Comments on: Morris: Infinite Products http://mathfactor.uark.edu/2009/06/infinite-products/ The Math Factor Podcast Site Sat, 21 Nov 2009 21:45:21 -0600 http://wordpress.org/?v=2.8.5 hourly 1 By: Mark http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-543 Mark Fri, 12 Jun 2009 18:13:29 +0000 http://mathfactor.uark.edu/?p=669#comment-543 a) 1 b) infinity c) 1/infinity a) 1
b) infinity
c) 1/infinity

]]> By: Stephen Morris http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-540 Stephen Morris Tue, 09 Jun 2009 00:44:38 +0000 http://mathfactor.uark.edu/?p=669#comment-540 Brian, thanks for getting the picture pun, it was tricky getting David Beckham and James Bond to pose for me.  I'm hoping for Victoria and Moneypenny next time. <div> <div>Actually what made me smile was the maths.  I love the fact that all of these products cancel completely [spoiler]and yet they all have completely different answers.[/spoiler]Unfortunately I can't afford prizes but anyone who gets this right should reward themselves with a Gucci bag or a Ferrari.  Bryan, you should buy yourself a cake shaped like a Gucci bag or a Ferrari!</div> </div> Brian, thanks for getting the picture pun, it was tricky getting David Beckham and James Bond to pose for me.  I’m hoping for Victoria and Moneypenny next time.

Actually what made me smile was the maths.  I love the fact that all of these products cancel completely Show Spoiler ▼

Unfortunately I can’t afford prizes but anyone who gets this right should reward themselves with a Gucci bag or a Ferrari.  Bryan, you should buy yourself a cake shaped like a Gucci bag or a Ferrari!

]]> By: Andy http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-539 Andy Mon, 08 Jun 2009 23:22:58 +0000 http://mathfactor.uark.edu/?p=669#comment-539 The first product is very clearly 1. If we write the second product as a partial product, then you get: P(n) = 1/2 * 2/3 * 3/4 * ... * (n-1)/n This cancels to get P(n) = 1/n, so taking the limit gives the infinite product as 0. The third product is the inverse of the second, so the partial product is P(n) = n. This doesn't converge, so the infinite product diverges. The first product is very clearly 1.

If we write the second product as a partial product, then you get:
P(n) = 1/2 * 2/3 * 3/4 * … * (n-1)/n
This cancels to get P(n) = 1/n, so taking the limit gives the infinite product as 0.

The third product is the inverse of the second, so the partial product is P(n) = n. This doesn’t converge, so the infinite product diverges.

]]> By: Brian Tristam Williams http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-538 Brian Tristam Williams Mon, 08 Jun 2009 23:12:41 +0000 http://mathfactor.uark.edu/?p=669#comment-538 Ok, never mind, lol I get it. "Infinite Products" picture pun. tyvm Ok, never mind, lol I get it. “Infinite Products” picture pun. tyvm

]]> By: Bryan http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-537 Bryan Mon, 08 Jun 2009 22:43:25 +0000 http://mathfactor.uark.edu/?p=669#comment-537 The first series is obviously equal to 1 because each term is equal to 1. For the second one, you can imagine sliding all the numbers in the denominator over one term to the right to obtain the top series, but with an extra term on the left and right of the series equal to 1/n where 'n' is however far you take the series. So this one is equal to zero because you can make 1/n arbitrarily small for a large enough n. In the third one, you can imagine sliding the numerators to the right to obtain the first series but now with an extra (n/1) term left over. Making 'n' as big as you want means that this series taken to infinity is infinitely large. Do I win a cake? :) Bryan South Africa The first series is obviously equal to 1 because each term is equal to 1.

For the second one, you can imagine sliding all the numbers in the denominator over one term to the right to obtain the top series, but with an extra term on the left and right of the series equal to 1/n where ‘n’ is however far you take the series. So this one is equal to zero because you can make 1/n arbitrarily small for a large enough n.

In the third one, you can imagine sliding the numerators to the right to obtain the first series but now with an extra (n/1) term left over. Making ‘n’ as big as you want means that this series taken to infinity is infinitely large.

Do I win a cake? :)
Bryan
South Africa

]]> By: Brian Tristam Williams http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-536 Brian Tristam Williams Mon, 08 Jun 2009 22:38:33 +0000 http://mathfactor.uark.edu/?p=669#comment-536 Yeah, so a) trends toward 1 :) and b) trends toward 0 and c) trends toward inifinity [<em>and beyond!</em>] being that c) is the reciprocal of b) but... Hmmm... I don't see where you get your smile from :D Yeah, so a) trends toward 1 :) and b) trends toward 0 and c) trends toward inifinity [and beyond!] being that c) is the reciprocal of b) but…

Hmmm… I don’t see where you get your smile from :D

]]> By: tjmathematica http://mathfactor.uark.edu/2009/06/infinite-products/comment-page-1/#comment-535 tjmathematica Mon, 08 Jun 2009 21:54:40 +0000 http://mathfactor.uark.edu/?p=669#comment-535 a) lim(n->infinity) PRODUCT(n/n) = 1 b) lim(n->infinity) PRODUCT(n/(n+1)) < 1 value starts at 1/2 and decreses c) lim(n->infinity) PRODUCT((n+1)/n) > 1 value starts at 2 and increses a) lim(n->infinity) PRODUCT(n/n) = 1
b) lim(n->infinity) PRODUCT(n/(n+1)) < 1
value starts at 1/2 and decreses
c) lim(n->infinity) PRODUCT((n+1)/n) > 1
value starts at 2 and increses

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