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notes: fix mistake in ex-2 computation

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This commit is contained in:
Michele Guerini Rocco 2020-07-06 16:29:22 +02:00
parent d8a00dbd10
commit 527f1fcfce
Signed by: rnhmjoj
GPG Key ID: BFBAF4C975F76450
1 changed files with 2 additions and 2 deletions
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notes/sections/2.md
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@ -249,8 +249,8 @@ $W$ both sides.
\begin{align*} \begin{align*}
\frac{5 \sqrt{2 \pi}}{12 \sqrt{x}} e^{-8x} = 10^{-D} \frac{5 \sqrt{2 \pi}}{12 \sqrt{x}} e^{-8x} = 10^{-D}
& \thus \left(\frac{12}{5}\right)^2 \frac{x}{2 \pi} e^{16x} = 10^{2D} \\ & \thus \left(\frac{12}{5}\right)^2 \frac{x}{2 \pi} e^{16x} = 10^{2D} \\
& \thus 16x e^{16x} = \left(\frac{5 \pi}{9}\right)^2 10^{2D + 1} \\ & \thus 16x e^{16x} = \frac{5 \pi}{9} 10^{2D + 1} \\
& \thus x = \frac{1}{16} W\left(\left(\frac{5 \pi}{9}\right)^2 10^{2D + 1}\right) & \thus x = \frac{1}{16} W\left(\frac{5 \pi}{9} 10^{2D + 1}\right)
\end{align*} \end{align*}
The smallest integer which satisfies the inequality is then The smallest integer which satisfies the inequality is then
$$ $$