Where results make sense

About us   |   Why use us?   |   Reviews   |   PR   |   Contact us

Bounded function - Wikipedia, the free encyclopedia An important special case is a bounded sequence, where X is taken to be the set N of natural numbers. The function f which takes the value 0 for x rational number and 1 for x irrational number is bounded. The set of all bounded functions defined on [0,1] is much bigger than the set of continuous functions on that interval. en.wikipedia.org /wiki/Bounded_function   (355 words)

Bounded variation Summary Functions of bounded variation are precisely those with respect to which one may find Riemann-Stieltjes integrals of all continuous functions. Another characterization states that the functions of bounded variation on a closed interval are exactly those f which can be written as a difference g − h, where both g and h are monotone. Functions of bounded variation have been studied in connection with the set of discontinuities of functions and differentiability of real functions, and the following results are well-known. www.bookrags.com /Bounded_variation   (449 words)

CMPSCI 601 Q&A for HW#5, Spring 2004 As you say, whether the function is total is in general a statement with unbounded foralls and thus in general not expressible by a bounded formula. The function takes a number n of a TM and gives a number f(n) of another TM, except that the way I defined Z I have to say that f(n) _is_ the TM rather than being just the number of the TM. You have three different characterizations of the general-recursive functions to work with: Floop, the original definition in terms of the unbounded mu-operator, and Turing machines (since we proved that general recursive and partial recursive functions are the same). www.cs.umass.edu /~barring/cs601/qa/5.html   (1728 words)

The Riemann Integral Is the Dirichlet function Riemann integrable on the interval [0, 1] ? Find a function that is not integrable, a function that is integrable but not continuous, and a function that is continuous but not differentiable. Suppose f is an continuous function defined on the closed, bounded interval [a, b], and F is a function on [a, b] such that F'(x) = f(x) for all x in (a, b). pirate.shu.edu /projects/reals/integ/riemann.html   (1676 words)

Bounded Distance Although D is not bounded whereas d is, when one is close to 0 so is the other.Thus both distances induce the same notion of nearness in the sense that sets closed in one are also closed in the other and vice versa. Let f be a 1-1 function from a set X into a metric space Y with the metric function D. With the assumption that it's easier to tell apart more distant numbers, we are looking for a distance function for which the distance between, say, 5 and 10 is greater than the distance between 55 and 60 which, in turn, exceeds the distance between 105 and 110. www.cut-the-knot.org /do_you_know/bounded_dist.shtml   (1018 words)

Springer Online Reference Works   (Site not responding. Last check: 2007-10-03) -means under the condition (*), play an important role in various problems of boundary properties of functions, in harmonic analysis, in the theory of power series, linear operators, random processes, and in the theory of extremal and approximation problems. Function of bounded characteristic); in particular, the functions of the Hardy classes have almost-everywhere on The characterization of Hardy classes in terms of the maximal function requires in a number of cases a recourse to probability concepts connected with Brownian motion (see [8]). eom.springer.de /h/h046320.htm   (1035 words)

7.4. Lebesgue Integral Just as step functions were used to define the Riemann integral of a bounded function f over an interval [a, b], simple functions are used to define the Lebesgue integral of f over a set of finite measure. Suppose f is a bounded function defined on a measurable set E with finite measure. Measurable functions that are bounded are equivalent to Lebesgue integrable functions. web01.shu.edu /projects/reals/integ/lebes.html   (1687 words)

All Elementary Mathematics - Study Guide - Functions and graphs - Basic notions and properties of functions... Now we can formulate a definition of a function more exactly: such a rule (law) of a correspondence between a set X and a set Y, that for each element of a set X one and only one element of a set Y can be found, is called a function. A graph of an even function is symmetrical relatively y-axis (Fig.5), a graph of an odd function is symmetrical relatively the origin of coordinates (Fig.6). Geometrically, a zero of a function is x-coordinate of a point of intersection of the function graph and x-axis. www.bymath.com /studyguide/fun/sec/fun6.htm   (742 words)

Bounded Linear Operator If f is bounded it has a norm, denoted f, which is the lower bound of all the constants k that make f a bounded operator. Clearly a function that is continuous everywhere is continuous at 0. Given a linear operator f on a normed vector space, f is continuous at a point, iff f is bounded, iff f is uniformly continuous everywhere. www.mathreference.com /top-ban,blo.html   (608 words)

T.J. Kaczynski: Boundary functions for bounded...   (Site not responding. Last check: 2007-10-03) A function p(e) defined on the unit circle is a boundary function for a function f(z) defined in the unit disk provided for each e, f(z) has the limit p(e) at e along some curve lying in the unit disk and having one endpoint at e. Any two boundary functions for the same function f differ at only countably many points by the ambiguous-point theorem of Bagemihl; and a boundary function for a continuous function differs from some function in the first Baire class at only countably many points. In answer to a question of Bagemihl and Piranian, the author constructs a bounded harmonic function having a boundary function that is not in the first Baire class. www.rpi.edu /~bulloj/tjk/tjk6.html   (140 words)

A function on a compact set is compact What I had handed in was that since the function is bounded on an epsilon neighborhood of every point, take the union of such neighborhoods to be a covering of E by open sets, call it K. Than F is bounded on K, since it is a union of sets on which F is bounded. The problem with this proof is that I am wrong in saying that F is bounded on K because it is a union of sets on which F is bounded. I guess from reading the post that the OP doesn't know that the image of a continuous function on a compact space is compact, which follows from the fact that f is continuous if and only if the inverse image of an open set is open. www.physicsforums.com /showthread.php?t=69026   (1620 words)

Riemann Integrable A function does not have to be continuous to be riemann integrable. note a function is discontinuois on a set if it has a discontinuity at any point of that set, not if it is discontinuous at all points of the set. A bounded function is Riemann integrable if and only if its set of discontinuities has zero area. www.physicsforums.com /showthread.php?t=79756   (782 words)

The Virial Theorem The asymptotic average of the derivative of a bounded function is zero. In words, this signifies that for any bound system of particles interacting by means of an inverse-square force, the average (negative) potential energy is twice the average kinetic energy. Kepler’s third law is essentially a special case of the virial theorem for gravitationally bound systems consisting of a small particle in circular orbit around a large massive body. www.mathpages.com /home/kmath572/kmath572.htm   (1253 words)

Boundedness - Wikipedia, the free encyclopedia In functional analysis, a subset A of a topological vector space is bounded if for every neighbourhood N of the zero vector there exists some scalar α to that A is a subset of αN A function is bounded if its range is a bounded set. A linear transformation is bounded if the image of the unit ball is a bounded set. en.wikipedia.org /wiki/Bounded   (284 words)

PlanetMath: bounded function is a bounded function if there exist a Cross-references: vector subspace, structure, continuous, topological space, compact, norm, satisfies, normed vector space, vector space, complex, scalar, function This is version 4 of bounded function, born on 2003-07-06, modified 2004-12-11. planetmath.org /encyclopedia/BoundedFunction2.html   (105 words)

[No title] Suppose you have an analytic function in the annulus bounded by the innermost and outermost circles, and you have an upper bound for its absolute value on the innermost circle and another bound on the outermost circle. Suppose you have a domain bounded by a Jordan curve and you cut out two regions from that domain bounded by Jordan curves inside the domain. We have an analytic function in the remaining triply connected domain and have various bounds - 3,5,7 - on the boundary components. www.princeton.edu /~missouri/Generals/generals/harcos_gergely   (1153 words)

JOT: Journal of Object Technology - The Theory of Classification Part 19: The Proliferation of Parameters, Anthony J H ... The polymorphic function appears to be “smart” because it somehow detects the type of its argument and returns a value of the exact same type. The outer function is a type-function and the inner function is a value-function. So, this kind of bounded type parameter captures exactly the sort of constraint we need when defining groups of functions that apply to all the types in a given class and only to those types. www.jot.fm /issues/issue_2005_07/column4   (3699 words)

Difference Between "a.e bounded function" and "a.e finite function" And sometimes they say if a function is in L^inf then it is a.e. And sometimes they say if a functions is in L^1 so it is a.e. The function f(x) = x^(-1/2), with f(0) = 0, is in L^1([0,1]). sci4um.com /about18116.html   (567 words)

Take a BrainSip   (Site not responding. Last check: 2007-10-03) The function f:'''R''' → R defined by f (''x'')=sin x is bounded. This function can be made bounded if one considers its domain to be for example The set of all bounded functions defined on [0,1] is a much bigger set than the set of continuous functions. bounded-function.mestskadoprava.sk   (295 words)

Real Analysis Proof Prove that if f : [a,b] ----> R is a bounded function that is continuous at all but finitely many points, then f is integrable over [a,b]. Real Anaylsis - Let f: [a,b] be mapped onto the Reals be a function that is integrable over [a,b] and let g: [a,b] be mapped onto the Reals be a function that agress with f except at finitely many points. Real Analysis Problem Prove a function is integrable over [a,b] - Let f:[a,b] mapped to the Reals be a function that is integrable over [a,b], and let g:[a,b] mapped to the Reals be a function that agrees with f except at two points. www.brainmass.com /homeworkhelp/math/other/10974   (278 words)

Liouville's theorem: a bounded analytic function is constant Liouville's theorem: a bounded analytic function is constant in the arc length of the surrounding circle, but that just repeats what we already know, that the function is bounded. Very well, take the limit, and arrive at the conclusion a bounded analytic function can only be a constant. delta.cs.cinvestav.mx /~mcintosh/comun/complex/node25.html   (147 words)

An Open Letter to Authors of Calculus Books A book covering the gauge integral might still include the characteristic function of the rationals, as an example of a bounded function that is gauge integrable but not Riemann integrable on a closed bounded interval. Ideally, one would like to also include a concrete example of a bounded function that is not gauge integrable on a closed bounded interval. That is both a strength of the theory (essentially all functions are gauge integrable) and a weakness of the theory (it is hard to convey the concept of the few functions that are not gauge integrable). www.math.vanderbilt.edu /~schectex/ccc/gauge/letter   (2241 words)

: Problem 4.52 The reason is that if S was not bounded, and we assume (for contradiction) that f is not bounded, we can not be sure that we can find a situation where (*) and (**) are satisfied. This function is uniformly continuous, even though it is unbounded. is bounded on S if there is a positive number M such that ||f(x)|| <= M for all x in S. That is, the image f(S) is a bounded set. home.cc.umanitoba.ca /cgi-bin/discus/board-auth.cgi?file=/10/50.html   (558 words)

Bounded Operators form a Vector Space   (Site not responding. Last check: 2007-10-03) Now let a and be be normed spaces, and note that the bounded operators from a into b form a vector space. The difference between two functions, on the unit sphere, is bounded by the norm of their difference, which is the "distance" between the two functions. For each x, the limit function g(x) has to be within ε of the functions beyond n. www.mathreference.com /top-ban,bhom.html   (408 words)

A Remarkable Monotonic Property of the Gamma Function   (Site not responding. Last check: 2007-10-03) and is positive because of convexity of the function 1/y. We showed that the derivative of the function P(x) is positive, which means that P(x) is increasing, which means that the function S(x) is increasing. Factorization of a polynomial, which defines values of sine function (angles n*pi/17). www.mathandcomp.com /mathcountry/analysis/gamma.htm   (611 words)

Springer Online Reference Works Apart from the bounded functions, similar product representations may be constructed for functions of bounded form and for Hardy classes [2]–[4] (cf. A solution was also found for the problem of constructing analogues of Blaschke products and Blaschke's theorem for doubly-connected domains [7] and, in general, finitely-connected [8] domains. The solution of the problem of constructing suitable analogues of the Blaschke product for holomorphic functions of several complex variables is rendered very difficult by the fact that the zeros of such functions cannot be isolated. eom.springer.de /B/b016630.htm   (502 words)

Theorems of Morera and Liouville and Extensions In this section we investigate some of the qualitative properties of analytic and harmonic functions. Theorem 6.18 shows that a nonconstant entire function cannot be a bounded function. is not constant, and hence it is not bounded. math.fullerton.edu /mathews/c2003/LiouvilleMoreraGaussMod.html   (251 words)

Remarks on Proving The Fundamental Theorem of Algebra The Extreme Value Theorem states that a continuous function from a compact set to the real numbers takes on minimal and maximal values on the compact set. We first prove the Bounded Value Theorem – the range of a continuous function on a compact set is bounded. Since the function is bounded, there is a least upper bound, say M, for the range of the function. www.cut-the-knot.org /fta/fta_note.shtml   (311 words)

Nathan S. Feldman's Home Page Even before the appearance of Thomson's Theorem (on the existence of analytic bounded point evaluations), the pure cyclic subnormal operators with finite rank self commutator were characterized by Olin, Thomson, and Trent; they showed, in particular, that the analytic bounded point evaluations of such an operator form a quadrature domain. The author defines a generalized quadrature domain to be a domain that has a generalized Schwarz function ; that is, a Nevanlinna class function on G that has boundary values almost everywhere on the boundary of G with respect to harmonic measure and these boundary values agree with z-bar a.e. In proving part (c) above, a general method is given for constructing bounded univalent functions that are smooth up to the boundary, have pseudocontinuations to the exterior of the unit disk and are not rational functions. home.wlu.edu /~feldmann/Papers/PhDThesis.html   (2289 words)

Try your search on: Qwika (all wikis)

About us   |   Why use us?   |   Reviews   |   Press   |   Contact us   Copyright © 2005-2007 www.factbites.com Usage implies agreement with terms.