A proof of mathematical analysis 30

I won't prove the first one, and I don't use it very often. But ideas can be provided. In general, the construction method is used to construct convergent subcolumns that conform to the topic.

First of all, because it is continuous in the closed interval, the points of h(x)=0 are limited, so it is better to set it to n. Then set it to x1 , xn

Let x0=0 ,xn+1=1 take (x0,x1) (x1,x2).xn,xn+1) (a total of n+1 points) to get n+1 intervals, and the first interval is not needed. Each of the remaining n intervals has a zero point.

Using the dichotomy, you can get a convergent subcolumn. To put it bluntly, this proof is that a bounded sequence must have a converging subcolumn.

The rest is up to you.

Question 2: The so-called open set means that there is another element in any neighborhood in the open set. It's enough to prove this, just prove it directly according to the definition.

Since h(x) is continuous within (0,1), there is, for element x0 in any u.

For any e (inverted e), there is o(delta) such that |x-x0|0 is known by the preserving sign that there is o(delta) such that x belongs to {x| |x-x0|0.

Apparently {x|.} |x-x0|0

For any x0 that belongs to u, it is true, so u is the open set.

The first h is a continuous function, and the precursor of the closed set is the closed set. The handover compact set [0,1] is a closed subset of the compact set, so it is tight.

The second, h is continuous, open set] 0, infinity['s preface is open set, open set is open set.

This is the first median theorem about integrals: the full description is: if the functions f(x) and g(x) are bounded and integrable in the interval [a,b], f(x) is continuous, and g(x) does not change in the interval [a,b], then there is at least one number in the district or interval [a,b< <

Detailed evidence is provided in general mathematical analysis textbooks. Proof idea: Let g(x)>0, firstly, use the maximum-value theorem of continuous functions on closed intervals to obtain the inequality, <>

The inequality is then obtained by using the valuation theorem of definite integrals.

Finally, the integral median theorem is applied to obtain the conclusion of the problem.

Take advantage of mean inequalities.

<> due to 1≠2≠3 ≠......n–1≠n

So you can't get an equal sign.

i.e. (n! )^1/n) ＜1＋2＋……n) n=(n 1) 2 so n! ＜[n＋1)/2]^n

This topic has just been seen on the headlines of someone ** to answer this question, method one you can try to use mathematical induction, method two is the mean inequality, first square both sides, and then use the deflation, so that you can also get a conclusion, of course, this mean inequality is an inequality of n yuan,

That is, to prove that the geometric mean is less than the arithmetic mean, there are several ways to do it.

Evidence 1:

It can be seen from the fact that the algebraic mean is greater than the geometric mean, (1+2+3+......n)/n>(1*2*3*……n) to the nth power, so n(n+1) 2n>(n!) to the nth power, so n!<[n+1)/2]^n。

Evidence 2: Evidence 2: Because 0<1*n<[(1+n) 2] 2,0<2*(n-1)<[1+n) 2] 2,......0<(n-1)*2<[(1+n)2] 2,0 multiplied by 1 2*2 2*......n^2<[(1+n)/2]^2n(n!n<[(1+n) 2] 2n, so n!<[n+1)/2]^n

As long as it is proved that there is a function value greater than 0 in (n m,sqrt(2)), it can be proved that (n m) 2 is not in (1,2). In this case, the interpolation method is adopted, using t to represent the value of the function of 2-x 2 at n m, and constructing r as an expression related to t.

Assuming r= tn m, then (n m+r) 2=(n m) 2+2nr m+r 2

n/m)^2+2αt(n/m)^2+α^2t^2(n/m)^2

T(n m) 2<2t, t 2(n m) 2<2t ((n m) 2<2 and t<1 were used

So(n m+r) 2<(2-t)+4 t+2 2t

To (n m+r) 2<2, (2-t)+4 t+2 2t<2 is required, i.e., 2 2+4 -1<0

-sqrt(6) 2-1< sqrt(6) 2-1 , so take =1 6 to complete the proof.

Let '=inf f(x), =inf g(x), =inf[f(x)g(x)],=sup f(x)

Defined by the lower definite, for >0, x d makes f(x)g(x)<

For 'and '' are the lower definites of f(x) and g(x) on d.

Therefore, for any x d, there is f(x) g(x) and f(x), g(x) is not negative, so f(x)g(x).

So + f(x)g(x) is determined by arbitrariness,

i.e. inf[f(x)g(x)] inf f(x) inf g(x).

Details: If < is arbitrary, it can be made = ' 0, then + = ' and the definition of + i.e. ' is a contradiction, so it can only be

By '' is the lower bound of g(x) on d, and for >0, x d, g(x)-

So 'g(x)- f(x)g(x)- where the last unequal sign is used to be the lower bound of f(x)g(x) on d.

It consists of arbitrariness and boundedness, i.e. inf[f(x)g(x)] sup f(x) inf g(x).

First, it is proved that (1+x) n 1+nx holds the proposition when n=1 holds for any x (-1, ) and the positive integer n.

Suppose n=k is true, i.e., (1+x) k 1+kx is due to x>-1, and the direction of the unequal sign does not change after multiplying (1+x) on both sides, and there is (1+x) (k+1) (1+kx)(1+x)=1+(k+1)x+kx

1+(k+1)x

That is, the proposition is true when n=k+1.

Therefore, (1+x) n 1+nx is true for any x (-1, ) and positive integer n, and the condition is kx = 0, i.e., x = 0

So if and only if x=0, there is (1+x) n=1+nx=1

The function derivative method is the most basic method for univariate inequality.

f（h)=（1+h）^n-（1+nh）

f'(h)=n（1+h）^（n-1）-n

When -1 h 0 f'＜0

When h 0 f'＞0

So f takes a minimum value of 0 at h=0

Hence f(h) 0

This is the mean inequality with (n+1) elements.

The geometric mean on the left and the arithmetic mean on the right.

Geometric mean Arithmetic mean.

Please refer to the web link for details.

How to take the n+1 numbers above has been written, where n is 1+1 n, and 1 is 1, and then just use the mean inequality.

The equation on the right is actually the following:

Proof: (1) Assuming that it is not an irrational number, then the difference between two rational numbers a+x-a=x is a rational number, which is contradictory. (2) Assuming that it is not an irrational number, then the quotient ax a=x of two rational numbers is a rational number, which is contradictory.

You may wish to set a0, so the original formula <1; (a+x) (b+x)-a b=x(b-a) b(b+x)>0, the same can be proved after (a>b, the conclusion is reversed).

Proof: (Counterproof) Assuming that it is not an irrational number, then the root number p can be expressed as a b, a, b are positive integers, then p=a2 b2, i.e., a2=p b 2

Push out P below and it's squared.

a=[x1,x2], x1 can be expanded into a set of orthogonal bases, and the scrambling beats are then re-unitized, i.e., there is an orthogonal matrix so that the He type t at= triangular matrix, then |λe-a|=|e-t¹at|, at are all real matrices, then slow shed|λe-a|=|e-t¹at|=（1）*（2）

So a has a real eigenvector.