2017年12月3日 · Lag=1 represents one hour. The autocorrelation function at lag=1 will experience a slight decrease in correlation. At lag=12 you will have the lowest correlation of the day, after what it will begin to increase. Move forward 6 month to 1 pm. Your time series is still somewhat correlated. Move lag to 6 months and 1 am.

To understand logistic growth, (say for example population growth) we employ a Verhulst-Pearl logistic equation and introduce a time lag to account for the fact that the system under study takes some time to respond to any perturbation and hence must also respond to the increasing population after a time lag. This time lag depends on the nature ...

2021年12月30日 · $\begingroup$ From the paper, and the paper's title even "ZERO LAG (well, almost)", the adaptive filter described in the paper is not exactly $0$ lag. The adaptive filter algorithm is designed to provide a compromise between reactivity (low delay) and smoothing.

The ° symbol means "degrees." Any answer marked with that is definitely in degrees. The technically correct thing to do is to assume that everything is in radians unless otherwise specified.

Note that $$\frac{1}{1+x}=\sum_{n \ge 0} (-1)^nx^n$$ Integrating both sides gives you \begin{align} \ln(1+x) &=\sum_{n \ge 0}\frac{(-1)^nx^{n+1}}{n+1}\\ &=x-\frac{x^2 ...

2017年10月24日 · i want to know how to calculate the autocorrelation of a markov chain (e.g for a simple random walk ). while i was searching online; i found a lecture with a two states {-1,1} markov chain with the

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$\begingroup$ I saw Henrik's comment above saying "in all likelyhood nobody is going to care" about a four-year old question, but this question has over 75,000 views and is on the first page of search results for common searches like "length of hypotenuse from angle", so I think it'd benefit from something that covers the options and jogs the memories of people (like me!) trying to …

2013年11月29日 · The Taylor series centered at $1$ can be easily derived with the geometric series $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$