How can we improve our mathematical analysis?

How can we improve our mathematical analysis?

Original mathematical forum for Dr. scibird
The requests from friends on the Forum tell me about my learning experience and experience in the score for your reference.
First, I declare that there is no omnipotent method in the world. Any method has its limitations and applicability.
The essence of what you say is a dialectical view. Similarly, if you see something similar in a book, "Let's see everything in the world"
You can basically think about throwing this book into the trash.

As the question is written, this article describes "how to improve the level of Mathematical Analysis". That is to say, this article is intended for friends who have learned a few points but suffer from a slow increase in the score level. I have some personal experiences and hope to provide guidance to those who need help. it is a logarithm, but it also has reference significance for other mathematical subjects. I am familiar with analysis, so the examples are mostly in analysis.

First of all, we should correct an attitude, that is, we should not treat a theorem or a question with the attitude of examination questions, but with the attitude of studying mathematics questions. try to explore new things, not limited to the conclusions in the problem. specifically, it is as follows:
In general, it is about existence, uniqueness, inadequate conditions, necessity, and necessity.
These general statements may be well known and justified, but I just don't know how to do it. I will explain in detail my experiences over the years for your reference.

1. Inspired by Geometric Intuition, bold imagination, and strict argumentation.
Currently, the analytics industry is prone to this kind of bad tendency. We think that the Geometric Intuition is not strict, So we reject the Geometric Intuition and substitute it for abstract analysis and demonstration. Some books do not even have a picture. it is true that one feature of college mathematics is high abstraction, and Geometric Intuition cannot replace rigorous proof. however, the emphasis on abstraction is easy to get lost, especially for beginners. in fact, Geometric Intuition inspires many analysis theorems. many theorems can be observed from Geometric Intuition and extracted. Finally, they are strictly proved to rise to the theorem. for example, considering the ferma's theorem, the derivative value at the extreme point of the function can be 0. intuitively, the tangent of a derivative function at the extreme point should be horizontal, and it does not necessarily require continuous function. Then our conjecture is proved through analysis.

But is the problem over? Can we leave the far point where the derivative of the extreme point of the derivative function is 0, then we can ask if the derivative is 0 is the extreme point? When is there an extreme point? The previous problem is negative. If the derivative is 0, it is not necessarily the Extreme Point. for the next question, there may be more than one condition. one of them is special. We know that continuous functions in the closed interval must have the maximum and minimum values. for a very number function, if the maximum value is obtained within the interval, it is also the extreme value. If F can be exported, f' (x0) = 0. so when can we get the most internal value (also the Extreme Value ). if a condition is not a constant and the points at both ends of the function are equal, the maximum value point must be in the interval. Therefore, the x0 in the interior point satisfies the f' (x0) = 0, so the Rolle Theorem is available. we asked again, is this condition necessary? We can give an inverse example. This shows that the conditions of the Role theorem are only sufficient conditions. there are still a lot of similar Geometric Intuition. For example, if an image is rotated, the Rolle Theorem becomes the Laplace theorem. If the parameter form is used to represent the Laplace theorem, it becomes the Poisson theorem. of course, the above is just an intuitive guess from ry, and we must strictly prove it next.

2. Questions can be considered from multiple perspectives.
After we solve a good problem, we don't have to go away immediately. You can explore it and see if there are any new discoveries. For example, I have adjusted the conditions and conclusions. Is the conclusion still true? The original question condition is P1. If I change the condition P2, is the conclusion still true? Or, if condition P1 is not met, is the conclusion still true?
The conditions of the original problem were too harsh. I weakened the conditions and concluded that the problem was true. The original problem was a three-dimensional problem. Is it true if it was changed to an n-dimensional problem? The original problem requires the function f to be continuous. After I replace it with the Riann product, what is the conclusion? Or the original problem is related to trigonometric functions (involving periodicity). After I change to a General periodic function, what is the conclusion? Or whether the original proposition has the possibility of promotion.
For example, to obtain the limit under an integral number (or the integral operation is interchangeable with the limit process), it is usually required to converge in a consistent manner. but the condition of consistent convergence is too strong. Can it be changed to a more general condition. as a result, the alzera theorem emerged, replacing consistent convergence with consistent bounded and point state convergence conditions. (Refer to Nankai mathematical analysis or Xie Huimin's book or calculus tutorial)

The so-called alzera theorem (also known as the control convergence theorem in the Riann integral theory) is in the following form:
The so-called uniformly bounded, that is, there is a positive number M> 0, so that any n, x in [a, B] has | fn (x) | <= m. alzera's theorem asserted that only the Product Function Column FN (x) points converge, that is, FN (x) → f (x), and are uniformly bounded,
And f (x) Riemann product, we can launch the lim limit [a, B] FN (x) dx = limit [a, B] f (x) DX (Limit operation and integral Operation Order)

Friends who are familiar with Lebesgue points will find that this theorem is a special case of Lebesgue's control convergence theorem in real changes. in contrast, the extra condition is that the "f (x) Riann product" is required, because the limit function is not necessarily the product of Riann. this requirement can be removed from Lebesgue's point theory, because the limit of measurable functions is also measurable. (This shows the superiority of L points over R points from a certain angle ).
In fact, there will be new gains from the perspective of real changes, such as ye guolov's theorem that reveals the relationship between point-state convergence and consistent convergence.

In another example, I would like to give the Riann theorem in the theory of Fourier series, that is, the extended form of the Fourier coefficient tends to 0.
When F (x) sin (λ x) dx = 0, when λ → ∞. we can guess what the result would be if we replaced SiNx with a more general periodic function g (x,
That is, limit [a, B] f (x) g (λ x) dx → 1/T limit [0, T] G (λ x) dx limit [, b] f (x) dx (T is the period of g (x)
This is what was later called the Riemann-Lebesgue theorem. in the 9th question of Peking University in, we investigated the extended Riemann-Lebesgue theorem. (In simple cases, we can use λ = N)
In fact, there are many wonderful theories of Fourier series. You can try to use a general periodic function instead of a trigonometric function for promotion. (This promotion is not necessarily acceptable, but only provides a possible idea)

My post is about how to improve the mathematics practice, rather than for the postgraduate entrance exam (although the postgraduate students can learn from it). This post is just a reference for the postgraduate students. frankly speaking, there are often conflicts between mathematics and postgraduate entrance exams. This is also one of the reasons why many masters or teachers despise postgraduate entrance exams. regarding postgraduate entrance exams, I talked about my opinion in the post posted by an outsider at Peking University (that is for postgraduate entrance exams ).

In addition, it is indeed time-consuming to improve the level of mathematics. unless you have a different talent in mathematics, you should spend more time on yourself. I feel that I am a beginner in calculus, but today I have worked hard in mathematics. for example, I often see some people complain that it is too difficult for Teacher Zhang zhusheng to write <new lecture on Mathematical Analysis>. Later, I was too lazy to reply to the debate. I bought a new lecture in my sophomore year. I have read it over and over again. Although I have not studied it carefully every time, there are several people like me. I am very familiar with the specific theorem (or even the page number) in the new lecture, so I have to memorize this set of books. I think anyone who looks at a score book for 20 times will not be afraid of improvement.

My creed is: repetition is the best way to remember.

If I did not read the new lecture 20 times, the score level of scibird is still half-hanging. I learned a few points by myself, and I have never been guided by anyone. Compared with the mathematics department, I took a lot of waste and wasted a lot of time. however, I never regret reading the new lecture for the past 20 times. Without those 20 times, I cannot lay a solid foundation, and I cannot use my spare time to complete the Real-Time Function Change within two months. my friends who have read the questions I have written will think that I have written a lot of pen and ink, but it is quite popular, and the method used is also very simple (so that some friends think that the method is unconventional, sigh !). This is because "new lecture" has a great impact on my style. To tell the truth, my style was totally different from the present one.

Tell a real story:
I used osai in the high school era. At that time, I admired the splendor and mystery of proof and liked playing skills! ----- I call it a romantic style. for some very simple methods, it seems that Luo Yun and the level is low. he often lives on his own with gorgeous skills and advanced theories (compared to high school, he is mainly engaged in competitions), emphasizing proof and light computing. the direct consequence is the mess of the College Entrance Examination mathematics examination. after going to college, especially after reading Dr. Zhang zhusheng's <new lecture>, my understanding has changed: it proves that simplicity is not profound, and it proves that skill is strong and therefore short, it does not mean it is general. writing less is not necessarily a good proof. now I have a taste of "heavy sword and no front". Haha, this is very similar to the new style.

As it takes time to explore new things, it is normal that you will be able to deeply understand it when you are a graduate student. your own things can be understood more deeply. it doesn't matter if the findings are the same as the previous crash. YY: Newton, Gauss, and Euler also discovered that I was the same as they were ......

After a while, I found that I couldn't even remember what I found myself. This is really depressing-you lack a summary. Many people do not like the summary or even despise the summary. This is wrong. when you sum up your knowledge (whether it's others' or your own), you will have a general understanding. I did not sum up my questions. Each time it was partial, I lost my way when I saw no trees in the forest.

Finally, I wish you and other friends the gold medal title! Bless all!

I feel more and more that it is not easy to write a mathematical book (why not write a mathematical article ). some people think it is difficult to write a little deeper; some people think it is too simple to write a little; some write a little professional, there are a lot of versions on the same as I am not a professional in mathematics, do not understand; when I write too much, some people think Luo Yun; when I write too little, some people tend to write too much. it is really difficult to take care of readers at different levels. It is indeed difficult to let others know what they are talking about!

After all, I am not a master. It is impossible to clarify the problem with just a few words. Therefore, it is better to say more than to say less. for example, if this method is used, I generally give two examples: one is hard and the other is simple. in simple cases, I try to limit it to a few minutes and give examples that are easier to understand. the example is intended and may not be the most appropriate.

3. Active manual computation, manual deduction, and regular identification in the computation examples.

At present, there is a bad phenomenon. The engineers prefer computing, and the question of proof is big. the preference of the mathematics department proves that they are not concerned with computing. the result is two extremes: the level of proof of engineering is relatively low, and the computing capability of mathematics is relatively poor. I remember when I was a graduate student in numerical analysis a, a mm complained that the teacher "told me so much about the theory, just tell me how to calculate it", and it was very confident and powerful. (I am sweating ). I was surprised to hear that some students could not calculate certain credits when studying mathematical analysis in an experimental class. I think it is necessary to reverse this bad phenomenon. the proof and plan are unified, and should not be separated manually.
The discovery or proof of many mathematical theorems is indeed calculated. Even if you plan to engage in basic mathematics in the future, it is also beneficial to strengthen your computing capability. you don't know how to calculate something yourself. you don't have to accumulate. How can this problem break out!

Looking back at history, many mathematicians are good at computing. For example, even the idol Gauss is engaged in astronomy for the rest of his life. At that time, there was no computer, and it was basically done by hand. astronomical numbers, is it easy to calculate? However, when a descendant sorted out the Gauss's manuscript, he found that he was rarely wrong. gauss himself said how he first discovered what was known as the "Prime Number Theorem" by future generations. He said that he had calculated all prime numbers within 3000000 (as if it was this number, but he could not remember it, then the result is guessed.

Prime Number Theorem: records π (X) to indicate the number of prime numbers not greater than X. Gauss conjecture Lim π (X)/(x/ln x) = 1.
This theorem is in the progressive (X → + ∞) sense, and the degree of approximation is not very high (not practical ). on the one hand, we are amazed at Gauss's amazing insights. At the same time, we also need to see that if Gauss did not calculate a large number of prime numbers in advance, he could not find out the distribution frequency of prime numbers.

Another example is familiar to everyone. For example, the levenitz formula of the n-order derivative of the product of two functions in micro-learning. this formula proves that it is not very complex and the conclusion is not hard to remember. I wonder if you have considered it?
Evaluate the n-order derivative (UV) ^ (n) of UV: First, we know that (UV) '= u'v + U ', this formula can be used repeatedly to obtain any order derivative. if you are patient and have many computations (such as 3, 4, 5 .....), after merging similar items in the result, you will find two rules. (1) Each coefficient is a binary coefficient (or Yang Hui triangle); (2) the sum of the Order of the u and v derivatives is N. if your memory is good enough or your high school is solid, you will immediately find that this is like a binary expansion. so you can boldly guess the result is (UV) ^ (n) = Σ C_n ^ K u ^ (k) V ^ (n-k), C_n ^ K is the binary coefficient. this is the later lavenitz formula. then you can use mathematical induction to strictly prove it.

Not all mathematical theorems are hidden deep, and it is difficult to find patterns. Mathematics is sometimes quite simple.

"I don't think it has much to do with reading Zhang zhusheng's book. It's mainly because your mathematics level has improved. "----- Well, I know myself better than myself. in fact, it must be related to my own improvement, because I have said that mathematics is a whole. the new lecture has had a huge impact on me. It has penetrated into my methods, ideas, and even spirit !! Therefore, the biggest influence of the new lecture on my level is not that of me.

As for the two-month self-study change, this is the case. I actually learned from myself during the Olympics, and I did not have much to do with it.
I would like to share some experiences with you.
At the beginning, I felt that I had encountered a bottleneck in learning scores from a traditional perspective. I heard that calculus was actually changed from a new perspective, and it was very essential. at that time, Tsinghua had the theory of actually changing functions, so I was very careful. I don't expect to be able to understand it again, because I am more interested in the part of the points theory. Let's take a look at the distance theory and remember the main nature.

I have said many times that mathematics is a whole and different disciplines are connected.
Let's measure this. Why do we study it? It was originally proposed to study the points, attention [a, B] f (x) dx =?
No matter how much the result is, the points are meaningful first. from the perspective of ry, The Geometric Intuition of integral is the area of the curve and the "Curved edge Trapezoid" enclosed by the X axis. so when will we propose "area "? Therefore, we must clarify the concept of "area" first, so that the promotion area will be measured. If you have a good score, you should have learned an approximate measurement, which initially explores the area.
In my opinion, an appointment measure is an external measure and an internal measure. The Lebesgue measure idea can be seen as an extension of an appointment measure. since measurement is a promotion of area, it should have compatibility, that is, the common rule image length, area, volume results are true. for example, the length of the interval [0, 1] is 1, and the rectangular area is S = AB.

To establish his own theory, Lebesgue should promote the approximate measurement. I think the general idea is as follows: 1) admit that the external measurement and the inner degree are still valid; 2) Promote the addition of the external measurement, from the finite addition to the infinite addition, why is this kind of promotion only traceable? In this case, first, the measurement of a single point set is 0. If it is an unmeasurable addition, you will get the measurement of the range [a, B] is also 0, this is in conflict with the compatibility idea of the initial design measure! Therefore, the value can be infinitely limited to only a few additions.

But is that OK? If you pay more attention to mathematics, you may find that some concepts are very strange and some conditions seem annoying. most of these are manually added to exclude some anti-BT examples. let's take a note that the approximate measure is J measure, and the Lebesgue measure is l measure.

J measure is defined by external pressure and internal squeeze (external measure = internal measure). It is similar to the upper and lower sum of dashboard, which is actually an approximation idea (the core idea of score ). but there is a problem. Some images have no interior, and you cannot approach them from the interior. do we only consider external measurement rows? No, because there is a counterexample: There are two non-cross sets a, B. the Outer Measure of the Union is not equal to the sum of outer measurements, which is contrary to our general understanding. therefore, the second is the internal measurement of "transformation. we define the lateral degree of the point set E (including in the range [a, B]), and consider the lower bounds of the covered (a heap range) area. The outer measure is m (e ), this can be done no matter whether e is internal or not. considering the external measure of E ^ c = [a, B]-E, the inner measure is defined as N (e) = (B-) -M (E ^ C ). when M (e) = N (E), e is measurable and its measurement is a common m (e ). this shows that the internal measure is induced by an external measure (indirect call of an external measure.

The introduction of external and internal measures above is still quite natural, and the basic properties of Lebesgue measures are also clear. however, these basic proofs are obscure. the method I used is to acknowledge the basic nature of these measures (which will be used), and I will prove it later. it is equivalent to avoiding the degree theory that fears beginners to some extent.

Let's talk about the Lebesgue point theory.

<Measurable Function>
Our goal is to study the points. Of course, we need to study the points that can "enclose" the area (measurement), so we can naturally go to the measurable function. in fact, the real change is not so mysterious. The key is whether you can grasp the main line. littlewood, a famous British mathematician, once proposed the famous "Three Principles of wood" in the theory of real-variable functions. The rough content is as follows:
1). Measure is similar to the Union set of a finite interval. (In the definition of an external measure, l measure is a subset of an external measure)
2). Measurable Functions are similar to continuous functions. (lujin theorem)
3). Consistent convergence is similar to Point Convergence (Ye Guoluo's theorem)
Note that the "and" on the right side of the above three principles are something we are familiar. if we acknowledge these three basic principles, many conclusions can be drawn from the "Three Principles of wood. I am afraid the most useful one belongs to ye guolov's theorem, because there are many conclusions about consistent convergence in the number points, the set of any small measures in the yeguolouf theorem (set of points with inconsistent convergence) can be controlled in the Integral Theory (the integral value can be any small ).
Of course, there are other theorems for measurable functions.

<Lebesgue points>
This is what I'm most interested in. familiar with the knowledge of Riemann points, the function changes in the study of Riemann points cannot be too drastic, and the continuity is better. we study that Riann integral is a defined domain, while Lebesgue integral is a value domain (to overcome the difficulties caused by drastic changes in functions ). however, when we follow the example of Riann, we will find that the set of X corresponding to y_ I to Y _ {I + 1} may not be some intervals, may be some point sets, and may be very complicated. fortunately, we have a measurement (the measurable set can be some point sets). In the past, we had to point in the range (or approximately when the measurable set). Now we can point in the L measurable set. of course there may be some very BT points, which are not in the number. A very Nb property of Lebesgue points is that it is compatible with Riann points, that is, all the Riann product functions must be Lebesgue product, and the integral value is equal.

The benefit of Lebesgue points is not only to expand the range of the product functions, but also to relax many limit conditions. This can be seen from Lebesgue's control convergence theorem, column-dimension theorem, and method graph theorem.
Interestingly:
Lebesgue's control convergence theorem corresponds to the alzera theorem in the number score.
The column-dimension theorem corresponds to the Dini Theorem in the number score.
By comparison, we can easily remember and deepen our understanding.

The following differential and indefinite integral can be seen as the deepening of the score:
Most of our common functions can be written in the piecewise monotonic form. From the perspective of real-time variation, we have obtained many profound conclusions through the study of monotonic functions. to some extent, the research on Bounded Variation Functions and absolute continuity functions solves many basic calculus problems. for example, the curve length and the application of NL formula.

The theory of L ^ 2 can be closely related to the theory of Fourier series in the number fraction, and there are many profound conclusions, such as: the convergence of Fourier series in L ^ 2 almost everywhere, then, the Fourier series of continuous functions almost converge everywhere.

4. Focus on multiple solutions.

Different from the previous one, here we are not exploring new problems from the old ones, but considering how to use different methods to prove the problem or solve multiple problems. in my opinion, a viewpoint, a concept, a method, and so on are all mathematical ideas. different methods reflect different mathematical ideas. every time we see a new method, we must learn to absorb something useful to ourselves. here, I would like to remind you that, for a problem, you should not only look at the concise method, but the method is long and cumbersome, so you will not read it. you need to know that conciseness is not profound. Some methods are long, but they may be more general or typical. Some methods are short, but they may only be effective for this question (some competition questions are like this ), not General.

The great mathematician Gauss paid special attention to multiple solutions. He provided four different proofs of the basic theorem of algebra in his life. These four proofs have different styles. I posted "several major issues in scibird analysis" on the forum, and I gave two of them. the old saying is "warm and new", learning mathematics is not a model catwalk, blindly pursuing the latest, fashionable, and turning yourself into a castle in the air. think about the painful lessons of the Great Leap Forward! Some valuable ideas in mathematics are inherited, and she will not be wiped out with time. The novelty and fashion of the moment may be like fireworks. after a short period of beauty, it will die out.

Here I want to talk about the famous Approximation Theorem of "weststras": that is, continuous functions in closed intervals can be uniformly approached by polynomials. this theory is very important and useful. It is generally used in the number of teaching materials. Most books have cited the elementary proof of bernstan. He constructed something called the bernstan polynomial, then it is proved that it converges in a continuous function. this constructive proof is very clever! For more information about how to figure it out, see the topic "function constructor guidance" or "several major problems in scibird analysis.

However, the other two proofs provided in the third book are clever! And more profound! One of the proof ideas is as follows: first prove that Y = | x | can be uniformly approached by polynomial, and then prove a certain deformation and combination of line functions (| x |) the continuous functions can be uniformly approached by polynomials. Finally, it is proved that the continuous functions can be uniformly approached by the line functions, and the continuous functions in the closed interval can be uniformly approached by polynomials. the idea is clear and natural, and it reflects the core idea of analysis-approaching idea. in fact, many theorems in the analysis are proved by the idea that simple situations approach complex situations.

In the new lecture, we also provide another proof-kernel function method. people who have seen the proof of the convergence of the Fourier series should be familiar with the kernel function method, where the first n terms and times of the Fourier series are discussed, there is a thing called the Dirichlet kernel (DN (t )), then, the first n terms of the Fourier series are combined into an integral. For the convergence, we only need to discuss the variation of the integral when n approaches infinity. the rest kernel also appeared in the rest summation method later. people who have learned mathematical equations may remember that when solving the Laplace equation again, poisson integral (bosonlet core) appears ). in short, the integral core related to the kernel function is an important idea of modern mathematics.

The above may be a bit deeper. Let's talk about a simple example with multiple solutions.
Limit! Think about the various methods for finding the limit. The law of robidda, The schunz theorem, the expansion of Taylor, and the coefficients of a certain power series. Each of these methods has its own characteristics, which can be said to be a hundred schools of Contention! This content may not be met by anyone who scores at the same time, but after learning the score, you need to sort out these scattered extreme methods, you will have a brand new overall understanding.

5. Pay attention to practicing your own counter examples. I usually look at some famous counter examples in mathematics.

Why do we propose this? This is an important part of mathematics. if you encounter a mathematical problem during actual research, you need to consider whether to prove it or construct an inverse example to deny it. frankly speaking, the inverse example is a weak link in Chinese Mathematics Teaching. Some students have hardly mentioned the inverse example. because we usually do problems designed by others, rather than discovering and solving problems by ourselves. when you prove that a problem cannot be solved, you should consider whether to create a counterexample to deny it. do not assume that this question is from the exercises after the XX book or the true questions of the XX University. it's not about where you come from that mathematical proof. be brave at the crucial moment, and prove that mathematics is the ultimate judge!

The traditional Chinese education model leads to a single direction of thinking for students. The problem is always considered positively, while the thinking adopted by constructing counterexamples is generally odd. due to the traditional influence, we always feel that weird things are not good or even crowded out. however, mathematics is not transferred by our will. Whether you think it is weird or not, it does exist objectively. I want to write more about this topic and talk about the "strange tricks" in mathematics "!

If we have carefully analyzed the proofs of some talented mathematical masters and read some of their growth stories, you will find that the thinking of Masters has a characteristic, that is, the so-called "not going on the ordinary road! "There is a legend about Gauss, saying that when he was a child in elementary school, one day the teacher was angry for some unknown reason and the consequences were very serious. This teacher made a relatively BT arithmetic problem,
1 + 2 + 3 + ...... + 99 + 100 =? If there is no time limit for this question, it is not very difficult, that is, there are too many numbers, and it is easy to calculate errors in the middle.
This is a hard calculation. if you have strong computing power and patience, stick to the promise. however, Gauss had a different path. He observed that (the inverted addition method)
1 2 3 4... 99 100
100 99 98 97... 2 1
The sum of all columns is 101.
101 101 101 101... 101 101
There are a total of 100, and the inverted back and unchanged, so 1 + 2 + 3 + ...... + 99 + 100 = 101 × 100 limit 2 = 5050.
This method, which is later called the inverted addition method, is a typical method for calculating the sum of the arithmetic difference series.
We can also think about it and use the geometric method to solve the above arithmetic problem. Consider the following "ladder chart", and use "×" to represent the unit square.
×
××
×××
××××
..........
××× ...... ××
The K layer has exactly K Square "X", a total of 100 layers. It is easy to see 1 + 2 + 3 + ...... + 99 + 100 = "×" number = Area of the tiered graph. from the upper left corner of the ladder graph to the lower right corner of the graph, a main diagonal line is connected, and the graph is just divided into a large isosceles right triangle and 100 small isosceles right triangle, we know the area formula of the isosceles right triangle, so the original image is divided into a 100x100 large isosceles right triangle, and 100 1x1 small isosceles right triangle.
So 1 + 2 + 3 + ...... + 99 + 100 = tiered graphic area = 100x100 + 1x1 drawing 2x100 = 5050!

Of course, if we think of a tiered image as a broken image, and suppose we make up the other half of it (in a rectangle), this is essentially the same as the Gaussian inverted addition method.

There are still many similar ways of thinking, such as "Reverse Thinking", such as the story of Sima Guang's cylinder and "Lenovo thinking", such as the ladder figure above. to learn mathematics well, you must make your thinking active so as to improve creativity. do not turn yourself into a "sophistry" or "regular ". you can't figure out how fast it is to learn mathematics, so you will eventually lose yourself.

After talking about the problem for a long time, I suggest you go back to the counterexample. in fact, the counterexamples can help us clarify many seemingly false things, deepen our understanding of concepts, and promote the perfection of mathematics to some extent. why do I suggest the famous antiexamples in Mathematics? Because these examples are hard to think about, and we will learn new ideas in the process of analyzing the antiexamples. I also want to learn more.

A classic example of the back examples in history is the continuous function that cannot be guided everywhere constructed by vilstras. Most mathematicians believed that continuous functions could be guided in most aspects. later, people used the bell's Theorem in functional analysis to prove that the continuous function (the second set) is much more than the continuous function, which is an amazing thing. mathematics is always full of surprises! While we are surprised, don't forget to bring their proof methodology.
By the way, f (x) is a continuous function with no monotonic place. the proof is very simple. The reverse proof method assumes that f (x) is monotonic in a cell, and the Lebesgue theorem in the Real Variable Function knows that "monotonic functions are almost everywhere", which leads to a conflict! In fact, it is estimated that many people think that continuous functions are traceable in most aspects, most of which are intuitive and misleading, because the functions that people generally imagine are segmented and monotonous. there is a beautiful curve in the fractal ry called "Snowflake curve" (also called the cohe curve). This curve is also inseparable everywhere. if you are interested, you can search for them online.

6. Be knowledgeable and be the best at the Expo!

I have seen some people learn mathematics. They think that learning mathematics is playing skills and fighting IQ. if a problem gets stuck, I think my mind is poor. I didn't expect "that skill" (is it really the reason for the skill ?). Never admit that (including subconscious) problems may not be due to poor knowledge ". this kind of learning about mathematics is mostly about leaving a book behind closed doors, rather than reading other reference books. the result may be like this. For example, if I scored 90 + (in percent) in our school, I couldn't even understand the question of Peking University, and I got 50 points at most. as a result, abstract difficulty theories like Peking University's question are too technical.

The difference between a math test and a math study is that the former limits time, compared with agile thinking. The latter generally does not limit time and allows you to access the literature! Mathematics has already entered the "modern" era. It is not the time in martial arts that no matter what the Foreign Affairs, closed-doors practice, and fight independently. when you give lectures in a prestigious school, a good teacher will usually pick an appropriate book as a teaching material, then specify a number of similar reference books, and tell you if you want to learn this course well, reading only teaching materials is far from enough. for everyone, the time is limited. No one can master every course. It would be nice to be proficient in two or three courses. after careful observation, you will find that a mathematical expert is knowledgeable. Many Masters have made achievements in several branches of mathematics. there is no way to learn mathematics. You can find another path. but the premise is that you know a lot. For example, if you only perform mathematical analysis, you are stuck in an analysis problem and you can only continue to try it using the analysis method. he may consider algebra, ry, topology, and other methods. in mathematics, the solution of many well-known theorems is often the result of the joint action of different branches. here I want to cite a famous theorem in mathematics in the 20th century ------ Atiyah-Singer's index theorem.

The description of Atiyah-singer index theorem is in the vernacular: Resolution Index = topology index.
At present, the theorem itself has many forms of representation and has been promoted. What I know is a simple and primitive form as follows:
Dim kerd-dim cokerd = KERNEL _ {t ^ * m} CH (σ (D) 1_td (m )-----(*)
Here, M is a compact, smooth, and oriented column-store, and D is an elliptical differential operator. (*) on the left side of the formula is an analytical indicator (it should not be unfamiliar to those who have learned functional functions, which has been defined in the fredham theory );
(*) The right side of the formula is the topological indicator, which is presented in the form of the above-mentioned simultaneous adjustment. The general idea is that the role of the Chen class and the Todd class is involved, and then the credits are made from the remainder.
The Atiyah-singer index theorem associates two seemingly unrelated things, namely the number of samples and the number of topologies. Moreover, this theorem can unify some other Nb theorem as its special case.

In fact, the Atiyah-singer index theorem tells us two things: 1) the parsing index (related to D) is a topological invariant; 2) how to calculate the parsing index. people familiar with the index theorem may know about
The fact that dim kerd-dim cokerd has nothing to do with the selection of D has long been discovered by GE fangde that dim kerd and dim cokerd are related to D, however, their difference is irrelevant to D, which is a topological invariant! This is an amazing conclusion! However, although the dim kerd and dim cokerd forms are simple, it is difficult to calculate the General D. although galentd knows that dim kerd-dim cokerd is a topological invariant, it does not know how to calculate it.
The road to analysis seems to be disconnected. Can it be solved through other ways? Yes! Atiyah later gave a method for calculating the topological index (see the (*) formula for a case), proving the Atiyah-singer index theorem.
Let's look back at the (*) method. Although the parsing index is simple, it is difficult to find it. Although the topology index is complex, it can be calculated. this is also a philosophical reflection. (Of course, complexity is relative. it is not difficult for mathematicians to calculate topological indexes .)

This incident enlightened us that, to solve the difficulties in a field, only the knowledge in this field is not enough. We can find a different path only when we are knowledgeable!

The index theorem has a special form for children, which is easier to understand.
If V and W are two finite dimensional linear spaces and D: V ---> W is a linear operator, the index theorem can be written:
Dim kerd-dim cokerd = dimv-dimw

7. learning mathematics should include mathematical discovery and mathematical proof!

As the title says, this is my long-term point of view. this is why I strongly recommend Mr. Zhang zhusheng on various occasions <new lecture on Mathematical Analysis>. I don't expect the mathematics book to write down how all the theorem can be found, but at least I should try to get as much as possible. because mathematical discovery is also an important part of mathematics!

I personally think that learning mathematics should actually include two parts: mathematical discovery + mathematical proof. however, it is a pity that the current teaching materials are mostly based on rigor and the discovery of mathematics is lost. the result is that the teaching material is probably written like this: Definition 1, Definition 2, Proof 1, Proof 2, Example 1, Definition 3, Definition 4 ,......, I call it a lexicographically written statement. in this way, there is no problem in mathematics logic, and it is very strict. however, the book is intended for people, most of whom are beginners. The consequences of the lexicographic formal writing are mostly confused. the result is likely to fear, dislike, or even dislike mathematics. as we all know, when mathematics is in college, it is almost impossible for a person to learn mathematics well if he is not interested or even excluded from mathematics. many people learn mathematics, but find that they only do well-designed questions. I am confused when I study mathematics. no idea, no direction, no inspiration, etc. as a result, most people sigh that their mathematical skills are too poor and their IQ is too low.

To be honest, is there a big gap between human IQ and human IQ except for a few days? The same family is closely related to each other. I think IQ is similar. the gap between math levels is not an order of magnitude. for scibird, he is not the smartest in his family. but none of my father's and my mother's relatives have the same level of mathematics as mine. in addition, I have established a far-ahead advantage in mathematics in junior high school. I never thought this mathematical advantage was born.

I have summarized my experience: diligence + attitude + method.
The first is diligence. If we say that a genius is born, we cannot change it. So diligence can change.
Second, attitude, low profile, modest, enterprising. don't be greedy for the speed of your tongue, and be pretentious. if you want to improve your mathematics proficiency, you must be "Pretentious. it's better to sit down and read more books than to make it easier.
The method may take a long time. I can only say one thing: learning mathematics should include mathematical discovery and mathematical proof.

Next I will talk about the discovery of mathematics:
I have met people in and out-of-mathematics departments and their ideas are confusing. it seems that he assumes that there must be a simple proof method (how to prove existence ?). Everyone wants a simple and beautiful method, but it cannot prove that this "simple and beautiful" method exists. Second, it is not easy to obtain a simple and beautiful method. what we need to do is first prove it and then find out if there is any better way.

Everyone who knows the history of mathematics knows that mathematics is very pioneering! However, the first strict proof of the mathematical theorem is often neither so simple nor so elegant. simple and elegant proofs are often provided by later people. Most of the classic proof methods included in textbooks are not the first one. They are probably processed by more than N people. it may be totally different from the original proof.

The discovery of mathematics is often not as elegant as many people think. The methods of discovery may be confusing, boring, not strict, or even violent! But this is the way it is created. It's not just common sense, it's not restricted, it's even possible! Solving the problem is the first priority. As for the strict, professional, and elegant issue, we can solve it later. I think the best example is how the Fourier series is discovered:

Generally, a major with high mathematical requirements will take the course of mathematical equations, that is, partial differential equations. there is a classic solution called the separation variable method. however, what you might not expect is that the Fourier series of the famous earthquake was found in the process of using the separation variable method to solve the heat conduction equation !!

Next I will give a brief introduction to the general discovery process. For details about solving heat conduction equations, refer to any mathematical equation teaching material.
Using the bottom mark method of Tex, U _ {x} = U _ {x} (x, t) is used to represent the partial derivative of the function u (x, t) to X, the second-order partial derivative is recorded as U _ {XX }. consider the heat conduction equation as follows:

Thermal equation: U _ {XX} = U _ {t} (0 <x <π, T> 0)
Initial Value Condition: u (x, 0) = f (x)
Edge Value Condition: U (0, T) = u (π, t) = 0

Fourier hard hypothesis u (x, t) can separate variables, that is, u (x, t) = u (x) T (t)
The heat equation is used to obtain U "(x) T (t) = u (x) t' (t)
So u "(X)/u (x) = t' (T)/T (t) = const: =-λ
Therefore, it can be converted into two ordinary differential equations.
U "(x) + λ u (x) = 0
U (0) = u (/PI) = 0
And t' (t) + λ T (t) = 0
Through discussion, we know that λ> 0
Solve the above equation and then combine the solution (For details, refer to the mathematical equation tutorial)
Obtain the level solution u (x, t) = Σ bn * (exp {wn * t}) * sin (nx)
Then, by u (x, 0) = f (x), f (x) = Σ bn * sin (nx )-----(*)
This seems to mean that f (x) can be expanded into a triangular series. Note that (*) each item on the right is a strange function (conflict ?)
The same program can export f (x) = A0/2 + Σ an * Cos (nx)

If we take the definition field of X as symmetric about the origin at the beginning, and consider that any function can be split into "why ?) ". So Fourier boldly guessed that there should be a general function.

F (x) = A0/2 + Σ an * Cos (nx) + BN * sin (nx)

This is the Fourier series of the famous earthquake qiangu. Fourier also finds the expressions of the coefficients {An} and {bn}, which is the expression on the current textbook.

Of course there are still many problems, such as when to expand, how to converge, and how to converge to f (x) itself. but that's the end. what I just want to say here is that the mathematical discovery needs to be bold, and over-emphasized rigor can easily suppress creation. "bold guesses, strict proof! "------ This is the correct way to learn mathematics !!

8. Focus on the relationship between different branches.

If you want to learn mathematics, you must not only cultivate your own self-cultivation skills, but also cultivate your mind and raise your ideological realm! What is the ideological realm? Here it refers to the long term of looking at the problem. It not only shows the content of the current branch, but also shows the connections between different branches and promote each other.
A typical representative of this field needs to push the geometric chart first! Recall that we learned Euclidean's classical plane ry in our high school. Those questions and proof methods are very beautiful. They can be said to reflect the beauty of mathematics! However, if we observe it carefully, we will find that most of its content is related to the circle. If we leave the circle, it seems that the method is not much. however, even if we limit the number of circles, it turns out to be very difficult. The most typical example is to connect the guides, which is quite skillful! Later, the introduction of analytic ry broke this technical situation and indeed provided general procedural methods. maybe the advantages of plane analytic ry are not so obvious. It is definitely a substantial improvement to use space analytic ry (including space vectors) to solve the three-dimensional ry problem. I believe many people have this experience. some people may say that the analytic ry is ugly and violent, which damages the geometric beauty. I think it is better to complain than this because the analytical method does not reject the traditional method. Both methods can be used, and sometimes the problem solving may be more important than the beauty of mathematics. therefore, both of them have meanings.

Tools that introduce other branches on the basis of one branch can often open up a new world, introduce the analytical method into the ry to create the analytic ry, and move the calculus to the ry to produce the differential ry, furthermore, many similar examples have been developed, such as the differential manifold.

The solution to many famous problems in history also relies on methods of other branches. for example, the simplest solution of "shortest line" applies the law of refraction of light. different branches also have internal relationships, such as the "proper form" in the differential form. Its physical prototype is "potential field ". de Rham theorem is used in the analysis for the parity between the above-mentioned tone and the topology. when studying the problem, we should not be confined to the problem itself, but be brave and look as far as possible. to see the links of different branches, we need to see similarity from similarity.

The following article was prepared during my undergraduate course and posted on the Forum (possibly sinking into the sea of data) mainly to solve the auxiliary function construction problem of the mean value theorem. (try to find a more general method)
As we all know, the differential mean value theorem is a tough bone, and its skill is quite strong. the most difficult part of most problems lies in the Construction of auxiliary functions. Once an auxiliary function is constructed, the problem is solved in half. unfortunately, few textbooks have carefully explained how to construct helper functions. Many people feel that helper functions are collected through observation or.

The mean-value theorem of a differential seems unrelated to a differential equation. However, if you think of the process of constructing an auxiliary function as "solving an equation", the relationship between them is still very subtle. please read the following articles carefully. You may feel something!

Consider a typical medium-value theorem problem first!
It is known that F and G are continuous on [a, B]. In (a, B), the second order can be deduced and there is an equal maximum value. F (a) = g (), F (B) = g (B)
Proof: the existence of ε (a, B) makes f "(ε)-f (ε) = g" (ε)-g (ε ).

Let me talk about the idea of this question. Generally, problems involving second-order Median Theorem require the three zeros of a function as the transition, for example, this question. I have considered the following forms:
F (x) is continuous on [a, B]. In (a, B), level 2 can be imported. F (a) = F (B) = F & copy; = 0, a <C <B
Proof: the existence of ε (a, B) makes f "(ε) + pf '(ε) + QF (ε) = 0, where P, Q is a constant and (△= P ^ 2-4q ≥ 0 ----- This is a post-addition condition, which is the default condition below ). I tried to find a more general proof of the problem (natural, smooth), but finally failed. I was inspired to find a general solution by turning to Zhang zhusheng's new lecture on Mathematical Analysis (my favorite score book), the first section of high-level ordinary coefficient differential equations.
Traditional high mathematics (not high school mathematics --- haha !) In the section y "+ py' + Qy = 0 (*) of the second-order ordinary differential equation, we can infer that it is (*) by observing the derivative properties of the exponent function exp (λ x (*) to obtain the expression formula of the general solution. However, this is more like the Inverse Solution with high requirements.
. Can I solve it directly by using points? ----- Of course.

The first-order linear equation has been obtained, and the second-order ordinary coefficient differential equation:
Note that the two roots of R ^ 2 + PR + q = 0 are λ 1 and λ 2. When a differential operator D = D/dx is introduced, (*) is equivalent to (d ^ 2 + PD + q) y = 0,
Then, we obtain (D-λ 1) (D-λ 2) y = 0, so that l = (D-λ 2) y ----- (1), then (D-λ 1) L = 0 ----- (2 ),
From (2) can solve L, and then into (1) can find y.

* ***** The mean value theorem is closely related to the differential equations. For example, the first-order form f' (ε) + N (ε) f (ε) = 0 ---- (3 ).
Possible differential equations f' (x) + n (x) f (x) = 0,
Separate the variable, and then point it, f (x) = exp [-limit n (x) dx]
If f (x) exp [limit n (x) dx] = 1 ----- (4) is obtained, (3) is obtained by performing (4) derivation ), in most cases, you can directly set the Helper function f (x) = f (x) exp [limit n (x) dx] ----- (5). Generally, you only need to verify f (x) there are 2 zeros.

Next we return to the original question (d-λ 1) (D-λ 2) f = 0. F has three zeros, which implies that two formulas (5) may be used ).
If l = (D-λ 2) F is used, application (5) can prove that l has two zeros.
Certificate: set f (x) = f (x) exp [-λ 2x]. The release of L has two zeros;
Then let's set PHI (x) = L (x) exp [-λ 1x]. Then, we can introduce the presence of ε (a, B), so that
F "(ε) + pf '(ε) + QF (ε) = 0. A similar method can be used for higher items.

Promotion: n times polynomial Pn (x) has n different solid roots, introducing differential operator D = D/dx, Pn (d) as operator polynomial, function f (x) in [a, B], Order N can be exported and there are n + 1 different zeros.
Proof: the presence of ε (a, B) makes Pn (d) f (ε) = 0.
For example, P2 (x) = x ^ 2 + 5x + 1, P2 (d) F = f "+ 5f' + F