I am having a hard time simplifying this product notation:
$$\prod_{x=1}^{(n+1)/2} (2x-1)$$
I would appreciate any help. Thank you!
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7Does this answer your question? Proving formula for product of first n odd numbers – Phil Sep 23 '20 at 16:47
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@Phil sort of, but I'm still a bit confused on how to deal with the upper bound of (n+1)/2 – curlypie99 Sep 23 '20 at 16:51
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2Just replace $n$ with $(n+1)/2$ in the other post. – Phil Sep 23 '20 at 16:52
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1oh ok, thank you! @Phil – curlypie99 Sep 23 '20 at 16:53
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In sums and products the variable (here $x$) is considered to be an integer.
Thus $\prod\limits_{x=a}^b f(x)$ means that $x\in[a,b]\cap\mathbb Z$ or $$a\le x\le b\text{ with }x\in\mathbb Z$$
In our case:
- when $n$ is odd then $n=2p+1\iff \frac{n+1}2=p+1$ and the involved indices are $x=1,2,3,\cdots,p+1$.
- when $n$ is even then $n=2p\iff \frac{n+1}2=p+\frac 12$ and the involved indices are $x=1,2,3,\cdots,p$ since $p$ is the last integer verifying $x\le p+\frac 12$.
E.g. $n=7$ then $x=1,2,3,4$ and for $n=10$ then $x=1,2,3,4,5$
Sometimes the bounds are made explicitly integers using floor or ceil function. In our case this would be $\lfloor\frac{n+1}2\rfloor$ but this is also equivalent to $\lceil\frac n2\rceil$.
zwim
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