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I am aware that if I want to write the product of a function $2N^2$ starting with numbers from $1$ to $n$, I write something like this $\prod_{N=1}^{n} 2N^2$, what if I want $N$ to start from 1 and step through odd numbers or even numbers? How do I express this? It is a simple question but I am getting somethings a bit mixed up.

Abdulhameed
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  • That product notation doesn't even make sense. Your question in general is also rather unclear. An explicit example would help a lot... – PrincessEev Dec 30 '18 at 04:37
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    $\prod_{k=0}^N 2(2k+1)^2$ – Zach Hunter Dec 30 '18 at 04:38
  • Its been edited. Thanks – Abdulhameed Dec 30 '18 at 04:38
  • @Zachary Hunter, I guess that is meant for the even function. Cant one use the $\prod$ sign? – Abdulhameed Dec 30 '18 at 04:40
  • You’re overloading the N variable – David Diaz Dec 30 '18 at 04:43
  • Yes. I hope this helps. Also, you notation is still ill defined as you have N as both a variable and a constant. – Zach Hunter Dec 30 '18 at 04:43
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    Your title refers to a sum while the body refers to a product. The question is the same, but they should be consistent. When you write $\prod_{N=1}^{N} 2N^2$ you are using $N$ in two different ways. The top one is the upper limit of the dummy variable while the other two are the dummy variable. It would be much better to write $\prod_{N=1}^{n} 2N^2$ The result will depend on $n$ but not $N$. – Ross Millikan Dec 30 '18 at 04:44
  • Thanks @Ross Millikan, I edited the question, I want to retain the product sign and express the product of $2N^2$ taking $N$ in steps of 2 for odd and even numbers – Abdulhameed Dec 30 '18 at 04:46

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If you want to step through the even numbers, you can write $\prod _{n=1}^k(2(2n)^2)$. If you want to step through the odd numbers, you can write $\prod _{n=0}^k(2(2n+1)^2)$. I don't know of a mathematical equivalent to the computer step as in for i=1 to 11 step 2 other than this.

I have seen the notation $$\sum_{\stackrel {p \le n}{p \text { prime}}}$$ for the sum of some expression over all the primes less than or equal to $n$.

Using the formula for even numbers makes it easier to deal with. In your example we can recognize the factorial and write $$\prod _{n=1}^k(2(2n)^2)=2^k(k!)^2\\ \prod _{n=0}^k(2(2n+1)^2)=2^{k+1}\left(\frac {(2n+1)!}{2^nn!}\right)^2$$

Ross Millikan
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  • @Thanks Ross!, but with the square you have just included, I guess it would not start from 1 and 2 this time. – Abdulhameed Dec 30 '18 at 04:48
  • We can just start with $n=0$ or use $2n-1$ if you want to start from $1$. I'll fix. You need to check the limits each time to make sure you get what you want and no more. The even one does start with $2$. – Ross Millikan Dec 30 '18 at 04:50
  • I want to start with 1, I guess it looks like $\prod _{n=1}^k(2(2n-1)^2)$ and for the even $\prod _{n=1}^k((2n))$ – Abdulhameed Dec 30 '18 at 04:54
  • I would think, as in my answer, that for the evens you want $2(2n)^2$ because you said you wanted the product of $2N^2$ for $N$ even. $2n$ will just give you the product of all the even numbers. That is fine if that is what you want. $2n$ starts $2,4,6,8$ while $2(2n)^2 starts $8,32,72$ – Ross Millikan Dec 30 '18 at 05:04
  • Thanks for the clarification @Ross Milikan – Abdulhameed Dec 30 '18 at 05:07