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The Dirichlet function is not Riemann-integrable on any segment of despite being bounded because the set of its discontinuity points is not negligible (for the Lebesgue measure). The Dirichlet function provides a counterexample showing that the monotone convergence theorem is not true in the context of the Riemann integral.
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Fixing an integer k ≥ 1, the Dirichlet L-functions for characters modulo k are linear combinations, with constant coefficients, of the ζ(s,a) where a = r/k and r = 1, 2, ..., k. This means that the Hurwitz zeta function for rational a has analytic properties that are closely related to the Dirichlet L-functions.
The name "Dirichlet's principle" is due to Bernhard Riemann, who applied it in the study of complex analytic functions. [1]Riemann (and others such as Carl Friedrich Gauss and Peter Gustav Lejeune Dirichlet) knew that Dirichlet's integral is bounded below, which establishes the existence of an infimum; however, he took for granted the existence of a function that attains the minimum.
If the sample space of the Dirichlet distribution is interpreted as a discrete probability distribution, then intuitively the concentration parameter can be thought of as determining how "concentrated" the probability mass of the Dirichlet distribution to its center, leading to samples with mass dispersed almost equally among all components, i ...
As an example, a bag of 100 real-world dice is a random probability mass function (random pmf)—to sample this random pmf you put your hand in the bag and draw out a die, that is, you draw a pmf. A bag of dice manufactured using a crude process 100 years ago will likely have probabilities that deviate wildly from the uniform pmf, whereas a bag ...
Of particular importance is the fact that the L 1 norm of D n on [,] diverges to infinity as n → ∞.One can estimate that ‖ ‖ = (). By using a Riemann-sum argument to estimate the contribution in the largest neighbourhood of zero in which is positive, and Jensen's inequality for the remaining part, it is also possible to show that: ‖ ‖ + where is the sine integral
Let σ 0 (n) be the divisor-counting function, and let D(n) be its summatory function: = = (). Computing D(n) naïvely requires factoring every integer in the interval [1, n]; an improvement can be made by using a modified Sieve of Eratosthenes, but this still requires Õ(n) time.