# Is there any way to find out $x! = n$ by hand?

I've given myself this question but I can't see how to get it. I'm actually not taking calculus I'm in high school taking Algebra 2 but I've learned a bit or two from this site.

I did:

$x! = 10$

Trying to solve for $x$ but I had trouble doing so. Can anyone give me a hand here?

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There is no integer $x$ whose factorial is $10$. Indeed, since factorials form a strictly increasing sequence and $4!=24$, if 10 were a factorial it would be one of $1!$, $2!$ or $3!$. But it isn't. – Mariano Suárez-Alvarez Apr 2 '12 at 23:31
However, one can extend the factorial function to a function of arbitrary reals (or even complex numbers) using the gamma function. There won't be a closed solution to this though (or I'll eat my shoe). – anon Apr 2 '12 at 23:32
@anon, the chances of that being the answer being sought are vanishingly small, don't you think? – Mariano Suárez-Alvarez Apr 2 '12 at 23:33
I suspect the question is $10!$, which is not a very difficult mental arithmetic exercise – Henry Apr 2 '12 at 23:44
@anon routine calculations show that $z = \frac{9 \pi^2 + 8 \pi - 45}{5 \pi}$ satisfies $\Gamma(z) = 10$. Bon appetit! – guy Apr 3 '12 at 1:07

Wolfram alpha gives the roots of $\Gamma(x+1)=10$ as $-4.995806, -4.016334, -2.947296, -2.097191, -1.095325, 3.390078.$

One root of $\Gamma(y)=x$ is approximated by $\frac{L(x)}{W(\frac{L(x)}e)}+\frac{1}2$ where $L(x)=\ln(\frac{x+c}{\sqrt{2\pi}})$, $c\approx0.036534$ and $W(x)$ is the principal branch of the inverse of $xe^x$. $W(x)$ can be approximated by selecting an intial $w_0$, and find succesive approximations $w_{j+1}=w_j-\frac{w_je^{w_j}-z}{e^{w_j}+w_je^{w_j}}$. $e^x$ and $\ln {1+x}$ can be found using $1+x+ \frac{x^2}/{2!}+\frac{x^3}{3!}...$ and $x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+...$ respectively. This can be done by hand but is tedious. http://mathforum.org/kb/message.jspa?messageID=342551&tstart=0

EDIT: The expansion for $\ln(x+1)$ only works if $|x|<1$. Otherwise use $\ln x\approx \frac{\pi}{2M(1,\frac{4}s}-m\ln 2$ with $M(1,\frac{4}s)=$ thearithmetic geometric mean of $1$ and $\frac{4}s$ ,$s=x2^m$ and $m$ any chosen integer (larger $m$ give a closer approximation).

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If one function doesn't work, you can always create a new one that does.

Choose a small number, say $d$ ($0\lt d \lt 1$), and define a factorial-like function $$F(k)=d(d+1)(d+2)\dots(k+d-1)(k+d).$$

After a bit of trial and error, if we take $d=\frac{16}{25} = 0.64$, we find that $F(3) = 0.64 \times 1.64 \times 2.64 \times 3.64 = 10.086$
to five significant digits, which is fairly close to 10. Other values of $d$ may get even closer.
It's not quite factorial, but is in the spirit of factorial.

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If you want to get a good numerical answer, get a calculator (such as the better HP calculators) that have both the Gamma function and a numeric solver.

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If you allow for approximate solutions, in other words, you interpret "=" as meaning "approximately equal to", then you might consider 3 as a solution, since 3!=6, and 4!=24.

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Then how about 4!! – The Chaz 2.0 Apr 3 '12 at 1:07
@TheChaz 4 might work also (but I don't see how 4!! will work). Do you want only an over-approximation, with an over-approximation satisfying (10-x)<0? Or do you want only an under-approximation with an under-approximation satisfying (10-x)>0? Do you want both over and under approximations? Which sort of metric do you think best suited here for determining how close a positive integer lies to n where n!=10 in say this graph en.wikipedia.org/wiki/File:Factorial_interpolation.png? – Doug Spoonwood Apr 3 '12 at 1:50
Oh I don't know. I was thinking of asking you the same question about metrics. Whatever you decide on, it should be that d(4!!, 10) = d(8, 10) < d(3!, 10), d(4!, 10) :) – The Chaz 2.0 Apr 3 '12 at 1:56