a periodically recursive sequence.

[@kokainum]

How about defining c_n such that c_{n+1} = c_n + e^(i*n). Easy to get a formula because it's sequence of partial sums of geometric series and then you find a_n as imaginary part of c_n. Seems more intuitive and less tricky (no guessing).

[@rogierbrussee3460]

An alternative way to doing this Is and dealing with trigonometric functions, is to use complex exponentials to make things easier. Set q = e^i then sin(n) = Im(q^n) so we get a_{n+1} - a_{n} = sin(n) = Im(q^n). Now we can use that a_n = a_{n-1} + Im(q^{n -1})= = a_{n-2} + Im (q^{n -1} + q^{n-2}) = a_0+ Im(q^0 + q^1 + .. + q^{n-1)) (by easy induction) = a_0 + Im((1 - q^{n})/(1 - q)) (summing a geometric sum) = a_0 + Im( (q^{-1/2} - q^{n -1/2})/(q^{- 1/2} - q^{1/2})) (dividing nominator and denominator by q^{1/2}.= e^{1/2i}) = a_0 + Im( (2i)^{-1} (q^{-1/2} - q^{n -1/2})/((q^{- 1/2} - q^{1/2})/2i)) (dividing nom and denom by 2i ) = a_0 - (1/2sin(1/2)) Re (q^{-1/2} - q^{n-1/2}) (using Im(iz) = - Re(z) and sin(x) = (q^x - q^{-x})/2i) = a_0 - 1/(2sin(1/2)) (cos(1/2) - cos(n-1/2)) or if you prefer a_n = C - cos(n-1/2)/2sin(1/2) = C - (cos(n)cos(1/2) + sin(n)sin(1/2))/2sin(1/2) = C - cos(n)cotan(1/2)/2 -sin(n)/2

[@titassamanta6885]

Can we use the follwoing method? a_(k+1)-a_k=sin(k). So a_(n+1) -a1= sin(1)+sin(2)+....+sin(n). Now we have a formula for the series on the lhs, which is sin(n/2)(sin((n+1)/2)÷sin(1/2)

[@TheEternalVortex42]

Yes, recursively defining a simple sum seems kinda pointless.

[@bjornfeuerbacher5514]

sin(n/2)(sin((n+1)/2)÷sin(1/2)? Where did you get that from?

[@titassamanta6885]

@@bjornfeuerbacher5514 Actually there is a formula for calculating the sums of the sines of angles in A.P. Theee are multiple ways to derive it.

[@bjornfeuerbacher5514]

@@titassamanta6885 What does A. P. mean?

[@titassamanta6885]

@@bjornfeuerbacher5514 arithmetic progression

[@SlipperyTeeth]

Alternatively, take the discrete derivative of sin(x) and of cos(x). They will each yield something of the form Asin(x)+Bcos(x). Then match coefficients.

[@SJ-ry6br]

There are many approaches to this problem, and this one is new to me. Thanks!

[@ralvarezb78]

this is often resolved using Z transform on signal processing

[@kkanden]

the fade-in at the start is a nice touch!

[@kkanden]

and the fade-out!

[@BikeArea]

A backflip-in and a backflip-out would impress me much more than these veeeery basic fadings. 😉

[@gat0tsu]

thanks alot for making the video. your presentation is very good!

[@insouciantFox]

sin(1)/(2(cos1-1)) = -1/2 cot(1/2) so a(n) can be further simplified into a(n) = -1/2 csc (1/2)cos (n-1/2) +C which I think is much prettier.

[@wolfwerewolf4754]

How do we know there could be no other sequence that satisfies the original equation? Without such justification the solution looks incomplete...

[@Noam_.Menashe]

There are infinitely many, as a_0 is left unknown.

[@shohamsen8986]

It's a linear recursion. It shouldn't be too hard to prove uniqueness of solution up to picking a_0. That's what the C does. If u can guess a solution then u are done.

[@TheEternalVortex42]

Proving uniquness is easy, say b_n is another solution that satisfies the same recurrence. Then a_{n+1} - b_{n+1} = a_n - b_n + sin(n) - sin(n). Thus a_{n+1} - b_n{n+1} = a_n - b_n, so we know a_n - b_n = C for all n. This proves that all solutions are unique up to a constant.

[@shohamsen8986]

@@TheEternalVortex42i was thinking more like given some seed a_0, if you plug it in, you will get a unique answer. If you construct the recursion relationship u wont get two answers cause the relationship is inherently single valued. There are no square roots type multivalued operations involved.

[@CM63_France]

You can built an infinite number of solutions by taking different values of the constant C. And this constant is nothing but the general solution of the equation without a second member.

[@DrR0BERT]

The matrix is glitching around 3:45.

[@rogerkearns8094]

Ah, thank you. I paused it just before that to come down and see if anyone else spotted anything wrong.

[@msli4882]

The core of the problem is the summation of sequence of sin(n), and can be easily calculated from the summation of sequence of e^(i n)

[@davidmelville5675]

That was off its chops!! Love it.

[@papanujian7758]

Thankyou, sir

[@19divide53]

Doe the method work if the RHS is replaced by sin(p(n)) where p is a polynomial?

[@tomholroyd7519]

How about a_n+1 = a_n + random(-1, 1) The random part averages to 0 like sin

[@TaladrisKpop]

The video only finds some solutions. I would be great to show there are no other solutions. It is not too hard: if a_{n+1}=a_n+f(n), then a_n = a_1 + \sum_{k=1}^{n-1} f(k) (proof by a simple induction). Here, the sum is easy to compute as f(k) = sin(k) = Im(e^{ik})

[@eduardomalacarne9024]

normally a would user complex equivalence and exponential function

[@元兒醬]

1-5-7-13-17-25-31-41-49............ I've found it an interesting number series , Do you mind introducing it ?

[@holyshit922]

I would probably use generating function (ordinary generating function is enough , no need for exponential generating function)

[@bjornfeuerbacher5514]

One could rewrite the coefficient of cos(n) a bit: 2 cos(n) - 2 = -4 ( 1/2 - 1/2 cos(n) ) = -4 sin²(1/2), so the coefficient becomes -sin(1)/4sin²(1/2).

[@stevenmellemans7215]

This one looks a bit handwaved to me.

[@theevermind]

It seems odd to call it "solving" the sequence. Finding a general form of a sequence isn't "solving" IMO.

[@soupisfornoobs4081]

Can you tell us what you think solving the sequence is? I genuinely don't know what you're referring to

[@Wielorybkek]

neat!