Show that A_n=\sum_{k=1}^n\sin k is bounded?

Alex Cervantes 2022-01-23 Answered
Show that An=k=1nsink is bounded?
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Addisyn Thompson
Answered 2022-01-24 Author has 16 answers
Note that
2(sink)(sin(0.5))=cos(k0.5)cos(k+0.5)
This is obtained by using the ordinary expression for the cosine of a sum.
Add up, k=1 to n. On the right, there is mass cancellation. We get
cos(0.5)cos(n+0.5)
Thus our sum of sines is
cos(0.5)cos(n+0.5)2sin(0.5)
We can now obtain the desired bound for |An|. For example, 2 works, but not by much.
We could modify the appearance of the above formula by using the fact that cos(0.5)cos(n+0.5)=2sin(n2)sin(n2+0.5)
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dikgetse3u
Answered 2022-01-25 Author has 10 answers
Hint: Since
sinx=eixeix2i
we can rewrite
n=1Ksinn=12in=1K(ei)n12in=1K(ei)n
and both of these are geometric series.
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RizerMix
Answered 2022-01-27 Author has 438 answers
A more general formula would be: An=k=1nsinkθ =sinθ+sinnθsin(n+1)θ2(1cosθ) So An is clearly bounded (just simply check the case where θ=1). The formula can be proved by induction using the trig identity: sinα+sinβ=2sin(α+β2)cos(αβ2).
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We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space C2. We denote by |0⟩ and |1⟩ the standard basis elements (1,0)T and (0,1)T. Let the Hamiltonian of the system in this basis be given by
H=ω0110=0-i-i0
and assume that for t=0 the state of the system is just given by ψ(t=0)=0. In the following, we also assume natural units in which h=1.
We expand the state at time t in the basis |0⟩, |1⟩ so:
Psi(t)=α0(t)0+α1(t)1
Problems: Use Schrödinger's equation in order to derive a differential equations for α0 and α1:
(i) Find a solution given the initial conditions.
(ii) What is the probability that the system can be measured in |1⟩ at some time t?