1.
Brian, there is no minima at 3pi/2 and 7pi/2.
2.
A function has minima at 'a' if f(x)>f(a) (not
f(x)>=f(a)) in a neighbourhood of 'a', x not equal to
'a'.
Sanjiva
--- Rob van Wijk <robvanwijk@gmx.net> wrote:
>
> http://mathworld.wolfram.com/Minimum.html
> The smallest value of a set, function, etc.
> -> Note that no requirement is made this smallest
> value is unique
>
> http://en.wikipedia.org/wiki/Maxima_and_minima
> [A] function has a local minimum point at x'
> if f(x') <= f(x) when |x âˆ' x'| < epsilon
> -> Note the use of smaller-or-equal (instead of
> strictly smaller)
>
> So, the function constructed by Sanjiva has a
> minimum
> {(4x-1)pi/2, for x in Integer} Union [3pi/2, 7pi/2]
> (the union of an infinite set of discrete points and
> a closed interval).
>
> While I haven't seen a formal prove yet, it
> certainly seems like
> the original proposition (the one Sanjiva's function
> was intended
> to disprove) holds.
>
> Grtz,
> Rob
>
>
> -------- Original-Nachricht --------
> > Datum: Wed, 3 Oct 2007 09:43:16 +0100
> > Von: "Stephen Tavener" <stephen.tavener@bbc.co.uk>
> > An: Math4u@yahoogroups.com
> > Betreff: RE: [Math4u] Re: Maxima & Minima
>
> > I think the point is more to do with definitions
> than anything else.
> >
> > If I understand correctly, according to Sanjeev:
> > At a local maximum, you can choose a small area
> around the point, where
> > that point is higher than every other point in the
> set.
> > At a local minimum, you can choose a small area
> around the point, where
> > that point is lower than every other point in the
> set.
> >
> > So, Sanjeev constructed a case where there is a
> plateau - no point is
> > lower than its' neighbour. So, while it is true
> that between two local
> > maxima, there are points with a minimum value,
> there does not need to be
> > a local minimum.
> >
> > However, as to whether this is a valid definition
> or not, I leave to the
> > reader. I reckon Wikipedia's and Mathworld's
> definitions would allow
> > all points in a plateau to be minima,
> contradicting Sanjeev, but in
> > maths, you start with a set of lies, and build
> elaborate structures, so
> > it just comes down to what lies we tell ourselves!
>
> >
> >
> > ________________________________
> >
> > From: Math4u@yahoogroups.com
> [mailto:Math4u@yahoogroups.com] On Behalf
> > Of Brian E. Jensen
> > Sent: 02 October 2007 20:06
> > To: Math4u@yahoogroups.com
> > Subject: Re: [Math4u] Re: Maxima & Minima
> >
> >
> > max at (pi/2,1)
> > min at (3pi/2,-1)
> > max at (5pi/2,1)
> > min at (7pi/2, -1)
> > max at 9pi/2,1)
> > It looks to me like the statement is proved true
> by Sanjiva's example.
> > regards, Brian
> >
> >
> > ----- Original Message -----
> > From: sanjiva dayal
> <mailto:sanjivadayal@yahoo.com>
> > To: Math4u@yahoogroups.com
> <mailto:Math4u@yahoogroups.com>
> > Sent: Tuesday, October 02, 2007 3:42 AM
> > Subject: RE: [Math4u] Re: Maxima & Minima
> >
> >
> > Hi,
> > 1.
> > The statement of my question is proved false by
> the
> > following example:-
> > f(x)=sinx, x<=3*pi/2
> > =-1, 3*pi/2<x<7*pi/2
> > =sinx, x>=7*pi/2.
> > This function f(x) is defined for all real x, is
> > continuous & differentiable for all real x and
> has
> > local maximas at x=pi/2 and x=9*pi/2, but there
> is no
> > local minima betweem pi/2 and 9*pi/2. Therefore,
> the
> > initial statement is FALSE.
> > 2.
> > Moral of the story:- "Never use common sense in
> > Mathematics, always think from the definitions".
> > Sanjiva
> >
> > --- Stephen Tavener <stephen.tavener@bbc.co.uk
> > <mailto:stephen.tavener%40bbc.co.uk> > wrote:
> >
> > > Hmmm... what about a function which has a
> plateau
> > > between two local
> > > maximae?
> > >
> > > At this late remove, I can't remember all my
> > > definitions, but I think we
> > > can get away with a function like:
> > >
> > > f(x) = {
> > > g(x) for x <= k1
> > > h(x) for k1 < x <= k2
> > > i(x) for k2 < x
> > > }
> > >
> > > define h(x) to be 0 at all values, and choose
> > > suitable cubic equations
> > > for g(x) and i(x) such that g(k1) = 0, i(k2) =
> 0,
> > > g'(k1) = 0, i'(k2) =
> > > 0, and g(x) and i(x) have suitable maximae.
> > >
> > > If memory serves, this should meet the
> definitions
> > > of continuous and
> > > differentiable at all points, but somebody who
> has
> > > done a maths degree
> > > more recently than 20 years ago (ouch) can tell
> me
> > > if I'm wrong!
> > >
> > >
> > > --
> > > CH3 Stephen Tavener
> > > | DigiText Programmer, BBC News,
> > > New Media.
> > > N
> > > / \ Room BC3 xtn (020 800)84739
> > > N----C C==O Broadcast Centre, 201 Wood Lane,
> > > London. W12 7TP
> > > || || |
> > > || || | mailto:Stephen.Tavener@bbc.co.uk
> > <mailto:Stephen.Tavener%40bbc.co.uk>
> > > (Work)
> > > CH C N--CH3 mailto:mrraow@gmail.com
> > <mailto:mrraow%40gmail.com> (Home)
> > > \ / \ /
> > > N C http://www.scat.demon.co.uk/
> > <http://www.scat.demon.co.uk/>
> > > (Games)
> > > | ||
> > > CH3 O Baby pictures:
> > >
> >
>
http://rainbot.pwp.blueyonder.co.uk/Katiefrog/group.jpg
> >
>
<http://rainbot.pwp.blueyonder.co.uk/Katiefrog/group.jpg>
>
> > >
> > >
> > >
> > >
> > > ________________________________
> > >
> > > From: Math4u@yahoogroups.com
> <mailto:Math4u%40yahoogroups.com>
> >
> > > [mailto:Math4u@yahoogroups.com
> > <mailto:Math4u%40yahoogroups.com> ] On Behalf
> > > Of Brian E. Jensen
> > > Sent: 01 October 2007 17:38
> > > To: Math4u@yahoogroups.com
> <mailto:Math4u%40yahoogroups.com>
> > > Subject: Re: [Math4u] Re: Maxima & Minima
> > >
> > >
> > >
> > > I don't have time to figure out what Jim
> FitzSimons,
> > > who is very
>
=== message truncated ===
Sanjiva Dayal, B.Tech.(I.I.T. Kanpur)
Address:A-602, Twin Towers, Lakhanpur, Kanpur-208024, INDIA.
Phones:+91-512-2581532,2581426.
Mobile:9415134052
Business email:sanjivadayal@yahoo.com
Personal email:sanjivadayal@hotmail.com
____________________________________________________________________________________
Take the Internet to Go: Yahoo!Go puts the Internet in your pocket: mail, news, photos & more.
http://mobile.yahoo.com/go?refer=1GNXIC
Yahoo! Groups Links
<*> To visit your group on the web, go to:
http://groups.yahoo.com/group/Math4u/
<*> Your email settings:
Individual Email | Traditional
<*> To change settings online go to:
http://groups.yahoo.com/group/Math4u/join
(Yahoo! ID required)
<*> To change settings via email:
mailto:Math4u-digest@yahoogroups.com
mailto:Math4u-fullfeatured@yahoogroups.com
<*> To unsubscribe from this group, send an email to:
Math4u-unsubscribe@yahoogroups.com
<*> Your use of Yahoo! Groups is subject to:
No comments:
Post a Comment