We show that the product 1/2 × 3/4 × 5/6 × ... × 9997/9998 × 9999/10000 is less than 1/100, and then generalise the result. 00:00 Solution 04:28 Generalisation
Yes, the upper bound with 2 can be improved. We should be able to do this by writing the product as a single fraction, then taking out a factor of 2^n so we can express the numerator as a falling factorial, and the denominator becomes n!, and then we can work with Stirling's approximation to get your approximation with pi in place of 2.
When you're doing the pairwise comparison of your odd/even and even/odd products, instead of adding the extra 10000/10001 at the end, could you not simply compare 9999/10000 to 1 and still get the same conclusion? That way you don't need to introduce an extra term in the product, nor do you need the 1/10001 < 1/10000 step.
@@jamesstrickland833 Actually I think I could have avoided adding the extra term for the more general result too, but I'd still need to deal with the (2n - 1)/2n term separately since it is no longer in a pair with a term from the second product.
You can also write the original term as 9999!!/10000!! Because !! is the double factorial, that skips every second factor: 10!! = 10 * 8 * 6 * 4 * 2 9!! = 9 * 7 * 5 * 3 * 1 So 10!! * 9!! = 10! or reversed, 10!/9!! = 10!! Using your method, it could be written as: 9999!! / 10000!! * 10000!!/10001!! = 10000!/10001! = 1/10001 And since 9999!! / 10000!! < 10000!!/10001!! we can square the first part and get: (9999!! / 10000!!)² < 9999!! / 10000!! * 10000!!/10001!! (9999!! / 10000!!)² < 1/10001 9999!! / 10000!! < 1/√10001 < 1/√10000 = 1/100
When you write the products with an ellipsis, you orally precise whether it's on odd or even numerators, but you omit to write it on the whiteboard. It's a bit confusing for those who watch your videos without sound.
Just throwing this out there, but would it be remotely possible to solve the problem using induction? At first I thought it wouldn't be feasible, but then the generalisation made me wonder. Again, not sure what I'm on about, just smells like there's a connection.
I thought it might be possible to go from the inductive step inequality 1/2 x 3/4 x ... x (2n-1)/(2n) < 1/sqrt{2n} to the desired new inequality 1/2 x 3/4 x ... x (2n-1)/(2n) x (2n+1)/(2n+2) < 1/sqrt{2n+2}. This is equivalent to proving that 1/sqrt{2n} * (2n+1)/(2n+2) < 1/sqrt{2n+2}, or equivalently (2n+1)/(2n+2) < sqrt{2n}/sqrt{2n+2}, but if we try to solve the inequality, it does not hold for any values of n. So it looks like a standard induction is not going to work.
Very neat indeed! I really love how you start with a seemingly specific, clever example and generalise it in your videos
Thank you!
I just love your videos Dr Barker; keep em coming
Thank you!
Really interesting video. Is there another easy way of proving this? I think there might be - I will give it a go.
Beautiful!!🤩🤩
Very nice! And now you can make a sequel to explain why the product approximates 1/√(πn) for large n :)
Yes, the upper bound with 2 can be improved. We should be able to do this by writing the product as a single fraction, then taking out a factor of 2^n so we can express the numerator as a falling factorial, and the denominator becomes n!, and then we can work with Stirling's approximation to get your approximation with pi in place of 2.
When you're doing the pairwise comparison of your odd/even and even/odd products, instead of adding the extra 10000/10001 at the end, could you not simply compare 9999/10000 to 1 and still get the same conclusion? That way you don't need to introduce an extra term in the product, nor do you need the 1/10001 < 1/10000 step.
That does work, I believe Dr. Barker was just trying to preserve the structure so that he can show the generalization at the end.
@@jamesstrickland833 Actually I think I could have avoided adding the extra term for the more general result too, but I'd still need to deal with the (2n - 1)/2n term separately since it is no longer in a pair with a term from the second product.
@@DrBarker-- That would be a (2n - 1)/(2n) term. The grouping terms are also required in the denominator because of the Order of Operations.
You can also write the original term as
9999!!/10000!!
Because !! is the double factorial, that skips every second factor:
10!! = 10 * 8 * 6 * 4 * 2
9!! = 9 * 7 * 5 * 3 * 1
So 10!! * 9!! = 10! or reversed, 10!/9!! = 10!!
Using your method, it could be written as:
9999!! / 10000!! * 10000!!/10001!! = 10000!/10001! = 1/10001
And since
9999!! / 10000!! < 10000!!/10001!!
we can square the first part and get:
(9999!! / 10000!!)² < 9999!! / 10000!! * 10000!!/10001!!
(9999!! / 10000!!)² < 1/10001
9999!! / 10000!! < 1/√10001 < 1/√10000 = 1/100
When you write the products with an ellipsis, you orally precise whether it's on odd or even numerators, but you omit to write it on the whiteboard. It's a bit confusing for those who watch your videos without sound.
Sir please make more videos 🙏🙏
Just throwing this out there, but would it be remotely possible to solve the problem using induction? At first I thought it wouldn't be feasible, but then the generalisation made me wonder. Again, not sure what I'm on about, just smells like there's a connection.
I thought it might be possible to go from the inductive step inequality 1/2 x 3/4 x ... x (2n-1)/(2n) < 1/sqrt{2n} to the desired new inequality 1/2 x 3/4 x ... x (2n-1)/(2n) x (2n+1)/(2n+2) < 1/sqrt{2n+2}.
This is equivalent to proving that 1/sqrt{2n} * (2n+1)/(2n+2) < 1/sqrt{2n+2}, or equivalently (2n+1)/(2n+2) < sqrt{2n}/sqrt{2n+2}, but if we try to solve the inequality, it does not hold for any values of n.
So it looks like a standard induction is not going to work.
Can this be related to approximating logarithms? Also, wolfram alpha code: Product[k/(1 + k), {k, 1, -1 + n}] = 1/n
Wow this was kind of hard. Need to do it by hand to understand better.