Postgraduate Mathematics Questions, Postgraduate Topics, Mathematics?

I just used the review book I have probably read Chen Wendeng's review guide and carefully read a little review of the whole book.

Cai Zihua's review is still good, but the question types of the example questions are basically comprehensive, but they are not completely comprehensive.

There is no comparison of infinitesimal scales like the first chapter, but the exercises involve yes.

Cai Zihua's review encyclopedia, some of the problem solving methods are stupid, and there are a few strange questions, but the overall is still good.

Cai Zihua's 1500 questions and 500 questions don't need to be bought The amount of questions is a little too big The selection of test questions is not good, it is just to make up the number of objective questions of 1500 and subjective questions of 500 There are too many and you don't have time to do it Using that book will outweigh the losses!

You are also more clear about the characteristics of Chen Wendeng and Li Yongle's books Don't say much The content of the review book is quite a lot more than Cai Zihua's review book It depends on your own preferences No matter which book you use The linear algebra part is Li Yongle's linear algebra tutorial handouts Cai Zihua's modern part can not be read Because that book is too classic!

In fact, you don't need to buy 660 questions There are many conceptual problems in this book that you can't use in graduate school entrance exams Anyone who does well in the exam will not emphasize 660 questions.

I suggest you buy a copy of Wu Zhongxiang's past past questions classification and analysis He arranged the 22-year past papers together according to chapters And the questions and analysis are separated Review Daquan (whole book) is mainly example questions It is difficult to cultivate your hands-on ability And repeatedly read the kind of big book and forget the front After reading the front and forgetting the back and forgetting the back It is difficult to grasp the key point Which book is a more effective method with review Many people have read the whole book n times and still failed to take the test, which is the reason.

Don't think that you won't have to do it in the future Past papers are not used to test yourself, but to study It is often difficult to really position yourself by using past papers to test yourself After all, many questions in reference books are selected from past papers.

It is necessary to master the basics and the ideas of the real questions well and prepare for the crazy set of questions in the future!

Li Yongle's review of the whole book is better, but Chen Wendeng's is more difficult!!

If you want to study mathematics for this postgraduate entrance examination, it is best to consider changing the yuan, and it should be easier to change the yuan.

Since ln(1+1 n)<1 n (n=1,2,3,...）

So the first n terms of the harmonic series are satisfied and satisfied.

sn=1+1/2+1/3+…+1/n>ln（1+1）+ln（1+1/2）+ln（1+1/3）+…ln（1+1/n)

ln2+ln（3/2）+ln（4/3）+…ln[(n+1）/n]

ln[2*3/2*4/3*…*n+1）/n]=ln(n+1）

Since lim sn(n)lim ln(n+1)(n)=+

So the limit of the SN does not exist, and the harmonic series diverges.

But the limit s=lim[1+1 2+1 3+....+1 n-ln(n)](n) ) exists, because.

sn=1+1/2+1/3+…+1/n-ln(n)>ln（1+1）+ln（1+1/2）+ln（1+1/3）+…ln（1+1/n)-ln(n)

ln(n+1）-ln(n)=ln（1+1/n)

Since lim sn(n)lim ln(1+1 n)(n)=0

Thus the SN has a nether.

And sn-s(n+1)=1+1 2+1 3+....+1/n-ln(n)-[1+1/2+1/3+…+1/(n+1）-ln(n+1）]

ln(n+1）-ln(n)-1/(n+1）=ln（1+1/n)-1/(n+1）

ln(1+1 n), take the first two terms, because the sum of the discarded terms is greater than 0.

ln（1+1/n)-1/(n+1）>1/n-1/（2n^2）-1/(n+1）=1/(n^2+n)-1/（2n^2）>0

i.e. ln(1+1 n)-1 (n+1)>0, so sn decreases monotonically. From the monotonic bounded series limit theorem, it can be seen that sn must have a limit, therefore.

s=lim[1+1/2+1/3+…+1 n-ln(n)](n) exists.

Provide a reference.

Hello, if you are a graduate student, there must be a second Li Quanshu, don't tell me you don't know the second Li Quanshu.

The first main line of this question is: monotonous and bounded do not need to converge.

When proving monotonicity, using one of the methods in the book (probably on page 10 of the book, method 9 of finding the limit), the monotonicity of the recursive series is related to the monotonicity of the function, and the monotonicity of the function is proved first, and then the monotony of the series.

Finally, it turns out that there is boundedness.

It's so hard, hehe, I've forgotten it.

1.Wrong. f(x) is continuous and not 0, it is not possible to tell whether f(x)>0 or f(x)<0;The discontinuity of the same g(x) does not know whether it is greater than zero, equal to zero, or less than zero.

If g(x) has a break at 0, g[f(x)] has no break.

2.Wrong. For example, if the discontinuity point is g(0)=-1 and lim[x->0+]g(x)=1, then the square is 1 and becomes continuous.

3.Wrong. For example: f(x)=1.

4.Right.

Derivative of the equation y=ax2+bx+c yields y'=2ax+b.Substituting the tangent slope k=y'=2ax0+bThe point-oblique equation for listing the straight line yields y-y0=(2ax0+b)(x-x0)

This is the tangent equation, which is substituted by x=0, y=0 to get y0=x0(2ax0+b).And since y0=ax02+bx0+c, so the bippial ax02=c, and since bx0 is reduced, b is an arbitrary real number. Deformed from ax02=c, c a=x02 x02 is equal to zero, so c a 0

Do you understand?