I am learning this MIT course, which gives this formula

$$O(n^2) = n^{1.5}$$

is there a table to calculate this? like $O(n^{1.5})$, $O(n^{5})$ ?

what x takes would have O(x) give $c \cdot n$ where c is a constant,

  • $\begingroup$ can you citate us teh definition of O? this will immediately give yiou an answer $\endgroup$ – Bulat May 19 at 8:40

The formula is wrong. The notation $f = O(g)$ is asymmetric, and has the meaning $f \in O(g)$. For more, check our reference question on Landau notation. Other relevant reference questions are this one and this one.

  • $\begingroup$ Just saying: The website gets it right, the poster swapped the sides. $\endgroup$ – gnasher729 May 19 at 11:19

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