughts about the 'cherry' reference in the article ('top' might be better), which I think would annoy mathematicians. I'm not happy about writing all this now rather than say I couple of weeks ago. Sorry.
I'd say best to hear from Oleg before doing anything. Or just leave as is. Its way good enough. My thx for your signal contributions. --Thomasmeeks 13:25, 19 January 2007 (UTC)
Well, I could do part of that. But then I'll have to start from scratch entirely (I'm quite busy these days). Also, I don't my graphing calculator has the capabilities. I hope you're not disappointed. But I could help you work on the text. --Freiddy 11:56, 26 January 2007 (UTC)
It was quite generous of to have made those figures, which are very good. They add another dimension to tha article. The best there is is, well, still the best. --Thomasmeeks 13:48, 26 January 2007 (UTC)

[edit] what are strict maxima and minima?

In some text, it says "strict local maximum", such as in http://stat-www.berkeley.edu/~peres/bmall.pdf, or the sentence "the origian can not be a proper local maximum" in http://www.springerlink.com/index/Q6164625N34P44Q3.pdf. Does it mean "strict local" or "strict maximum"?

In this article http://www.emis.de/journals/EJP-ECP/EcpVol5/paper11.pdf, it also says "strict fine maximum". does it mean "strict fine" or "strict maximum"?

[edit] Do minima and maxima belong to the range or to the domain?

Comparing with mathworld's definition [1], I wonder if our definition could be too informal. Mathworld says: "A global maximum of a function is the largest value in the entire range of the function, and a local maximum is the largest value in some local neighborhood." Are we confusing the maximum with the point in the domain where the function takes such a maximum? Can anyone clarify this, please? Another Wikipedian 04:00, 12 May 2007 (UTC)

  • =

YES = the point that you raise is valid. = It is clear to me that "our" definition is imprecise and should be re-worded along the lines: "A function f has a local (or relative) maximum at a point x* if ..." Then go on to point out: inasmuch as f HAS a maximum at x*, that maximum is f(x*). = OR (take a vote?): don't say "f has a local maximum at x* ..." but say instead "f has a local maximum VALUE at x* ..." and the local maximum value is f(x*) L P Meissner 01:44, 15 May 2007 (UTC)L P Meissner

[edit] Local max/min when domain is restricted

The problem is how sophisticated to make this discussion. OK, explain everything at first in terms as simple as possible, and only for functions from "all reals" to reals. BUT for such functions it is true that every global extremum is also a local extremum; and it needs to be explained that such an assumption is not universal - especially for optimization problems (a couple of paragraphs down) on a finite domain where the possibilities to be considered are (1) global extrema at endpoints of closed intervals of the domain, and (2) local extrema at interior points. So, after introducing the simple case where the domain is "all reals" and global extrema are always local extrema, it is NECESSARY to point out that the definition of local (but not global) extrema involves some kind of "neighborhood" concept which does not apply at the endpoints of finite portions of the domain. And it follows that on some kinds of restricted domains it is NOT TRUE that global extrema are always local extrema, in particular when they occur at endpoints. L P Meissner 03:51, 20 May 2007 (UTC)LPMeissner

[edit] Can a function whose range is NOT "reals" have a max/min?

Main page says: "In order to be able to define local maxima and local minima, the function needs to take real values,..." = Seems like the requirement is only that the range be ORDERED in some definite way, so that the words "maximum" and "minimum" make sense. For example, the set of all SUBSCRIBERS' NAMES, in a phone book that covers multiple towns in separate sections, might have a local max at the end of each town's section, assuming names within each town are ranked in increasing alphabetical order (whatever that means, for a phone book). L P Meissner 02:02, 15 May 2007 (UTC)L P Meissner

  • I edited this to the following: "One can talk about global maxima and global minima for functions whose domain is any set, and whose range is an ordered set." I am thinking of an example such as a "Miss America" beauty pageant: the domain is the set of contest entrants, and the entrant who wins is globally "most beautiful" as determined by the judges or whatever. L P Meissner 17:19, 16 May 2007 (UTC)L P Meissner
I restricted the text back to functions with real values, since is far and away the most important case. Also, for a ordered set you have other issues, such as whether the order is total. None of the analysis works there either. I think this is an elementary enough article that going beyond functions whose range is the reals is probably not practical. Oleg Alexandrov (talk) 01:36, 17 May 2007 (UTC)
Yeah, the words, taken literally, can have a multitude of meanings, but for anything except real-valued functions one might as well just use a dictionary definition of maximum and minimum, instead of trying to cover all bases in a technical article.
Even global min/max is not a very technical term - only the local case really neads a mathematical definition - pretty simple, really: a local extremum has to be the extreme value for some neighborhood. And then it is necessary to point out that there can be points, such as endpoints in restricted domains, where neighborhoods in the required sense don't exist. L P Meissner 15:47, 22 May 2007 (UTC) L P Meissner

So a method of finding a global maximum (or minimum) is to look at all the local maxima (or minima) in the interior, and also look at the maxima (or minima) of the points on the boundary; and take the biggest (or smallest) one.

Is there an efficient way to find the global maximum/minimum? Take for example the sine integral. It has an infinite number of local maxima and minima. So how can one decide which one is the global maximum/minimum? --Abdull 17:04, 17 May 2007 (UTC)

[edit] Definition of the local maximum / minimum

I believe the definition for the local maximum / minimum is incorrect. Take some point x1 which is a point of removable discontinuity of this function, e.g. let f(x1) "jump" higher then all the neighbor points forming such a discontinuity. Following the current definition, there exists έ>0 such that |x-x1|<έ implies f(x1)≥f(x). x1 is a local maximum then, but is it really so? If f(x1) jumps up from the very bottom of the hill upside down (where we would normally try to find a local minimum), the it forms a local maximum even that it does not look like one at all. How about the requirement for the function f to be continuous at x1? <br />Dmitry Dmitriev

[edit] Anybody speak Polish?

http://pl.wikipedia.org/wiki/Ekstremum is so much better than ours.. Miserlou (talk) 16:47, 3 March 2008 (UTC)

[edit] Finding global extrema

I'm curious about algorithms to find global extrema. I think there is theory out there about it such as use of scale space for optical flow problems. I did some Googling and found this short discussion, along with other hits. Can anyone point me to other ways of searching for global extrema in an efficient way when there are many local extrema? —Ben FrantzDale (talk) 02:02, 28 May 2008 (UTC)