Solve $n < e^{6 \sqrt{n}}$

Question

Find for which values of $n \in \mathbb{N}$ it holds that $$n < e^{6 \sqrt{n}}.$$

I tried to use the inequality $(1 + x) \leq e^x$, but from this, I can only find that the inequality holds for $n > 36$. But I need to get $n$ as small as possible.

I also tried the induction on $n$, but I stucked in the induction step. In particular, in showing that $e^{6\sqrt{n}} + 1 \leq e^{6\sqrt{n+1}}$.

I appreciate any help and suggestions.

try deriving. You have already $1<e^6$, so I suspect it is true for every $n$ — Exodd, Oct 09 '20 at 10:38
I will give you a blind guess, which may be the wrong approach; it is just an idea that may or may not lead anywhere. I would let $\displaystyle f(x) = e^{(6\sqrt{x})} ~-~ x$, and look at the behavior of $f'(x).$ — user2661923, Oct 09 '20 at 10:39
Are you allowed to take one more term in the power series, i.e. $e^x \ge 1 + x + \frac{x^2}2$? — player3236, Oct 09 '20 at 10:43
Thank you to all of you. I start focus on the hint from @user2661923 and this give me also the solution (so I do not try the other hints). — Kapur, Oct 09 '20 at 11:18

score 3 · Accepted Answer · answered Oct 09 '20 at 11:31

3

We have $e^{3x} > 3x \ge x$ for $x\ge 0$. Apply this to $x = \sqrt{n}$ to get $e^{3\sqrt{n}} > \sqrt{n}$. Square both sides and you have $e^{6\sqrt{n}} > n$ for all $n\ge 0$.

answered Oct 09 '20 at 11:31

jjagmath

18,214

score 0 · Answer 2 · answered Oct 09 '20 at 11:17

0

(Based on hint from user2661923, see the comments above.)

I set $f(x) = e^{6\sqrt{x}} - x$. The first derivation of $f(x)$ is $$f'(x) = 3 \frac{e^{6\sqrt{x}}}{\sqrt{x}} - 1.$$

We note that $f(1) = e^6 - 1 > 0$. Moreover, it is not hard to check that $f'(x) > 0$ for $x \in [1, +\infty)$. Thus $f$ is increasing on $[1, +\infty)$ and $f(1) > 0$. Therefore $f(x) > 0$ for all $x \in [1, +\infty)$. It implies that $x < e^{6\sqrt{x}}.$

And my question easily follows.

answered Oct 09 '20 at 11:17

Kapur

489

For completeness, I suggest that you edit your answer to demonstrate that $f'(x) > 0$, for all $x > 1.$ One way of doing this is to interpret $f'(x)$ as $3 \times e^{[g(x)]} - 1,$ and explicitly examine the behavior of $g(x) = 6\sqrt{x} - (\log x/2).$ Again, a blind guess is to examine $g'(x)$. – user2661923 Oct 09 '20 at 11:32

score 0 · Answer 3 · answered Oct 09 '20 at 11:33

The inequality holds for all positive $x$, so effectively that a proof seems overkill :-)

(Notice that $0<e^{6\sqrt0}$ and $1<\dfrac6{2\sqrt x}e^{6\sqrt x}$.)

Or using Taylor,

$$n<1+6\sqrt n+18n+\cdots$$

Solve $n < e^{6 \sqrt{n}}$

3 Answers3