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01-Hasznos_matematikai_formulak.lyx
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01-Hasznos_matematikai_formulak.lyx
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#LyX 2.2 created this file. For more info see http://www.lyx.org/
\lyxformat 508
\begin_document
\begin_header
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theorems-std
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\end_header
\begin_body
\begin_layout Part
Néhány hasznos matematikai formula
\end_layout
\begin_layout Section
\begin_inset CommandInset href
LatexCommand href
name "Gauss-integrál"
target "https://en.wikipedia.org/wiki/Gaussian_integral"
\end_inset
\end_layout
\begin_layout Standard
Vezessük le a következő integrált:
\end_layout
\begin_layout Standard
\begin_inset Box Shaded
position "t"
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thickness "0.4pt"
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framecolor "black"
backgroundcolor "none"
status open
\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
I=\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\mathrm{d}x=\sqrt{\pi}.\label{eq:gauss-0}
\end{equation}
\end_inset
\end_layout
\end_inset
Célszerű a kifejezés négyzetét vizsgálni!
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
I^{2}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}-y^{2}}\mathrm{d}x\mathrm{d}y.\label{eq:gauss-levezet-1}
\end{equation}
\end_inset
Az integrált írjuk át polárkoordinátákba!
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
x & = & r\cos\varphi,\label{eq:gauss-polar}\\
y & = & r\sin\varphi,\nonumber \\
\mathrm{d}x\mathrm{d}y & = & r\mathrm{d}\varphi\mathrm{d}r.\nonumber
\end{eqnarray}
\end_inset
A keresett integrál a következő alakot ölti:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
I^{2}=\int_{0}^{\infty}\int_{0}^{2\pi}\mathrm{e}^{-r^{2}}r\mathrm{d}\varphi\mathrm{d}r=2\pi\int_{0}^{\infty}\mathrm{e}^{-r^{2}}r\mathrm{d}r.\label{eq:gauss-levezet-2}
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
\emph on
\color blue
Megjegyzés:
\emph default
Vegyük észre, hogy míg a polárkoordinátáknál
\begin_inset Formula $0$
\end_inset
-tól
\begin_inset Formula $\infty$
\end_inset
-ig integráltunk, az eredeti integrálunk
\begin_inset Formula $-\infty$
\end_inset
-től
\begin_inset Formula $\infty$
\end_inset
-ig megy.
Itt nem követtünk el semmilyen matematikai hibát ugyanis ezt a szögfüggéssel
figyelembe vettük a polárkoordináták segítségével.
\color inherit
\begin_inset Newline newline
\end_inset
Végezzünk el még egy változó cserét!
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
u & = & r^{2},\label{eq:gauss-rtou}\\
\frac{\mathrm{d}u}{\mathrm{d}r} & = & 2r\rightarrow\mathrm{d}r=\frac{\mathrm{d}u}{2r}.\nonumber
\end{eqnarray}
\end_inset
Így már elemi integrációs szabályokkal kiértékelhető összefüggésre jutunk:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
I^{2}=2\pi\int_{0}^{\infty}\frac{\mathrm{e}^{-u}}{2}\mathrm{d}u=\pi\left[-\mathrm{e}^{-u}\right]_{0}^{\infty}=\pi.\label{eq:gauss-levezet-3}
\end{equation}
\end_inset
A keresett integrál tehát:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\int_{-\infty}^{\infty}\mathrm{e}^{-x^{2}}\mathrm{d}x=\sqrt{\pi}.\label{eq:gauss-eredmeny}
\end{equation}
\end_inset
Egy egyszerű változó cserével lássuk be a Gauss-integrál egyszerű általánosításá
t:
\end_layout
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position "t"
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\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
\int_{-\infty}^{\infty}\mathrm{e}^{-ax^{2}}\mathrm{d}x=\sqrt{\frac{\pi}{a}}.\label{eq:gauss-scaled}
\end{equation}
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{align}
ax^{2} & =t^{2},\\
\frac{\mbox{d}t}{\mbox{d}x} & =\sqrt{a}\rightarrow\mbox{d}t=\sqrt{a}\mbox{d}x.
\end{align}
\end_inset
Kapjuk tehát, hogy
\begin_inset Formula
\begin{equation}
\int_{-\infty}^{\infty}\mathrm{e}^{-ax^{2}}\mathrm{d}x=\int_{-\infty}^{\infty}\mathrm{e}^{-t^{2}}\frac{\mbox{d}t}{\sqrt{a}}=\sqrt{\frac{\pi}{a}}.
\end{equation}
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Box Shadowbox
position "t"
hor_pos "c"
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framecolor "black"
backgroundcolor "none"
status open
\begin_layout Exercise
Lássuk be, hogy ha
\begin_inset Formula $a>0$
\end_inset
valós szám akkor:
\end_layout
\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
\int_{-\infty}^{\infty}\mathrm{e}^{-ax^{2}+bx+c}\mathrm{d}x=\sqrt{\frac{\pi}{a}}e^{\frac{b^{2}}{4a}+c}\label{eq:gauss-alltalanos}
\end{equation}
\end_inset
\end_layout
\begin_layout Plain Layout
Útmutatás:
\end_layout
\begin_layout Itemize
Alakítsuk teljes négyzetté a kitevőben szereplő polinomot!
\end_layout
\begin_layout Itemize
A Gauss-integrál invariáns az integrandus
\begin_inset Quotes eld
\end_inset
eltolására
\begin_inset Quotes erd
\end_inset
!
\end_layout
\begin_layout Itemize
A négyzetes tag együtthatójától egy alkalmas változó cserével szabadulhatunk
meg.
\end_layout
\end_inset
\end_layout
\begin_layout Section
A
\begin_inset CommandInset href
LatexCommand href
name "Gamma-függvény"
target "https://hu.wikipedia.org/wiki/Gamma-f%C3%BCggv%C3%A9ny"
\end_inset
néhány tulajdonsága
\end_layout
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filename Fig/Gamma.png
lyxscale 40
width 50text%
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\end_layout
\begin_layout Plain Layout
\begin_inset Caption Standard
\begin_layout Plain Layout
A Gamma függvény
\begin_inset CommandInset label
LatexCommand label
name "fig:A-Gamma-fuggveny"
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\end_layout
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\begin_layout Plain Layout
\series bold
A Gamma-függvény:
\end_layout
\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
\Gamma(x)=\int_{0}^{\infty}\mathrm{e}^{-t}t^{x-1}\mathrm{d}t,\quad\mathrm{Re}(x)>0,\label{eq:gamma-def}
\end{equation}
\end_inset
\end_layout
\end_inset
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\begin_layout Standard
A fenti definíció segítségével lássuk be, hogy
\end_layout
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\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
\Gamma(x+1)=x\Gamma(x).\label{eq:gamma-plusone}
\end{equation}
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\Gamma(x+1) & = & \int_{0}^{\infty}\underbrace{\mbox{e}^{-t}}_{v'}\underbrace{t^{x}}_{u}\mbox{d}t\label{eq:gamma-levezet-1}\\
& = & \underbrace{[t^{x}(-\mbox{e}^{-t})]_{0}^{\infty}}_{0}-\int_{0}^{\infty}\underbrace{-\mbox{e}^{-t}}_{v}\underbrace{xt^{x-1}}_{u'}\mbox{d}t\\
& = & x\int_{0}^{\infty}\mbox{e}^{-t}t^{x-1}\mbox{d}t\\
& = & x\Gamma(x)
\end{eqnarray}
\end_inset
ahol kihasználtuk a parciális integrálás szabályát
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{equation}
\int u(t)v'(t)\mbox{d}t=u(t)v(t)-\int u'(t)v(t)\mbox{d}t
\end{equation}
\end_inset
és az alábbi két ismert összefüggést
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\int\mbox{e}^{\alpha t}\mbox{d}t & = & \frac{\mbox{e}^{\alpha t}}{\alpha},\\
\partial_{t}t^{\alpha} & = & \alpha t^{\alpha-1}.
\end{eqnarray}
\end_inset
Lássuk be a következő két összefüggést is!
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\begin_layout Plain Layout
\begin_inset Formula
\begin{eqnarray}
\Gamma(1) & = & 1\label{eq:gamma-one}\\
\Gamma\left(\frac{1}{2}\right) & = & \sqrt{\pi}\label{eq:gamma-half}
\end{eqnarray}
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\Gamma(1) & = & \int_{0}^{\infty}\mbox{e}^{-t}t^{1-1}\mbox{d}t\\
& = & \int_{0}^{\infty}\mbox{e}^{-t}\mbox{d}t\\
& = & [-\mbox{e}^{-t}]_{0}^{\infty}=1
\end{eqnarray}
\end_inset
Felhasználva a (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:gamma-plusone"
\end_inset
) és (
\begin_inset CommandInset ref
LatexCommand ref
reference "eq:gamma-one"
\end_inset
) összefüggéseket a
\begin_inset Formula $\Gamma(x)$
\end_inset
függvényt tetszőleges pozitív egész számokra meghatározhatjuk:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\Gamma(2) & = & 1\Gamma(1)=1\\
\Gamma(3) & = & 2\Gamma(2)=2\cdot1\\
\Gamma(4) & = & 3\Gamma(3)=3\cdot2\cdot1
\end{eqnarray}
\end_inset
\end_layout
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status open
\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
\Gamma(n)=(n-1)!\quad n\in\mathbb{N}\label{eq:gamma-factorial}
\end{equation}
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\Gamma(1/2) & = & \int_{0}^{\infty}\mbox{e}^{-t}t^{1/2-1}\mbox{d}t\\
& = & \int_{0}^{\infty}\frac{\mbox{e}^{-t}}{t^{1/2}}\mbox{d}t
\end{eqnarray}
\end_inset
Hajtsuk végre a következő változó cserét:
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
u & = & t^{1/2},\\
\frac{\mbox{d}u}{\mbox{d}t} & = & \frac{1}{2}\frac{1}{t^{1/2}}.
\end{eqnarray}
\end_inset
Így kapjuk, hogy
\end_layout
\begin_layout Standard
\begin_inset Formula
\begin{eqnarray}
\int_{0}^{\infty}\frac{\mbox{e}^{-t}}{t^{1/2}}\mbox{d}t & = & \int_{0}^{\infty}\mbox{e}^{-u^{2}}2\mbox{d}u,\\
& = & \int_{-\infty}^{\infty}\mbox{e}^{-u^{2}}\mbox{d}u=\sqrt{\pi}.
\end{eqnarray}
\end_inset
\begin_inset Float figure
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\align center
\begin_inset Float figure
wide false
sideways false
status open
\begin_layout Plain Layout
\begin_inset Graphics
filename Fig/Stirling.png
lyxscale 40
width 40text%
\end_inset
\begin_inset Caption Standard
\begin_layout Plain Layout
A Stirling-formula levezetése során alkalmazott közelítés
\begin_inset CommandInset label
LatexCommand label
name "fig:Stirling formula-integrand"
\end_inset
\end_layout
\end_inset
\end_layout
\end_inset
\begin_inset Formula $\quad\quad\quad\quad$
\end_inset
\begin_inset Float figure
wide false
sideways false
status open
\begin_layout Plain Layout
\begin_inset Graphics
filename Fig/Stirling-approx.png
lyxscale 40
width 40text%
\end_inset
\begin_inset Caption Standard
\begin_layout Plain Layout
A Stirling-formula és a
\begin_inset Formula $\Gamma$
\end_inset
-függvény
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\end_layout
\end_inset
\end_layout
\begin_layout Plain Layout
\begin_inset Caption Standard
\begin_layout Plain Layout
Stirling közelítés
\end_layout
\end_inset
\end_layout
\end_inset
\end_layout
\begin_layout Standard
Termodinamikai határesetek vizsgálata során sokszor fogunk találkozni olyan
esetekkel amikor a
\begin_inset Formula $\Gamma(n)$
\end_inset
függvényt a
\begin_inset Formula $n\gg1$
\end_inset
értékekre kell kiértékelnünk.
\end_layout
\begin_layout Standard
Lássuk be a következő hasznos közelítő formulát:
\end_layout
\begin_layout Standard
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\begin_layout Plain Layout
\series bold
\begin_inset CommandInset href
LatexCommand href
name "Stirling-formula:"
target "https://hu.wikipedia.org/wiki/Stirling-formula"
\end_inset
\series default
\begin_inset Formula
\begin{equation}
n!=n^{n}\mbox{e}^{-n}\sqrt{2\pi n}\left(1+\mathcal{O}\left(\frac{1}{n}\right)\right)\label{eq:stirling-formula}
\end{equation}
\end_inset
\end_layout
\end_inset
\begin_inset Formula
\begin{align}
n! & =\Gamma(n+1)=\int_{0}^{\infty}\mbox{e}^{-t}t^{n}\mbox{d}t\label{eq:stirling-init}\\
& =\int_{0}^{\infty}\mbox{e}^{-t+n\ln t}\mbox{d}t\nonumber \\
& =\int_{0}^{\infty}\mbox{e}^{-f_{n}(t)}\mbox{d}t,\nonumber \\
f_{n}(t) & =t-n\ln t.
\end{align}
\end_inset
Fejtsük sorba az
\begin_inset Formula $f_{n}(t)$
\end_inset
függvényt a minimuma körül!
\begin_inset Formula
\begin{equation}
\partial_{t}f_{n}(t)=1-\frac{n}{t},
\end{equation}
\end_inset
\begin_inset Formula
\begin{equation}
\partial_{t}f_{n}(t_{0})=0\rightarrow t_{0}=n.
\end{equation}
\end_inset
Elegendő elvégezni a sorfejtést másod rendig.
Azaz a következő közelítéssel élünk:
\begin_inset Formula
\begin{equation}
f_{n}(t)\approx f_{n}(t_{0})+\partial_{t}f_{n}(t_{0})\left(t-t_{0}\right)+\frac{1}{2}\partial_{t}^{2}f_{n}(t_{0})\left(t-t_{0}\right)^{2},
\end{equation}
\end_inset
\begin_inset Formula
\begin{equation}
\partial_{t}^{2}f_{n}(t)=\frac{n}{t^{2}}\rightarrow\partial_{t}^{2}f_{n}(t_{0})=\frac{n}{n^{2}}=\frac{1}{n},
\end{equation}
\end_inset
\begin_inset Formula
\begin{equation}
f_{n}(t)\approx n-n\ln n+\underbrace{\partial_{t}f_{n}(t_{0})\left(t-t_{0}\right)}_{\text{by def}=0}+\frac{1}{2}\left(\frac{1}{n}\right)\left(t-n\right)^{2}.
\end{equation}
\end_inset
Visszaírva ezt a
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:stirling-init"
\end_inset
kifejezésbe:
\begin_inset Formula
\begin{align}
n! & =\int_{0}^{\infty}\mbox{e}^{-t}t^{n}\mbox{d}t\\
& =\int_{0}^{\infty}\mbox{e}^{-f_{n}(t)}\mbox{d}t\nonumber \\
& \approx\mbox{e}^{-\left(n-n\ln n\right)}\int_{0}^{\infty}\mbox{e}^{-\frac{\left(t-n\right)^{2}}{2n}}\mbox{d}t.\nonumber
\end{align}
\end_inset
A kifejezésben szereplő integrál alsó határát kiterjeszthetjük
\begin_inset Formula $-\infty$
\end_inset
-ig hiszen feltettük, hogy
\begin_inset Formula $n\gg1$
\end_inset
:
\begin_inset Formula
\begin{align}
n! & \approx\mbox{e}^{-\left(n-n\ln n\right)}\int_{-\infty}^{\infty}\mbox{e}^{-\frac{\left(t-n\right)^{2}}{2n}}\mbox{d}t\\
& =\mbox{e}^{-\left(n-n\ln n\right)}\sqrt{2\pi n}.\nonumber
\end{align}
\end_inset
Ahol felhasználtuk a Gauss-integrálra vonatkozó
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:gauss-scaled"
\end_inset
azonosságot.
A kapott eredmény pedig nem más mint maga a
\begin_inset CommandInset ref
LatexCommand eqref
reference "eq:stirling-formula"
\end_inset
Stirling-formula.
Sokszor fogunk találkozni a Stirling-formula logaritmusával:
\end_layout
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backgroundcolor "none"
status open
\begin_layout Plain Layout
\begin_inset Formula
\begin{equation}
\ln n!\approx n\ln n-n+\frac{1}{2}\ln\left(2\pi n\right)\label{eq:stirling-log}
\end{equation}
\end_inset