Solving $ 2^\frac{n}{8} – n < 0$

Question

While working on a question on algorithm the equation reduces to

$$2^{n/8} - n < 0$$

How do I go from there ?

Ans : $2≤n≤43$

Edit :

$n$ is an integer, the question can be found here

Answer 1

$2^{n/8}-n<0\implies 2^{n/8}

$2^n < n^8 = 2^{8\log_2 n } $

$n < 8*\log_2 n $

So for $n=1;1 <0$ is false.

For $2^i \le n <2^{i+1} $ verify that $2^i < 2^3*i \le 8\log_2 n $.

That requires $2^{i-3}

Obviously true for $i \le 4$. For $i=5$ we have $2^2 <5$ but this will fail for any higher $i $.

So somewhere between $n=32$ and $n=63$ this fails.

If that was was opaque I really just calculated:

For $i=1: n=2,3 < 8$ is true.

For $i=2:n=4-7 <8*2$ is true.

For $i=3:n=8-15 < 8*3$is true.

$16-31 < 8*4=32$

$32-63 <|> 8*5 =40\le 8*\log_2 n$

So it breaks down in there somewhere.

But $32 < 8*\log_2 32 = 40$

So $n $ must be over $40$ when it breaks.

$41 < 8\log_2 41 = 8*5.3\approx 42.4$ so true for $n $ up to $42$

Try $n=43$

$43 < 8\log_2 43 \approx 43.4$ still true but barely!

$44 > 8\log_2 44 \approx 43.67$

So is true for $2\le n\le 43$.

Answer 2

We have: $$2^{n/8}-n<0$$ There exists only 2 roots to: $$2^{n/8}=n$$ Since there does not exist a closed form solution for $n$ in terms of elementary functions, we can find these roots using a numerical method. I will use the Newton-Raphson method. You can apply the method with a spreadsheet or a more sophisticated software such as MATLAB.

The process is as follows: $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}\tag{1}$$

We have our function: $$f(x)=2^{x/8}-x$$ Evaluating its derivative: $$f'(x)=2^{\frac{x}{8}-3}\cdot \ln(2)-1$$ Substituting into $(1)$: $$x_{n+1}=x_n-\frac{2^{x_n/8}-x_n}{2^{\frac{x_n}{8}-3}\cdot \ln(2)-1}\tag{2}$$ Now, we will apply this method with reasonable initial values $x_0$ (An educated guess of the solution).

First, I tried $x_0=5$, and did multiple iterations. Note that as the iteration number $n\to \infty$, $x_n\to x$. $$\begin{array}{c|c}n&x_n\\\hline0&5\\1&1.00891\\2&1.09996\\3&1.099997\\4&1.099997\end{array}$$ $$x\approx 1.099997030237609$$ Then, I tried $x_0=50$ and got the second root: $$\begin{array}{c|c}n&x_n\\\hline0&50\\1&45.3329\\2&43.7269\\3&43.5609\\4&43.5593\\5&43.5593\end{array}$$ $$x\approx 43.55926043688166$$ From this, you can deduce the signs of the inequality, and obtain: $$1.099997

Answer 3

Okay a smart brute force method.

$n=1: 2 < 1^8=1$ is not true.

$n=2: 2^2=4 < 2^8$ is true and will obviously be true for $n\le 8$ (because $2^n \le 2^8 < n^8$.

So jump to $n= 9$

$2^9 < 9^8 $ as $9^8>8^8=2^{24} $.

So now we can jump to $n=25$

$2^{25} < 25^8$??? Well, $25^8>16^8=2^{32} $ so we can jump to $n=33$.

$2^{33} < 33^8$??? Yes, because $33^8 > 32^8 = 2^{40} $.

So we can jump to $n = 41$

$2^{41 } < 41^8$?? Well...calculator time.

$41^8 =7,984,925,229,121$

$ =7,984,925,229,121=2^{42.86} $

So we can jump to $n=43$

$2^{43}= 8,796,093,022,208$

$43^8= 11,688,200,277,601$

And $ 11,688,200,277,601=2^{43.4} $.

If we test for $n=44$ we get a failure.

And... well, once $2^n >n^8$ then as $n $ increases $2^n$ increases faster than $n^8$ does. So this will not hold for any larger $n$.

As someone pointed out in the comments, it's concave.

Answer 4

Okay. The simplest answer.

Graph $y = 2^{x/8} $. This is a nice bowed exponential curve we are all used to.

Graph the line $y=x $. As one is straight and the other bowed they intersect at 2 points. $2^{n/8} - n < 0$ for $n $ between the points of intersection.

It is pretty clear from the article you cited, you are to take it at face value the points of intersections are between $(1,2) $ and $(43,44) $

This is equivalent to $n < 8\log_2 n $ which can be approximated be trial and error. Especially when one realizes $32 < 8*\log_2 32 =8*5 <40 < 8*6 = 8*\log_2 64=48 <64$.

Answer 5

Considering this problem from an algebraic point of view, let $$f(x)=2^{x/8}-x\qquad f'(x)=2^{\frac{x}{8}-3} \log (2)-1\qquad f''(x)=2^{\frac{x}{8}-6} \log ^2(2)$$ The first derivative cancels for $$x_*=24-\frac{8 \log (\log (2))}{\log (2)}\approx 28.2301$$ $$f(x_*)=\frac{8 \left(1+\log \left(\frac{\log (2)}{8}\right)\right)}{\log (2)}\approx-16.6886$$ SInce the second derivative is always positive, equation $f(x)=0$ shows two roots $x_1$ and $x_2$ between which the function is negative.

Sooner or later, you will learn about Lambert function and the roots are explicitely given by $$x_1=-\frac{8 }{\log (2)}W\left(-\frac{\log (2)}{8}\right)\approx 1.099997030$$ $$x_2=-\frac{8}{\log (2)} W_{-1}\left(-\frac{\log (2)}{8}\right)\approx 43.55926044$$ The Wikipedia page gives series expansions for an "easy" computation of these values.