[Keywords:] mathematical thinking problem-solving method construction method
In mathematics learning, how to find a method to solve problems is an important problem that is often encountered. To solve some complex problems, we often need to re-create thinking by decomposing, combining, exchanging, analogizing, limiting and popularizing existing knowledge and methods, and construct new formulas or graphs to help solve the problems. This is the so-called construction method.
Euclid, an ancient Greek mathematician, was not only the founder of Euclid geometry, but also the founder of mathematical construction methods. In the Elements of Geometry, he skillfully proved for the first time the basic theorem named after him in number theory, "The number of prime numbers is infinite". Many mathematicians in history, such as Gauss, Euler and Lagrange, have successfully solved mathematical problems by using the construction method. The ingenious method of solving problems by construction method is to solve the original problem by constructing an auxiliary problem related to the original problem. If the auxiliary question is simpler and more intuitive than the original question, this way of thinking will be successful. The following examples illustrate the concrete application of construction method in solving mathematical problems in secondary vocational schools.
First, construct a proposition.
When some propositions are difficult to prove, we can solve the problem by constructing their equivalent propositions, lemmas or auxiliary propositions.
1. Construct an equivalent proposition.
Example 1: It is proved that a triangle with an area equal to 1 cannot be covered by a parallelogram with an area less than 2.
Analysis: This proposition is expressed in concise mathematical language as "If S△PQR= 1, SABCD"
It is proved that the parallel lines with P as AD intersect AB and CD at N and M (as shown in figure 1). Obviously:
S△PQK≤ S？ ANMD，
S△PPK≤ S？ Money, so:
S△PQK+S△PPK≤ S？ ABCD, that is
S△PQK≤ S？ Accelerated business collection and delivery system (adopted by the United States post office)
2. Construct auxiliary propositions.
Example 2: Known
Analysis: this question is to prove that the above two inequalities are established at the same time.
Namely "and",
Proving that the inequality ￠ holds, can be transformed into proving that all the following auxiliary propositions are true: x b1= a1,x b2 = a2, x bn = an, which is known and easy to know, and all the above auxiliary inequalities hold, then the original inequality is proved to hold.
Proof: ∫ …, (All letters are positive numbers)
∴×b 1=a 1，×B2 \u a2，…，×bn \u an .
Add the above inequalities to get:
(b 1+B2+…+bn)﹤a 1+a2+…+an，
∴﹤
The same principle can also be proved:, so the original inequality holds.
3. Structural lemma.
Example 3: Let the elliptic equation be += 1, and find the trajectory equation of the symmetric figure whose center trajectory is about point M (- 1, 1).
Lemma: The curve equation f(x, y) is known, and the curve equation symmetric about point M(x0, y0) is f(2x0-x, 2y0-y)=0 (the proof is abbreviated).
Solution: Let the center of the ellipse be (x, y). According to the meaning of the question, there are
x=2t
y=-t2
The ellipse center trajectory equation for eliminating parameters is:
f(x，y)=x2+4y=0
According to the lemma, the trajectory equation of m (- 1, 1) symmetric graph is f(-2-x, 2-y)=0.
Namely: (-2-x)2+4(2-y)=0,
Converted into: (x+2)2=4(2-y) is the ballistic equation.
Example 4: The straight line L passing through point A (1, 2) intersects with hyperbola x2- = 1 at two points p 1, p2, and the trajectory equation of point P in line segment p 1p2 is found.
Lemma: The midpoint locus equation of the chord intersecting the moving straight line L passing through the fixed point P(x0, y0) and the quadratic curve C: f (x, y)=0 is:
F(x, y)=F'(x0, y0)(x, y), (the proof is abbreviated).
Solution: According to lemma, the trajectory equation of point P is:
x2- - 1=2x- - 1
It is concluded that 2x2-y2-4x+y=0 is the trajectory equation.
Second, the structural model
When solving mathematical problems, if the conditions or conclusions are similar to the theorems, formulas and identities we have learned before, we can solve the problems through association and analogy with the help of theorems, formulas and identities. This method is called pattern construction. There are some conclusions (such as formulas, inequalities, etc. ) In mathematics, it can be used as a "model" to solve other mathematical problems with similar "shapes", and trigonometric identities are a widely used model.
Example 5: Find the range of the function y=.
Analysis: borrowing the famous Cauchy inequality as a "model";
(Abby) 2≤( ai2)( bi2)
The equal sign is true if and only if = = … =.
The characteristic of "mode" is that Abby is divided into the sum of squares of ai and bi, and "one bracket" is converted into "two brackets" (and vice versa).
To find the range of y2, try the combination of sin2x and cos2x.
Rewrite y=( 1-y)sinx+(3-2y)cosx,
According to Cauchy inequality are:
y2≤[( 1-y)2+(3-2y)2](sin2x-cos2x):
2y2-7y+5≥0, y≤ 1 or y≥
Third, the construction of special cases.
When solving the "at most" (or "at least") "existence" problem, a special case (special value or algebraic expression) can often be constructed to solve the problem.
Example 5: Let a column number AI (I = 1, 2, …, n), in which the sum of any three consecutive terms is positive and the sum of any five consecutive terms is negative. Verification: n≤6.
It is proved that the original proposition holds when the inverse is n≥7, which depends on the situation of seven numbers. If seven numbers are proved, it is proved to be n>7 points, and the proposition is not valid. Construct the following permutation table:
a 1，a2，a3
a2、a3、a4
a3、a4、a5
a4、a5、a6
a5、a6、a7
According to the conditions, fifteen numbers are positive in the horizontal direction and negative in the vertical direction, so the proposition of n≥7 does not hold.
And when n=6, the construction sequence: 3, -5, 3, 3, -5, 3 meets the requirements.
Fourth, the structure of graphics
For some mathematical problems, according to the known conditions or conclusions of the topic, appropriate analysis and association are carried out to construct graphs related to or satisfying the conditions, so as to achieve the purpose of solving problems. This method is called structural graph method. The essence of structural graphic method is to "turn numbers into shapes" and solve problems with the help of graphics. With the help of form, make full use of the intuitive characteristics of graphics, enter the problem situation as soon as possible, grasp the key to solving the problem, understand and analyze the problem at the height of the close combination of image thinking and logical thinking, and finally solve the problem. Constructing graphics is a traditional basic equation in classical geometry (like adding auxiliary lines is also a kind of construction). Graphics are not only the object of geometric problems, but also can be used to solve various problems that seem to have nothing to do with geometry at first glance.
Example 6: given 0 < x < 1, 0 < y < 1, verify:
+ + + ≥2 。
Analysis: If we use the knowledge of algebraic inequality to prove it, it is obviously complicated. As shown in Figure 2, the mathematical intuitive model unit square ABCD is established.
Prove that p is any point in the unit square ABCD, then
PD=，
PA=
PC=
PB=
∴ PA+PC≥AC，
PD+PB≥BD，
∴ PA+PB+PC+PD≥AC+BD=2. That is
+ + + ≥2 。
Example 7: Known geometric series:,,,, …
Seeking to be n? When Joan ∞, the limit of the sum Sn of the first n terms of this series.
Solution: Build a unit square as shown in Figure 3. Imagine the limit of the sum of geometric series as the limit of the sum (shaded part) of the areas of small rectangles and small squares in the graph. Obviously, this limit is equal to the area of the unit square 1. Namely: = 1.
V. Construction Equation
When encountering some problems of dealing with equivalence or some calculation problems, if a quantity cannot or is difficult to obtain directly, try to derive the equation it satisfies, and the method of solving the problem by solving the equation is called structural equation method. According to the needs of the problem, we can use the methods of finding roots, Vieta theorem, discriminant method, equation discussion method, basic theorem of algebra and so on to solve the problem.
Example 8: If a+b+c=m,+=, A, B and C are not equal, it is proved that one of A, B and C must be equal to M.
Analysis: If A, B and C are regarded as unknown quantities, it can be known from the conditions that their sum is M, and the sum of pairwise is ab+bc+ca=, so that after abc is set, the equation for finding the root can be constructed according to Vieta's theorem of cubic equation.
It is proved that if abc=n, then ab+bc+ca=, then A, B and C are three roots of the equation t3-mt2+ t-n=0, and the equation (t-m)(t2+ )=0 has a root t 1=m, that is, one of A, B and C must be equal to M.
Sixth, the builder
When solving some mathematical problems, an appropriate auxiliary function is constructed by using the concept and properties of the function. The method to study the properties of this auxiliary function is called the constructor method. Function is one of the centers of mathematical knowledge. An equation can be regarded as a case where the function value is zero, and an inequality can be regarded as an unequal relationship between two functions. So equations and inequalities are special forms of functions. The premise and foundation of constructing function is to be familiar with the concept of function and firmly grasp the properties of various elementary functions. The process of constructing functions needs our keen observation, correct judgment, reasonable selection of suitable functions and accurate use of the properties of functions. Some mathematical problems can be solved by connecting some variable quantities to construct functions, and then using the properties of functions; Some problems are essentially related to a certain property of the function, which can be summed up as studying the properties of the related function and then constructing an auxiliary function to solve the problem.
Example 9: Verification: For all real numbers x, there are:
≤ ≤7
Prove: constructor y=, now you only need to prove: ≤y≤7.
This is actually the problem of finding the function value domain, using the discrimination method.
y(x+3x+4)=x2-3x+4
∴(y- 1)x2+y+ 1+3x+4(y- 1)= 0
X∈R, so △≥0, i.e.
9(y+ 1)2-4(y- 1)2？ 6? 14 ≥ 0, simplified, and obtained:
7y2-50y+7≤0, that is, (7y- 1)(y-7)≤0.
So ≤y≤7.
Example 10: A, b, c, d, e∈R are known and satisfy:
a+b+c+d+e=8，a2+b2+c2+d2+e2= 16
Try to determine the maximum value of e (the seventh American middle school mathematics competition);
Solution: Because a+b+c+d=8-e, a2+b2+c2+d2= 16-e2.
According to the above formula, a quadratic function with coefficients a, b, c and d is constructed as an auxiliary tool, from which the inequality of e is transformed. Construct a quadratic function:
f(x)= 4x 2+2(a+b+c+d)x+(a2+B2+C2+D2)
=(x+a)2+(x+b)2+(x+c)2+(x+d)2≥0
Since the quadratic coefficient of the quadratic function is 4 > 0, and f(x)≥0, △≤ 0: 4 (A+B+C+D) 2-16 (A2+B2+C2+D2) ≤ 0 is obtained.
According to the known conditions: 4(8-e)2≤ 16( 16-e2).
Solution: 0≤e≤, when a=b=c=d, it is given by emax=
Seven, the construction formula
Example 1 1: verification: cot? Zhuo -8cot? Cut = Tan? Zhuo +2tan 2? Cut = 4? chisel
Proof: Try to construct a recurrence formula of double angles, which is easy to prove.
A cot? Zhuo tanqie =2cot 2? Shear (1)
Recursion: 2cot? Zhuo tanqie =4cot 4? Shear (2)
4cot 4？ Zhuo -4tan 4? Cut =8cot 8？ Shear (3)
The above formulas (1), (2) and (3) are summed up: cot? Zhuo -8cot 8? Cut = Tan? Zhuo +2tan 2? Cut = 4? Zhuo was established.
Eight, structural analytical formula
The method of constructing appropriate relations to help explore the idea of solving problems is called the method of constructing analytical expressions. This method can often bring great convenience. The general mode of constructing analytical formula is: according to the characteristics of the problem, construct a related relationship to replace or simplify the original problem, so as to promote the original problem to be completely solved.
Example 12: Verification:++…+= 2n
It is proved that it is not difficult to construct the expression by association because the left side of the equation is the sum of the coefficients of binomial expansion.
(a+b)2=an+an- 1b +…+bn
The equation to be proved is a special case of expansion when a=b= 1 If a=b= 1, the expansion becomes 2n =++…+,which proves the original formula. In this paper, the binomial theorem is used to transform the problem into a special case of a=b= 1, which greatly simplifies the problem.
IX. Structural sequence
When solving a problem, the method of constructing an appropriate sequence to solve the problem according to the known conditions of the topic is called sequence construction method. The premise of constructing sequence method is to flexibly use the concept and nature of sequence, find out the relationship between the known conditions or conclusions of the topic and sequence, and then use the knowledge of sequence to solve the problem.
Example 13: known crime? Z +Cos？ We =? Z ∈(0，？ What about Kurt? The value of this is _ _ _ _. (1999 college entrance examination questions)
Solution: by conditional crime? Z +Cos？ Z =, structural arithmetic progression: Sin? z，Cos？ Therefore, let its tolerance d be: sin? Z = -d，Cos？ Z = +d
By Sin2? Z +Cos2？ Based on this = 1, we can get:
(-d)2+( +d)2= 1, and the solution is d=+.
∵ 0