Suppose $$f(x)=x!$$ for all $x$ in $N$,and $f$ is also continuous in $x>0$,does this mean $f(x)=\Pi (x)$ for all $x>0$?
Asked
Active
Viewed 149 times
0
-
What would that even mean if $x$ wasn't a natural number? – Randall Sep 20 '19 at 14:17
-
The Pi function? – Sasuke_Bird Sep 20 '19 at 14:19
-
Then that should be in your question: does that mean $f(x) = \Pi(x)$ for all $x > 0$? – Randall Sep 20 '19 at 14:21
-
See also https://en.wikipedia.org/wiki/Bohr%E2%80%93Mollerup_theorem – J.G. Sep 20 '19 at 14:21
-
If that's your question, then no. You can take $\Pi(x)$ and wrinkle it continuously between integers and still get a continuous function but it's not $\Pi(x)$ anymore. – Randall Sep 20 '19 at 14:22
-
1Even smoothness, concavity, and the functional equation are not sufficient to define the Gamma function, as pointed out in the above duplicate. – Simply Beautiful Art Sep 20 '19 at 14:31
3 Answers
1
Do you see why the following question has a negative answer?
If $f$ is continuous on $(0,+\infty)$ and $f(n)=n$ for all natural numbers $n$, must $f(x)=x$ for all $x$?
Of course not, because $f(x)=\cos(2\pi x)x$ has exactly the same property, but surely isn't equal to $x$ everywhere.
Your question has a negative answer for the same reason: knowing only what it does on integers and nothing else (beyond continuity) is not going to make your function unique.
Randall
- 18,542
- 2
- 24
- 47
1
There are infinitely many continuous functions $\mathbb{R} \rightarrow \mathbb{R}$ that satisfy $f(x)=x! \space \forall x \in \mathbb{N}$.
Start with one such function $g(x)$ and then add any continuous function that is zero at all integers. For example you can create an infinite family $g_n(x)$ as follows:
$g_n(x) = g(x) + \sin (n \pi x)$
gandalf61
- 15,326
0
Strictly speaking, the question doesn't make sense, as the factorial operator is only defined for natural numbers. However: You might want to look up the Gamma function.
PMar
- 1