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Because the function is continuous and monotonically increasing over the interval, a right Riemann sum overestimates the integral by the largest amount (while a left Riemann sum would underestimate the integral by the largest amount).
Riemann's explicit formula for the number of primes less than a given number states that, in terms of a sum over the zeros of the Riemann zeta function, the magnitude of the oscillations of primes around their expected position is controlled by the real parts of the zeros of the zeta function.
The sum over prime powers then gets extra factors of χ(p m), and the terms Φ(1) and Φ(0) disappear because the L-series has no poles. More generally, the Riemann zeta function and the L-series can be replaced by the Dedekind zeta function of an algebraic number field or a Hecke L-series .
One popular restriction is the use of "left-hand" and "right-hand" Riemann sums. In a left-hand Riemann sum, t i = x i for all i, and in a right-hand Riemann sum, t i = x i + 1 for all i. Alone this restriction does not impose a problem: we can refine any partition in a way that makes it a left-hand or right-hand sum by subdividing it at each t i.
The trapezoidal rule may be viewed as the result obtained by averaging the left and right Riemann sums, and is sometimes defined this way. The integral can be even better approximated by partitioning the integration interval, applying the trapezoidal rule to each subinterval, and summing the results. In practice, this "chained" (or "composite ...
The entire function ξ(s), related to the zeta function through the gamma function (or the Π function, in Riemann's usage) The discrete function J(x) defined for x ≥ 0, which is defined by J(0) = 0 and J(x) jumps by 1/n at each prime power p n. (Riemann calls this function f(x).) Among the proofs and sketches of proofs:
From the previous fact, Riemann integrals are at least as strong as Darboux integrals: if the Darboux integral exists, then the upper and lower Darboux sums corresponding to a sufficiently fine partition will be close to the value of the integral, so any Riemann sum over the same partition will also be close to the value of the integral.
The Riemann–Stieltjes integral admits integration by parts in the form () = () () ()and the existence of either integral implies the existence of the other. [2]On the other hand, a classical result [3] shows that the integral is well-defined if f is α-Hölder continuous and g is β-Hölder continuous with α + β > 1 .