How should I simplify this product expression?

AI Thread Summary
The discussion focuses on simplifying a product expression that varies based on whether n is even or odd. For even n, the product can be expressed as 2n/(n+1), while for odd n, the denominator changes, complicating the expression. Participants suggest analyzing the product by writing out terms and identifying patterns, with one noting that many factors cancel out. The use of factorials is proposed to further simplify the expression, leading to a final solution that retains only the outer terms. Ultimately, the conversation emphasizes the importance of recognizing patterns and using mathematical induction for proof.
musicgold
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Homework Statement
I need to simplify the following product expression and show it in factorial terms.
Relevant Equations
As shown below I have calculated the value of bn for 10 terms.
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I see that when n is an even number, the product can be represented as ## \frac {2n} {(n+1)} ##. When n is an odd number, the denominator seems to be changing and I am not able to define an expression for it.

How should I go about solving this?

Thanks
 
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The equality also seems to hold for odd n:

##b_3 = 2.3/(3+1) = 6/4 = 3/4##
##b_5 = 2.5/(5+1) = 10/6 = 5/3##

and so on.
 
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Math_QED said:
The equality also seems to hold for odd n:

##b_3 = 2.3/(3+1) = 6/4 = 3/4##
##b_5 = 2.5/(5+1) = 10/6 = 5/3##

and so on.

Oh..I totally missed it. Thanks.
How would a trained mathematician go about analyzing this kind of question? Would they do what I did - analyzing a first few steps?
 
Math_QED said:
The equality also seems to hold for odd n:

##b_3 = 2.3/(3+1) = 6/4 = 3/4##
6/4 = 3/2
 
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musicgold said:
How would a trained mathematician go about analyzing this kind of question?
Write a few of the terms. If I start with ##P_4## (the product of the first three factors), I get
##\frac 4 3 \frac 9 8 \frac {16}{15} = \frac{2^23^34^2}{3\cdot8\cdot15}##
The numerator can already be expressed in terms of factorials, but I don't think the denominator can.
 
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Mark44 said:
Write a few of the terms. If I start with ##P_4## (the product of the first three factors), I get
##\frac 4 3 \frac 9 8 \frac {16}{15} = \frac{2^23^34^2}{3\cdot8\cdot15}##
The numerator can already be expressed in terms of factorials, but I don't think the denominator can.
Yes, the denominator can ... well almost.

Factor the denominator. ##(k-1)(k+1)##Write out a few terms.
 
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musicgold said:
Oh..I totally missed it. Thanks.
How would a trained mathematician go about analyzing this kind of question? Would they do what I did - analyzing a first few steps?
There are many factors that cancel numerator and denominator. I would write down some factors and see what cancels and what is left. Probably a lot cancels and if you are lucky, all you would be left with are at the beginning and the end.
 
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I would do it the way you did, and hope I saw a pattern. Of course, one then has to prove this pattern is correct. For this, you can use induction.
 
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if we take 3 successive terms we ll have
$$\frac{(k-1)(k-1)}{(k-2)k}\frac{kk}{(k-1)(k+1)}\frac{(k+1)(k+1)}{k(k+2)}$$ and after simplifications we are left with the two terms in the edges ##\frac{k-1}{k-2}## and ##\frac{k+1}{k+2}## which will get simplified by the previous and next terms so the total product will have all intermediate terms simplified to 1 and only the two outer-outer terms will survive which are the terms ##\frac{2}{1}## and ##\frac{n}{n+1}## giving the final solution $$\frac{2n}{n+1}$$
 
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  • #10
SammyS said:
Yes, the denominator can [be written as factorials] ... well almost.

Factor the denominator. ##(k-1)(k+1)##

Write out a few terms.
Supposing you are interested in solving this with factorials, you might find it helpful to do the following prior to writing out a few terms.

Rewrite the product as follows.

##\displaystyle b_n = \prod_{k=2}^n {\ \dfrac{k^2}{k^2-1}} ##

## = \dfrac{\displaystyle \left( \prod_{k=2}^n k \right) \left( \prod_{k=2}^n k \right) }{\displaystyle \left( \prod_{k=2}^n (k-1) \right) \left( \prod_{k=2}^n (k+1) \right) } ##​

Write out a few terms, or do the general expanded expression using ellipses (...) .
 
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  • #11
SammyS said:
Supposing you are interested in solving this with factorials, you might find it helpful to do the following prior to writing out a few terms.

Rewrite the product as follows.

##\displaystyle b_n = \prod_{k=2}^n {\ \dfrac{k^2}{k^2-1}} ##

## = \dfrac{\displaystyle \left( \prod_{k=2}^n k \right) \left( \prod_{k=2}^n k \right) }{\displaystyle \left( \prod_{k=2}^n (k-1) \right) \left( \prod_{k=2}^n (k+1) \right) } ##​

Write out a few terms, or do the general expanded expression using ellipses (...) .

Cool!

So if I understand this correctly, all terms except ## \left( \prod_{k=2}^n (k+1) \right) ## can be converted into factorials, because the lowest term of that product will be 3. Is that correct?
 
  • #12
musicgold said:
Cool!

So if I understand this correctly, all terms except ## \left( \prod_{k=2}^n (k+1) \right) ## can be converted into factorials, because the lowest term of that product will be 3. Is that correct?
That can be converted to factorial too, it is equal to ##(n+1)!/2##
 
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  • #13
Delta2 said:
That can be converted to factorial too, it is equal to ##(n+1)!/2##
Great! Thanks
 
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