A little understanding of the consistent convergence of series Σ (x n-x n-1)

 

A.The consistent convergence of the Σ (x n-x n-1) series is somewhat interesting. It does not converge on the open interval (0, 1). However, if a positive number r <1 is given, it converges on the closed interval [0, R. It sounds like a general loop. Of course, the Weierstrass theorem can be easily used to determine the consistent convergence of series (I think this theorem proves that a series is a good choice for consistent convergence, and it will be a little difficult if it proves that it is not consistent convergence ), but here I am talking about how to understand it according to definition.

 

B.First, let's look at the definition of consistent convergence.: Set function item level ΣUN (x). For any given positive number ε, there is a natural number n that only depends on ε.So that when n> NTime, interval IEverything on X, All have inequalities

| RN (x) | = | S (X)-Sn (x) | <ε

Function level.Sigma uN (x)In the interval IOn the same side, it converges on and S (X), Also known as function sequence {sN (x )}In the interval IIt converges on S (X).Call, okay, okay ....

 

C.It is clear that the original determination of consistent convergence depends on the absolute values of other items.Σ(XN-X n-1)To try:

∵ S (x) = limsn (x) = Lim (x n) = 0 (N tends to be infinite)

Records | RN (x) | = | S (x)-Sn (x) | = x n

This value does not satisfy the definition. That is to say, when I give a ε, it corresponds to a N, but when n> N, it is not true for all X in the range, specifically, in the range I, there are always points that do not satisfy the inequality. For example, if I take a vertex x = A in the interval, that is, 0 <A <1, then obviously 0 <A1/n <1 is in the interval, therefore, x = A1/N is also a point in the interval and should be tested. Now, if I assume that this series is uniformly converged, then I will take an ε <A/2 (can I take any ε), which is easy to obtain. According to the definition, there should be a corresponding N so that when n> N, all X satisfies x n <ε <A/2, but obviously, X = A1/n does not meet this condition. Therefore, this assumption is incorrect.

 

D.Just now, we have to come up with a more incredible conclusion: although this series is inconsistent in (), if any number is given to a positive number r <1, in this case, it converges in the closed range [0, R. What does this mean? If I use the same method as I did, will this conclusion be broken if I find a point in [0, R] that does not meet the conditions? Okay, let's play it again: I take a bit of X = A in the range [0, R,SoX = A1/nIt should also be in the range [0, R]Internal, Then x = A1/n needs to be tested. Now I take a ε <A/2. Similarly, there should be a corresponding N so that when n> N, all X satisfies x n <ε <A/2, however, the point X = A1/n does not meet this condition. Therefore, the series in [0, R] are not uniformly converged, so the conclusion is wrong.

 

E.If you have a bit of reverence for truth, you will not doubt the classic theorem that our predecessors have summed up. In fact, there is no error in the conclusion. What is wrong is that the inference process is biased. However, reasoning is based on the correct reasoning above. Everything is done step by step. Where is the error? Let's take a look: You can pay attention to that line.ItalicsIn fact, this judgment is unfounded. x = A is in the range [0, R] and does not mean that X = A1/n must be in the range [0, R. The correct situation is: it may or may not be in it, for example, to make r = 0.8, A = 0.04, when n = 2, A1/N = 0.2, within 0.8, however, if r = 0.1, it is no longer in it (R can be arbitrary). Therefore, it cannot be used as a basis for judgment. In fact, I guess if the final conclusion is correct, then x = A1/N should be out of the range [0, R. Although I don't know what it actually means (too abstract), I can use the logic to prove it as follows:

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Proposition: knownAny given positive number r <1, In [0, R]The upper order of this closed interval is Σ (xN-X n-1)Consistent convergence, that is, for [0, R]Any point in XMakeX n <If ε is set up,(0 <A <r),1/nIn [0, R].

 

Proof: Use the reverse verification method.

 

Assume 0 <A1/n <R. Take an ε to make the ε <A <R, then any X has

X n <ε

Derivative of x n

(X n) '= NX n-1> = 0

Therefore, this function increases monotonically in [0, R], so max {x n} = R n,

Therefore

R n <ε

=> R n <ε <

=> R <A1/n

This is in conflict with assumptions, so assumptions are incorrect. A1/n> r, that is, A1/N is out of [0, R. Pass.Pai_^!

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F.This series is so special because of the form of the remainder and the selected range, because the remainder is x n and the interval is (0, 1 ), this leads to when x tends to 1, no matter how many remainder N is, it will inevitably tend to 1. Therefore, when x infinitely approaches 1, x n is bound to infinitely close to 1, in this way, the condition close to 0 (ε is small enough) is not met. This makes it better to understand why it is not consistent convergence. For the series of consistent convergence, the points in each interval will eventually reach the end point at the same time, however, for non-consistent convergence functions (as in the example discussed today), although we can ensure that the vertices in each interval eventually reach the destination, it is impossible to have all vertices reach the destination at the same time, in more detail, when the very distant points you expect finally reach the end point, there are still many points that are still earlier than the end point, but they are also approaching the end point, in the end, they will also reach the destination, but after them, there will still be a previous point approaching the destination ...... Today's example also has a name called "point-by-point convergence ". "Consistent convergence" is also called "Uniform Convergence ". For more information about point-by-point convergence and even convergence, see Wikipedia.

The following is the graph of Y = x n I made with Flash as2. X is taken in (), and N is obtained in 1, 2, 3, 10, 20, 30 respectively. (The longer the N is, the longer it takes, in order to make these curves appear at the same time, the machine asked me if I could continue five times .......).

 

 

 

 

 

 

 

G.My recent feelings about mathematics have found that the more I look at it, the less I can understand it with common sense and intuitive image thinking. It is built on an abstract set, the correct logic leads to further rational and abstract reasoning, so that we do not know what he is talking about due to the abstract relationship, we can understand that this theorem is correct and useful, but we do not know what he is talking about...