Question

Any hints on how to process product notation?

Answer 1

\[\prod_{something}^{something} something\]

Answer 2

Uhmmm.... take the product of all the somethings on the right, run over the something from to the bottom to the something on the top? LOL... I'm not sure what you're asking... sorry...

Answer 3

\[\Large\prod_{n=1}^{10} 2n=2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16\cdot18\cdot20\]

Answer 4

\[\prod_{n=1}^{4} n^2 = 1^2\times2^2\times3^2\times4^2\]
I'm looking for a way to find out what they are without multiplying everything out manually or entering the product into anywhere, maybe a way to simplify it to a sum?

Answer 5

OK, this is interesting. I bet I learned this somewhere in the deep, dark past and have forgotten... lol.... but this is the product:

\[\prod_{n=1}^{k} n^2 = 1^2\times2^2\times...(k-2)^2\times(k-1)^2\times k^2\]

Which I entered on Wolfram and tried just playing around, and have discovered that:

\[\prod_{n=1}^{k} n^2 = (k!)^2\]

Cool, huh? Now.... how to do about proving that, or putting some intuition behind it..........? I'd have to think about it some more.

Just try it with different values of k:

http://www.wolframalpha.com/input/?i=product+k^2+for+k%3D1+to+8%2C+%288!%29^2

Answer 6

Oh... duh..... of course it does... it's not that hard. It's just the rules of exponents.

Answer 7

Maybe that's not what you're looking for....? But that seems like a pretty easy way to evaluate it.

Answer 8

<-- needs more coffee.

Answer 9

thanks, that helps

Answer 10

is there a way to express a product as a summation?

Answer 11

possible but impractical?

Answer 12

Well, if you look at it as a polynomial in k...

\[\Large \prod_{n=1}^{k} n^2 = k^2(k-1)^2(k-2)^2....1\]
\[\Large = k^2(k-2k+1)(k-4k+4)....1\]
etc... you can expand that out, for any particular value of k....

E.g., if k=3 you get \(\Large k^6-6 k^5+13 k^4-12 k^3+4 k^2\)

If k=4 you get \(\Large k^8-12 k^7+58 k^6-144 k^5+193 k^4-132 k^3+36 k^2\)

(thank you Wolfram, lol)

But good Lord, I don't think that's easier to evaluate then \((k!)^2\), lol... nor do I know what the algorithmic way is to cook up that polynomial (although I suspect there is one... something related to binomial coefficients, maybe?)

Answer 13

how about a more complex product like \[\prod_{n=1}^{10} n^2+1\]

Answer 14

Of course there is :)
Let\[\Large P = \prod_{n=1}^ka_n \]
Then...
\[\Large P = e^{\ln P}\]
\[\Large \ln P = \ln \left(\prod_{n=1}^ka_n \right)\]\[\Large = \sum_{n=1}^k\ln(a_n)\]

Answer 15

Then just raise e to that sum.

Answer 16

Logarithms: They turn products into sums...
XD

Answer 17

wow, I had not thought of that.

Answer 18

But then again, this'd probably only work out right if every \(a_n\)is positive...
Might work out for negatives, but then again, nothing's perfect ^_^

Answer 19

it seems like it would work out as long as both the product and the sum are negative