Benutzer:Daniel AT/Meine FS

pq-Formel:

${\displaystyle x^{2}+p\cdot x+q=0}$

${\displaystyle x_{1,2}=-{p \over 2}\pm {\sqrt {{p^{2} \over 4}-q}}}$

${\displaystyle a\cdot x^{2}+b\cdot x+c=0}$

${\displaystyle x_{1,2}={\frac {-b\pm {\sqrt {b^{2}-4\cdot ac}}}{2\cdot a}}}$

${\displaystyle \left.{\begin{matrix}{\mbox{Zerlegung in Linearfaktoren:}}&x^{2}+px+q=(x-x_{1})\cdot (x-x_{2})=0\\{\mbox{Satz von Vieta:}}&p=-(x_{1}+x_{2})\qquad q=x_{1}\cdot x_{2}\end{matrix}}\right\}x_{1},x_{2}{\mbox{ sind Lösungen}}}$

Weiterhin sind die Additionstheoreme nützlich:

${\displaystyle \sin(x+y)=\sin x\;\cos y+\sin y\;\cos x}$

${\displaystyle \sin(x-y)=\sin x\;\cos y-\sin y\;\cos x}$

${\displaystyle \cos(x+y)=\cos y\;\cos x-\sin x\;\sin y}$

${\displaystyle \cos(x-y)=\cos y\;\cos x+\sin x\;\sin y}$

${\displaystyle \tan(x+y)={\frac {\tan x+\tan y}{1-\tan x\;\tan y}}={\frac {\sin(x+y)}{\cos(x+y)}}}$

${\displaystyle \tan(x-y)={\frac {\tan x-\tan y}{1+\tan x\;\tan y}}={\frac {\sin(x-y)}{\cos(x-y)}}}$

${\displaystyle \cot \left(x+y\right)={\frac {\cot x\cot y-1}{\cot x+\cot y}}={\frac {\cos(x+y)}{\sin(x+y)}}}$

${\displaystyle \cot \left(x-y\right)={\frac {-\left(\cot x\cot y+1\right)}{\cot x-\cot y}}={\frac {\cos(x-y)}{\sin(x-y)}}}$

${\displaystyle \sin(x+y)\sin(x-y)=\cos ^{2}y-\cos ^{2}x}$

${\displaystyle \cos(x+y)\cos(x-y)=\cos ^{2}y-\sin ^{2}x}$

Für x = y folgen hieraus die Doppelwinkelfunktionen.

${\displaystyle \sin(2\;x)=2\sin x\;\cos x={\frac {2\tan x}{1+\tan ^{2}x}}}$

${\displaystyle \cos(2\;x)=\cos ^{2}x-\sin ^{2}x=1-2\sin ^{2}x=2\cos ^{2}x-1={\frac {1-\tan ^{2}x}{1+\tan ^{2}x}}}$

${\displaystyle \tan(2\;x)={\frac {2\tan x}{1-\tan ^{2}x}}={\frac {2}{\cot x-\tan x}}}$

${\displaystyle \cot(2\;x)={\frac {\cot ^{2}x-1}{2\cot x}}={\frac {\cot x-\tan x}{2}}}$

Aus den Additionstheoremen lassen sich Identitäten ableiten, mit denen die Summe zweier trigonometrischer Funktionen als Produkt aufgefasst werden kann:

${\displaystyle \sin x+\sin y=2\sin {\frac {x+y}{2}}\cos {\frac {x-y}{2}}}$

${\displaystyle \sin x-\sin y=2\cos {\frac {x+y}{2}}\sin {\frac {x-y}{2}}}$

${\displaystyle \cos x+\cos y=2\cos {\frac {x+y}{2}}\cos {\frac {x-y}{2}}}$

${\displaystyle \cos x-\cos y=-2\sin {\frac {x+y}{2}}\sin {\frac {x-y}{2}}}$

${\displaystyle \tan x+\tan y={\frac {\sin(x+y)}{\cos x\cos y}}}$

${\displaystyle \tan x-\tan y={\frac {\sin(x-y)}{\cos x\cos y}}}$

${\displaystyle \cot x+\cot y={\frac {\sin(x+y)}{\sin x\sin y}}}$

${\displaystyle \cot x-\cot y={\frac {-\sin(x-y)}{\sin x\sin y}}}$

${\displaystyle \sin x\;\sin y={\frac {1}{2}}\left[\cos(x-y)-\cos(x+y)\right]}$

${\displaystyle \cos x\;\cos y={\frac {1}{2}}\left[\cos(x-y)+\cos(x+y)\right]}$

${\displaystyle \sin x\;\cos y={\frac {1}{2}}\left[\sin(x-y)+\sin(x+y)\right]}$

${\displaystyle \tan x\;\tan y={\frac {\tan x+\tan y}{\cot x+\cot y}}=-{\frac {\tan x-\tan y}{\cot x-\cot y}}}$

${\displaystyle \cot x\;\cot y={\frac {\cot x+\cot y}{\tan x+\tan y}}=-{\frac {\cot x-\cot y}{\tan x-\tan y}}}$

${\displaystyle \tan x\;\cot y={\frac {\tan x+\cot y}{\cot x+\tan y}}=-{\frac {\tan x-\cot y}{\cot x-\tan y}}}$

${\displaystyle \sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\;-\;\dots \;=\;\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\quad f{\ddot {u}}r\quad |x|<\infty }$

${\displaystyle \cos(x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+\;-\;\dots \;=\;\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}\qquad \quad f{\ddot {u}}r\quad |x|<\infty }$

${\displaystyle \tan(x)=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+{\frac {62}{2835}}x^{9}+\;\dots \;=\;\sum _{n=0}^{\infty }{\frac {2^{2n}(2^{2n}-1)B_{n}}{(2n)!}}\quad f{\ddot {u}}r\quad |x|<{\frac {\pi }{2}}}$

${\displaystyle \cot(x)={\frac {1}{x}}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-{\frac {1}{4725}}x^{7}-\;\dots \qquad f{\ddot {u}}r\quad 0<|x|<\pi }$

Originalfunktion f(t)	Bildfunktion F(s)
${\displaystyle \delta (t)\,}$	${\displaystyle 1\,}$
${\displaystyle 1\,}$	${\displaystyle {\frac {1}{s}}}$
${\displaystyle t\,}$	${\displaystyle {\frac {1}{s^{2}}}}$
${\displaystyle e^{-at}\,}$	${\displaystyle {\frac {1}{s+a}}}$
${\displaystyle t^{n}e^{-at}\,}$	${\displaystyle {\frac {n!}{(s+a)^{n+1}}}}$
${\displaystyle \sin(\omega t)\,}$	${\displaystyle {\frac {\omega }{s^{2}+\omega ^{2}}}}$
${\displaystyle \cos(\omega t)\,}$	${\displaystyle {\frac {s}{s^{2}+\omega ^{2}}}}$
${\displaystyle \sinh(\omega t)\,}$	${\displaystyle {\frac {\omega }{s^{2}-\omega ^{2}}}}$
${\displaystyle \cosh(\omega t)\,}$	${\displaystyle {\frac {s}{s^{2}-\omega ^{2}}}}$
${\displaystyle f(t)\cdot \sin(at)\,}$	${\displaystyle {\frac {1}{2i}}\cdot (F(s-ia)-F(s+ia))}$
${\displaystyle f(t)\cdot \cos(at)\,}$	${\displaystyle {\frac {1}{2}}\cdot (F(s-ia)-F(s+ia))}$
${\displaystyle {\frac {t^{n-1}}{(n-1)!}}\cdot e^{\pm at}\,}$	${\displaystyle (s\mp a)^{-n}}$
${\displaystyle {\frac {1}{T}}\cdot e^{-t/T}\,}$	${\displaystyle {\frac {1}{1+Ts}}}$
${\displaystyle {\frac {e^{-t/T_{1}}-e^{-t/T_{2}}}{T_{1}-T_{2}}}\,}$	${\displaystyle {\frac {1}{(1+T_{1}s)(1+T_{2}s)}}}$
${\displaystyle {\frac {1}{\omega }}e^{-\delta t}\sin(\omega t)\,}$	${\displaystyle {\frac {1}{s^{2}+2\delta s+(\delta ^{2}+\omega ^{2})}}}$

${\displaystyle {\mathcal {L}}\{a_{1}f_{1}(t)\ +a_{2}f_{2}(t)\}=a_{1}{\mathcal {L}}\{f_{1}(t)\}+a_{2}{\mathcal {L}}\{f_{2}(t)\}}$

1) Verschiebung nach rechts

${\displaystyle {\mathcal {L}}\{f(t-a)\}=\mathrm {e} ^{-as}{\mathcal {L}}\{f(t)\}=\mathrm {e} ^{-as}F(s)\qquad (t\geq a>0)}$

2) Verschiebung nach links

${\displaystyle {\mathcal {L}}\{f(t+a)\}=\mathrm {e} ^{as}\left(F(s)-\int _{0}^{a}f(t)\mathrm {e} ^{-st}\mathrm {d} t\right)\qquad (t\geq a>0)}$

${\displaystyle {\mathcal {L}}\{f(at)\}={\frac {1}{a}}F\left({\frac {s}{a}}\right)\qquad (a>0)}$

${\displaystyle {\mathcal {L}}\{\mathrm {e} ^{-at}f(t)\}=F(s+a)\qquad (a\in \mathbb {C} )}$

${\displaystyle {\mathcal {L}}\{t^{n}f(t)\}=(-1)^{n}F^{(n)}(s)\qquad (n=1,2,\dots )}$

${\displaystyle {\mathcal {L}}\left\{{\frac {1}{t}}f(t)\right\}=\int _{s}^{\infty }F(q)\mathrm {d} q}$

${\displaystyle {\mathcal {L}}\{f'(t)\}=sF(s)-f(+0)}$

${\displaystyle {\begin{matrix}{\mathcal {L}}\{f^{(n)}(t)\}&=&s^{n}F(s)-s^{n-1}f(+0)-s^{n-2}f'(+0)-\dots -f^{(n-1)}(+0)\\&=&s^{n}F(s)-\sum _{v=1}^{n}s^{n-v}f^{(v-1)}(+0)\end{matrix}}}$

${\displaystyle {\mathcal {L}}\left\{\int _{0}^{t}f(q)\mathrm {d} q\right\}={\frac {1}{s}}F(s)}$

${\displaystyle {\mathcal {L}}\{f_{1}(t)*f_{2}(t)\}=F_{1}(s)\cdot F_{2}(s)={\mathcal {L}}\{\int _{0}^{t}f_{1}(u)f_{2}(t-u)\}\mathrm {d} u}$

${\displaystyle {\mathcal {L}}\{f_{1}(t)\cdot f_{2}(t)\}={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }F_{1}(\sigma )F_{2}(s-\sigma )\mathrm {d} \sigma }$

${\displaystyle {\mathcal {L}}\{p(t)\}={\frac {1}{1-e^{-sT}}}\int _{0}^{T}p(\tau )\cdot e^{-s\tau }d\tau }$

wobei T die Periode der Funktion p(t) darstellt.

${\displaystyle \lim _{s\to 0}F(s)={\mathcal {L}}\{\int _{0}^{\infty }f(t)dt\}}$

${\displaystyle \lim _{s\to 0}s\cdot F(s)={\mathcal {L}}\{\lim _{t\to \infty }f(t)dt\}}$