How to solve the equation $\sqrt{\frac{\sqrt{x!}\times \sqrt{x!!}\times \sqrt{x!!!}}{\sqrt{\left( x-2 \right)!+x}}}=12$

Question

Consider the following equation,

$$\sqrt{\frac{\sqrt{x!}\times \sqrt{x!!}\times \sqrt{x!!!}}{\sqrt{\left( x-2 \right)!+x}}}=12$$ I tried first to eliminates all radicals using squring both sides to get the following , $$\frac{\sqrt{x!}\times \sqrt{x!!}\times \sqrt{x!!!}}{\sqrt{\left( x-2 \right)!+x}}={{12}^{2}}\Leftrightarrow \frac{\sqrt{x!\times x!!\times x!!!}}{\sqrt{\left( x-2 \right)!+x}}=\frac{x!\times x!!\times x!!!}{\left( x-2 \right)!+x}={{12}^{4}}$$

Now how will continue attacking this problem?

I suppose that $x$ must be $\geq 2$ since $(-1)!=\Gamma(0)$ is undefined. Also: Does $x!!!$ denote a multifactorial? — Maximilian Janisch, Feb 22 '20 at 10:39
Okay so since your function is increasing in $x$, you can just try out all integers until your function is greater than $12^4$ — Maximilian Janisch, Feb 22 '20 at 10:49
Already for $x=3$, we have $x!!!=720!$, which has many sizeable prime factors that will not be cancelled by the denominator, and it's much too big anyway. — John Bentin, Feb 22 '20 at 11:04
@JohnBentin $x!!!$ is a multifactorial It doesn't mean $((x!)!)!$. — saulspatz, Feb 22 '20 at 11:36
You can try $x = 10$ which gives a value of nearly $10^8$, much larger than $12^4$. There are only a few positive integers to try that are less than $10$, from which you can quickly get $x = 6$. — Toby Mak, Feb 22 '20 at 11:39
I did it this way. $x=4$ is too small. $x=5$ gives a factor of $5$ in the numerator, but not the denominator. $x=6$ works. — saulspatz, Feb 22 '20 at 11:43

score 1 · Answer 1 · answered Feb 22 '20 at 16:23

Let's say we have next to no intuition for the size of a potential solution, and don't want to try out small numbers by hand. Then we can use the Gamma function in an extension of the multifactorial to make this a continuous function so that a computer can more-easily help us hunt for the root.

By the calculations in the OP, with the equation with $12^4$, we can see at the very least that the solution probably isn't bigger than $12^2$, and likely much less. Let's just start looking around $150$: Wolfram|Alpha says that $45.1$ would be a better guess. If you start at $45$, it says that something around $6.2$ would be better. If you start at $6$, it's still thinking around $6.2$ (but these continuous functions are hard to approximate), so we should probably check around $6$ or $7$.

If we happen to choose $7$ first, it says the value is about $18.5$. And if we ask for the value at 6, we find that it is indeed $12$.

If you add the option MaxIterations -> 1000, then it will instantly give out $6.211$. Interestingly, the Gamma function "rewriting" of the original function is not $=0$ at $6$, but indeed $=0$ at a number strictly larger than $6$ (see plot) — Maximilian Janisch, Feb 23 '20 at 10:40

How to solve the equation $\sqrt{\frac{\sqrt{x!}\times \sqrt{x!!}\times \sqrt{x!!!}}{\sqrt{\left( x-2 \right)!+x}}}=12$

1 Answers1