In this case this happens to work because the recursion is based on a Tchebychev polynomial. Now what if it were another random third degree polynomial with 0 as a fixed point? Could this be generalized?
I think one interesting thing is it takes aₙ beyond local maximum at x=1. There is a corresponding local mimimum at x=-1 as well of course. It would be interesting to see 'path' of aₙ in ℝ² where values and inputs are based on aₙ and aₙ₋₁ as aₙ follows a path to its point of convergence. As it seems aₙ decreases on its path until reaching 2pi. The ratio of a(n-1)/a(n) must make some interesting changes. On a different level the video shows an interesting link between some discrete things (sequences) and continuous functions
I initially assumed convergence to zero but was mistaken. It would be interesting to see how aₙ bounces around in ℝ² towards its point of convergence. What is interesting is that f(x) ~ -x³ is increasing on [0, 1] and at same time aₙ is initially decreasing. This masks the obvious path to converence point of the sequence. So aₙ must bounce about a bit on its path to point of convergence at a guess
It makes sense to have that restriction. Set lim(a_n)=L, then solving L=L(3-L²), either L=0 or 3-L²=1 iff L=√2. If L=0, then since 3^n→∞, lim(3^n a_n) might be interesting and finite. If L=√2, then 3^n a_n→±∞ depending on the choice of √2. Either way, it's clear and not that interesting what would happen in that case.
There is, first, notice given by the question we see a(n) = 3a(n+1) - a(n+1)^3 and extracting the right hand side, we have something like 3y - y^3, which just like Michael mentioned is exactly the form of the sin formula for triple angle. Second, as we see a(n) is between 0 and 1 for all n >= 3, we may hope that a(n) can be parametrized as some other function of unit length, and the kind of most obvious candidates are the sin and cos functions (as they are between -1 and 1). Both of the rationale above suggests and motivates the possibility of introducing the sin function into the play.
@ This is one of those problems that is tortuously reversed engineered. I would be very interested to know if Michael generated the solution by himself or it’s just one of those “gotcha” solutions no one solves on their own. Did HE suddenly see the triple angle formula (sort of) in the beginning of an alternating sign polynomial of odd exponents?
I keep watching this video over and over again, because I do not know where a good place to stop is
No
ice
:(
No WHAT?
In this case this happens to work because the recursion is based on a Tchebychev polynomial. Now what if it were another random third degree polynomial with 0 as a fixed point? Could this be generalized?
8:25 Isn't alpha_n = alpha_2/3^{n-2} and not alpha_2/3^{n-1}?
That would make the final result require an extra factor of 3
@@minamagdy4126 Yes, exactly!
Yes, the correct answer is 6*pi - there's a mistake in the video
What is the role of a1=2?
Yeah there's no point, should have written the qn with the indexes shifted down by one.
I think one interesting thing is it takes aₙ beyond local maximum at x=1. There is a corresponding local mimimum at x=-1 as well of course.
It would be interesting to see 'path' of aₙ in ℝ² where values and inputs are based on aₙ and aₙ₋₁ as aₙ follows a path to its point of convergence.
As it seems aₙ decreases on its path until reaching 2pi. The ratio of a(n-1)/a(n) must make some interesting changes.
On a different level the video shows an interesting link between some discrete things (sequences) and continuous functions
In fact a_1 = 0 is more natural
Love this trigonometry trick
Would be nice to have common theory of how and when one should be using it
Perhaps because there's a well-known trigonometric formula for the depressed cubic x^3+p*x+q=0 when (q/2)^2+(p/3)^3
Thanks for sharing!
I have no idea why the a_1 term was given. Is it just there to mess with people who didn't read the problem statement properly?
No good place to stop?
0:30 doesnt work for n=2 yet 2>=2 hehehe
If a_1 = 2, then a_2 = 1 or a_2 = -2
Where can i access those math magzines ? From which these problems are taken.... ??
Please tell anyone
....
I think this is from Mathemtics Magazine published by the MAA. If you are a member of that association, you will be able to find those problems
infinity
So it's equal to 360°
Well, not really
Never been this early
I initially assumed convergence to zero but was mistaken.
It would be interesting to see how aₙ bounces around in ℝ² towards its point of convergence.
What is interesting is that f(x) ~ -x³ is increasing on [0, 1] and at same time aₙ is initially decreasing.
This masks the obvious path to converence point of the sequence. So aₙ must bounce about a bit on its path to point of convergence at a guess
Nice
It makes sense to have that restriction. Set lim(a_n)=L, then solving L=L(3-L²), either L=0 or 3-L²=1 iff L=√2. If L=0, then since 3^n→∞, lim(3^n a_n) might be interesting and finite. If L=√2, then 3^n a_n→±∞ depending on the choice of √2. Either way, it's clear and not that interesting what would happen in that case.
No motivation for the sin substitution. Pulled out of thin air. Unimpressive.
There is, first, notice given by the question we see a(n) = 3a(n+1) - a(n+1)^3
and extracting the right hand side, we have something like 3y - y^3, which just like Michael mentioned is exactly the form of the sin formula for triple angle.
Second, as we see a(n) is between 0 and 1 for all n >= 3, we may hope that a(n) can be parametrized as some other function of unit length, and the kind of most obvious candidates are the sin and cos functions (as they are between -1 and 1).
Both of the rationale above suggests and motivates the possibility of introducing the sin function into the play.
@ This is one of those problems that is tortuously reversed engineered. I would be very interested to know if Michael generated the solution by himself or it’s just one of those “gotcha” solutions no one solves on their own. Did HE suddenly see the triple angle formula (sort of) in the beginning of an alternating sign polynomial of odd exponents?