Unifying the product of odd powers of function with the product of even powers of the same function in one product.

Question

Assume that \begin{equation} \begin{split} f_n(x)&= \begin{cases} \big(g(x)\big)\cdot \big(g(x)\big)^3\cdot \big(g(x)\big)^5\cdots \big(g(x)\big)^{n-1},& n \ \text{even},\\ \big(g(x)\big)^2\cdot \big(g(x)\big)^4\cdot \big(g(x)\big)^6\cdots \big(g(x)\big)^{n-1},& n \ \text{odd}, \end{cases}\\ &=\begin{cases}\displaystyle\prod^{n-1}_{k\ \ odd} \big(g(x)\big)^{k} ,& n \ \text{even},\\ \displaystyle\prod^{n-1}_{k\ \ even}\big(g(x)\big)^{k},& n \ \text{odd}, \end{cases} \end{split} \end{equation} where $g(x)$ is any polynomial of order $m$(with real coefficient). I beleive that it's possible to unify the two cases($n$ odd and even) in one product like $\prod^{n-1}_{k=1}(\cdots)$ or $\prod^{n-2}_{k=0}(\cdots)$ $\big($or like $\prod^{n/2}_{k=1}(\cdots)$ or $\prod^{(n/2)-1}_{k=0}(\cdots)$$\big)$ which gives the desired $f_n(x)$ for each $n$. I hope someone helps me to figure it out.

Answer 1

How about $g(x)^{\lfloor n^2/4\rfloor}$?

Answer 2

How about something like: $$ \prod_{k=1}^{\lfloor \frac n2 \rfloor} g(x)^{n+1-2k} $$

Answer 3

We can actually simplify this considerably. It turns out that if $n$ is even, then $$1+3+...+n-1=\frac{n^2}{4}\tag{1}$$ and if $n$ is odd, then $$2+4+...+n-1=\frac{n^2-1}{4}\tag{2}$$ Now, notice that the function $$\frac{n^2-n}{4}+\frac{1}{2}\bigg\lfloor\frac{n}{2}\bigg\rfloor$$ is equal to $(1)$ for even $n$ and $(2)$ for odd $n$. Thus, we have $$f_n(x)=g(x)^{\frac{n^2-n}{4}+\frac{1}{2}\lfloor\frac{n}{2}\rfloor}$$

Answer 4

you can do much better actually. Let me start with $n=2N+1$ odd, then $$ f_n(x)=g(x)^{\sum_{k=0}^{N}(2k)}=g(x)^{2\sum_{k=0}^{N-1}k}=g(x)^{N(N+1)} $$ and for $n=2N$ even $$ f_n(x)=g(x)^{\sum_{k=0}^{N-1}(2k+1)}=g(x)^{2\sum_{k=0}^{N-1}k+N}=g(x)^{N(N+2)} $$ There are many ways to rewrite this in a single line, one of them is $$ f_n=g(x)^{\lfloor n/2\rfloor\left(\lfloor n/2\rfloor+1+\frac{1+(-1)^n}{2}\right)} $$