How should I simplify this product expression?

In summary, the conversation discusses a pattern in a product involving even and odd numbers, where the product can be represented as ## \frac {2n} {(n+1)} ## for even numbers and the denominator seems to change for odd numbers. The conversation also discusses how a trained mathematician might approach analyzing this question, including using induction and expressing the product in terms of factorials. Ultimately, it is determined that all terms in the product can be converted to factorials.
  • #1
musicgold
304
19
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.
1569876566334.png

1569877136336.png


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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  • #2
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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  • #3
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?
 
  • #4
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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  • #5
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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  • #6
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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  • #7
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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  • #8
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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  • #9
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
 

1. How do I know if a product expression can be simplified?

A product expression can be simplified if it contains like terms, which are terms with the same variables and exponents. If there are no like terms, the expression cannot be simplified.

2. What is the first step in simplifying a product expression?

The first step is to factor out the greatest common factor (GCF) of all the terms in the expression. This will help to reduce the number of terms and make the expression easier to work with.

3. Can I combine terms with different variables in a product expression?

No, only terms with the same variables and exponents can be combined. If there are terms with different variables, they cannot be simplified and will remain separate in the expression.

4. How do I simplify a product expression with exponents?

To combine terms with exponents, you need to use the rules of exponents. For example, when multiplying two terms with the same base, you can add the exponents. If the exponents are different, you can use the power rule to simplify the expression.

5. Is it necessary to simplify a product expression?

In some cases, simplifying a product expression can make it easier to work with and understand. However, if the expression is already in its simplest form and cannot be further simplified, then it is not necessary to simplify it further.

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