Točno
3. svibnja 2016. 20:35 (8 godine, 2 mjeseci)
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pretpostavimo suprotno, da postoje takve
i
.
tada ocito postoji
za koji
(inace bi npr. uvrstavanje
dovelo na kontradikciju).
uvrstavajuci
, nalazimo
![f(x) = \frac{(x - c) e^{-(x - c)^2} + f(c)g(x)}{g(c)} \qquad \forall x \in \mathbb{R}](/media/m/c/3/b/c3bfa808d0c207275fbb66cb9bb994be.png)
uvrstavajuci to nazad u pocetnu dobivamo
![\frac{(x - c) e^{-(x - c)^2} + f(c)g(x)}{g(c)}g(y) - \frac{(y - c) e^{-(y - c)^2} + f(c)g(y)}{g(c)}g(x) = (x - y) e^{-(x - y)^2} \qquad \forall x, y \in \mathbb{R}](/media/m/c/1/2/c127da5806923443d4aa7b8ec7be3a4b.png)
odnosno nakon mnozenja s
i sredivanja,
![g(y) (x - c) e^{-(x - c)^2} - g(x) (y - c) e^{-(y - c)^2} = g(c)(x - y) e^{-(x - y)^2} \qquad \forall x, y \in \mathbb{R}](/media/m/1/5/7/157da2facfe8e236df0dc0f4119bd0f4.png)
uvrstavajuci
i
te izrazavajuci
, zbog
dobivamo
![g(y) = \frac{g(2c) (y - c) e^{-(y - c)^2} + g(c)(2c - y) e^{-(2c - y)^2}}{c e^{-c^2}} \qquad \forall y \in \mathbb{R}](/media/m/9/6/2/9621d0913bf320567786705bfdcfe02d.png)
![g(y) = \frac{- g(0) (y - c) e^{-(y - c)^2} + g(c)y e^{-y^2}}{c e^{-c^2}} \qquad \forall y \in \mathbb{R}](/media/m/c/6/0/c60fd2a991bcf26ff6252c64f9f8f09f.png)
pa sada izjednacavanjem slijedi
![(g(2c) + g(0)) (y - c) e^{-(y - c)^2} + g(c)((2c - y) e^{-(2c - y)^2} - y e^{-y^2}) = 0 \qquad \forall y \in \mathbb{R}](/media/m/5/1/0/5107d17061f448d2219f7cbae493638e.png)
uvedimo supstituciju
zbog simetrije. takoder, neka je
sto je opravdano zbog
. prethodna jednakost postaje
![Aze^{-z^2} - (z-c)e^{-(z-c)^2} - (z+c)e^{-(z+c)^2} = 0 \qquad \forall z \in \mathbb{R}](/media/m/4/e/7/4e787c5065f9edb84a70a235fd98ef45.png)
nekako je jasno da je vrlo tesko da ce ovaj identitet vrijediti
, ako je
, ali pokazimo to. podijelimo sve s
i s
te oznacimo jos
. takoder, neka je
, te podijelimo sve s
(opet, sve je opravdano zbog
). dobivamo
![Bw - (w-1)e^{2wc^2} - (w+1)e^{-2wc^2} = 0 \qquad \forall w \in \mathbb{R}](/media/m/5/7/8/578478eff33c9deed634e7d9e8cae655.png)
ako je
, tada je
, pa ovo ne moze biti istina
, jer za dovoljno velike
,
raste linearno a
eksponencijalno, dok zadnji clan,
tezi u nulu, pa cijeli izraz tezi u
.
![f](/media/m/9/9/8/99891073047c7d6941fc8c6a39a75cf2.png)
![g](/media/m/9/5/8/958b2ae8c90cadb8c953ce50efb9c02a.png)
tada ocito postoji
![c \in \mathbb{R}, c \neq 0](/media/m/b/e/2/be2849c8aeb9bed9e4f71392f520f8c7.png)
![g(c) \neq 0](/media/m/0/b/8/0b8edf354947128f7213ccdc7f1e7268.png)
![x = 2, y = 1](/media/m/3/0/4/3040b3978bac1bfdbec354d19cb983bf.png)
uvrstavajuci
![(x, c)](/media/m/1/5/a/15a5cfc943009e3bc81ee8a942f02cde.png)
![f(x) = \frac{(x - c) e^{-(x - c)^2} + f(c)g(x)}{g(c)} \qquad \forall x \in \mathbb{R}](/media/m/c/3/b/c3bfa808d0c207275fbb66cb9bb994be.png)
uvrstavajuci to nazad u pocetnu dobivamo
![\frac{(x - c) e^{-(x - c)^2} + f(c)g(x)}{g(c)}g(y) - \frac{(y - c) e^{-(y - c)^2} + f(c)g(y)}{g(c)}g(x) = (x - y) e^{-(x - y)^2} \qquad \forall x, y \in \mathbb{R}](/media/m/c/1/2/c127da5806923443d4aa7b8ec7be3a4b.png)
odnosno nakon mnozenja s
![g(c) \neq 0](/media/m/0/b/8/0b8edf354947128f7213ccdc7f1e7268.png)
![g(y) (x - c) e^{-(x - c)^2} - g(x) (y - c) e^{-(y - c)^2} = g(c)(x - y) e^{-(x - y)^2} \qquad \forall x, y \in \mathbb{R}](/media/m/1/5/7/157da2facfe8e236df0dc0f4119bd0f4.png)
uvrstavajuci
![x = 2c](/media/m/a/3/5/a35f3f0c29e6eba519c637a8a91ce9ea.png)
![x = 0](/media/m/8/2/6/8268205131b210915acd475f4122137c.png)
![g(y)](/media/m/b/b/f/bbfa4d549421dc358e2294af5eb10227.png)
![ce^{-c^2} \neq 0](/media/m/8/4/7/847f783cb7bc469bfe4ad46343a54125.png)
![g(y) = \frac{g(2c) (y - c) e^{-(y - c)^2} + g(c)(2c - y) e^{-(2c - y)^2}}{c e^{-c^2}} \qquad \forall y \in \mathbb{R}](/media/m/9/6/2/9621d0913bf320567786705bfdcfe02d.png)
![g(y) = \frac{- g(0) (y - c) e^{-(y - c)^2} + g(c)y e^{-y^2}}{c e^{-c^2}} \qquad \forall y \in \mathbb{R}](/media/m/c/6/0/c60fd2a991bcf26ff6252c64f9f8f09f.png)
pa sada izjednacavanjem slijedi
![(g(2c) + g(0)) (y - c) e^{-(y - c)^2} + g(c)((2c - y) e^{-(2c - y)^2} - y e^{-y^2}) = 0 \qquad \forall y \in \mathbb{R}](/media/m/5/1/0/5107d17061f448d2219f7cbae493638e.png)
uvedimo supstituciju
![z = y-c](/media/m/4/3/1/431ec6730b1805ce0016fa7f860f20e9.png)
![A = \frac{g(2c)+g(0)}{g(c)}](/media/m/8/a/c/8acef21dbd28528e58f56a79c29ec68d.png)
![g(c) \neq 0](/media/m/0/b/8/0b8edf354947128f7213ccdc7f1e7268.png)
![Aze^{-z^2} - (z-c)e^{-(z-c)^2} - (z+c)e^{-(z+c)^2} = 0 \qquad \forall z \in \mathbb{R}](/media/m/4/e/7/4e787c5065f9edb84a70a235fd98ef45.png)
nekako je jasno da je vrlo tesko da ce ovaj identitet vrijediti
![\forall z \in \mathbb{R}](/media/m/a/9/6/a968f6c2db0b44a73f9127cf8853ec2a.png)
![c \neq 0](/media/m/b/a/d/badc550248e46fce5223b73584f6398a.png)
![e^{-z^2}](/media/m/6/f/3/6f39a18e474ab45fb5b1ff0cbe22f7fa.png)
![e^{-c^2}](/media/m/8/4/3/8438a6a10d07a036737e72046c73e17a.png)
![B = \frac{A}{e^{-c^2}}](/media/m/0/d/1/0d1028f4a7708af507a1f2eebd3983d5.png)
![w = \frac{z}{c}](/media/m/f/6/9/f69de04f372772cd861d6d5df5820063.png)
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
![c \neq 0](/media/m/b/a/d/badc550248e46fce5223b73584f6398a.png)
![Bw - (w-1)e^{2wc^2} - (w+1)e^{-2wc^2} = 0 \qquad \forall w \in \mathbb{R}](/media/m/5/7/8/578478eff33c9deed634e7d9e8cae655.png)
ako je
![c \neq 0](/media/m/b/a/d/badc550248e46fce5223b73584f6398a.png)
![c^2 > 0](/media/m/9/8/f/98f916f7577bdbd518da6b5996e9faa0.png)
![\forall w \in \mathbb{R}](/media/m/c/f/c/cfca8be0948e06f298c322946cb0c8ea.png)
![w](/media/m/a/7/a/a7abf250ebf14efa424fde966849d5f9.png)
![Bw](/media/m/2/a/9/2a9c036c61a737b658c12bc230cdec48.png)
![(w-1)e^{2wc^2}](/media/m/2/3/2/23232479dd34df4e4d1962e7ec5d1548.png)
![(w+1)e^{-2wc^2}](/media/m/3/3/c/33c0dddd16c15ea67d2d907017d72536.png)
![- \infty](/media/m/c/8/6/c86ba3f9d68242d2ff2aa5b922245ae9.png)