a periodically recursive sequence.
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- Опубликовано: 10 фев 2025
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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).
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)
Yes, recursively defining a simple sum seems kinda pointless.
sin(n/2)(sin((n+1)/2)÷sin(1/2)? Where did you get that from?
@@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.
@@titassamanta6885 What does A. P. mean?
@@bjornfeuerbacher5514 arithmetic progression
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
There are many approaches to this problem, and this one is new to me. Thanks!
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.
the fade-in at the start is a nice touch!
and the fade-out!
A backflip-in and a backflip-out would impress me much more than these veeeery basic fadings. 😉
thanks alot for making the video. your presentation is very good!
Thankyou, sir
How do we know there could be no other sequence that satisfies the original equation? Without such justification the solution looks incomplete...
There are infinitely many, as a_0 is left unknown.
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.
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.
@@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.
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.
That was off its chops!! Love it.
this is often resolved using Z transform on signal processing
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.
The matrix is glitching around 3:45.
Ah, thank you. I paused it just before that to come down and see if anyone else spotted anything wrong.
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)
1-5-7-13-17-25-31-41-49............
I've found it an interesting number series ,
Do you mind introducing it ?
Doe the method work if the RHS is replaced by sin(p(n)) where p is a polynomial?
How about a_n+1 = a_n + random(-1, 1)
The random part averages to 0
like sin
normally a would user complex equivalence and exponential function
I would probably use generating function (ordinary generating function is enough , no need for exponential generating function)
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).
neat!
This one looks a bit handwaved to me.
It seems odd to call it "solving" the sequence. Finding a general form of a sequence isn't "solving" IMO.
Can you tell us what you think solving the sequence is? I genuinely don't know what you're referring to
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})