Cesaro summable proofs in real analysis?
a) If ∑a(sub n) is convergent to a (in usual sense), show that ∑a(sub n) is Cesaro summable to a. b) Show that in the sense of Cesaro ∑(-1)^n is summable to 1/2. c) Let a(sub n)=k when n=k^3 for some k∈N, otherwise let a(sub n)=0. Show that ∑a(sub n) diverges (in usual sense, partial sums grow without bound), but it is Cesaro summable to 0 (seems paradoxical at first sight).
Let [math]S[/math] be a set, possibly uncountable. For any [math]p\in [1,\infty)[/math] we may form the space [math]\ell_p(S)[/math] comprising all scalar-valued functions [math]f[/math] for which the number [math]\|f\|^p = \sum_{s\in S} |f(s)|^p[/math] is finite. When [math]S=\mathbb N[/math] it is customary to view such functions as sequences, however the order of summation does not matter as we are summing non-negative numbers.For every [math]s\in S[/math] let us define the function [math]e_s[/math] by [math]e_(s)=1[/math] and [math]e_s(t) = 0[/math] otherwise, that is for all [math]t[/math] in [math]S[/math] different from [math]S[/math]. Clearly, [math]e_s\in \ell_p(S)[/math].The family [math]\{e_s\colon s\in S\}[/math] forms an unconditional Schauder basis of [math]\ell_p(S)[/math] for any [math]p\in [1,\infty)[/math]. (Note that the notion of an unconditional basis makes sense in non-separable spaces too.)When [math]p=2[/math], [math]\{e_s\colon s\in S\}[/math] is an orthonormal basis of [math]\ell_2(S)[/math].The family [math]\{e_s\colon s\in S\}[/math] is sometimes referred as the standard unit vector basis of [math]\ell_p(S)[/math].
Convergent series are easy to handle and one can perform many kinds of operation on them, like adding and subtracting two convergent series, differentiating or integrating (under suitable conditions), or even extending them in some sort of regular manner. Contrary to convergent series, the ones which are divergent cannot be handled in a satisfactory manner. They may come in two types -a) the partial sums either diverges to either positive or negative infinity - an example may be 1+2+3+...b) the partial sum do not converge to a single finite number (oscillating behavior) - an example may be 1-1+1-1+...A natural question arises which asks if there is a more general formulation of convergence which also includes some of the divergent series? Cesaro summation is one of the simplest technique to do so. It allows one to assign a sum to many divergent series and has the benefit that it yields the same answer when applied to convergent series. To get the Cesaro sum of a series one just replaces the usual sequence of partial sums by the sequence of the average of the sum of first n partial sums. In other words, the Cesaro sum of an infinite series is the limit of the arithmetic mean (average) of the first n partial sums of the series, as n goes to infinity.For the two examples given above, one can readily verify that the Cesaro sum of 1-1+1-1+... is 1/2, which is a bit intuitive to expect. On the other hand the series 1+2+3+... is not even summable using Cesaro method (it again diverges to infinity).I am not sure of any wider application of the Cesaro summability, but there are other methods of summability of divergent series such as zeta function regularization and Abel summation, among which, from what I have heard, zeta function regularization is used in some aspects of theoretical physics. Zeta function regularization is also used to assign a definite meaning to -1+2+3+... = -1/12, which evades common sense. To end I would remark that it is really difficult to tame divergent series and reproduce one of the quotes by Abel -"The divergent series are the invention of the devil, and it is a shame to base on them any demonstration whatsoever."(I may refine this answer later, preferably latex the equations.)
How to estimate fraction sums?
I assume you means sums of fractions. There are many tricks you can use depending on the sizes of the fractions and how many fractions you have (if there are 30 or 40 fractions I will use different tricks than if there are 3 fractions). It generally boils down to "pick nearby numbers that are easier to add" e.g. 3/8 + 2/3 would be easier if 3/8 was 3/9: 3/9 + 2/3 = 1/3 + 2/3 = 1, so the real answer is a bit bigger than 1 e.g. 1/2 + 1/4 + 1/7 + 1/8 would be easy if the 1/7 was 1/8 (because then the two 1/8 would add to 1/4, the two 1/4s add to 1/2 and the two 1/2s add to give 1). So the actual answer is just a tiny bit bigger than 1. If I have a *lot* of fractions, I would break it up into ranges, count how many fractions in each range and replace each range by a convenient value near the middle of the range, or just below the middle, particularly on a range starting from 0. e.g. "I have five values between 1/6 and 1/3 and two values between 1/3 and 2/3" So that's roughly 5 x 1/4 and 2 x 1/2 = 2 1/4. There are a bunch of other tricks but it just depends on what the situation is
Prove or disprove that series ∑(tan k)/k is Cesàro summable and it's sum is 0?
Some perhaps useless observations… Observation 1: The following sequence might be helpful: tan(n+1) = (tan(n) + tan(1))/(1 - tan(n)tan(1)) *** Observation 2: Claim: Fix delta>0. For almost all real numbers z in [0, pi], tan(zn)/n^(1+delta) converges to 0 as n goes to infinity over the integers. Proof: Fact 1 (Theorem 1.6 in "Probability and Measure" by Billingsley): Fix epsilon>0. For almost all real numbers x in [0,1], the following equation holds for all but a finite number of positive integers p, q: |x - p/q| >= 1/q^(2+epsilon). Fact 2: d*tan(pi/2 + d) converges to 1 as d goes to 0. From fact 2 we get: Fact 3: There is a constant C such that for all non-zero d in (-pi/2, pi/2) we have |tan(pi/2+d)| <= |C/d|. Now define epsilon = delta/2. For each n, define k[n] as the integer such that |zn - pi/2 - k[n]*pi | is smallest. Define d[n] = zn - pi/2 - k[n]*pi. Then: |tan(zn)| = |tan(d[n] + pi/2 + k[n]*pi)| =|tan(d[n] + pi/2)| <= |C/d[n]| [From fact 3] However, d[n] = zn - pi/2 - k[n]*pi, so: |z/pi - (2k[n] + 1)/(2n)| = |d[n]/(pi*n)| Now Fact 1 implies that for almost all z in [0, pi]: |z/pi - (2k[n] + 1)/(2n)| >= 1/(2n)^(2+epsilon) for all but a finite number of integers n. Thus (for almost all z in [0,pi]): |d[n]/(pi*n)| >= 1/(2n)^(2+epsilon) for all but a finite number of integers n. Define A = pi/2^(2+epsilon). Then: |d[n]| >= A/n^(1+epsilon) Thus (for almost all z in [0,pi]): |tan(zn)| <= |C/d[n]| <= |C/A|*n^(1+epsilon) = |C/A|*n^(1+delta/2). Thus, for almost all z in [0,pi] and all but a finite number of positive integers n: |tan(zn)|/n^(1+delta) <= |C/A|/n^(delta/2) and this converges to zero as n goes to infinity. *** Does anyone know any results, like Theorem 1.6 from Fact 1, that bound the error in rational approximations of pi? Such as a result that says p/q must be a distance at least f(q) away from pi, for some function f(q)...
WRESTLING : TRUE or FALSE.....Without HEYMEN, CESARO is a nobody?
Not true at all my friend! Cesaro by himself is one of the finest wrestlers we've seen. Very few young (and old) talent can match skills with him. The "King Of Swing" is truly one of the men that's going to be the future of this company. However, Paul Heyman benefits him. As amazing as Cesaro is in the ring.. he's garbage on the mic. Covering that part up by adding Heyman to the mix makes him an all round character. To answer your question, even without Heyman, Cesaro is something... A damn fine wrestler!
Ceasro? Nah. I don't see it happening in the near future, nor should it. The image of lesnar right now is that of an indomitable force. There is only one way Cesaro may win,and that is if wwe itself books it. But wwe writers won't do it. Cesaro may execute 20–30 swings, some uppercuts and display his strength via the suplexes he delivers. But lesnar will win. Cesaro possess amazing talent and many, including me, feels that he's being wasted in the 7-match series against Sheamus, when he can be booked for US or Intercontinental championships. But defeating Lesnar? Not by a long shot.Out of the current roster, I don't think there's anyone who can take him on,that's why there are rumors that Goldberg is gonna take him on. But if a current superstar has to be pitted against the beast, I would go with Kevin Owens.
I WILL not answer anymore question by are old friend k k k, your exactly right, if we stick together and give?
I NEVER reply to him, he is the most anoying i have seen on here, i dont even find him amusing, i find him down-right anoying, offencive, rude and with no life, who has nothing else to do but pick on other people. He reminds me of those bullies at high schools and you know those are the most messed up ones? They are so INSECURE about themselves that they pick on other people to make themselves feel big. Look at him, he asked a question why gays think he is in the closet when he "hates" them. What reason does he have to hate them. I dont understand how anyone can hate someone else because of who they choose to love. You can hate their charactereistics, their personality, if they've did you wrong, then you can hate them, but id like to see his reasons, my only conclusion is that maybe he was molested by a gay man or something so he now hates every person thats gay, what other explenetion can he give me? Im with you.
Why not all function stems from form? Sounds to me as if you could argue that either way since both are interdependent. Not sure that either has to do with “the moral end of philosophy”, but I'm not a professional philosopher of course.