Is the growth on an exponential function faster than on a power function? Is this explanation valid?

Question

I'm trying to find simple mathematical proof that lengthy passwords are more secure than complex (with multiple symbols) ones. I don't want to get into the concept of entropy - it is not useful to me atm. I've found this explanation at Security Stack, which it seemed pretty reasonable to me. My question is: is this explanation valid and, if so, is there a limit to it's validity?. My doubt comes from testing it out considering a password with $n$ digits, but restrict to a group of 26 characters (the alphabet) vs a password with fixed 10 digits, with $n$ possible characters:

$$f(n) = 26^n \qquad\text{vs}\qquad f(n) = n^{10}.$$

For these two functions it just doesn't work out as the answer in Security Stack explained. So, how does this work?

I guess the title may be a bit misleading, because this question markes as duplicate does not answer my question! I get that exponential growth is faster than polynomial growth, but why doesn't it apply to these two functions I'm comparing? I mean, what is wrong in my line of thought?

Possible duplicate of How to prove that exponential grows faster than polynomial? — caverac, Jun 21 '18 at 17:00
An exponential has a power series representation $\sum \frac {a^n}{n!}.$ What ever degree of polynomial you choose, the exponential has a higher degree polynomial embedded inside of it. — Doug M, Jun 21 '18 at 17:05
perhaps graphing the functions might help you better understand — The Integrator, Jun 21 '18 at 17:14
I did, @TheIntegrator! That's what actually got me confused, because the results of the power function were higher than the results of the exponential function, but I guess that's because the $x$s I checked weren't large enough, as J.G. further explained. — yyyyyyyan, Jun 21 '18 at 17:19

J.G. · Accepted Answer · 2022-05-09T21:18:49.550

1

It's certainly true that $a^x,\,a>1$ eventually grows faster than $x^b,\,b>0$. (Perhaps the easiest proof is the fact that $\int_0^\infty xe^{-cx}dx=\frac{1}{c^2},\,c>0$ is finite. Thus $\lim_{x\to\infty}xe^{-cx}=0$, so $x^b$ can't keep up with $e^{bcx}$; now take $c=b^{-1}\ln a$.) So $26^x>x^{10}$ for all sufficiently large $x$.

edited May 09 '22 at 21:18

answered Jun 21 '18 at 17:12

J.G.

115,835

score 1 · Answer 2 · answered Jun 21 '18 at 17:31

1

Look at what happens when $x$ increases from $1\text{ million}$ to $1\text{ million}+1.$

$3^x$ gets multiplied by $3.$ And does so every time $x$ increases by $1.$

But $x^{10}$ gets multiplied by something that's very small by comparison to $(1\text{ million})^{10}$ (even though it's very large in absolute terms).

answered Jun 21 '18 at 17:31

Michael Hardy

1

Just curious, but there was another user with either the same name or very similar name. That user had more than 100K reputation points. Are you that former user? If so, may I ask what happened to your reputation points? – Mark Viola Dec 07 '18 at 19:42

Is the growth on an exponential function faster than on a power function? Is this explanation valid?

2 Answers2