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Old 15-03-2003, 00:01   #1
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je majka svih nauka. Ajnstajn je tvrdio da se sve smislene stvari na svijetu mogu staviti u matematicki izraz. <img border="0" title="" alt="[Smile]" src="smile.gif" />

http://lavica.fesb.hr/mat1/osnovna.html
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Old 15-03-2003, 01:21   #2
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The following is from a book whose title I don't recall. The book is in German
but the article is actually a translation from the original by H. Petard which
appared in the American Monthly 54, 466 (1938). Unfortunately our library is
lacking some years of this journal around WW 2, so I had to re-translate the
stuff into English. (That will make you people share the experience of reading
German translations of books on Einstein which also usually re-translate
Einstein's words :-) ).

A Contribution to the Mathematical Theory of Big Game Hunting
================================================== ===========

Problem: To Catch a Lion in the Sahara Desert.

1. Mathematical Methods

1.1 The Hilbert (axiomatic) method

We place a locked cage onto a given point in the desert. After that we
introduce the following logical system:
Axiom 1: The set of lions in the Sahara is not empty.
Axiom 2: If there exists a lion in the Sahara, then there exists a lion in
the cage.
Procedure: If P is a theorem, and if the following is holds:
"P implies Q", then Q is a theorem.
Theorem 1: There exists a lion in the cage.

1.2 The geometrical inversion method

We place a spherical cage in the desert, enter it and lock it from inside. We
then performe an inversion with respect to the cage. Then the lion is inside
the cage, and we are outside.

1.3 The projective geometry method

Without loss of generality we can view the desert as a plane surface. We
project the surface onto a line and afterwards the line onto an interiour point
of the cage. Thereby the lion is mapped onto that same point.

1.4 The Bolzano-Weierstrass method

Divide the desert by a line running from north to south. The lion is then
either in the eastern or in the western part. Lets assume it is in the eastern
part. Divide this part by a line running from east to west. The lion is either
in the northern or in the southern part. Lets assume it is in the northern
part. We can continue this process arbitrarily and thereby constructing with
each step an increasingly narrow fence around the selected area. The diameter
of the chosen partitions converges to zero so that the lion is caged into a
fence of arbitrarily small diameter.

1.5 The set theoretical method

We observe that the desert is a separable space. It therefore contains an
enumerable dense set of points which constitutes a sequence with the lion as
its limit. We silently approach the lion in this sequence, carrying the proper
equipment with us.

1.6 The Peano method

In the usual way construct a curve containing every point in the desert. It has
been proven [1] that such a curve can be traversed in arbitrarily short time.
Now we traverse the curve, carrying a spear, in a time less than what it takes
the lion to move a distance equal to its own length.

1.7 A topological method

We observe that the lion possesses the topological gender of a torus. We embed
the desert in a four dimensional space. Then it is possible to apply a
deformation [2] of such a kind that the lion when returning to the three
dimensional space is all tied up in itself. It is then completely helpless.

1.8 The Cauchy method

We examine a lion-valued function f(z). Be \zeta the cage. Consider the integral

1 [ f(z)
------- I --------- dz
2 \pi i ] z - \zeta

C

where C represents the boundary of the desert. Its value is f(zeta), i.e. there
is a lion in the cage [3].

1.9 The Wiener-Tauber method

We obtain a tame lion, L_0, from the class L(-\infinity,\infinity), whose
fourier transform vanishes nowhere. We put this lion somewhere in the desert.
L_0 then converges toward our cage. According to the general Wiener-Tauner
theorem [4] every other lion L will converge toward the same cage.
(Alternatively we can approximate L arbitrarily close by translating L_0
through the desert [5].)

2 Theoretical Physics Methods

2.1 The Dirac method

We assert that wild lions can ipso facto not be observed in the Sahara desert.
Therefore, if there are any lions at all in the desert, they are tame. We leave
catching a tame lion as an execise to the reader.

2.2 The Schroedinger method

At every instant there is a non-zero probability of the lion being in the cage.
Sit and wait.

2.3 The nuclear physics method

Insert a tame lion into the cage and apply a Majorana exchange operator [6] on
it and a wild lion.

As a variant let us assume that we would like to catch (for argument's sake) a
male lion. We insert a tame female lion into the cage and apply the Heisenberg
exchange operator [7], exchanging spins.

2.4 A relativistic method

All over the desert we distribute lion bait containing large amounts of the
companion star of Sirius. After enough of the bait has been eaten we send a
beam of light through the desert. This will curl around the lion so it gets all
confused and can be approached without danger.

3 Experimental Physics Methods

3.1 The thermodynamics method

We construct a semi-permeable membrane which lets everything but lions pass
through. This we drag across the desert.

3.2 The atomic fission method

We irradiate the desert with slow neutrons. The lion becomes radioactive and
starts to diintegrate. Once the disintegration process is progressed far enough
the lion will be unable to resist.

3.3 The magneto-optical method

We plant a large, lense shaped field with cat mint (nepeta cataria) such that
its axis is parallel to the direction of the horizontal component of the
earth's magnetic field. We put the cage in one of the field's foci. Throughout
the desert we distribute large amounts of magnetized spinach (spinacia
oleracea) which has, as everybody knows, a high iron content. The spinach is
eaten by vegetarian desert inhabitants which in turn are eaten by the lions.
Afterwards the lions are oriented parallel to the earth's magnetic field and
the resulting lion beam is focussed on the cage by the cat mint lense.

[1] After Hilbert, cf. E. W. Hobson, "The Theory of Functions of a Real
Variable and the Theory of Fourier's Series" (1927), vol. 1, pp 456-457
[2] H. Seifert and W. Threlfall, "Lehrbuch der Topologie" (1934), pp 2-3
[3] According to the Picard theorem (W. F. Osgood, Lehrbuch der
Funktionentheorie, vol 1 (1928), p 178) it is possible to catch every lion
except for at most one.
[4] N. Wiener, "The Fourier Integral and Certain of itsl Applications" (1933),
pp 73-74
[5] N. Wiener, ibid, p 89
[6] cf e.g. H. A. Bethe and R. F. Bacher, "Reviews of Modern Physics", 8
(1936), pp 82-229, esp. pp 106-107
[7] ibid "
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Old 15-03-2003, 04:18   #3
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a = b
a^2 = a * b
a^2 - b^2 = a b - b^2
(a - b)(a + b) = b (a - b)
a + b = b
b + b = b
2b = b
[b:250a247ca9]2 = 1[/b:250a247ca9]
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Old 19-03-2003, 15:03   #4
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ne mozes ovaj dio [b:8ce526f608](a - b)(a + b) = b (a - b)[/b:8ce526f608] tako da kratis jer je a-b=0.
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Old 19-03-2003, 15:10   #5
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tacno. Osim toga rjesenje b=2b je sasvim validno ako je dio rjesenja nekoliko formula. Na taj nacin dobivas jedno rjesenje da je b=0.
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Old 24-03-2003, 04:19   #6
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padoh matematiku.
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Old 24-03-2003, 10:20   #7
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Veljkom u klub. <img border="0" title="" alt="[Smile]" src="smile.gif" />
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Old 24-03-2003, 10:22   #8
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Ah da, velkome, velkome. <img border="0" title="" alt="[Frown]" src="frown.gif" />
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Old 24-03-2003, 14:16   #9
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je li to klub 1200?
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Old 24-03-2003, 20:38   #10
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To je zato sto visis na forumu umjesto da vjezbas zadatke iz Matematiskopa.
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Old 24-03-2003, 20:53   #11
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Evo jedan stari, ali dobar skroz. Ko zna, nek' cuti da ne kvari.

Zamislite jedan veliki kais koji je obmotan oko ekvadora. Sada taj kais produzite za jedan metar i obmotajte ga opet oko ekvadora, ali tako da je svugdje isto udaljen od zemlje. Koliko je sada kais udaljen od zemlje? Bar otprilike, moze li se macka provuci ispod kaisa?

Napomena: Podrazumijeva se da je zemlja savrsena lopta.

<small>[ 24.03.2003, 20:55: Message edited by: sch ]</small>
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Old 24-03-2003, 21:01   #12
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</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by sch:
Ko zna, nek' cuti da ne kvari.

</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">OK, javi nam kad budemo mogli poceti odgovarati na ovo tvoje pitanje. ccc <img border="0" title="" alt="[Smile]" src="smile.gif" />
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Old 24-03-2003, 22:20   #13
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fomula za obim kruga je 2r puta Pi, ekvator je cca. 40.000 metara dugacak (nazovimo ga E).

formula je
2r x Pi = E,

prvo cemo izracunati originalni r po formuli izvedenoj iz prethodne:
E / 2 x Pi = r
a onda cemo primjeniti formulu
(E + 1) / 2 x Pi = r + y

y ce pokazati kolika je razlika u "ekvatoru" tj. koliko se povisio. sumnjam da ce macka proc tuda, ovako od oka.mozda zohar <img border="0" title="" alt="[Smile]" src="smile.gif" />

p.s. nisam nista izracunao, samo sam pokazao da bih znao kad bih htio. znaci ne znam odgovor (iako sam ga pretpostavio) - ovo se zove "educated guess" tako da mislim da sam zasluzio jednu bajaderu. ili bar plus u teci (moze li iko sapnut mom profesoru da sam pametan i da znam ovako ove folove-golove-trikove?)

<small>[ 24.03.2003, 22:23: Message edited by: Lucky ]</small>
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Old 31-03-2003, 17:00   #14
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Lucky, pokusaj nastaviti:

Dakle, dobio si: (E + 1) / 2 x Pi = r + y

odnosno:

E/2xPi + 1/2xPi = r + y

posto je: E/2xPi = r, to se skrati i dobijes:

y = 1/2xPi

Dakle, oko 16 cm, sto znaci da macka moze proci. Ali, to i nije toliko bitno. Primjecujes li jednu drugu stvar:

y uopce ne zavisi od r ! Dakle, isti je razmak ako za METAR povecas obim zemlje ili ping-pong loptice, uvijek je oko 16 cm!

Meni je to fenomenalno.
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Old 31-03-2003, 19:27   #15
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nisam skontao i ne razumijem te najbolje, mozda jer nemam kad da procitam i razmislim opet - ali ali ali - tvrdis neovisno od toga koliki je obim, ako se poveca za 100 cm, poluprecnik se uveca za 16 cm? nema logike doode? ili ja nisam najbolje shvatio te omjere?

da probamo primjerom

krug precnika 1 m ima obim = precnik puta pi = 1 x 3.14 = 3.14 m.

uvecajmo obim kruga od 3.14 m, za jedan metar, koliko se povecava precnik? precnik je 4.14 / pi; rezultat je 1.31 m. nisam ti skonto tih 16 cm profesore
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Old 31-03-2003, 20:08   #16
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ah, da vidim poluprecnik se uvecao za cca. 16 cm (malo manje), ali da li ovo radi za sve? vidi, trebalo bi da radi, jer je uvijek jedan metar vise... ili ne treba raditi... racunacu opet. u svakom slucaju dobro, 16 cm, sto znaci - moze macka ako se sagne. ne moze ako joj se digne.
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Old 31-03-2003, 23:40   #17
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Da, POLUPRECNIK se UVIJEK poveca za cca. 16 cm, neovisno od toga koliki je prvobitno taj precnik. Formule su jednostavne, ali je rezultat nevjerovatan. Meni je to toliko bilo fascinantno da sam uzimao kanap, pa ga obmotavao oko ping-pong loptice i oko sestrinog hula-hopa. Jedna od stvari koje toliko nestvarno djeluju cak i kad se svojim ocima uvjeris u to. Matematika! Zato je kraljica nauka.
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Old 01-04-2003, 04:08   #18
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jeste, u stvari se poluprecnik uveca za polovinu od 1m/Pi, sto si - dobro si - oko 16 cm i interesantno je jako
phunny
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Old 08-04-2003, 22:17   #19
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Primate li novog clana u klub "matematickih luzera"?
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Old 09-04-2003, 19:45   #20
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Cuj, luzera?! Gdje ti ovdje vidis luzere? ccc, ma, garant za Celik ovaj navija
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Old 10-04-2003, 11:49   #21
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Quote:
Originally posted by sch
Cuj, luzera?! Gdje ti ovdje vidis luzere? ccc, ma, garant za Celik ovaj navija

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Old 09-05-2003, 23:27   #22
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Imam par pitanja za vas, matematicari!
Koliko je beskonacno kroz beskonacno?
Sta je O?
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Old 10-05-2003, 07:38   #23
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Beskonacno kroz beskonacno se moze rijesiti uz pomoc Lopitelovog pravila gdje izvadis derivat iz obadva izraza u razlomku i onda rijesis uz pomoc limesa. To naravno vazi samo ako se radi o izrazu sa jednom nepoznatom ili vise.

O je slovo na latinici, nalazi se prije P i odmah poslije N
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Old 10-05-2003, 15:09   #24
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O je notacija prilikom analiziranja T(n) funkcije za performansu algoritama(npr.pretraga niza od n elemenata). Analiza uzima u obzir samo najvaznije informacije.

Malo detaljnije:

Uzmimo npr funkciju T(n)=3n^2 + 9n. Kada god je n>=5 onda je 3n^2 + 9n <= 4n^2.

Dakle funkcija T(n) je ustvari O(F(n)) gdje je F(n) = n^2.

Dokaz mozda drugom prilikom.

Pozdrav.
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Old 10-05-2003, 15:30   #25
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Sorry, los sam primjer dao gore.

T(n) = n^2 + 3n + 4

T(n) < 2n^2 za n > 10

T(n) <= k* F(n), T(n) je O(n^2) funkcija.
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