For those who are interested: the binomial coefficient can be generalized so that the upper index in the symbol is actually any complex number, not just an integer. In fact, it can be generalized to be anything you want it to be, as long as it is part of a ring. How does it work? Well, again, notice that the binomial coefficient is equal to n!/[k!·(n - k)!]. Let us just focus on the symbol n!/(n - k)! for a second. This is called the falling factorial of order k, of n. What is nice about the falling factorial is that it can be expressed as a polynomial with respect to n: this polynomial has degree k, and is called the kth Stirling polynomial. Now, why is this important? It is important because polynomials can have complex inputs, matrix inputs, or whatever input you want them to have, as long as you are not changing the degree, and as long as the thing you are inputting belongs to a ring (so that addition and multiplication are well-defined and associative and addition is commutative). So if I substitute n for a complex number z in the kth Stirling polynomial, and I divide by k!, then that results in the binomial coefficient z choose k. This is extremely useful, as it helps express a generalization of the binomial theorem.
Thank you for this comment. Just realized r is a ring and this fits perfectly well with the binomial series(learning about this in calc 2 currently). A bit interested about the complex case now
@@Spacexioms The complex case is just a direct generalization of the real case, nothing too special happens. In the complex case, an expression such as (1 + z)^α would be written instead as exp[α·ln(1 + z)], assuming z is real and α is complex. It becomes more complicated if z is complex too, as you can no longer write it in such a manner.
Hi, there is a little mistake in your notation of the binomial coefficient at 2:46, the multiplication series above goes on from (n) till and including (n-k-1), not till (n-k+1)
@@brightsideofmaths you're completely right! I saw it as (n-(k+1)). In which case it would be true that (n-(k-1)) is the correct answer. But (n-(k-1)) has the same outcome as (n-k+1).
The way you can think about the final equation is for each arrangement of the numbers you did choose, you're also going to have every arrangement of the numbers of you didn't choose, so the denominator becomes k!(n-k)!. This actually presents an interesting insight: n choose k is the same as n choose (n - k), because multiplication is commutative.
![Matt Explains: Binomial Coefficients [featuring: choose function, pascal's triangle]](http://i.ytimg.com/vi/Pcgvv6T_bD8/mqdefault.jpg)
PDF versions and quizzes: tbsom.de/s/aoms
For those who are interested: the binomial coefficient can be generalized so that the upper index in the symbol is actually any complex number, not just an integer. In fact, it can be generalized to be anything you want it to be, as long as it is part of a ring. How does it work? Well, again, notice that the binomial coefficient is equal to n!/[k!·(n - k)!]. Let us just focus on the symbol n!/(n - k)! for a second. This is called the falling factorial of order k, of n. What is nice about the falling factorial is that it can be expressed as a polynomial with respect to n: this polynomial has degree k, and is called the kth Stirling polynomial. Now, why is this important? It is important because polynomials can have complex inputs, matrix inputs, or whatever input you want them to have, as long as you are not changing the degree, and as long as the thing you are inputting belongs to a ring (so that addition and multiplication are well-defined and associative and addition is commutative). So if I substitute n for a complex number z in the kth Stirling polynomial, and I divide by k!, then that results in the binomial coefficient z choose k. This is extremely useful, as it helps express a generalization of the binomial theorem.
Thank you for this comment. Just realized r is a ring and this fits perfectly well with the binomial series(learning about this in calc 2 currently). A bit interested about the complex case now
@@Spacexioms The complex case is just a direct generalization of the real case, nothing too special happens. In the complex case, an expression such as (1 + z)^α would be written instead as exp[α·ln(1 + z)], assuming z is real and α is complex. It becomes more complicated if z is complex too, as you can no longer write it in such a manner.
Fantastic explanation, clearing up my own misunderstanding the day before a final exam in university. Dear thanks to you!
You're very welcome!
Awesome. Very well explained in just under 4 minutes. Thank you very much
Best video i could fint on this subject, thanks man :D
A very important symbol! Thanks for all your videos! 😃
Glad you think so!
Great explanation! Thanks, will help me with a math proofs course
Nice :)
Hi, there is a little mistake in your notation of the binomial coefficient at 2:46, the multiplication series above goes on from (n) till and including (n-k-1), not till (n-k+1)
My definition is correct. Maybe you think about setting parentheses like (n-(k-1))?
@@brightsideofmaths you're completely right! I saw it as (n-(k+1)). In which case it would be true that (n-(k-1)) is the correct answer. But (n-(k-1)) has the same outcome as (n-k+1).
This explained a concept my biostats professor has had me confused on for 3 weeks….in 3 minutes…. Omg
I love how you visualize everything with the colours.
Makes it more clear for me :)
What wrong with you bro
Thank you :)
Also is it any different than Combinations( C(n, r))?
Thanks! It's the same :)
understood it very well
The way you can think about the final equation is for each arrangement of the numbers you did choose, you're also going to have every arrangement of the numbers of you didn't choose, so the denominator becomes k!(n-k)!.
This actually presents an interesting insight: n choose k is the same as n choose (n - k), because multiplication is commutative.
very well explained!! Thnx !!
You are welcome :)
Makes me want to take private lessons with this guy, just for fun.
Nice!
It is also a mini basic combinatorics lesson!
Thanks ❤
Where is the generalized binomial theorem to comppex numbers?
I love linear algebra now but Probability and statics is too stressful :( thank you for explaining !!!!
You can do it! You can check out my probability theory course: tbsom.de/s/pt
Is it possible to find a question with negative n
It's possible.
Tnx a lot
How did you get (n-k+1)?
Maybe try to write the expression for some chosen natural numbers :)
Impressive ❤️
Is it the same as nCr?
Yes!
amazing
♥
Du bist safe Deutsch 😂
Yes!