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I have two functions $f(n) = n^{1.001}$ and $g(n) = nlog_2n$ and I want to determine if the inequality $f(n) < c*g(n)$ is true, where $f(n), g(n), c > 0$.

To do this, I am using the following property:

$f(n) < c*g(n)$ if and only if $log(f(n)) < logc + log(g(n))$

So:

$1.001*log_2n < log_2c + log_2{(nlog_2n)}$

By setting c to 1:

$1.001*log_2n < log_2n(nlog_2n)$

I am not sure how can I solve this inequality so that I can find what the value of 'n' is.

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  • $\begingroup$ It's not true (no matter the $c$), and the reason is simple. $f(n)/g(n)\to \infty$ when $n\to \infty$ $\endgroup$
    – Jakobian
    Mar 11 '19 at 22:45
  • $\begingroup$ Would it be correct to say that g(n)/f(n) → 0 when 𝑛→∞? Since f(n) grows faster than g(n)? $\endgroup$
    – ceno980
    Mar 12 '19 at 0:43
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\begin{align} \lim_{n \to \infty} \frac{f(n)}{g(n)} &=\lim_{n \to \infty} \frac{n^{1.001}}{n \log_2 n} \\ &= \lim_{n \to \infty} \frac{n^{0.001}}{\log_2 n} \\ &= \ln 2 \lim_{n \to \infty} \frac{0.001 n^{-0.999}}{\frac1n} \\ &= 0.001 \ln 2 \lim_{n \to \infty } n^{0.001} \\ &= \infty \end{align}

Hence $\frac{f(n)}{g(n)}$ can't be bounded by a constant.

Also for your question in the comment, yes, $\lim_{n \to \infty}\frac{g(n)}{f(n)}=0$

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