Michael can you share HOW.and WHY anyone would even come up with this problem? Who would think to mix a fsctorial.with e and a floor I mean? Thanks for.sharing and hope to hear from you.
This makes it really easy to find even numbers, especially extremely large ones. Just plug in a big value for n. Also odd numbers if you choose the ceiling function. What a helpful tool! 😉
@@jorgekennedy3241 that sounds great, but I'm not sure @MichaelPennMath has done a video proving your hypothesis. Guess we'll have to wait for a new video where he tackles this important topic! P.S. whoosh
@@TimMaddux its not an hypothesis, n! is always even because by definition its divisible by every number less or equal to n (in particular, divisible by 2). So n!+1 is always odd.
Fun fact. Ignoring the floor, n!/e actually approximates the subfactorial of n (denoted !n) which represents the number derrangements of n items (the permutations where no item goes back to its original position. The exact formula for !n is n! multiplied by the taylor series evaluating e^-1, but we only use the first n+1 terms of the infinite series. Thus, the approximation gets better the larger n is.
Specifically, for n odd, we have !n = floor(n!/e), so they are exactly the 0, 2, 44, 1854, 133496, ... in the video (note he skipped n=6 which should have been 264), and for n even, !n is one more than floor(n!/e), so we get 1, 9, 264, 14833, 133461, .... Rather than using the floor, it makes more sense to use the nearest integer. In the video the tail is always in (0,1/2) for n odd and (-1/2,0) for n even, so rounding to the nearest integer means rounding down for odds and rounding up for evens, agreeing with floor for odd and floor+1 for even. In other words !n = near(n!/e) for n at least 1.
Yes but if you’re dividing by a different number and taking the floor (rounding down) it’s possible that for some combination ann odd number is spat out. For example, the ceiling function instead of the floor here will always give you an odd number despite n! Being even for n>=2
Also the number of chaotic permutations of a set with n elements. This can be found using recurrence relations and generating function, that Michael loves for sure!
An interesting but hard problem will be: find all real numbers x such that floor of n!/x has the same parity. Don’t know if there will be any other solutions other than e.
I made some playlists of your videos on my channel Michael. New ones like viewer suggested, Lie algebras and the answer is... and extended some of your existing ones.
well e is irrational which means there is no integer p such that p/e∈ℤ. Given that n! is is always positive, n!/e will always be a positive real number which is not an integer, and as such the target ⌈n!/e⌉=⌊n!/e⌋+1, and as such is always odd. i know you meant this as a joke but i wanna feel like my pea brain can do some of the math on this channel, and this is as much as it'll process, i'm weirdly proud:)
In min 11, in the homework, you Forget to say that the -1 in the begining of the rightside of the equality came from appliying floor function to the sum where m goes from n+1 to infinity (in purple)
Won't be easier. You have to essentially prove a more restricted bounds on n!/e than being between an even number and its successor as your induction hypothesis, or else you can not conclude the case for (n+1)!. Those "more restricted bounds" are essentially the proof you are looking for. So, I don't think induction is of any use here.
It's just a convention that varies from person to person/region to region/field to field. There's no real reason to include or exclude 0 from the natural numbers, any statement that assumes one of them is trivially easy to modify to a statement that assumes the other, so there's no real need for a universal standard there.
For clarity my professors taught us to use N to exclude 0 and N_0 (subscript) to include 0. It depends on the context whether you need/want 0 in there or not.
@@gregoryford2532Is this specific to a particular subfield, out of curiosity? My grad work was in combinatorics in the US, and we do include 0 in the natural numbers by convention. Same from what I'm familiar with in algebra and group theory. And I don't think I've ever seen "whole numbers" used as a formally-defined term; not in any of the papers I read, at least. I've only ever heard the term in the more casual sense of "a number that isn't a fraction" in math education. (Honestly, if anything, my intuition feels like "whole number" would be synonymous with "integer"; I'd still call, say, -3 a whole number.)
Can it be that if you substitute some other number in a neighbourhood of e that this still works? Or maybe all the numbers are odd for some other values of the denominator?
That was my first thought, but I think it's not good enough--the error in the factorial approximation is proportionately small but it gets bigger than 1.
I seem to have stumbled across a fallacy when trying to prove this by induction. Let g(n) = n!/e, f(n) = ⌊g(n)⌋ Base case - f(0) is even: f(0) = 0 Recursive case - Assume f(n) is even. - Separate g(n) into integer & fractional components: g(n) = f(n) + d(n). - Divide both sides by 2: g(n)/2 = f(n)/2 + d(n)/2 - Floor both sides: ⌊g(n)/2⌋ = ⌊f(n)/2 + d(n)/2⌋ - Because f(n) is even: ⌊f(n)/2 + d(n)/2⌋ = f(n)/2 - Multiply both sides by 2: 2 ⌊g(n)/2⌋ = f(n) - Increment by 1: f(n+1) = 2 ⌊g(n+1)/2⌋ - RHS is 2 * integer, thus f(n+1) is even. This proof is clearly wrong because it also applies for ⌊n!/2⌋, which fails when n=3. Can anyone spot where the mistake lies?
I think the issue with it was incrementing like was done in the second to last line, as it hasn’t been shown that you can do that. For instance… Claim: f(n)=n equals 1 for all natural numbers n Proof: - This is true for n=1. - Proceeding inductively, assume f(n)=1. - Increment by 1: f(n+1)=1. - Therefore, f(n)=1 for all natural numbers n To do a proof by induction, I would expect having to use g(n+1) = (n+1)*g(n), and multiply both sides of line 2 by n+1… But this makes things complicated, because the floor of (n+1)*d(n) is no longer guaranteed to be zero
@Jack Bellamy second sum in the video is based on the taylor series expansion for e^x. i'm sure your logic holds but you need a more in-depth examination.
AI like chatgpt is very bad at proofs, it's almost the polar opposite of its designed purpose. When it gets stuck or doesn't know something it imagines an answer and prints it very confidently, like we want it to do when we ask it to be creative, write a song or summarize a book, but it's not at all what we want for proofs.
Thanks again to our sponsor Brilliant! To get started for free, visit brilliant.org/MichaelPenn/
Great observation. Michael, What happens if you replace e with π? Do you get a similar result? Keep doing what you are doing.
Michael can you share HOW.and WHY anyone would even come up with this problem? Who would think to mix a fsctorial.with e and a floor I mean? Thanks for.sharing and hope to hear from you.
This makes it really easy to find even numbers, especially extremely large ones. Just plug in a big value for n. Also odd numbers if you choose the ceiling function. What a helpful tool!
😉
I wonder what the largest known even number is and if this can be used to compute it
n! is always even and larger and easier to calculate than n!/e. Also n!+1 is always odd. Maybe you confuse with the search of big primes numbers.
@@jorgekennedy3241 😄😆😂🤣 come on man. U math nerds don't get sarcasm huh lol.
@@jorgekennedy3241 that sounds great, but I'm not sure @MichaelPennMath has done a video proving your hypothesis. Guess we'll have to wait for a new video where he tackles this important topic!
P.S. whoosh
@@TimMaddux its not an hypothesis, n! is always even because by definition its divisible by every number less or equal to n (in particular, divisible by 2). So n!+1 is always odd.
Fun fact. Ignoring the floor, n!/e actually approximates the subfactorial of n (denoted !n) which represents the number derrangements of n items (the permutations where no item goes back to its original position.
The exact formula for !n is n! multiplied by the taylor series evaluating e^-1, but we only use the first n+1 terms of the infinite series. Thus, the approximation gets better the larger n is.
Specifically, for n odd, we have !n = floor(n!/e), so they are exactly the 0, 2, 44, 1854, 133496, ... in the video (note he skipped n=6 which should have been 264), and for n even, !n is one more than floor(n!/e), so we get 1, 9, 264, 14833, 133461, ....
Rather than using the floor, it makes more sense to use the nearest integer. In the video the tail is always in (0,1/2) for n odd and (-1/2,0) for n even, so rounding to the nearest integer means rounding down for odds and rounding up for evens, agreeing with floor for odd and floor+1 for even. In other words !n = near(n!/e) for n at least 1.
w
@@burk314 "near" is lame you could've used a big o notation
also you can rewrite it as !n = floor(n!/e + 1/2)
@1:42 dang that’s a nice looking exclamation point!
wouldn't you know that n! is even anyway since n> 2 always contains a 2 in the product ?
But you divide by e before flooring. The irrationality of e makes it slightly more intriguing
@@tanchienhao Hi Chien Hao :0
@@advaypakhale5254 lmao hi
Yes but if you’re dividing by a different number and taking the floor (rounding down) it’s possible that for some combination ann odd number is spat out. For example, the ceiling function instead of the floor here will always give you an odd number despite n! Being even for n>=2
Nop, try this with π instead of e, and notice how for n=6 the result is odd!
Also the number of chaotic permutations of a set with n elements. This can be found using recurrence relations and generating function, that Michael loves for sure!
From the thumbnail, I was expecting the argument to be from the point of view of combinatorics...
Actually, the number of Derangements (or chaotic permutations) is the nearest integer to this fraction, not the floor.
Unfortunately, there are 9 derangement of 4 objects…!
@@cabrazarado Oh, yeah mb.
@@cabrazarado what the heck are you guys talking about sorry?
10:41 Homework
12:01 Good Place To Stop
LMAO
Engineer Michael doesn't exist
Engineer Michael: 1:06
Thanks!
I mean, 2x where x
An interesting but hard problem will be: find all real numbers x such that floor of n!/x has the same parity. Don’t know if there will be any other solutions other than e.
you could use the same taylor series expansion for e^x given that n!/x = n!exp(-ln(x)).
3.35×2 is 6.7, which isnt even
I made some playlists of your videos on my channel Michael. New ones like viewer suggested, Lie algebras and the answer is... and extended some of your existing ones.
thanks for sharing your playlists
@@yoursleepparalysisdemon1828 You're welcome, that's kind of you to say :)
@3:56
To be rigorous and precise, that all good professional mathematicians should do, it should be written that
n >= 2
where should that be written, at the top of the sigma sum notation?
@@unflexian
No, not at the top of the sigma, but somewhere before the sigma notation.
Otherwise, we would have (-1)! and (-2)! which are undefined.
Can this be done by some version of Stirling's formula?
Good question. One reason why this is true is indeed the Stirling's formula.
I had the same thought, but it's not obvious to me how to do it. n! ~ sqrt(2 pi n) (n/e)^n doesn't look like dividing by e would simplify anything.
@@RobbieRosati You're right, it's not at all obvious.
As we've recently covered power series in calculus class, this is a really cool application of how useful a tool they can be!
n! is always even for all integers anyways unless you use 0! or 1!
Interesting corollary: the ceiling is always odd 😂
well e is irrational which means there is no integer p such that p/e∈ℤ. Given that n! is is always positive, n!/e will always be a positive real number which is not an integer, and as such the target ⌈n!/e⌉=⌊n!/e⌋+1, and as such is always odd.
i know you meant this as a joke but i wanna feel like my pea brain can do some of the math on this channel, and this is as much as it'll process, i'm weirdly proud:)
@@unflexian Oh nice didn’t expect any replies. It is just because if the input is not an integer, then the ceiling is 1 more than the floor of it.
This blew my freaking mind, without having to see the proof.
Wait, it's almost the same as !n (derangement of n)
In min 11, in the homework, you Forget to say that the -1 in the begining of the rightside of the equality came from appliying floor function to the sum where m goes from n+1 to infinity (in purple)
MIND BLOWN
Great video! Small error at 1:00: you're missing floor(6!/e) = 264.
I’m very intrigued that the floor function is one of your favourite things
Great problem!
Could you prove this by induction? Do you think it would be any easier?
Won't be easier. You have to essentially prove a more restricted bounds on n!/e than being between an even number and its successor as your induction hypothesis, or else you can not conclude the case for (n+1)!. Those "more restricted bounds" are essentially the proof you are looking for. So, I don't think induction is of any use here.
Seems to work with Pi also
You are a talented mathematician.
Heh, that's a really *odd* situation
This problem is mind-blowingly tricky
You could have used 'e' in the denominator for the word 'even' in preview
Why isn’t 0 a natural number ? I’m French and we consider it to be a natural number
It's just a convention that varies from person to person/region to region/field to field. There's no real reason to include or exclude 0 from the natural numbers, any statement that assumes one of them is trivially easy to modify to a statement that assumes the other, so there's no real need for a universal standard there.
For clarity my professors taught us to use N to exclude 0 and N_0 (subscript) to include 0. It depends on the context whether you need/want 0 in there or not.
@@gregoryford2532Is this specific to a particular subfield, out of curiosity? My grad work was in combinatorics in the US, and we do include 0 in the natural numbers by convention. Same from what I'm familiar with in algebra and group theory. And I don't think I've ever seen "whole numbers" used as a formally-defined term; not in any of the papers I read, at least. I've only ever heard the term in the more casual sense of "a number that isn't a fraction" in math education. (Honestly, if anything, my intuition feels like "whole number" would be synonymous with "integer"; I'd still call, say, -3 a whole number.)
Can it be that if you substitute some other number in a neighbourhood of e that this still works? Or maybe all the numbers are odd for some other values of the denominator?
I had this same identical question. Now that we have the solution for exactly e using the power series, can we somehow derive for e +/- epsilon?
it's a shame that its id number on math Stackexchange is odd
Hi micheal i have an intresting question ...find posible digits for which 2^n ,5^n start with same digit n>0
Example 1234 ,198 both start with 1 ...so 1 may be possible digit
Can you prove floor(n!/(3*e)) is also even for all n? Is it?
Pretty closely related to derangement counting.
Had no idea about this result , it’s unexpected for me. I will have guessed that is was random.
Can we use Stirling's approximation?
That was my first thought, but I think it's not good enough--the error in the factorial approximation is proportionately small but it gets bigger than 1.
I seem to have stumbled across a fallacy when trying to prove this by induction.
Let g(n) = n!/e, f(n) = ⌊g(n)⌋
Base case
- f(0) is even: f(0) = 0
Recursive case
- Assume f(n) is even.
- Separate g(n) into integer & fractional components: g(n) = f(n) + d(n).
- Divide both sides by 2: g(n)/2 = f(n)/2 + d(n)/2
- Floor both sides: ⌊g(n)/2⌋ = ⌊f(n)/2 + d(n)/2⌋
- Because f(n) is even: ⌊f(n)/2 + d(n)/2⌋ = f(n)/2
- Multiply both sides by 2: 2 ⌊g(n)/2⌋ = f(n)
- Increment by 1: f(n+1) = 2 ⌊g(n+1)/2⌋
- RHS is 2 * integer, thus f(n+1) is even.
This proof is clearly wrong because it also applies for ⌊n!/2⌋, which fails when n=3.
Can anyone spot where the mistake lies?
I think the issue with it was incrementing like was done in the second to last line, as it hasn’t been shown that you can do that. For instance…
Claim: f(n)=n equals 1 for all natural numbers n
Proof:
- This is true for n=1.
- Proceeding inductively, assume f(n)=1.
- Increment by 1: f(n+1)=1.
- Therefore, f(n)=1 for all natural numbers n
To do a proof by induction, I would expect having to use g(n+1) = (n+1)*g(n), and multiply both sides of line 2 by n+1… But this makes things complicated, because the floor of (n+1)*d(n) is no longer guaranteed to be zero
Sir Linear Diophantine Equations 11x+y=11 please
that's interesting maybe someone (not me) could attempt finding all real x where floor(n!/x) is even for all natural n
@Jack Bellamy second sum in the video is based on the taylor series expansion for e^x. i'm sure your logic holds but you need a more in-depth examination.
BTW, The series expansion of e^(-1) is used in a very simple proof that e is irrational.
Great observation. Michael, What happens if you replace e with π? Do you get a similar result? Keep doing what you are doing.
[3!/pi] = 1.
no pattern with pi because it does not have a series expansion
@@AayushSrivastava0307 It does have series expansions, but these do not fit nicely with the factorial the way e's does.
@@russellsharpe288 Consequently, you can see that the floor of n!/π does change parity. For example, 5!/π = 38.1... and 6!/π = 229.1...
If s(n)=[n!/e] then both s(2n) and s(2n+1) are divisible by 2n so both are even. I might investigate further.
I find the thumbnail hard to believe. I guess I'll have to watch the video.
Well I'll be damned. This means something.
I have another one i learned in my calculus 6 class at harvard, 2n will always be even
no way
I have something better, 2k is always an even number (k an integer). Great video!
small type: you missed n=6 in your examples
Why can’t you just show it’s 0 for n=0,1 and n! is even for n>1 so 2floor(k/e) is even? Or is it cuz you can’t necessarily pull out the 2?
you cannot distribute the integer function over multiplication
[4 * ¾] is 3 while 4*[¾] is 0
🤓
What an extraordinary coincidence
fun fact! floor(e) = 2 ....
You should have an AI learn to prove unproven math theorems.
AI like chatgpt is very bad at proofs, it's almost the polar opposite of its designed purpose.
When it gets stuck or doesn't know something it imagines an answer and prints it very confidently, like we want it to do when we ask it to be creative, write a song or summarize a book, but it's not at all what we want for proofs.
@@unflexian I'm not talking about current AI. I agree that current AI can't learn to prove theorems.
Why is n!/(n-1!) = n?