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The restrictions on the function in the statements are the following:

  • The functions are positive
  • They are monotonically increasing
  • $f,g: [1,\infty) \to [1, \infty)$
  • They go to $+\infty$ when $x \to +\infty$
  • All of the statements below imply $x \to +\infty$

The statements are:

  1. If $f(x) \sim g(x)$, then $g(x) \sim f(x)$
  2. If $f(x) = o(g(x))$, then $g(x)=o(f(x))$
  3. If $f(x) = O(g(x))$, then $g(x)=O(f(x))$
  4. If $f(x) \sim g(x)$, then $(f(x))^2 \sim (g(x))^2$
  5. If $f(x) \sim g(x)$, then $\ln (f(x)) \sim \ln (g(x))$
  6. If $f(x) = O(g(x))$, then $\ln(f(x)) = O(\ln(g(x))$
  7. If $f(x) \sim g(x)$, then $2^{f(x)} \sim 2^{g(x)}$
  8. If $f(x) = o(g(x))$, then $2^{f(x)} = o(2^{g(x)})$
  9. If $f(x) = O(g(x))$, then $2^{f(x)} = O(2^{g(x)})$

The options that are given in this exercise change all the time, so here are the options that I selected to be correct from my previous answer that are not in the above list:

  1. If $f(x)=o(g(x))$, then $(f(x))^2 = o((g(x))^2)$
  2. If $f(x) = O(g(x))$, then $(f(x))^2 = O((g(x))^2)$

My initial answer, which was incorrect, is that the following set has the correct statements: $\{1, 4, 10, 11, 5, 8, 9\}$. It also, looks like for the current options, $2$ and $3$ are correct, as they are equivalent to $10$ and $11$. So really, just the options $6$ and $7$ are false in my view. But that is an incorrect answer.

For $7$ and $6$ I have found the counterexamples to be: $x$ and $x+1$, $2x$ and $x$, respectively.

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1 Answer 1

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Always assume $x \to +\infty$:

(6) is true: $\frac{f}g < M$ means $\ln f - \ln g < \ln M$, so $\frac{\ln f}{\ln g} < \frac{M}{\ln g} +1 < 2$ once $\ln g$ gets big enough.

(9) is not true: Take $f(x)=x+\log_2 x, g(x) = x$, then you get $\frac{2^f}{2^g}=x$.

The others are correct, IMO

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