anonymous
  • anonymous
Any hints on how to process product notation?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\prod_{something}^{something} something\]
DebbieG
  • DebbieG
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...
DebbieG
  • DebbieG
\[\Large\prod_{n=1}^{10} 2n=2\cdot4\cdot6\cdot8\cdot10\cdot12\cdot14\cdot16\cdot18\cdot20\]

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anonymous
  • anonymous
\[\prod_{n=1}^{4} n^2 = 1^2\times2^2\times3^2\times4^2\] 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?
DebbieG
  • DebbieG
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
DebbieG
  • DebbieG
Oh... duh..... of course it does... it's not that hard. It's just the rules of exponents.
DebbieG
  • DebbieG
Maybe that's not what you're looking for....? But that seems like a pretty easy way to evaluate it.
DebbieG
  • DebbieG
<-- needs more coffee.
anonymous
  • anonymous
thanks, that helps
anonymous
  • anonymous
is there a way to express a product as a summation?
anonymous
  • anonymous
possible but impractical?
DebbieG
  • DebbieG
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?)
anonymous
  • anonymous
how about a more complex product like \[\prod_{n=1}^{10} n^2+1\]
terenzreignz
  • terenzreignz
Of course there is :) Let\[\Large P = \prod_{n=1}^ka_n \] Then... \[\Large P = e^{\ln P}\] \[\Large \ln P = \ln \left(\prod_{n=1}^ka_n \right)\]\[\Large = \sum_{n=1}^k\ln(a_n)\]
terenzreignz
  • terenzreignz
Then just raise e to that sum.
terenzreignz
  • terenzreignz
Logarithms: They turn products into sums... XD
anonymous
  • anonymous
wow, I had not thought of that.
terenzreignz
  • terenzreignz
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 ^_^
anonymous
  • anonymous
it seems like it would work out as long as both the product and the sum are negative

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