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update uebungen

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christian
2020-05-05 16:16:41 +02:00
parent b33c550fbb
commit fb7f3aad6c
6 changed files with 13 additions and 14 deletions
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sose2020/ana/uebungen/ana2.pdf
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sose2020/ana/uebungen/ana2.tex
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@@ -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}
\frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x
\int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
\frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x
.\end{align*}
Zu zeigen:
\[
\int_{-1}^{1} \frac{\mathrm{d}}{\d x^{n}}(x^2-1)^{n}
\frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x = 0
\int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
\frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x = 0
.\] Mit $(*)$ und $k = m+1 \le n$ folgt
\begin{align*}
\int_{-1}^{1} \frac{\mathrm{d}}{\d x^{n}}(x^2-1)^{n}
\frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x
\int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
\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}
\frac{\mathrm{d}}{\d x^{m}}(x^2-1)^{m} \d x
\int_{-1}^{1} \frac{\mathrm{d}^{n}}{\d x^{n}}(x^2-1)^{n}
\frac{\mathrm{d}^{m}}{\d x^{m}}(x^2-1)^{m} \d x
&= 0
.\end{align*}
\end{proof}