Find product of a fraction

[wolly]

Hi I want to calculate this product:


[math]\prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}[/math]
How do You Express this product as a factorial?If not do You use double factorial in these 2 fractions?
Also If You apply the double factorial îs there a proof for [math]n!![/math]. ?

 

[mario99]

Hi I want to calculate this product:


[math]\prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}[/math]
How do You Express this product as a factorial?If not do You use double factorial in these 2 fractions?
Also If You apply the double factorial îs there a proof for [math]n!![/math]. ?

Yes, you can express it as factorial, but look at the denominator: when [imath]k = 1, 2(1)+2 = 4[/imath], not [imath]2[/imath].

Do you mean:

[imath]\displaystyle \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k)}[/imath]

?

 

[wolly]

Yes, you can express it as factorial, but look at the denominator: when [imath]k = 1, 2(1)+2 = 4[/imath], not [imath]2[/imath].

Do you mean:

[imath]\displaystyle \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k)}[/imath]

?

No the denominator is [math]2k+2[/math] and not [math]2k[/math]!
Also how can the numerator be used as factorial?

 

[wolly]

@mario99 I know that the bottom fraction can be written as [math]2*(1*2*3*...*(k+1))[/math].Is this correct?
What about the numerator?Is there a formula ?

 

[wolly]

This is what I found in the internet but I don't know where the 2k term comes from!

 

[mario99]

No the denominator is [math]2k+2[/math] and not [math]2k[/math]!
Also how can the numerator be used as factorial?

Let us first solve this problem:

Show me how the product of [imath]2k + 2[/imath] will give us [imath]2*4*6*\cdots *(2k + 2)[/imath]

 

[wolly]

[math]2∗(1∗2∗3∗...∗(k+1))[/math]
Isn't this one?

 

[mario99]

The product of [imath]2k + 2[/imath] means:

when [imath]k = 1, 2(1)+2 = 4[/imath]

when [imath]k = 2, [2(1)+2]*[2(2)+2] = 4*6[/imath]

when [imath]k = 3, [2(1)+2]*[2(2)+2]*[2(3)+2] = 4*6*8[/imath]

If [imath]k[/imath] starts at zero, we have:

when [imath]k = 0, 2(0)+2 = 2[/imath]

when [imath]k = 1, [2(0)+2]*[2(1)+2] = 2*4[/imath]

when [imath]k = 3, [2(0)+2]*[2(1)+2]*[2(2)+2] = 2*4*6[/imath]

But in the OP, you said that [imath]k[/imath] starts at [imath]1[/imath].

 

[wolly]

@mario99 That's how the exercise appeared in my textbook!

Did I applied the factor of 2k+2 correctly?

 

[mario99]

[imath]\displaystyle \prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}[/imath]

is wrong!

It should be:

[imath]\displaystyle \prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k)}[/imath]

@mario99 That's how the exercise appeared in my textbook!

Did I applied the factor of 2k+2 correctly?

How can I answer your question when the problem is written wrong?

In fact, it should be written like this:

[imath]\displaystyle \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k)}[/imath]

Or

[imath]\displaystyle \prod_{k=1}^{n} \frac{2k-1}{2k}[/imath]

Because

[imath]\displaystyle \prod_{k=1}^{n} \frac{2k-1}{2k} = \frac{1*3*5*...*(2n-1)}{2*4*6*...*(2n)}[/imath]

 

[Dr.Peterson]

@mario99 That's how the exercise appeared in my textbook!

Please show us an image of the problem as given to you, so we can be sure you copied it correctly (and also so we can know what you were asked to do with this expression).

What you wrote does have a meaning, and could be what someone wants, but it is odd both in the way @mario99 has pointed out, and in another way that hasn't been mentioned.

Hi I want to calculate this product:


[math]\prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}[/math]
How do You Express this product as a factorial?If not do You use double factorial in these 2 fractions?
Also If You apply the double factorial îs there a proof for [math]n!![/math]. ?

My concern is that you are multiplying at two levels, and I suspect you don't really mean to do that.

First, [imath]\frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}[/imath] itself is a product, namely
[math]\frac{1}{2\cdot4}\cdot\frac{3}{6}\cdot\frac{5}{8}\cdots\frac{2k-1}{2k+2}[/math]
(For example, the first factor, with [imath]k=1[/imath], is [imath]\frac{1}{2\cdot4}[/imath] since the denominator ends with [imath]2(1)+2=4[/imath], and the second, with [imath]k=2[/imath], is [imath]\frac{2(2)-1}{2(2)+2}=\frac{3}{6}[/imath], and so on.)

But then you are multiplying together n of these:
[math]\prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}=\left[\frac{1}{2\cdot4}\right]\cdot\left[\frac{1}{2\cdot4}\cdot\frac{3}{6}\right]\dots\left[\frac{1}{2\cdot4}\cdot\frac{3}{6}\cdot\frac{5}{8}\cdots\frac{2n-1}{2n+2}\right][/math]
Moreover, for [imath]n=3[/imath], the last factor in the outer product would be [imath]\left[\frac{1}{2\cdot4}\cdot\frac{3}{6}\cdot\frac{5}{8}\right][/imath], which doesn't appear to be what one would expect from the way it's written in the problem.

Do you really want such a product of products? And do you really want the first factor of each product to have 2*4 as its denominator? That's what the given expression seems to mean. But the work you quote in #5 doesn't include the outer product; do you see that? I suspect that it is answering the problem I think you would really have been given, namely something like this:

Write the following expression using product notation [or, using factorials]: [math]\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}[/math]​

 

[mario99]

Now this:

[imath]\displaystyle \frac{1*3*5*...*(2n-1)}{2*4*6*...*(2n)}[/imath]

Can be converted to factorial easily. First, let us focus on the numerator.

[imath]\displaystyle 2n-1[/imath]

When [imath]\displaystyle n = 3[/imath] you should get [imath]\displaystyle 1*3*5[/imath]. (Odd integers.)

Now choose a factorial. Let us choose [imath]\displaystyle n![/imath].

[imath]\displaystyle 3! = 1*2*3[/imath], this does not give me [imath]\displaystyle 5[/imath].

Now choose another factorial. Let us choose [imath]\displaystyle (2n)![/imath]

[imath]\displaystyle (2*3)! = 1*2*3*4*5*6[/imath]. Yes, I got [imath]\displaystyle 1*3*5[/imath], but I don't want [imath]\displaystyle 2*4*6[/imath], so I can get rid of them by division.

[imath]\displaystyle \frac{1*2*3*4*5*6}{2*4*6} = 1*3*5[/imath] the result that I want.

So I can write:

[imath]\displaystyle\frac{(2*3)!}{2*4*6} \rightarrow[/imath] this is only valid for [imath]n = 3[/imath].

Can I write the denominator also in factorial? Yes.

[imath]\displaystyle\frac{(2*3)!}{2*4*6} = \frac{(2*3)!}{(2\times 1)*(2\times 2)*(2\times 3)} = \frac{(2*3)!}{2^3*1*2*3} = \frac{(2*3)!}{2^3*3!} \rightarrow[/imath] this is only valid for [imath]n = 3[/imath].

Now anywhere you see three, replace it by [imath]n[/imath].

[imath]\displaystyle \prod_{k=1}^{n} 2k-1 = \frac{(2n)!}{2^n \ n!} \rightarrow[/imath] this is valid for all [imath]n \geq 1[/imath].

Now let us go the denominator.

[imath]2n[/imath]

When [imath]\displaystyle n = 3[/imath] you should get [imath]\displaystyle 2*4*6[/imath]. (Even integers.)

I have already calculated the factorial for even integers above [imath]\displaystyle 2*4*6 = 2^3*3! \rightarrow[/imath] this is only valid for [imath]n = 3[/imath].

Now anywhere you see three, replace it by [imath]n[/imath].

[imath]\displaystyle \prod_{k=1}^{n} 2k = 2^n \ n! \rightarrow[/imath] this is valid for all [imath]n \geq 1[/imath].

Now combine the numerator and the denominator.

[imath]\displaystyle \prod_{k=1}^{n} \frac{2k-1}{2k} = \frac{\frac{(2n)!}{2^n \ n!}}{2^n \ n!} = \frac{(2n)!}{2^{2n} \ (n!)^2} \rightarrow[/imath] this is valid for all [imath]n \geq 1[/imath].

 

[mario99]

Please show us an image of the problem as given to you, so we can be sure you copied it correctly (and also so we can know what you were asked to do with this expression).

What you wrote does have a meaning, and could be what someone wants, but it is odd both in the way @mario99 has pointed out, and in another way that hasn't been mentioned.

My concern is that you are multiplying at two levels, and I suspect you don't really mean to do that.

First, [imath]\frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}[/imath] itself is a product, namely
[math]\frac{1}{2\cdot4}\cdot\frac{3}{6}\cdot\frac{5}{8}\cdots\frac{2k-1}{2k+2}[/math]
(For example, the first factor, with [imath]k=1[/imath], is [imath]\frac{1}{2\cdot4}[/imath] since the denominator ends with [imath]2(1)+2=4[/imath], and the second, with [imath]k=2[/imath], is [imath]\frac{2(2)-1}{2(2)+2}=\frac{3}{6}[/imath], and so on.)

But then you are multiplying together n of these:
[math]\prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)}=\left[\frac{1}{2\cdot4}\right]\cdot\left[\frac{1}{2\cdot4}\cdot\frac{3}{6}\right]\dots\left[\frac{1}{2\cdot4}\cdot\frac{3}{6}\cdot\frac{5}{8}\cdots\frac{2n-1}{2n+2}\right][/math]
Moreover, for [imath]n=3[/imath], the last factor in the outer product would be [imath]\left[\frac{1}{2\cdot4}\cdot\frac{3}{6}\cdot\frac{5}{8}\right][/imath], which doesn't appear to be what one would expect from the way it's written in the problem.

Do you really want such a product of products? And do you really want the first factor of each product to have 2*4 as its denominator? That's what the given expression seems to mean. But the work you quote in #5 doesn't include the outer product; do you see that? I suspect that it is answering the problem I think you would really have been given, namely something like this:

Write the following expression using product notation [or, using factorials]: [math]\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}[/math]​

Professor Dave, I have an extra thought and I just want to share it. Since the OP is insisting that the problem is correct, I think that the problem would be equivalent to be written like this:

[imath]\displaystyle \prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)} = \frac{1}{2}\prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{4*6*8*...*(2k+2)}[/imath]

Now, by using the same method in my post #12, I get this answer:

[imath]\displaystyle \frac{1*3*5*...*(2k-1)}{4*6*8*...*(2k+2)} = \prod_{k=1}^{n} \frac{2k-1}{2k+2} = \frac{(2n)!}{2^{2n}n!(n+1)!}[/imath]

And the original problem will be:

[imath]\displaystyle \prod_{k=1}^{n} \frac{1*3*5*...*(2k-1)}{2*4*6*...*(2k+2)} = \prod_{k=1}^{n}\frac{(2k)!}{2^{2k+1}k!(k+1)!}[/imath]

where the main goal was to write the product of the product in factorial form. And this factorial form is giving the same results as yours!

Just to confirm it, let us compare some products. You have:

[imath]k = 1[/imath]

[imath]\displaystyle \left[\frac{1}{2\cdot4}\right] = \frac{1}{8}[/imath]


[imath]k = 2[/imath]

[imath]\displaystyle \left[\frac{1}{2\cdot4} \cdot \frac{3}{6}\right] = \frac{1}{16}[/imath]


I have:

[imath]k = 1[/imath]

[imath]\displaystyle \frac{(2(1))!}{2^{2(1)+1}1!(1+1)!} = \frac{1}{8}[/imath]


[imath]k = 2[/imath]

[imath]\displaystyle \frac{(2(2))!}{2^{2(2)+1}2!(2+1)!} = \frac{1}{16}[/imath]


It seems that my factorial formula is working properly! And I hope that this is what the OP wants to achieve.

 

[wolly]

@mario99 How did you get [math]\frac{(2n)!}{2^{2n}*(n)!*(n+1)!}[/math] ?

 

[wolly]

@mario99 If 2n-1 has n=3 where does the factorial come from? n=3 2*3-1???????? I'm confused!

 

[mario99]

@mario99 How did you get [math]\frac{(2n)!}{2^{2n}*(n)!*(n+1)!}[/math] ?

In the same way as post #12. What you did not understand in that post?

@mario99 If 2n-1 has n=3 where does the factorial come from? n=3 2*3-1???????? I'm confused!

When I say [imath]2n - 1[/imath], I mean the product of [imath]2n - 1[/imath]. In other words:

[imath]\displaystyle \prod_{k=1}^{3}2n-1 = [2(1) - 1]*[2(2) - 1]*[2(3) - 1] = 1*3*5 \rightarrow[/imath] (So, you get these numbers when [imath]n = 3[/imath].)

Follow the process in post #12 and tell me where you are stuck!

 

[lookagain]

@mario99 That's how the exercise appeared in my textbook!

A textbook can have an error in the desciption of the problem. An author
could post something ignorant, such as including a wrong symbol. The
more likely scenario here is that the problem is supposed to be just the
ratio of two products.

You should speak to your instructor about what the problem shows with
its particular symbols, versus the comments made by the mathematics
helpers here.