Graduate School Entrance Examination High Mathematics Problems Solved by Masters Next 10

Problem 1: I don't understand much. Does it mean that "an+1 an<= 1" is introduced from "positive series convergence"? According to the definition of ratio discriminant, when n tends to infinity, if the ratio of the posterior term to the previous term of the positive term series is less than 1, it converges.

Greater than 1, divergent. When it is equal to 1, the ratio method is invalid and other methods are needed to judge, so the equal sign cannot be concluded by the ratio judgment method.

The so-called inverse proposition.

No, it does not mean that the ratio discriminant method is not true, but that the convergence of any series cannot be judged by "an+1 an< 1", because this judgment is not valid without knowing whether the series is a positive series.

Question 2, there is no special rule, it depends on whether the formula can be set. Derivation or integration is premised on summing the series, so if you find that n=0 cannot be simplified by an existing formula, but n=1 can be simplified, you can separate n=0 from the whole, and the rest of the sum starts from n=1. The same is true for other cases, such as the addition of splitting into two sums.

First question: Can positive series convergence deduce an+1 an<= 1 ?

The answer is no (even when n tends to infinity, the conclusion an+1 an<= 1) cannot be obtained

Counterexample: when n is odd, an=1 n 2, when n is even, an=1 n 3 then an is constant and the series converges, but an does not decrease monotonically.

Reason: "an+1 an<= 1" and "an+1 an>1 since a certain term" are not mutually negative, and it is possible that neither is true.

The second question, which you asked too vaguely, may be n=0, an=0If you want to know more details, please ask the question to be more explicit.

1. First of all, it can be guided continuously.

So in x=x. A decentric neighborhood can be derived at x=x. of a decentric neighborhood in a continuous; The title is back:

f(x) in the answer x=x. This point is continuous. Well, the whole x = x.

's neighborhoods are all continuous. Since they are all continuous, then this conclusion can be proved by the definition of derivatives.

2. The discontinuity point of the derivative function can only be the second type of discontinuity, and the absolute value of f(x)=x is not a derivative function.

3. Original = i (n+1) Subscript i=1 Superscript n+1 = 1 (n+1) i Subscript i=1 Superscript n+1 = (definite integral) xdx The upper limit is 0 to 1. In the postgraduate entrance examination, there is a problem such as finding the limit, which is to be solved by definite integrals, and there is a similar problem here, which the landlord can take a look at.

Question 1 Do you put f'(x0) and limf'(x) x x0 confused copy bai, or you send a picture, I can't understand what you say du

Question 2 If there is a discontinuity in the derivative function of the zhi number, it must be the discontinuity of the second DAO class, and the example you gave is a non-derivative function.

Question 3 First of all, you need to understand that the definition of a definite integral is the limit of the sum of infinite multinomials, for the sake of convenience, the integral interval is 0 to 1 (so that b-a=1 can make the situation simple), and dx is 1 n, and the part of i n in the general term is x, so that it becomes a definite integral.

Alas, I took the exam 2 years ago, and I forgot about it.

The integration field is a triangle with o(0, 0), a(1, 0), b(1, 1) as vertices, and the integration order is exchanged.

i = ∫<0, 1>(tanx/x)dx∫<0, x>dy = ∫<0, 1>tanxdx

-lncosx]<0, 1> = 1 - lncos1

0 < x -3 = (x n-3) [sqrt(x n+6)+ 3] Add: No, isn't it enough to write like this, then I'll write two more sentences.

According to the previous ones, it can be summarized.

x_n-3|Then whatever you use the entrapment criterion or the definition, there is no hope for any more.

Assuming that the limit of the series is a, then there is a+1= (6+a), and the value of a can be obtained by solving the equation, and then the definition of the limit is analyzed: when n tends to infinity, abs(xn-a) is an infinitesimal quantity.

Let the limit be a then substitute a into the upper test, a = (6 + a) to find a is to assume that the limit exists and find the method, and then prove according to the limit definition, the landlord himself substitute it.

Let y f(x), then g(y) x, so g(f(x)) g(y) x, and in the same way let y g(x), then f(y) x, so f(g(x)) f(y) x, so f(g(x)·g(f(x)) x·x, or the same as below.

I don't know what x you are asking.

If you ask about the eigenvalue of a eigenvector, this eigenvalue is calculated and not taken arbitrarily.

Eigenvalues:

1. Write the characteristic equation|λe-a|=0；Bihao or a is the coefficient matrix of the original equation, and e is the unit matrix;

2. Solution. This is the eigenvalue. Bend.

If you ask how to assign the basic solution of the eigenvector, the position of the unknown can be assigned at will, generally one unknown is 1, and all other unknowns are 0, which is convenient to solve.

<> is like Qiao, and Kaibu is noisy and filial piety.

<> such as the hail of Yuliang sail slag Qitu town.