SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\lim  _ {z \rightarrow \infty} z \cot \alpha = \frac {K _ {1}}{1 + ^ {4} / 3 K _ {2} ^ {2}}.
$$
Proof. The expression for  $z \cot \alpha$  or  $z / \tan \alpha$  follows from Lemma 24:
$$
\lim  _ {z \rightarrow \infty} z \cot \alpha = \frac {z K _ {1}}{p + \sqrt {p ^ {2} + K _ {1} ^ {2}}},
$$
where  $p = \frac{K_1^2}{z} \left( 1 + \frac{1}{3d_{\mathrm{in}}} \right) + \frac{\sigma_{\mathrm{in}}^2}{3} \left( 1 - \frac{1}{d_{\mathrm{in}}} \right) - \left( \frac{z}{2} + \frac{2K_2^2z}{3} + \frac{1}{6} \right)$ . Applying L'Hôpital's (relevant expressions are continuous in  $z$ ) rule we get:  $\lim_{z \to \infty} z \cot \alpha = \frac{K_1}{1 + 4/3K_2^2}$ .
Lemma 28 (asymptotic behavior of  $z / \tau^2$ ). For  $\gamma = K_1 / \sqrt{z}$ ,  $\sigma_{\mathrm{sp}} = K_2\sqrt{z}$  following the expressions in Lemma 24,
$$
\lim _ {z \rightarrow \infty} z / \tau^ {2} = \frac {2 \sigma_ {\mathrm {i n}} ^ {2} / 3 (1 - 1 / d _ {\mathrm {i n}})}{1 + 4 / 3 K _ {2} ^ {2}}.
$$
Proof. For  $\tau = \lambda_1 / \lambda_2$ , substituting the relevant expressions from Lemma 24, we get:
$$
\begin{array}{l} z / \tau^ {2} = \frac {z \lambda_ {2}}{\lambda_ {1}} \\ = z \cdot \frac {2 K _ {1} ^ {2} / _ {z} (1 + 1 / _ {3 d _ {\mathrm {i n}}}) + 2 \sigma_ {\mathrm {i n}} ^ {2} (1 - 1 / _ {d _ {\mathrm {i n}}}) + p - \sqrt {K _ {1} ^ {2} + p ^ {2}}}{2 K _ {1} ^ {2} / _ {z} (1 + 1 / _ {3 d _ {\mathrm {i n}}}) + 2 \sigma_ {\mathrm {i n}} ^ {2} (1 - 1 / _ {d _ {\mathrm {i n}}}) + p + \sqrt {K _ {1} ^ {2} + p ^ {2}}}, \\ \end{array}
$$
where  $p = z / 2 + 2K_2^2 z / 3 + 1 / 6$ . Applying L'Hôpital's (relevant expressions are continuous in  $z$ ) rule we get:  $\lim_{z\to \infty}z / \tau^2 = \frac{2\sigma_{\mathrm{in}}^2 / 3(1 - 1 / d_{\mathrm{in}})}{1 + 4 / 3K_2^2}$ .
Lemma 29 (0-1 error of a classifier on target). Assume a classifier of the form  $w = l_{1} \cdot w_{\mathrm{in}} + l_{2} \cdot w_{\mathrm{sp}}$  where  $l_{1}, l_{2} \in \mathbb{R}$  and  $w_{\mathrm{in}} = [w^{\star}, 0, \dots, 0]^\top$ , and  $w_{\mathrm{sp}} = [0, \dots, 0, \mathbf{1}_{d_{\mathrm{sp}}} / \sqrt{d_{\mathrm{sp}}}]^\top$ . Then the target accuracy of this classifier is given by  $0.5 \cdot \operatorname{erfc}\left(-\frac{l_{1} \cdot \gamma}{\sqrt{2} \cdot l_{2} \cdot \sigma_{\mathrm{sp}}}\right)$ .
Proof. Assume  $(x,y)\sim \mathrm{P}_{\mathsf{T}}$ . Accuracy of  $w$  is given by  $\mathbb{E}_{\mathrm{P_T}}[(\mathrm{sign}(w^\top x) = y)]$ .
$$
\begin{array}{l} \mathbb {E} _ {\mathrm {P} _ {\tau}} \left[ \operatorname {s i g n} \left(w ^ {\top} x\right) = y \right] = \mathbb {E} _ {\mathrm {P} _ {\tau}} \left[ y \cdot \operatorname {s i g n} \left(w ^ {\top} x\right) = 1 \right] \\ = \mathbb {E} _ {\mathrm {P} _ {\top}} \left[ y \cdot \left(w ^ {\top} x\right) > 0 \right] \\ = \mathbb {E} _ {\mathrm {P} _ {\mathrm {T}}} \left[ y \cdot \left(x ^ {\top} \left(l _ {1} \cdot w _ {\text {i n}} + l _ {2} \cdot w _ {\text {s p}}\right)\right) > 0 \right] \\ = \mathbb {E} _ {\mathrm {P} _ {\mathrm {T}}} \left[ y \cdot \left(\gamma \cdot l _ {1} \cdot y + l _ {2} \cdot \sigma_ {\mathrm {s p}}\right) > 0 \right] \\ = \mathbb {E} _ {z \sim \mathcal {N} (0, 1)} \left[ \left(\gamma \cdot l _ {1} + y \cdot l _ {2} \cdot \sigma_ {\mathrm {s p}} \cdot z\right) > 0 \right] \\ = \mathbb {E} _ {z \sim \mathcal {N} (0, 1)} \left[ y \cdot l _ {2} \cdot \sigma_ {\mathrm {s p}} \cdot z > - \gamma \cdot l _ {1} \right] \\ = \mathbb {E} _ {z \sim \mathcal {N} (0, 1)} \left[ l _ {2} \cdot \sigma_ {\mathrm {s p}} \cdot z > - \gamma \cdot l _ {1} \right] \\ = \mathbb {E} _ {z \sim \mathcal {N} (0, 1)} \left[ z > - \frac {\gamma \cdot l _ {1}}{l _ {2} \cdot \sigma_ {\mathrm {s p}}} \right] \\ \end{array}
$$
Using the definition of erfc function, we get the aforementioned accuracy expression.
Lemma 30. For  $\sigma > 0$  and  $\mu \in \mathbb{R}$ , we have
$$
\begin{array}{l} g (\mu , \sigma) := \mathbb {E} _ {z \sim \mathcal {N} (0, \sigma)} [ \exp (- | \mu + z |) ] (200) \\ = \frac {1}{2} \left(\exp \left(\sigma^ {2} / 2 - \mu\right) \cdot \operatorname {e r f c} \left(- \mu / \sqrt {2} \sigma + \sigma / \sqrt {2}\right) + \exp \left(\sigma^ {2} / 2 + \mu\right) \cdot \operatorname {e r f c} \left(\mu / \sqrt {2} \sigma + \sigma / \sqrt {2}\right)\right) (201) \\ \end{array}
$$
Proof. The proof uses simple algebra and the definition of erfc function.
$$
\begin{array}{l} g (\mu , \sigma) := \mathbb {E} _ {z \sim \mathcal {N} (0, \sigma)} [ \exp (- | \mu + z |) ] \\ = \frac {1}{\sqrt {2 \pi}} \int_ {z} \exp (- | \mu + z |) \cdot \exp \left(- \frac {z ^ {2}}{2 \sigma^ {2}}\right) d z \\ = \frac {1}{\sqrt {2 \pi}} \int_ {- \infty} ^ {\infty} \exp (- | \mu + z |) \cdot \exp \left(- \frac {z ^ {2}}{2 \sigma^ {2}}\right) d z \\ = \frac {1}{\sqrt {2 \pi}} \int_ {- \mu} ^ {\infty} \exp (- \mu + z) \cdot \exp \left(- \frac {z ^ {2}}{2 \sigma^ {2}}\right) d z + \frac {1}{\sqrt {2 \pi}} \int_ {- \infty} ^ {- \mu} \exp (\mu + z) \cdot \exp \left(- \frac {z ^ {2}}{2 \sigma^ {2}}\right) d z \\ = \exp \left(\sigma^ {2} / 2 - \mu\right) \int_ {\frac {- \mu}{\sqrt {2} \sigma} + \frac {\sqrt {2} \sigma}{2}} ^ {\infty} \exp (- z ^ {2}) d z + \exp \left(\sigma^ {2} / 2 + \mu\right) \int_ {- \infty} ^ {\frac {- \mu}{\sqrt {2} \sigma} - \frac {\sqrt {2} \sigma}{2}} \exp (- z ^ {2}) d z \\ = \frac {1}{2} \left(\exp \left(\sigma^ {2} / 2 - \mu\right) \cdot \operatorname {e r f c} \left(- \mu / \sqrt {2} \sigma + \sigma / \sqrt {2}\right) + \exp \left(\sigma^ {2} / 2 + \mu\right) \cdot \operatorname {e r f c} \left(\mu / \sqrt {2} \sigma + \sigma / \sqrt {2}\right)\right) \\ \end{array}
$$