MHB Determining the growth rate of a function

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The discussion focuses on determining the growth rate of a function, specifically evaluating the expression involving \(\frac{3(n+1)^{\frac{2}{3}}}{2}-\frac{3(1)^{\frac{2}{3}}}{2}\). The conclusion reached is that the growth rate is \(\Theta(n^{\frac{2}{3}})\), but the initial poster is criticized for not providing a clear explanation of their calculations. It is noted that while the final result is correct, the lack of context and detail in the explanation is problematic. Additionally, it is suggested that the first line may be part of a larger problem that should have been shared for better clarity. Overall, the conversation emphasizes the importance of thorough explanations in mathematical discussions.
DanSlevin
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I'm trying to figure out the growth rate of a function. Below is what I believe to be the solution, but I'm wondering if I've properly taken into account all the factors necessary, so I wanted to see if this appears correct.

$$\Large\frac{3(n+1)^{\frac{2}{3}}}{2}-\frac{3(1)^{\frac{2}{3}}}{2}$$

$$\Large\frac{3}{2}((n+1)^{\frac{2}{3}}-1) $$

$$\Large\Theta(n^{\frac{2}{3}}) $$
 
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DanSlevin said:
I'm trying to figure out the growth rate of a function. Below is what I believe to be the solution, but I'm wondering if I've properly taken into account all the factors necessary, so I wanted to see if this appears correct.

$$\Large\frac{3(n+1)^{\frac{2}{3}}}{2}-\frac{3(1)^{\frac{2}{3}}}{2}$$

$$\Large\frac{3}{2}((n+1)^{\frac{2}{3}}-1) $$

$$\Large\Theta(n^{\frac{2}{3}}) $$

It does not appear correct because you provide no explanation of how you get from the first line to the last or indeed what the relationship between the expression on the first line is with that on the last.

It is indeed the case that

\[\large \left[ \frac{3(n+1)^{2/3}}{2}-\frac{3(1)^{2/3}}{2}\right]\in \Theta(n^{2/3}) \]

but I won't say your explanation is inadequate because it is not an explanation at all. Also the form of your first line suggests that this is part of a larger problem, which you really should have posted.

CB
 

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