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In section 1.2.11.2 of Donald Knuth's first volume of TAOCP, he gives an exercise at the end to prove that $g(n) = \Omega (f(n)) \iff f(n) = O(g(n))$ (exercise 13). His solution is simply: "we may take $L = \frac{1}{M}$ in the definitions of $O$ and $\Omega$." However, earlier in the book he states, "Every appearance of $O(f(n))$ means precisely this: There are positive constants $M$ and $n_0$ such that the number $x_n$ represented by $O(f(n))$ satisfies the condition $|x_n|\leq M|f(n)|$ for all integers $n\geq n_0$." Similarly, he says, "The statement $$g(n) = \Omega (f(n))$$ means that there are positive constants $L$ and $n_0$ such that $$|g(n)| \geq L |f(n)| \text{ for all }n\geq n_0."$$

However, if $L$ and $M$ have to be positive constants, then we can't say that $L=\frac{1}{M}$ because the reciprocal of a positive constant is not a positive constant. You can adjust the proof to make it work when $L$ and $M$ have to be positive constants, but I'm wondering if Knuth made an oversight by stating that $L$ and $M$ have to be positive constants because that makes his proof for exercise 13 not work. Additionally, I see no reason that these must be positive constants (and Wikipedia agrees with me, calling the constants positive and real).

If this is in fact an error, how should I report it to claim a hexadecimal dollar in a fictional bank?

Eli Yablon
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  • Exact conditions are $M,L>0$ and $n_0\in\mathbb{N}$, so no problems here. As to last question, then you can simply directly write to author, but be prepared that the answer will be late. I received mine, for example, after about 8 months. – zkutch Apr 03 '22 at 20:38

1 Answers1

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The reciprocal of a positive constant is indeed a positive constant.

Just like $2>0$ and the reciprocal of $2$ is $\frac{1}{2}>0$.

So Knuth didn't make a mistake

nir shahar
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    For some reason I saw positive constant and I thought "integer constant" lmao. Yeah, I was very confused. – Eli Yablon Apr 03 '22 at 23:31