Ox Hugo
Here’s a link to the Circle.
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git checkout
is deprecated.
git checkout --<file> will discard working tree changes.
recall that your working-tree is working-directory. similarly there is a staging area and a place to commit things.
finally, there is also the upstream repository that contains a copy of the code should shit go to fuck.
que
q3
LaTeX formatted equation: \( E = -J \sum_{i=1}^N s_i s_{i+1 }\)
not nested
markdown
Q4
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Which is larger asymptotically as \(n \rightarrow \infty\)? \[2^n \ll n!\] OR \[2^n \gg n!\] Give a proof by induction.
We claim that \[ \boxed{2^n \ll n!} \] for sufficiently large \(n\).
Proof by induction:
We will prove that \(n! > 2^n\) for all \(n \geq 4\).
Base case: \(n = 4\) \[ 4! = 24 > 16 = 2^4 \]
Inductive step: Assume \(k! > 2^k\) for some \(k \geq 4\). We must show \((k+1)! > 2^{k+1}\).
\begin{align*} (k+1)! &= (k+1) \cdot k!\\ &> (k+1) \cdot 2^k \qquad\text{(by inductive hypothesis)}\\ &> 2 \cdot 2^k \qquad\text{(since }k+1 > 2\text{ for }k \geq 4\text{)}\\ &= 2^{k+1} \end{align*}
Therefore, by induction, \(n! > 2^n\) for all \(n \geq 4\).<br><br>
Asymptotic analysis: The ratio \(\frac{n!}{2^n}\) grows without bound: \[ \frac{n!}{2^n} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdots n}{2^n} = \frac{1}{2} \cdot \frac{2}{2} \cdot \frac{3}{2} \cdot \frac{4}{2} \cdots \frac{n}{2} \]
For \(n \geq 5\), each factor \(\frac{k}{2}\) for \(k \geq 3\) is \(> 1\), and we have infinitely many such factors as \(n \to \infty\). Thus \(\frac{n!}{2^n} \to \infty\), confirming \(n! \gg 2^n\).
He puts his claw against the divider. “Fist my bump.”
Andy Weir, Project Hail Mary
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