update uebungen

sose2020/ana/uebungen/ana2.tex

@@ -29,25 +29,25 @@
                 \begin{align*}
                     \int_{-1}^{1} P_n(x) P_m(x) \d x
                     &= \frac{1}{2^{n} 2^{m} n! m!}
-                    \int_{-1}^{1} \frac{\mathrm{d}}{\d x^{n}}(x^2-1)^{n}
+                    \int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
-                    \frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x
+                    \frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x
                 .\end{align*}
                 Zu zeigen:
                 \[
-                 \int_{-1}^{1} \frac{\mathrm{d}}{\d x^{n}}(x^2-1)^{n}
+                 \int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
-                    \frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x = 0
+                    \frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x = 0
                 .\] Mit $(*)$ und $k = m+1 \le n$ folgt
                 \begin{align*}
-                    \int_{-1}^{1} \frac{\mathrm{d}}{\d x^{n}}(x^2-1)^{n}
+                    \int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
-                        \frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x
+                        \frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x
                     &= (-1)^{m+1} \int_{-1}^{1} \frac{\mathrm{d}^{n-m-1}}{\d x^{n-m-1}}
                     (x^2-1)^{n} \frac{\mathrm{d}^{2m+1}}{\d x^{2m+1}}(x^2-1)^{m}\d x
                 .\end{align*}
-                    Wegen $\text{deg } (x^2-1)^{m} = 2^{m}$ folgt
+                    Wegen $\text{deg } (x^2-1)^{m} = 2m$ folgt
                     $\frac{\mathrm{d}^{2m+1}}{\d x^{2m+1}}(x^2-1)^{m} = 0$. Damit folgt
                 \begin{align*}
-                    \int_{-1}^{1} \frac{\mathrm{d}}{\d x^{n}}(x^2-1)^{n}
+                    \int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
-                        \frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x
+                        \frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x
                         &= 0
                 .\end{align*}
             \end{proof}

sose2020/la/uebungen/la1.tex

@@ -41,7 +41,7 @@
             \end{proof}
         \item $\Z / 4 \Z$ ist wegen $\overline{2} \cdot \overline{2} = \overline{4} = \overline{0}$ und
             $\overline{2} \neq 0$ nicht nullteilerfrei. Hier ist
-            $\overline{1} + \overline{2}t \in (\Z / 4 \Z)^{\times }$, wegen
+            $\overline{1} + \overline{2}t \in ((\Z / 4 \Z)[t])^{\times }$, wegen
             \[
                 (\overline{1} + \overline{2}t) (\overline{1} + \overline{2}t)
                 = \overline{1} + \underbrace{\overline{4}t + \overline{4}t^2}_{= 0}

sose2020/theo/uebungen/theo2.tex

@@ -107,11 +107,10 @@
                 \intertext{Zusammen folgt}
                 &\Delta t = \int_{x_0}^{x_E} \frac{\sqrt{1 + f'(x)^2}}{\sqrt{\frac{2}{m} (E - V(\vec{x}))}} \d x
             .\end{align*}
-        \item Mit $V(z) = mgz$, $E = 0$, $x_0 = 0$, $x_E = 1$ und $f(x) = x$ folgt
+        \item Mit $V(z) = mgz$, $E = \frac{m}{2} v_0^2$, $x_0 = 0$, $x_E = 1$ und $f(x) = x$ folgt
             \[
-            \Delta t = \int_{0}^{1} \frac{\sqrt{2} }{\sqrt{\frac{2}{m} mgx} } \d x
+                \Delta t = \int_{0}^{1} \frac{\sqrt{2} }{\sqrt{\frac{2}{m} (\frac{m}{2}v_0^2 + mgx)} } \d x
-            = \int_{0}^{1} \frac{\sqrt{2} }{\sqrt{2 g x} } \d x = \frac{2}{\sqrt{g} } \sqrt{x}
+            = \int_{0}^{1} \frac{\sqrt{2} }{\sqrt{v_0^2 + 2 g x} } \d x = 2 \cdot \frac{1}{2g} \cdot \sqrt{2} \sqrt{v_0^2 + 2gx} \Big|_{0}^{1}  \d x 
-            \Big|_{0}^{1} = \frac{2}{\sqrt{g} }
             .\] 
     \end{enumerate}
 \end{aufgabe}