Mathematical proof of mathematical progress

Let a(x) and b(x) be even functions, then a(-x)=a(x), b(-x)=b(x);

c(x) and d(x) are odd functions, then c(-x) = -c(x) and d(-x) = -d(x).

The sum of even functions f(x) = a(x) + b(x), f(-x) = a(-x) + b(-x) = a(x) + b(x) = f(x), that is, the sum of two even functions is an even function;

The sum of odd functions g(x) = c(x) + d(x), g(-x) = c(-x) + d(-x) = -c(x)-d(x) = -g(x), that is, the sum of two odd functions is an odd function;

The product of even functions m(x)=a(x)b(x), m(-x)=a(-x)b(-x)=a(x)b(x)=m(x), that is, the product of two even functions is an even function;

The product of odd functions n(x)=c(x)d(x),n(-x)=c(-x)d(-x)=(-c(x))(d(x))=c(x)d(x)=n(x), that is, the product of two odd functions is an even function;

The product of even and odd functions y(x) = a(x) c(x), y(-x) = a(-x) c(-x) = a(x) (-c(x)) = -a(x) c(x) = -y(x), i.e., the product of even and odd functions is an odd function. Certification.

It can only be proved by definition.

1. Let f(x) be an even function and g(x) be an even function, then f(x)+g(x) is an even function, because f(-x)=f(x) and g(-x)=g(x).

So f(-x)+g(-x)=f(x)+g(x), so f(x)+g(x) is an even function.

Let f(x) be an odd function and g(x) be an odd function, then f(x) + g(x) be an odd function.

Because f(-x)=-f(x) g(-x)=-g(x), then f(-x)+g(-x)=-f(x)-g(x)=-[f(x)+g(x)].

So f(x) + g(x) are odd functions.

The second question is the same, which is proved according to the definition of the parity function.

1.Comparative method Comparative method is one of the most basic and important methods to prove inequality, it is a direct application of the order of the magnitude of two real numbers and the nature of the operation, the comparative method can be divided into difference comparison method (referred to as difference method) and quotient value comparison method (referred to as business method).

2.The comprehensive method uses known facts (known conditions, important inequalities or proven inequalities) as the basis, with the help of the properties of inequalities and relevant theorems, and finally introduces the inequalities to be proved after step-by-step logical reasoning3.

Analytical method The analytical method refers to starting from the inequality that needs to be proven, analyzing the sufficient conditions for the establishment of this inequality, and then transforming it into determining whether the condition is met4.The proof of some inequalities in the counterproof method is difficult to explain clearly from the positive evidence, and it can be considered from the perspective of positive and difficult and negative, that is, to prove the inequality a>b, first assume a b, and deduce the contradiction from the question and other properties, so as to affirm a>b

Counter-proof may be considered when the proof inequality involved is a negative proposition, a unique proposition, or contains words such as "at most", "at least", "non-existent", "impossible", etc.

5.The substitution method is to introduce one or more variables for substitution of inequalities that are more complex in structure, with many variables and unclear relationships between variables, so as to simplify the original structure or achieve some transformation and adaptation, and bring new enlightenment and methods to the proof. There are two main forms of substitution.

1) Triangular substitution filial piety fan rough method: mostly used for the proof of conditional inequality, when the given conditions are more complex, a variable is not easy to be represented by another variable, then triangular substitution can be considered, and the two variables have the same parameter representation. If this method is used properly, it can communicate the connection between trigonometry and algebra, and transform complex algebraic problems into trigonometric problemsAccording to specific problems, the trigonometric substitution methods implemented are:

If x2+y2=1, you can set x=cos and y=sin; If x2+y2 1, x=rcos ,y=rsin (0 r 1); For inequalities contained, due to |x|1. x=cos can be set; If x+y+z=xyz, known from tana+tanb+tanc=tanatan-btanc, we can set x=taaa, y=tanb, z=tanc, where a+b+c= .2) Incremental commutation method: In the inequality of symmetry (arbitrary exchange of two letters, algebraic formula unchanged) and given alphabetical order (such as a>b>c, etc.), consider using the incremental method to commute, the purpose of which is to achieve the reduction of the element through the commutation, so that the problem is difficult to change to easy, and the complex is simplified.

For example, a+b=1, you can use a=1-t, b=t or a=1 2+t, b=1 2-t for commutation.

6.The deflation method is to prove inequality a

First, the derivative is used to find the maximum value of the hand n n. Then magnify the left side of the above equation to (n 1) and multiply it by the maximum value. Now it is enough to prove that the formula to the right of the above equation is larger than the formula [(n 1) times the maximum].

However, the numerator on the right side of the above equation can be factored into (n-1) (2n+1) and can therefore be reduced. All you need is (2n+1) (4n+4) >. From the derivative, it can be seen that (2n+1) (4n+4) increases monotonically.

Therefore, it is only necessary to prove that the minimum value is greater than. This easy to prove is therefore the original form. The above gait postures can be carried out and verified by myself.

A series of deductive reasoning confirms that a conclusion is true.

Proof of congruence.

If you don't put the isosceles jujube shoot abc, ab=ac, the rock to be proved is abc= acb

Over a as ad bc on d, then.

In RT ADB and RT ADC, there are.

ad=ad,ab=ac

adb≌△adc （hl）

abc=∠acb

Evidence of the envy. Thank you.

Inverse theorem: The two angles of the bucket wheel nucleus on the same bottom are equal in an isosceles trapezoidal hollow dig.

Make a perpendicular line from the two ends of the other bottom edge to this base, and the perpendicular line, the waist and the bottom line form two right-angled triangles.

Since the waist is equal, the perpendicular length is equal, and the two triangles are equal. then the two corners on the bottom are equal.

Let sn= k=(0,n) and (k-1) 2 k= (0-1) 2 0 + 1-1) 2 1+...n-2)/2^(n-1)+ n-1)/2^n

2sn= (0-1)/2^(0-1)+(1-1)/2^(1-1)+ 2-1)/2^(2-1)+.n-1)/2^(n-1)

Subtract to get sn=(0-1) 2 (0-1)+1 2 0+1 2 1+.1/2^(n-1)-(n-1)/2^n

2+(1-(1/2)^n)/(1-1/2)-(n-1)/2^n

2(1/2)^n-(n-1)/2^n

n-> infinite sn->0

sn=(2+1)x^2+(2*2+1)x^4 +.2n+1)x^(2n)

x^2sn = 2+1)x^4+ (2*2+1)x^6+..2n-1)x^(2n)+ 2n+1)x^(2n+2)

Subtraction. (1-x^2)sn=(2+1)x^2+2x^4+2x^6+..2x^(2n)- 2n+1)x^(2n+2)

sn=[ 2+1)x^2+2x^4+2x^6+..2x^(2n)- 2n+1)x^(2n+2) ]1-x^2)

x^2+2x^2(1-(x^2)^n)/(1-x^2)- 2n+1)x^(2n+2) ]1-x^2)

If |x|> = 1, n-> infinite, divergent.

If |x|<1, n-> infinity.

sn->[x^2+2x^2/(1-x^2)]/1-x^2)=(3x^2-x^4)/(1-x^2)^2

Let f(x)= (x+1)- x, f(x)=1 ((x+1)+ x), it can be found that f(x) is subtracted, so f(x) >f(x+1). Therefore, (x+1)- x> (x+2)- x+1).