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战斗力 鹅
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本帖最后由 RookieTnT 于 2026-10-14 11:58 编辑
Try to solve this problem
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For $x \in \mathbb{R}$, define
\[
W(x) = 2 \sum_{n=1}^{\infty} \left( 2\pi^2 n^4 e^{\frac{9}{2}|x|} - 3\pi n^2 e^{\frac{5}{2}|x|} \right) e^{-\pi n^2 e^{2|x|}}.
\]
Let
\[
C = \int_{\mathbb{R}} W(x) \, dx, \quad p(x) = \frac{W(x)}{C}, \quad F(x) = \int_{-\infty}^x p(y) \, dy.
\]
On the complex Hilbert space $L^2(\mathbb{R})$, define the integral operator
\[
(Bf)(x) = \frac{i}{\sqrt{p(x)}} \left[ \int_{-\infty}^x \sqrt{p(y)} f(y) \, dy - F(x) \int_{\mathbb{R}} \sqrt{p(y)} f(y) \, dy \right].
\]
The following known properties may be used:
\begin{enumerate}
\item $W$ is strictly positive, smooth, and an even function, satisfying $0 < C < \infty$.
\item $p$ is a probability density function, and for every $a > 0$,
\[
\int_{\mathbb{R}} p(x) e^{a|x|} \, dx < \infty.
\]
\item $B$ is a Hilbert--Schmidt operator, and is therefore a compact operator.
\end{enumerate}
\textbf{Prove:} For any $\lambda \in \mathbb{C} \setminus \{0\}$, if there exists a non-zero function $f \in L^2(\mathbb{R})$ such that
\[
Bf = \lambda f,
\]
then
\[
\operatorname{Im} \lambda = 0.
\]
可以用上面的来测试ai. 答案是做不出来的. |
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楼主: 绕指流光
沪公网安备 31010702007642号 )
发表于 2026-10-14 10:54
现在国内模型是不是都在适配国产卡训练和推理啊？感觉上面的态度就是趁现在差距不大，宁可拼着落后一年也要把国内生态搞完整。
