Extending $2^n > n $ from set of natural to set of real numbers

Question

I was given a task to prove that $2^n>n$ for any $n \in N \cup \{0\}$. I am aware that this can be solved by induction and that the solution is pretty easy but instead of meddling with induction and loosing time, is it valid to solve this graphically. The problem here is that $2^n$ is not a valid exponential function and $n$ is not a valid linear function since their domains are discrete. Is it possible to somehow extend the domain and prove that the inequality holds for $n \in \mathbb{R^+\cup \{0\}}$. This is probably more time-demanding than induction itself but I would still like to know if it's possible.

Answer 1

Some elementary analysis.

If $f(x) =2^x-x $, then $f'(x) =2^x \ln 2 - 1 $ and $f''(x) =2^x (\ln 2)^2 $.

From this, $f''(x) > 0 $ for all $x$.

$f'(x) = 0$ when $0 = 2^x \ln 2 - 1$ or $2^x = 1/\ln 2$ or $x =(\ln(1/\ln 2))/\ln 2 =(-\ln(\ln 2)/\ln 2 \approx 0.5287663729 = x_0 $.

Therefore, for $x > x_0$, $f'(x) > 0$.

$f(x_0) =2^{x_0}-x_0 =1/\ln 2 -(-\ln(\ln 2)/\ln 2 =\frac{1+\ln(\ln 2)}{\ln 2} \approx 0.9139286679 $.

Therefore, $f(x) > 0$ for $x \ge x_0$.

Answer 2

HINT: Show that the derivative of $2^x-x$ is positive for $x>?$.

Answer 3

One fairly obvious fact about any reasonable extension of $2^n$ to real numbers is that it ought to be strictly increasing; then, if $n+1>x>n$ so must $2^x>2^n$ so if $2^n\geq n+1$, then $$2^x>2^n\geq n+1>x$$ as desired. So you only need $2^n\geq n+1$, which holds for all naturals (and 0)

Answer 4

For a real number $A > 1,$ we always have $A^x \geq x$ as long as $$ A \geq e^{1/e} \approx 1.44466786. $$

Note that $$ \left( e^{1/e} \right)^e = e^1 = e, $$ so $x=e$ is giving equality.

Same fact, we always have $A^x > x$ as long as $$ A > e^{1/e} \approx 1.44466786. $$

Proof: the maximum of $$ \frac{\log x}{ x} $$ occurs when $x=e.$