Alex Cervantes
2022-01-23
Answered

Show that ${A}_{n}=\sum _{k=1}^{n}\mathrm{sin}k$ is bounded?

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Addisyn Thompson

Answered 2022-01-24
Author has **16** answers

Note that

$2\left(\mathrm{sin}k\right)\left(\mathrm{sin}\left(0.5\right)\right)=\mathrm{cos}(k-0.5)-\mathrm{cos}(k+0.5)$

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

$\mathrm{cos}\left(0.5\right)-\mathrm{cos}(n+0.5)$

Thus our sum of sines is

$\frac{\mathrm{cos}\left(0.5\right)-\mathrm{cos}(n+0.5)}{2\mathrm{sin}\left(0.5\right)}$

We can now obtain the desired bound for$\left|{A}_{n}\right|$ . For example, 2 works, but not by much.

We could modify the appearance of the above formula by using the fact that$\mathrm{cos}\left(0.5\right)-\mathrm{cos}(n+0.5)=2\mathrm{sin}\left(\frac{n}{2}\right)\mathrm{sin}(\frac{n}{2}+0.5)$

This is obtained by using the ordinary expression for the cosine of a sum.

Add up,

Thus our sum of sines is

We can now obtain the desired bound for

We could modify the appearance of the above formula by using the fact that

dikgetse3u

Answered 2022-01-25
Author has **10** answers

Hint: Since

$\mathrm{sin}x=\frac{{e}^{ix}-{e}^{-ix}}{2i}$

we can rewrite

$\sum _{n=1}^{K}\mathrm{sin}n=\frac{1}{2i}\sum _{n=1}^{K}{\left({e}^{i}\right)}^{n}-\frac{1}{2i}\sum _{n=1}^{K}{\left({e}^{-i}\right)}^{n}$

and both of these are geometric series.

we can rewrite

and both of these are geometric series.

RizerMix

Answered 2022-01-27
Author has **438** answers

A more general formula would be:
${A}_{n}=\sum _{k=1}^{n}\mathrm{sin}k\theta $
$=\frac{\mathrm{sin}\theta +\mathrm{sin}n\theta -\mathrm{sin}(n+1)\theta}{2(1-\mathrm{cos}\theta )}$
So ${A}_{n}$ is clearly bounded (just simply check the case where $\theta =1$ ).
The formula can be proved by induction using the trig identity: $\mathrm{sin}\alpha +\mathrm{sin}\beta =2\mathrm{sin}(\frac{\alpha +\beta}{2})\mathrm{cos}(\frac{\alpha -\beta}{2})$ .

asked 2022-06-28

Since everyone freaked out, I made the variables are the same.

$\sum _{x=1}^{n}{2}^{x-1}$

$\sum _{x=1}^{n}{2}^{x-1}$

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Using a change in variables, Show that the boundary value problem:

$-\frac{d}{dx}\left(p\left(x\right){y}^{\prime}\right)+q\left(x\right)y=f\left(x\right),$

$a\le x\le b,$

$y\left(a\right)=\alpha ,$

$y\left(b\right)=\beta$

Can be transformed to the form:

$-\frac{d}{dw}\left(p\left(w\right){z}^{\prime}\right)+q\left(w\right)z=F\left(w\right),$

$0\le w\le 1,$

$z\left(0\right)=0,$

$z\left(1\right)=0$

Can be transformed to the form:

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Evaluate the following integrals. Include absolute values only when needed.

$\int}_{e}^{{e}^{2}}\frac{dx}{x{\mathrm{ln}}^{3}x$

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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 $\mathbb{C}}^{2$. 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

$H=\omega \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)$

and assume that for $t=0$ the state of the system is just given by $\psi (t=0)=\mid 0\u27e9$. In the following, we also assume natural units in which $h\stackrel{\u2015}{=}1$.

We expand the state at time t in the basis |0⟩, |1⟩ so:

$\mid Psi\left(t\right)\u27e9={\alpha}_{0}\left(t\right)\mid 0\u27e9+{\alpha}_{1}\left(t\right)\mid 1\u27e9$

Problems: Use Schrödinger's equation in order to derive a differential equations for $\alpha}_{0}\text{}\text{and}\text{}{\alpha}_{1$:

(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?