# Thread: Product of arithmetic sequence in term of x.

1. ## Product of arithmetic sequence in term of x.

My question is:there is an arithmetic sequence that start from 2x-3 ,and followed by 2x-5,2x-7 and so on, so the diffrence between two terms is -2, and finally what is the product of the terms when reaching the (x-1)th term, in term of x.

My classmate ask me this question.I try to surf the internet but cannot understand or have few resources. Please help me to do this as elementary as possible (step by step) so I can more easily understand. Thanks.

2. ## Re: Product of arithmetic sequence in term of x.

product[(2x-1-2n),{n,1,infinity}] - Wolfram|Alpha Results

This is not a special product that I know of. Wolframalpha doesn't know it either.

4. ## Re: Product of arithmetic sequence in term of x.

Please show me step by step as Wolfram results that you have given seems to have the final product only.

5. ## Re: Product of arithmetic sequence in term of x.

Let's look for a special case. Assume $x\in \mathbb{Z}$ and $x>1$.

Let's start with the $x-1$ term. It is $2x-2(x-1)-1 = 1$

So, you have $(2x-3)(2x-5)\cdots 3\cdot 1$

This is the product of all odd integers less than $2x-1$.

Let's consider the product:

$(2x-2)(2x-4)\cdots 4\cdot 2$

Each of these terms are even. We can factor out a 2 from every term, and we know there are $x-1$ terms. That gives:
$2^{x-1}(x-1)!$

Now, the product you want is given by

$\dfrac{(2x-2)!}{2^{x-1}(x-1)!}$

6. ## Re: Product of arithmetic sequence in term of x.

Thanks!!!Is any products of arithmetic sequence can be solved using this method?Also can I ask you another question:can a factorial of x can be said that is a product of arithmetic sequence for the first x terms starting from x with diffrence of -1?Thanks again!

7. ## Re: Product of arithmetic sequence in term of x.

Your sequence represents odd numbers from 1 to 2x-3;
as example, if x = 5: 1,3,5,7.
Product = 1*3*5*7 = 105

This is known as the "double factorial function" and is
represented using symbol !!
In other words, 7!! = 105

Google "double factorial" and all will become clear.

In doubt? Go to Wolfram and enter f=7!!

8. ## Re: Product of arithmetic sequence in term of x.

Thanks for your information. But I hope to expand it without using double factorial. It is possible the answer above can be presented in a form without factorials?

9. ## Re: Product of arithmetic sequence in term of x.

Originally Posted by lai001
Thanks for your information. But I hope to expand it without using double factorial. It is possible the answer above can be presented in a form without factorials?
Yes, it is possible. Do you like the $\Gamma$ function better than factorials? Which form do you like best?

10. ## Re: Product of arithmetic sequence in term of x.

Isn't gamma function also actually in terms of factorials?If so, what is the difference between 2 forms?

11. ## Re: Product of arithmetic sequence in term of x.

Originally Posted by lai001
But I hope to expand it without using double factorial.
Expand what?

You started off with "help classmate" solving an
obscure representation of "1st n odd numbers";
how can that be "expanded"?

Supply an example.

12. ## Re: Product of arithmetic sequence in term of x.

Originally Posted by lai001
Isn't gamma function also actually in terms of factorials?If so, what is the difference between 2 forms?
No, the $\Gamma$ function satisfies the property that $\Gamma(x) = (x-1)!$ for all positive integers $x$. But, it is a continuous function, not representable by factorials (which are only defined over nonnegative integers).

13. ## Re: Product of arithmetic sequence in term of x.

I meant that using factorial for the final form of solution, not using double factorial.

14. ## Re: Product of arithmetic sequence in term of x.

Oh I see.... so how the above answer express in form of gamma function?Thanks.

15. ## Re: Product of arithmetic sequence in term of x.

Originally Posted by lai001
Oh I see.... so how the above answer express in form of gamma function?Thanks.
One way is

$\dfrac{\Gamma(2x-1)}{2^{x-1}\Gamma(x)} = \dfrac{2^{x-1}\Gamma\left(x-\dfrac{1}{2}\right)}{\sqrt{\pi}}$

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