Generating functions are already powerful on their own. Your restrictions to multiples and extensions to partitions (different video) have made them even more general and powerful.
Around 18:30 they should have been positive etas since you were multiplying -x^(1/3) by itself but they still would have added to zero so you still got the answer.
It looks like the F(3n) series converges for abs(x) < phI^-3 = sqrt(5) - 2 which is about .236, which makes sense as phi^3 would be the limiting ratio of every third term of the Fibonacci sequence.
I have a question, if someone could help me give it an answet I´d be grateful. If you have a generating function, for example, for the positive integers (1/(1-x)^2), could you substitute that in the formulas given in this video to obtain the generating functions of the multiples of 2 or any other natural number? Thanks in advance for the answer
Since that part was general, did not depend on any property of the Fibonacci numbers or their generating function, you can use complex roots as coefficients to get any subsequence of a sequence whose generating function you already have. Notice how in the beginning he mentioned that he only chose the Fibonacci numbers and their generating function because he likes the Fibonacci numbers.
Good news: I'm not the only one who love these kind of toys. Bad news: my result (for F_kn over a^n to find the minimum real a while keep convergence) does not match your... and I believe your... D'OH!
Generating functions are already powerful on their own. Your restrictions to multiples and extensions to partitions (different video) have made them even more general and powerful.
Amazing video.... The only thing I do not understand is why you have only 24.4 K subscribers, instead of their number tending to infinity....
It’s ok, Michael finally reindexed the sum so he has 100,000 subs
@@tomatrix7525 A further 35k now!
Great, a new video on generating functions! 🙂
Around 18:30 they should have been positive etas since you were multiplying -x^(1/3) by itself but they still would have added to zero so you still got the answer.
It looks like the F(3n) series converges for abs(x) < phI^-3 = sqrt(5) - 2 which is about .236, which makes sense as phi^3 would be the limiting ratio of every third term of the Fibonacci sequence.
Many thanks. Is there a video on the "way" to determine a generating function, as that would be helpful indeed?
with this method we obtained a formula for a_(3n), what about a_(3n+1)? any ideas?
You could maybe shift the generating function to the left and re-do the original procedure.
i.e. start with A_n(x) --> A_n(x)/x
Wow, these gemerators are crazy
... this was the vid I was supposed to be watching!
I have a question, if someone could help me give it an answet I´d be grateful. If you have a generating function, for example, for the positive integers (1/(1-x)^2), could you substitute that in the formulas given in this video to obtain the generating functions of the multiples of 2 or any other natural number? Thanks in advance for the answer
Since that part was general, did not depend on any property of the Fibonacci numbers or their generating function, you can use complex roots as coefficients to get any subsequence of a sequence whose generating function you already have.
Notice how in the beginning he mentioned that he only chose the Fibonacci numbers and their generating function because he likes the Fibonacci numbers.
thanks
Good news: I'm not the only one who love these kind of toys.
Bad news: my result (for F_kn over a^n to find the minimum real a while keep convergence) does not match your... and I believe your...
D'OH!