Finding an $a$, such that $\forall x(x^x-a\cdot x!=0).$

Question

I've been finding myself wondering about this equation for a long time, however due to my limited math knowledge, I can't solve or even determine if there is a solution to that equation.

So I ask: is there an equation or number that can satisfy that?

you can approximate it with Stirling's. No reason to expect closed form. https://en.wikipedia.org/wiki/Stirling%27s_approximation — Will Jagy, Sep 22 '15 at 19:09
Are you asking for an $a$ such that the equation is $0$ for all $x$? If no, then $a=1,x=1$ works. — YoTengoUnLCD, Sep 22 '15 at 19:12
For the first case, an a such that the equation is 0 for all x, after reading the other responses though, I must say that I'm in the last high school year, so my math knowledge is indeed limited, but I will try to follow along the responses — Henke, Sep 22 '15 at 19:14

score 1 · Answer 1 · answered Sep 22 '15 at 19:54

1

$\mathbf{Proposition}$

$\not \exists a\in \Bbb R\forall x(x^x=a\cdot x!$).

$\mathbf{Proof:}$

We are looking for an $a$ that works for all $x$'s. Let's look at two different particular cases: $x=2$, $x=1$.

Case 1: $x=2\implies x^x=4; a\cdot x!=2a \implies a=2$.

Case 2: $x=1\implies x^x=1; a\cdot x!=a\implies a=1$. As these two particular cases require different values of $a$, an $a$ that satisfies all the cases does not exist.

answered Sep 22 '15 at 19:54

YoTengoUnLCD

13,384

But is there an equation that can satisfy all the cases? – Henke Sep 22 '15 at 21:42
What do you mean? I showed that for both cases you had only one choice of $a$, and these were distinct. – YoTengoUnLCD Sep 22 '15 at 21:46
Yeah, I got it now, thanks for the help – Henke Sep 22 '15 at 21:57

Will Jagy · Accepted Answer · 2015-09-22T19:45:03.087

from https://en.wikipedia.org/wiki/Stirling%27s_approximation#Speed_of_convergence_and_error_estimates we get an approximation that gets pretty good as $x$ gets large, $$ \frac{x^x}{x!} \approx \frac{e^x}{\sqrt{2 \pi x} \left( 1 + \frac{1}{12 x} + \frac{1}{288 x^2} - \frac{139}{51840 x^3} - \frac{571}{2488320 x^4} \right)} $$

Quite good, even with small values:

x   x^x                          x!      x^x / x!
1   1                             1       1                   approx 1.00050078212357
2   4                             2       2                   approx 2.000042039543573
3   27                            6       4.5                 approx 4.500013477631439
4   255.9999999999999            24      10.66666666666666   approx 10.66667448781807
5   3124.999999999999           120      26.04166666666666   approx 26.0416730195439
6   46656.00000000003           720      64.80000000000004   approx 64.80000640706673
7   823542.9999999994          5040     163.4013888888888   approx 163.401396402745
8   16777215.99999998         40320     416.1015873015867   approx 416.1015971490262
9   387420489.0000001        362880    1067.627008928572   approx 1067.62702298249
10   10000000000            3628800    2755.73192239859    approx 2755.731943854943
11   285311670610.9999     39916800    7147.658895778216   approx 7147.658930376819
12   8916100448255.988    479001600   18613.92623376621   approx 18613.92629213758
13   302875106592253.4   6227020800   48638.84613847017   approx 48638.84624076476

Finding an $a$, such that $\forall x(x^x-a\cdot x!=0).$

2 Answers2