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## New answers tagged time-series

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Written as you did, this seems periodic to me. As you observed, the answer depends on the parity of $n$. You can rewrite $n=2\nu_n+\epsilon_n$, with $\nu_n$ integer ($\nu_n = \lfloor n/2\rfloor$), and $\epsilon_n = n-\lfloor n/2\rfloor$ is zero if $n$ is even, and $\epsilon_n$ is one if $n$ is odd. The latter series $\epsilon_n$ is $2$-periodic. Thus, $x[... 1 You're right, the given$x[n]$is clearly periodic. You can show this by simply checking if $$x[n]\stackrel{?}{=}x[n+N]\tag{1}$$ is satisfied for some positive integer$N$. For the given$x[n]you get \begin{align}(-1)^{n^2}&\stackrel{?}{=}(-1)^{(n+N)^2}\\&=(-1)^{n^2}(-1)^{N(N+2n)}\tag{2}\end{align} From(2)$it follows that for$(1)$(and ... 0 I think you're missing a critical component in your drawing/calculation Hint: Look at the exponent EDIT: Yep I was wrong, my bad 0 For measuring the similarity between two temporal signals, you can try using Dynamic Time Warping (DTW). DTW constructs a distance matrix between the two signals and tries to find minimum distance the two signals. If the two signals are identical, then distance is zero. The answer to your second questions depends on the signal model used to generate these ... 1 Hi: Testing for H0:$\phi_2 = \phi_1$in an AR(1) where$\phi_1$can be done using the offset function in R. Do you use R and, if so, are you familiar with the offset function ? If so, I can explain how to do this in more detail. Note that what I will do here is definitely an approximation ( econometricians might cringe. I'm not sure) because$\phi_1$... 1 You could try to Fourier transform,$F(k) = \int_{-\infty}^{\infty} x(t) e^{-2\pi i kt} dt$which can be expressed in terms of magnitude and phase in the following way$F(k) = |F(k)|e^{i\Phi(k)} = a(k) + ib(k)$with magnitude$|F(k)|$and phase$\Phi(k) = \tan^{-1}(\frac{b}{a})$for each frequency component$k$. Since it looks like you have a dominant$k\$...

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