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anamech
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37
S2/AnaMech/VL/AnMeVL6.typ
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37
S2/AnaMech/VL/AnMeVL6.typ
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// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 6,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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Das dritte Kepler Gesetz zeigt, dass $pi a b = T L/(q mu) ==> T = (2 mu a b pi) / (L) $.
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= Das $1/r$ Potential
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Es gilt $alpha>0$ und $r_("min") <r<r_("max") $ und $V_0 <= E <0$.
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Unser Ziel ist $1 + epsilon cos phi = r_0 /r)$.
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Wir wissen
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$
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d phi = plus.minus L/r^2 (d r) / [(d mu (E - V_("eff") )]^(1/2).
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$
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Das kann dann durch Substitution geloest werden.
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@@ -4,7 +4,7 @@
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#let rot = math.op("rot")
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#let grad = math.op("grad")
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#let conf(num: none, date: "", type: none, body) = {
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#let conf(num: none, date: "", type: none, body, ueb: false) = {
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// Global settings
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show: default
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141
S2/DiffII/VL/DiIIVL7.typ
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141
S2/DiffII/VL/DiIIVL7.typ
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// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 7,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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Ist $f$ diffbar in $x_0 $, so gilt
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$
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exists "lineare Abbildung" L: RR^n -> CC, "sodass" \
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lim_(n -> 0) (f (x_0 + h) - f (x_0 ) - L h) / (norm(h)) = 0, L = d f (x_0 ).
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$
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#theorem[
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Sei $U subset RR^n $ offen und $f: U -> CC$ in $a in U$ differenzierbar. Dann ist $f$ in $a$ in jede Richtung differenzierbar und es gilt
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$
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d f (a) h = partial / (partial n) f (a) = sum_(i = 1)^(n) partial / (partial i) f (a) h_i .
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$
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] <tem2>
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#proof[
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Fuer $h$ hinreichend klein schreibe
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$
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f (a + h) = f (a) + d f (a) h + R (h) "mit" lim_(h -> 0) (R (h)) / (norm(h)) = 0.
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$
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Fuer $h in RR^n $ und $t in RR$ hinreichend klein folgt
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$
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f (a + t h) = f (a) + d f (a) (t h) + R (t h).
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$
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Fuer $t != 0$ folgt
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$
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d f (a) h = (f (a + t h) - f (a) - R (t h)) / (t),
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$ <gl1>
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da $f$ in $a$ diffb. ist, gilt $lim_(t -> 0) (R (t h)) / (t) < norm(h) (R (t h)) / (t norm(h)) = 0$
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Mit @gl1 folgt, dass der Grenzwert $lim_(t -> 0) (f (a + t h)- f (a)) / (t) $ existiert und dass gilt
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$
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d f (a) h = partial / (partial h) f (a).
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$ <gl2>
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Das ist eine wichtige Gleichheit zwischen Differenzial und Richtungsableitung
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Da $ d f (a) h$ eine lineare Abbildung ist, gilt
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$
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d f (a) h = d f (a) (sum_(k = 1)^(n) e_k h_k ) = sum_(k = 1)^(n) h_k underbrace(d f (a) e_k, #[@gl2] ).
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$
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Also $d f (a) h = sum_(k = 1)^(n) h_k * partial / (partial n) f (a)$.
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]
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#remark[
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Sei $s in RR$.
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$
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lim_(t -> 0) (f (a + t * s h) - f (a)) / (s t) = partial / (partial s h) f (a), \
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s lim_(t' -> 0) (f (a + t' h)-f (a)) / (t') = s partial / (partial h) f (a).
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$
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Betrachte
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$
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f: U subset RR^n -> CC, d f (a): RR^n -> CC, h |-> d f (a) h = sum_(k = 1)^(n) h_k partial / (partial k) f (a).
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$
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]
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#definition[
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Ist $f: U -> CC$ in $a in U$ partiell Differenzierbar, so definieren wir den Gradienten von $f$ in $a$ als
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$
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grad(f) (a) := vec(partial / (partial 1) f (a), partial / (partial 2) f (a), dots.v , partial / (partial n) f (a) ).
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$
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Wir schreiben auch
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$
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arrow(nabla) f (a) = grad(f) (a) = f' (a).
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$
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]
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#example[
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Ist $f (x_1, x_2 )= x_1 ^2 + x_2 ^2 $ so gilt in $(a_1, a_2)$
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$
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arrow(nabla) f (a) = vec(2 a_1 , 2 a_2 ).
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$
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]
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"Der Gradient gibt die Richtung des starksten Anstiegs an."
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#theorem[
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Sei $U subset RR^n $ offen und $f: U -> CC$ in $a in U$ differenzierbar. Ist $arrow(nabla) f (a) = 0$ so gilt
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$
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partial / (partial h) f (a) = 0, forall h in RR^n .
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$
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Ist $arrow(nabla) f (a) != 0$, so gilt $abs(partial / (partial n) f (a)) <= norm(arrow(nabla) f (a))_(2) forall h in RR^n "mit" norm(h)_(2) = 1$ und
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Gleichheit fuer $h = (arrow(nabla) f (a)) / (norm( arrow(nabla) f (a))_(2) ) $.
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]
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#proof[
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Ist $arrow(nabla) f (a) = 0 "und" h in RR^n $, so folgt nach @tem2, $partial / (partial n) f (a) = sum_(k = 1)^(n) partial / (partial n) f (a) h_(k) = arrow(nabla) f (a) = 0 $.
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Sei nun $h in RR^n $ mit $norm(h)_(2) = 1$. Mit Cauchy-Schwarz erhalten wir
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$
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abs(partial / (partial n) f (a)) = abs( sum_(k = 1)^(n) h_k partial / (partial k) f (a)) = underbrace(norm(h)_(2), =1) norm( arrow(nabla) f (a))_(2)
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$
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und Gleichheit fuer
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$
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h = (arrow(nabla) f (a)) / (norm( arrow(nabla) f (a))_(2) )
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$
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]
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$
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f: U -> CC, a in U
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$
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$
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f "diffbar" ==>^(#[@tem2]) f "ist partiell diffbar".
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$
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#example[
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Sei $f: RR^2 -> RR$ gegeben durch
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$
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f (x,y) = cases((x ^2 y) / (x^2 + y^2 ) ,0).
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$
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Dann ist $f$ stetig in 0 und fuer jedes $h = (h_1, h_2 ) in RR^2$ existiert die Richtungsableitung.
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$
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(f (h_1, h_2 ) - f (0,0) - d f (0) h)/(norm((h_1, h_2 ))_(2) ) = (h_1 ^2 h_2 ) / (norm((h_1, h_2 ))_(2) ^3) = (h_1 ^3 ) / (( 2 h_1 ^2 )^(3/2) ) = +- (1) / (2^(3/2) ) != 0
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$
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Also ist $f$ nicht diffbar in dem Punkt $(0,0)$.
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]
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#theorem[
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Sei $U subset RR^n $ offen, $f: U -> CC$ in jedem Punkt in U partiell differenzierbar und die partiellen Ableitungen $partial / (partial k) f: U -> CC, 1 <= k <= n "im" a in U$ stetig. Dann ist $f$ in $a$ differenzierbar.
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]
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