How would I write a function for the following pattern?

Question

\begin{align} Y(0) ={}& 1\\ Y(1) ={}& 2.5\\ Y(2) ={}& 2.5\cdot2.3\\ Y(3) ={}& 2.5\cdot 2.3\cdot 2.1\\ Y(4) ={}& 2.5\cdot 2.3\cdot 2.1\cdot 1.9\\ \vdots\,\,\, \end{align}

How would I solve for something like $Y(1.3)$ or $Y(2.7)$? How would a function for $Y(x)$ be defined?

Answer 1

We may use the Pochhammer symbol (falling factorial ) once $Y(x)$ multiplied by $5^x$ : \begin{align} Y(x)\;5^x&=\left(\frac {25}2\right)_{x}=\frac{\Gamma\left(\frac {27}2\right)}{\Gamma\left(\frac {27}2-x\right)}\\ \\ \end{align}

Answer 2

Notice that \begin{align*} Y(x) &= \frac{25}{10} \cdot \frac{23}{10} \cdot \frac{21}{10} \cdots \frac{27 - 2x}{10} \\[5pt] &= \frac{1}{10^x} \cdot \frac{(26)(25)(24)(23)(22)(21) \cdots (28-2x)(27-2x)}{(26)(24)(22)\cdots (28-2x)} \\ &= \frac{1}{10^x} \cdot \frac{(26)(25)(24)(23)(22)(21) \cdots (2)(1)}{(26)(24)(22)\cdots (4)(2)} \cdot \frac{(26-2x)(24-2x)(22-2x)\cdots (4)(2)} {(26-2x)(25-2x)(24-2x)(23-2x)\cdots (2)(1)} \\ &= \frac{1}{10^x} \cdot \frac{26!}{2^{13} 13!} \cdot \frac{2^{13-x} (13-x)!} {(26-2x)!} \\ &= \boxed{\frac{1}{20^x} \cdot \frac{26!}{13!} \cdot \frac{(13-x)!}{(26-2x)!}}. \end{align*}

In particular, this gives you a closed form for integer $x$ when $x = 0, 1, 2, 3, \ldots, 13$.

But we'd like this to work for $x > 13$ as well (when the above formula is undefined), and for real $x$ instead of just integer $x$. In fact, the factorial function extends to a function $\Pi(z)$ for all complex $z$ except negative integers ($z! = \Pi(z) = \Gamma(z+1)$ where $\Gamma$ is the Gamma function).

Then we get $$ Y(x) = \frac{1}{20^x} \cdot \frac{26!}{13!} \cdot \frac{\Pi(13-x)}{\Pi(26-2x)} $$ We still have a problem when $x > 13$ is an integer: $\Pi(13-x)$ and $\Pi(26-2x)$ are undefined but we expect $Y(x)$ to still have a formula. We can employ the duplication formula which says that $$ \Pi(2z) = \frac{1}{\sqrt{\pi}} 4^z \Pi(z) \Pi(z - \tfrac12) $$ with $z = 13-x$ to obtain \begin{align*} Y(x) &= \frac{1}{20^x} \cdot \frac{26!}{13!} \cdot \frac{\Pi(13-x)}{(\tfrac{1}{\sqrt{\pi}})4^{13-x} \Pi(13-x)\Pi(\tfrac{25}{2} - x)} \\ &= \boxed{\left( \frac{26! \sqrt{\pi}}{13! \;4^{13}} \right) \left(\frac{1}{5^x}\right) \left(\frac{1}{\Pi(\tfrac{25}{2} - x)}\right)}. \end{align*}

The constant term in front is of course $\Pi(\tfrac{25}{2}) = \Gamma(\tfrac{27}{2})$, so this agrees with Raymond Manzoni's answer.