I'm told to prove or disprove that $4^{\sqrt{n}}$ grows faster than $\sqrt{4^n}$ As n tends to infinity.

From my Previous years Calculus I know that if I take the derivative of two functions, and one is bigger, than it must grow faster.

$f(n) = 4^{\sqrt{n}}$

$g(n) = \sqrt{4^n}$

if $f'(n)$ is > $g'(n)$ Then $f(n)$ grows faster than $g(n)$, as $n\to\infty$

My Attempt:

$f'(n) = (\ln(4)\times 4^{\sqrt n)}) /2^{\sqrt n}$

$g'(n) = (\ln(4)\times 4^{n/2}) /2$

But Now I'm not sure if f(n) does grow faster, how would I know that it's first derivative is bigger here, just by plugging in numbers?

  • 2
    $\begingroup$ Hint: $\sqrt{4^n}=4^{n/2}$, and $4^x$ is increasing, so it's enough to compare the growth of $\sqrt{n}$ to the growth of $n/2$. $\endgroup$ – Ian Feb 5 '16 at 16:08
  • $\begingroup$ And then taking the derivative of those two? Where it would be 1/2(sqrt(x) and 1/2, and g(n) being greater so the answer would be false. $\endgroup$ – TTEd Feb 5 '16 at 16:10
  • $\begingroup$ There are different ways; an easier way would be to compute $\lim_{n \to \infty} \frac{\sqrt{n}}{n/2}$, which is not hard. (L'Hopital's rule will work, but there is a still-simpler way by simplifying first...) $\endgroup$ – Ian Feb 5 '16 at 16:14
  • $\begingroup$ Sounds good, thanks. $\endgroup$ – TTEd Feb 5 '16 at 16:15

Hint: Look at the ratio $\dfrac{4^{\sqrt{n}}}{\sqrt{4^n}}$. This is equal to $$\left(\frac{4}{4^{\sqrt{n}/2}}\right)^{\sqrt{n}}.$$


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