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Subject: "product of fractions"     Previous Topic | Next Topic
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Conferences The CTK Exchange High school Topic #379
Reading Topic #379
ke_45
Member since May-5-08
May-05-08, 03:08 PM (EST)
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"product of fractions"
 
   Hi

In a book on sequences, there's this product:

(1 - 1/4 ) · (1 - 1/9) · (1 - 1/16) · ... · (1 - 1/nē) = (n + 1) / 2n

How do we derive this product? What general methods do we have to derive products?

(I've tried Google, but the word "product" returns too many irrelevant pages.)

Please note that I'm not asking for you to solve this. Only suggest methods (preferably, with partial demonstration -- like the one shown in the thread on "sum of squares"), and any URLS you know that contains relevant material.

Thanks.


KE


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alexb
Charter Member
2346 posts
May-05-08, 03:15 PM (EST)
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1. "RE: product of fractions"
In response to message #0
 
   First, note that

1 - 1/k² = (k - 1)(k + 1)/k².

Then, write a few terms consecutive terms next to each other and observe that most of the factors involved cancel out. Do the cancelling.


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ke_45
guest
May-06-08, 08:34 AM (EST)
 
2. "RE: product of fractions"
In response to message #1
 
   Thanks, Alex. This did the trick, and I figured it out in a breeze.

However...

...when I naively tried to exptrapolate the idea to a variant of this sequence, I failed.

This time, I used "2" as the numerator of the second term:

(1 - 2/4 ) · (1 - 2/9) · (1 - 2/16) · ... · (1 - 2/nē) = ?

In the first sequence, because of the nice properties of the number 1, this doesn't become obvious, but if you try the same technique, you first rush to a quadratic like so:

1st seq.: 1 - (1/nē) = (nē - 1)/nē = ((n - 1) · (n + 1))/nē

2nd seq.: 1 - (2/nē) = (nē - 2)/nē = ((n - sqrt(2)) · (n + sqrt(2)))/nē

Unfortunately, the terms do not cancel each other this time. (And/or I simply do not have the *algebraic imagination* to express this in a way that'll be conducive to that technique -- a greater possibility ;)).

I didn't want to spend hours in trial-and-error with complex numbers so I came back -- to ask for you to kindly suggest material (on this or other sit's)discussing this issue in further detail.

--

BTW Alex, I absolutely LOVED cut-the-knot.org. Fantastic site! It's been such a pleasure for someone like me (a 40-something math-challenged enthusiast) to discover it. Congratulations! :)


Cheers
KE


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alexb
Charter Member
2346 posts
May-06-08, 08:37 AM (EST)
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3. "RE: product of fractions"
In response to message #2
 
   Thank you for the kind words.

The fact is that very few series or infinite products can be computed in closed form. Those that admit'some kind of a nice formula are exceptions rather than a rule.


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jack202
Member since Sep-30-08
Mar-28-09, 10:46 AM (EST)
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4. "RE: product of fractions"
In response to message #2
 
   You can use the Weierstrass product for the sine function:

sin(x)/x = prod(j=1..+inf) (1 - x^2/(pi^2 j^2))

and evaluate it in x = sqrt(2) pi to have:

(1-2/4)(1-2/9)... = - sin(sqrt(2) pi) / (sqrt(2) pi)


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