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chore: add maths6

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Jan Meyer
2026-06-04 19:29:51 +02:00
parent d91ccb84b9
commit 7432f09be9
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11 Homework/Mathe II/Homework 6.pdf
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11 Homework/Mathe II/Homework 6.typ
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@@ -14,29 +14,29 @@
= Problem 2 = Problem 2
== (a) == (a)
$ f_n (x) = (sin(n x) + n) / (3n + 1) = (sin(n x) / n + 1) / (3 + 1/n) $ $f_n (x) = (sin(n x) + n) / (3n + 1) = (sin(n x) / n + 1) / (3 + 1/n)$
Da $ lim_(n -> oo) sin(n x) / n = 0 $, ist der punktweise Grenzwert: Da $lim_(n -> oo) sin(n x) / n = 0$, ist der punktweise Grenzwert:
$ f(x) = lim_(n -> oo) f_n (x) = (0 + 1) / (3 + 0) = 1/3 $ $f(x) = lim_(n -> oo) f_n (x) = (0 + 1) / (3 + 0) = 1/3$
== (b) == (b)
Zu zeigen: $ f_n (x) = (n x) / (n x^2 + 1) $ konvergiert auf $ [r, 1) $ mit $ r > 0 $ gleichmäßig gegen $ f(x) = 1/x $. Zu zeigen: $f_n (x) = (n x) / (n x^2 + 1)$ konvergiert auf $[r, 1)$ mit $r > 0$ gleichmäßig gegen $f(x) = 1/x$.
$ |f_n (x) - f(x)| = |(n x) / (n x^2 + 1) - 1/x| = |(n x^2 - (n x^2 + 1)) / (x(n x^2 + 1))| = 1 / (n x^3 + x) $ $|f_n (x) - f(x)| = |(n x) / (n x^2 + 1) - 1/x| = |(n x^2 - (n x^2 + 1)) / (x(n x^2 + 1))| = 1 / (n x^3 + x)$
Da $ x >= r > 0 $, gilt $ n x^3 + x > n r^3 $. Daraus folgt für das Supremum: Da $x >= r > 0$, gilt $n x^3 + x > n r^3$. Daraus folgt für das Supremum:
$ sup_(x in [r, 1)) |f_n (x) - f(x)| <= 1 / (n r^3) $ $sup_(x in [r, 1)) |f_n (x) - f(x)| <= 1 / (n r^3)$
Wegen $ lim_(n -> oo) 1 / (n r^3) = 0 $ konvergiert die Funktionenfolge gleichmäßig. Wegen $lim_(n -> oo) 1 / (n r^3) = 0$ konvergiert die Funktionenfolge gleichmäßig.
= Problem 3 = Problem 3
Sei $ cal(E) = (bold(m)_0, bold(m)_1) $ die Standardbasis. Die Basiswechselmatrizen von $ cal(B) $ und $ cal(C) $ in $ cal(E) $ sind: Sei $cal(E) = (bold(m)_0, bold(m)_1)$ die Standardbasis. Die Basiswechselmatrizen von $cal(B)$ und $cal(C)$ in $cal(E)$ sind:
$ M_cal(B) = mat(3, 2; 3, 1), quad M_cal(C) = mat(4, 1; 0, 1) $ $M_cal(B) = mat(3, 2; 3, 1), quad M_cal(C) = mat(4, 1; 0, 1)$
Der Koordinatenvektor von $ bold(p) = bold(m)_0 + bold(m)_1 $ in $ cal(E) $ ist $ bold(p)^cal(E) = vec(1, 1) $. Der Koordinatenvektor von $bold(p) = bold(m)_0 + bold(m)_1$ in $cal(E)$ ist $bold(p)^cal(E) = vec(1, 1)$.
== (a) == (a)
$ bold(p)^cal(B) = M_cal(B)^(-1) bold(p)^cal(E) = 1/(3-6) mat(1, -2; -3, 3) vec(1, 1) = mat(-1/3, 2/3; 1, -1) vec(1, 1) = vec(1/3, 0) $ $bold(p)^cal(B) = M_cal(B)^(-1) bold(p)^cal(E) = 1/(3-6) mat(1, -2; -3, 3) vec(1, 1) = mat(-1/3, 2/3; 1, -1) vec(1, 1) = vec(1/3, 0)$
== (b) == (b)
$ T_(cal(C) <- cal(B)) = M_cal(C)^(-1) M_cal(B) = 1/4 mat(1, -1; 0, 4) mat(3, 2; 3, 1) = mat(1/4, -1/4; 0, 1) mat(3, 2; 3, 1) = mat(0, 1/4; 3, 1) $ $T_(cal(C) <- cal(B)) = M_cal(C)^(-1) M_cal(B) = 1/4 mat(1, -1; 0, 4) mat(3, 2; 3, 1) = mat(1/4, -1/4; 0, 1) mat(3, 2; 3, 1) = mat(0, 1/4; 3, 1)$
$ T_(cal(B) <- cal(C)) = (T_(cal(C) <- cal(B)))^(-1) = 1/(0 - 3/4) mat(1, -1/4; -3, 0) = -4/3 mat(1, -1/4; -3, 0) = mat(-4/3, 1/3; 4, 0) $ $T_(cal(B) <- cal(C)) = (T_(cal(C) <- cal(B)))^(-1) = 1/(0 - 3/4) mat(1, -1/4; -3, 0) = -4/3 mat(1, -1/4; -3, 0) = mat(-4/3, 1/3; 4, 0)$
== (c) == (c)
$ bold(p)^cal(C) = T_(cal(C) <- cal(B)) bold(p)^cal(B) = mat(0, 1/4; 3, 1) vec(1/3, 0) = vec(0, 1) $ $bold(p)^cal(C) = T_(cal(C) <- cal(B)) bold(p)^cal(B) = mat(0, 1/4; 3, 1) vec(1/3, 0) = vec(0, 1)$