How to determine the periodicity of a function. Properties of periodic functions Rules for finding the period of periodic functions

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Algebra and the beginning of analysis, grade 10 (profile level) A.G. Mordkovich, P.E. Semenov Teacher Volkova S.E.

Definition 1 A function y = f (x), x ∈ X is said to have period T if for any x ∈ X the equality f (x - T) = f (x) = f (x + T) is true. If a function with a period T is defined at a point x, then it is also defined at the points x + T, x - T. Any function has a period equal to zero at T = 0, we get f (x - 0) = f (x) = f ( x + 0) .

Definition 2 A function that has a non-zero period T is called periodic. If a function y = f (x), x ∈ X, has a period T, then any multiple of T (i.e., a number of the form kT, k ∈ Z) is also its period.

Proof Let 2T be the period of the function. Then f(x) = f(x + T) = f((x + T) + T) = f(x + 2T), f(x) = f(x - T) = f((x - T) -T) = f(x - 2T). Similarly, it is proved that f(x) = f(x + 3 T) = f(x - 3 T), f(x) = f(x + 4 T) = f(x - 4 T), etc. So f(x - kT) = f(x) = f(x + kT)

The smallest period among the positive periods of a periodic function is called the main period of this function.

Features of the graph of a periodic function If T is the main period of the function y \u003d f (x), then it is enough: to build a branch of the graph on one of the intervals of length T, to perform a parallel translation of this branch along the x axis by ±T, ±2T, ±3T, etc. . Usually choose a gap with ends at points

Properties of periodic functions 1. If f(x) is a periodic function with period T, then the function g(x) = A f(kx + b), where k > 0, is also periodic with period T 1 = T/k. 2. Let the function f 1 (x) and f 2 (x) be defined on the entire real axis and be periodic with periods T 1 > 0 and T 2 >0. Then, for T 1 /T 2 ∈ Q, the function f(x) = f(x) + f 2 (x) is a periodic function with period T equal to the least common multiple of the numbers T 1 and T 2 .

Examples 1. The periodic function y = f(x) is defined for all real numbers. Its period is 3 and f(0) =4 . Find the value of the expression 2f(3) - f(-3). Solution. T \u003d 3, f (3) \u003d f (0 + 3) \u003d 4, f (-3) \u003d f (0–3) \u003d 4, f (0) \u003d 4. Substituting the obtained values ​​into the expression 2f (3) - f(-3) , we get 8 - 4 =4 . Answer: 4.

Examples 2. The periodic function y = f(x) is defined for all real numbers. Its period is 5, and f(-1) = 1. Find f(-12) if 2f(3) - 5f(9) = 9. Solution T = 5 F(-1) = 1 f(9) = f(-1 +2T) = 1⇨ 5f(9) = 5 2f(3) = 9 + 5f(9) = 14 ⇨f(3)= 7 F(-12) = f(3 – 3T) = f (3) = 7 Answer: 7.

References A.G. Mordkovich, P.V. Semyonov. Algebra and the beginnings of analysis (profile level), Grade 10 A.G. Mordkovich, P.V. Semyonov. Algebra and beginnings of analysis (profile level), Grade 10. Methodological guide for the teacher

On the topic: methodological developments, presentations and notes

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Repeating its values ​​at some regular interval of the argument, that is, not changing its value when some fixed non-zero number is added to the argument ( period functions) over the entire domain of definition.

More formally, the function is said to be periodic with period T ≠ 0 (\displaystyle T\neq 0), if for each point x (\displaystyle x) from its point definition area x + T (\displaystyle x+T) and x − T (\displaystyle x-T) also belong to its domain of definition, and for them the equality f (x) = f (x + T) = f (x − T) (\displaystyle f(x)=f(x+T)=f(x-T)).

Based on the definition, the equality also holds for a periodic function f (x) = f (x + n T) (\displaystyle f(x)=f(x+nT)), where n (\displaystyle n)- any integer.

However, if a set of periods ( T , T > 0 , T ∈ R ) (\displaystyle \(T,T>0,T\in \mathbb (R) \)) there is a smallest value, it is called main (or main) period functions.

Examples

Sin ⁡ (x + 2 π) = sin ⁡ x , cos ⁡ (x + 2 π) = cos ⁡ x , ∀ x ∈ R . (\displaystyle \sin(x+2\pi)=\sin x,\;\cos(x+2\pi)=\cos x,\quad \forall x\in \mathbb (R) .)

• The Dirichlet function is periodic; its period is any non-zero rational number. It also does not have a main period.

Some features of periodic functions

• The sum of two functions with incommensurable periods is not always a non-periodic function.

Purpose: to generalize and systematize students' knowledge on the topic "Periodicity of functions"; to form skills in applying the properties of a periodic function, finding the smallest positive period of a function, plotting periodic functions; promote interest in the study of mathematics; cultivate observation, accuracy.

Equipment: computer, multimedia projector, task cards, slides, clocks, ornament tables, folk craft elements

“Mathematics is what people use to control nature and themselves”
A.N. Kolmogorov

During the classes

I. Organizational stage.

Checking students' readiness for the lesson. Presentation of the topic and objectives of the lesson.

II. Checking homework.

We check homework according to samples, discuss the most difficult points.

III. Generalization and systematization of knowledge.

1. Oral frontal work.

Questions of theory.

1) Form the definition of the period of the function
2) What is the smallest positive period of the functions y=sin(x), y=cos(x)
3). What is the smallest positive period of the functions y=tg(x), y=ctg(x)
4) Use the circle to prove the correctness of the relations:

y=sin(x) = sin(x+360º)
y=cos(x) = cos(x+360º)
y=tg(x) = tg(x+18 0º)
y=ctg(x) = ctg(x+180º)

tg(x+π n)=tgx, n ∈ Z
ctg(x+π n)=ctgx, n ∈ Z

sin(x+2π n)=sinx, n ∈ Z
cos(x+2π n)=cosx, n ∈ Z

5) How to plot a periodic function?

oral exercises.

1) Prove the following relations

a) sin(740º) = sin(20º)
b) cos(54º ) = cos(-1026º)
c) sin(-1000º) = sin(80º )

2. Prove that the angle of 540º is one of the periods of the function y= cos(2x)

3. Prove that the angle of 360º is one of the periods of the function y=tg(x)

4. Transform these expressions so that the angles included in them do not exceed 90º in absolute value.

a) tg375º
b) ctg530º
c) sin1268º
d) cos(-7363º)

5. Where did you meet with the words PERIOD, PERIODICITY?

Students' answers: A period in music is a construction in which a more or less complete musical thought is stated. The geological period is part of an era and is divided into epochs with a period of 35 to 90 million years.

The half-life of a radioactive substance. Periodic fraction. Periodicals are printed publications that appear on strictly defined dates. Periodic system of Mendeleev.

6. The figures show parts of the graphs of periodic functions. Define the period of the function. Determine the period of the function.

Answer: T=2; T=2; T=4; T=8.

7. Where in your life have you met with the construction of repeating elements?

Students answer: Elements of ornaments, folk art.

IV. Collective problem solving.

(Problem solving on slides.)

Let us consider one of the ways to study a function for periodicity.

This method bypasses the difficulties associated with proving that one or another period is the smallest, and also there is no need to touch on questions about arithmetic operations on periodic functions and about the periodicity of a complex function. The reasoning is based only on the definition of a periodic function and on the following fact: if T is the period of the function, then nT(n? 0) is its period.

Problem 1. Find the smallest positive period of the function f(x)=1+3(x+q>5)

Solution: Let's assume that the T-period of this function. Then f(x+T)=f(x) for all x ∈ D(f), i.e.

1+3(x+T+0.25)=1+3(x+0.25)
(x+T+0.25)=(x+0.25)

Let x=-0.25 we get

(T)=0<=>T=n, n ∈ Z

We have obtained that all periods of the considered function (if they exist) are among integers. Choose among these numbers the smallest positive number. it 1 . Let's check if it is actually a period 1 .

f(x+1)=3(x+1+0.25)+1

Since (T+1)=(T) for any T, then f(x+1)=3((x+0.25)+1)+1=3(x+0.25)+1=f(x ), i.e. 1 - period f. Since 1 is the smallest of all positive integers, then T=1.

Task 2. Show that the function f(x)=cos 2 (x) is periodic and find its main period.

Task 3. Find the main period of the function

f(x)=sin(1.5x)+5cos(0.75x)

Assume the T-period of the function, then for any X the ratio

sin1.5(x+T)+5cos0.75(x+T)=sin(1.5x)+5cos(0.75x)

If x=0 then

sin(1.5T)+5cos(0.75T)=sin0+5cos0

sin(1.5T)+5cos(0.75T)=5

If x=-T, then

sin0+5cos0=sin(-1.5T)+5cos0.75(-T)

5= - sin(1.5T)+5cos(0.75T)

Adding, we get:

10cos(0.75T)=10

2π n, n € Z

Let's choose from all numbers "suspicious" for the period the smallest positive one and check whether it is a period for f. This number

f(x+)=sin(1.5x+4π)+5cos(0.75x+2π)= sin(1.5x)+5cos(0.75x)=f(x)

Hence, is the main period of the function f.

Task 4. Check if the function f(x)=sin(x) is periodic

Let T be the period of the function f. Then for any x

sin|x+T|=sin|x|

If x=0, then sin|T|=sin0, sin|T|=0 T=π n, n ∈ Z.

Suppose. That for some n the number π n is a period

considered function π n>0. Then sin|π n+x|=sin|x|

This implies that n must be both even and odd at the same time, which is impossible. Therefore, this function is not periodic.

Task 5. Check if the function is periodic

f(x)=

Let T be the period f, then

, hence sinT=0, T=π n, n € Z. Let us assume that for some n the number π n is indeed the period of the given function. Then the number 2π n will also be a period

Since the numerators are equal, so are their denominators, so

Hence, the function f is not periodic.

Group work.

Tasks for group 1.

Tasks for group 2.

Check if the function f is periodic and find its main period (if it exists).

f(x)=cos(2x)+2sin(2x)

Tasks for group 3.

At the end of the work, the groups present their solutions.

VI. Summing up the lesson.

Reflection.

The teacher gives students cards with drawings and offers to paint over part of the first drawing in accordance with the extent to which, as it seems to them, they have mastered the methods of studying the function for periodicity, and in part of the second drawing, in accordance with their contribution to the work in the lesson.

VII. Homework

one). Check if function f is periodic and find its main period (if it exists)

b). f(x)=x 2 -2x+4

c). f(x)=2tg(3x+5)

2). The function y=f(x) has a period T=2 and f(x)=x 2 +2x for x € [-2; 0]. Find the value of the expression -2f(-3)-4f(3,5)

Literature/

1. Mordkovich A.G. Algebra and the beginning of analysis with in-depth study.
2. Maths. Preparation for the exam. Ed. Lysenko F.F., Kulabukhova S.Yu.
3. Sheremetyeva T.G. , Tarasova E.A. Algebra and beginning analysis for grades 10-11.

Application No. 7

Municipal educational institution

secondary school No. 3

Teacher

Korotkov

Asya Edikovna

Kurganinsk

2008

CONTENT

Introduction ………………………………………………… 2-3

Periodic functions and their properties ……………. 4-6

Tasks ………………………………………………… 7-14

Introduction

Note that problems on periodicity in the educational and methodological literature have a difficult fate. This is explained by a strange tradition - to allow one or another negligence in the definition of periodic functions that lead to controversial decisions and provoke incidents in exams.

For example, in the book “Explanatory Dictionary of Mathematical Terms” - M, 1965, the following definition is given: “a periodic function is a function

y = f(x), for which there is a number t > 0, which for all x and x + t from the domain f(x + t) = f(x).

Let us give a counter-example showing the incorrectness of this definition. According to this definition, the function is periodic with period t = 2π

с(x) = Cos(√x) 2 – Cos(√4π - x) 2 with a limited domain of definition, which contradicts the generally accepted point of view about periodic functions.

Similar problems arise in many of the newest alternative textbooks for the school.

A. N. Kolmogorov’s textbook gives the following definition: “Speaking of the periodicity of the function f, it is believed that there is such a number T ≠ 0 that the domain of definition D (f) together with each point x contains points that are obtained from x by parallel translation along axis Ox (right and left) by a distance T. The function f is called periodical with period T ≠ 0, if for any of the domain of definition the values ​​of this function at the points x, x - T, x + T are equal, i.e. f (x + T) \u003d f (x) \u003d f (x - T) ". Further in the textbook it is written: “Since the sine and cosine are defined on the entire number line and Sin (x + 2π) = Sin x,

Cos (x + 2π) \u003d Cos x for any x, sine and cosine are the period of a function with a period of 2π.

For some reason, this example does not check what is required in the definition of the condition that

Sin (x - 2π) \u003d Sin x. What's the matter? The point is that this condition is superfluous in the definition. Indeed, if T > 0 is the period of the function f(x), then T will also be the period of this function.

I want to give one more definition from the textbook by M.I. Bashmakov "Algebra and the beginning of analysis in 10-11 cells." “The function y \u003d f (x) is called periodic if there is such a number T ≠ 0 that the equality

f(x + T) = f(x) holds identically for all values ​​of x.

The above definition does not say anything about the scope of the function, although it means x from the scope of the definition, not any real x. According to this definition, the function y \u003d Sin (√x) can be periodic 2 , defined only for x ≥ 0, which is not true.

In the unified state exam there are tasks for periodicity. In one scientific periodical journal, as a training for section C of the USE, the solution of the problem was given: “is the function y (x) \u003d Sin 2 (2 + x) - 2 Sin 2 Sin x Cos (2 + x) periodic?

The solution shows that y (x - π) \u003d y (x) in the answer - an extra entry

"T = π" (after all, the question of finding the smallest positive period is not raised). Is it really necessary to carry out a complex trigonometric formation to solve this problem? After all, here you can focus on the concept of periodicity, as the key in the condition of the problem.

Solution.

f1 (x) \u003d Sin x - a periodic function with a period T \u003d 2π

f2 (x) = Cos x is a periodic function with period T = 2π, then 2π is a period and for functions f 3(x) = Sin(2+x) and f 4 (x) = Cos (2 + x), (this follows from the definition of periodicity)

f5 (x) = - 2 Sin 2 = Const, its period is any number, including 2π.

Because the sum and product of periodic functions with a common period T is also T-periodic, then this function is periodic.

I hope that the material presented in this work will help in preparing for the unified state exam in solving problems for periodicity.

Periodic functions and their properties

Definition: a function f(t) is called periodic if for any t from the domain of definition of this function D f there is a number ω ≠ 0 such that:

1) numbers (t ± ω) є D f ;

2) f(t + ω) = f(t).

1. If the number ω = the period of the function f (t), then the number kω, where k = ±1, ±2, ±3, … are also the periods of the function f(t).

EXAMPLE f(t) = Sint. The number T = 2π is the smallest positive period of this function. Let T 1 = 4π. Let us show that T 1 is also the period of this function.

F (t + 4π) = f (t + 2π + 2π) = Sin (t + 2π) = Sin t.

So T 1 is the period of the function f (t) = Sin t.

2. If the function f(t) - ω is a periodic function, then the functions f (at), where a є R, and f (t + c), where c is an arbitrary constant, are also periodic.

Find the period of the function f(аt).

f(аt) = f(аt + ω) = f (а(t + ω/а)), i.e. f (аt) = f (а(t + ω/а).

Therefore, the period of the function f(аt) – ω 1 = ω/а.

EXAMPLE 1. Find the period of the function y = Sin t/2.

Example 2. Find the period of the function y \u003d Sin (t + π / 3).

Let f(t) = Sin t; y 0 \u003d Sin (t 0 + π / 3).

Then the function f(t) = Sin t will also take the value y 0 for t = t 0 + π/3.

Those. all the values ​​that the function y takes are also taken by the function f(t). If t is interpreted as time, then each value of y 0 function y \u003d Sin (t + π / 3) is taken π / 3 units of time earlier than the function f (t) "shift" to the left by π / 3. Obviously, the period of the function will not change from this, i.e. T y \u003d T 1.

3. If F(x) is some function, and f(t) is a periodic function, and such that f(t) belongs to the domain of the function F(x) – D F , then the function F(f (t)) is a periodic function.

Let F(f (t)) = φ.

Φ (t + ω) = F(f (t + ω)) = F(f (t)) = φ (t) for any t є D f.

EXAMPLE Investigate the function for periodicity: F(x) = ℓ sin x .

The scope of this function D f coincides with the set of real numbers R. f (x) = Sin x.

The set of values ​​of this function is [-1; one]. Because segment [-1; 1] belongs to D f , then the function F(x) is periodic.

F(x+2π) = ℓ sin (x + 2π) = ℓ sin x = F(x).

2 π is the period of this function.

4. If the functions f 1 (t) and f 2 (t) periodic, respectively, with periods ω 1 and ω 2 and ω 1 / ω 2 = r, where r is a rational number, then the functions

С 1 f 1 (t) + С 2 f 2 (t) and f 1 (t) f 2 (t) are periodic (~ 1 and C 2 are constants).

Note: 1) If r = ω 1 /ω 2 = p/q, because r is a rational number, then

ω 1 q = ω 2 p = ω, where ω is the least common multiple of the numbers ω 1 and ω 2 (LCM).

Consider the function C 1 f 1 (t) + C 2 f 2 (t).

Indeed, ω = LCM (ω 1 , ω 2 ) - the period of this function

С 1 f 1 (t) + С 2 f 2 (t) = С 1 f 1 (t+ ω 1 q) + С 2 f 2 (t+ ω 2 p) + С 1 f 1 (t) + С 2 f 2 (t) .

2) ω is the period of the function f 1 (t) f 2 (t), because

f 1 (t + ω) f 2 (t + ω \u003d f 1 (t + ω 1 q) f 2 (t \u003d ω 2 p) \u003d f 1 (t) f 2 (t).

Definition: Let f 1 (t) and f (t) are periodic functions with periods, respectively, ω 1 and ω 2 , then two periods are said to be comparable ifω 1 / ω 2 = r is a rational number.

3) If the periods ω 1 and ω 2 are not commensurable, then the functions f 1 (t) + f 2 (t) and

f 1 (t) f 2 (t) are not periodic. That is, if f 1 (t) and f 2 (t) are different from a constant, periodic, continuous, their periods are not commensurate, then f 1 (t) + f 2 (t), f 1 (t) f 2 (t) are not periodic.

4) Let f(t) = С, where С is an arbitrary constant. This function is periodic. Its period is any rational number, which means that it does not have the smallest positive period.

5) The statement is also true for more functions.

Example 1. Investigate the periodicity of the function

F(x) = Sin x + Cos x.

Solution. Let f 1 (x) = Sin x, then ω 1 = 2πk, where k є Z.

T 1 = 2π is the smallest positive period.

f 2 (x) \u003d Cos x, T 2 \u003d 2π.

Ratio T 1 /T 2 = 2π/2π = 1 is a rational number, i.e. periods of functions f 1 (x) and f 2 (x) are commensurate. So this function is periodic. Let's find its period. By definition of a periodic function, we have

Sin (x + T) + Cos (x + T) = Sin x + Cos x,

Sin (x + T) - Sin x \u003d Cos x - Cos (x + T),

2 Cos 2x + π / 2 Sin T / 2 \u003d 2 Sin 2x + T / 2 Sin T / 2,

Sin T / 2 (Cos T + 2x / 2 - Sin T + 2x / 2) \u003d 0,

√2 Sin T / 2 Sin (π / 4 - T + 2x / 2) \u003d 0, therefore,

Sin T/2 = 0, then T = 2πk.

Because (х ± 2πk) є D f , where f(x) = Sin x + Cos x,

f(х + t) = f(х), then the function f(х) is periodic with the least positive period 2π.

Example 2. Is the periodic function f (x) \u003d Cos 2x Sin x, what is its period?

Solution. Let f 1 (x) \u003d Cos 2x, then T 1 \u003d 2π: 2 \u003d π (see 2)

Let f 2 (x) = Sin x, then T 2 = 2π. Because π/2π = ½ is a rational number, then this function is periodic. Its period T = LCM

(π, 2π) = 2π.

So, this function is periodic with a period of 2π.

5. Let the function f(t), which is not identically equal to a constant, be continuous and periodic, then it has the smallest positive period ω 0 , any other period of its ω has the form: ω= kω 0 , where k є Z.

Note: 1) Two conditions are very important in this property:

f(t) is continuous, f(t) ≠ C, where C is a constant.

2) The converse is not true. That is, if all periods are commensurable, then it does not follow from this that there is a smallest positive period. Those. a periodic function may not have the smallest positive period.

EXAMPLE 1. f(t) = C, periodic. Its period is any real number, there is no smallest period.

Example 2. Dirichlet function:

D(x) =

Any rational number is its period, there is no smallest positive period.

6. If f(t) is a continuous periodic function and ω 0 is its smallest positive period, then the function f(αt + β) has the smallest positive period ω 0 /‌‌/α/. This statement follows from item 2.

Example 1. Find the period of the function y \u003d Sin (2x - 5).

Solution. y \u003d Sin (2x - 5) \u003d Sin (2 (x - 5/2)).

The graph of the function y is obtained from the graph of the Sin x function, first by “compressing” twice, then by “shifting” to the right by 2.5. “The shift does not affect the periodicity, T = π is the period of this function.

It is easy to get the period of this function using the property of item 6:

T \u003d 2π / 2 \u003d π.

7. If f (t) - ω is a periodic function, and it has a continuous derivative f "(t), then f" (t) is also a periodic function, T \u003d ω

EXAMPLE 1. f(t) = Sin t, T = 2πk. Its derivative f "(t) = Cos t

F "(t) \u003d Cos t, T \u003d 2πk, k є Z.

EXAMPLE 2. f(t) = Cos t, T = 2πk. Its derivative

F "(t) \u003d - Sin t, T \u003d 2πk, k є Z.

Example 3. f(t) =tg t, its period is Т = πk.

F "(t) \u003d 1 / Cos 2 t is also periodic by property item 7 and has period T = πk. Its smallest positive period is T = π.

TASKS.

№ 1

Is the function f(t) = Sin t + Sin πt periodic?

Solution. For comparison, we solve this problem in two ways.

First, by the definition of a periodic function. Assume that f(t) is periodic, then for any t є D f we have:

Sin (t + T) + Sin π (t + T) = Sin t + Sin πt,

Sin (t + T) - Sin t \u003d Sin πt - Sin π (t + T),

2 Cos 2t + T/2 Sin T/2 = -2 Cos 2 πt + πt/2 Sin πt/2.

Because this is true for any t є D f , then, in particular, for t 0 , at which the left side of the last equality vanishes.

Then we have: 1) Cos 2t 0 + T/2 Sin T/2 = 0. Resolve with respect to T.

Sin Т/2 = 0 at Т = 2 πk, where k є Z.

2) Cos 2πt 0 + πt 0 /2 Sin πТ/2 = 0. Resolve with respect to Т.

Sin πТ/2 = 0, then Т = 2πn/ π = 2n, n≠0, where n є Z.

Because we have an identity, then 2 πk = 2n, π = 2n/2 k = n/ k, which cannot be, because π is an irrational number, and n/ k is rational. That is, our assumption that the function f(t) is periodic was not correct.

Secondly, the solution is much simpler if we use the above properties of periodic functions:

Let f 1 (t) = Sin t, Т 1 = 2 π; f 2 (t) = Sin πt, Т 2 - 2π/π = 2. Then, Т 1 /Т 2 = 2π/2 = π is an irrational number, i.e. periods T 1 , T 2 are not commensurate, so f(t) is not periodic.

Answer: no.

№ 2

Show that if α is an irrational number, then the function

F(t) = Cos t + Cos αt

is not periodic.

Solution. Let f 1 (t) = Cos t, f 2 (t) = Cos αt.

Then their periods are respectively T 1 \u003d 2π, T 2 = 2π//α/ - the smallest positive periods. Let's find, T 1 /T 2 = 2π/α//2π = /α/ is an irrational number. So T 1 and T 2 are incommensurable, and the function

f(t) is not periodic.

№ 3

Find the smallest positive period of the function f(t) = Sin 5t.

Solution. By property item 2 we have:

f(t) is periodic; T = 2π/5.

Answer: 2π/5.

№ 4

Is F(x) = arccos x + arcsin x a periodic function?

Solution. Consider this function

F(x) \u003d arccos x + arcsin x \u003d π - arcsin x + arcsin x \u003d π,

those. F(x) is a periodic function (see property item 5, example 1.).

Answer: yes.

№ 5

Is a periodic function

F (x) \u003d Sin 2x + Cos 4x + 5?

solution. Let f 1 (x) = Sin 2x, then T 1 = π;

F 2 (x) \u003d Cos 4x, then T 2 \u003d 2π / 4 \u003d π / 2;

F 3 (x) \u003d 5, T 3 - any real number, in particular T 3 we can assume equal to T 1 or T 2 . Then the period of this function is T = LCM (π, π/2) = π. That is, f(x) is periodic with period Т = π.

Answer: yes.

№ 6

Is the function f(x) = x - E(x) periodic, where E(x) is a function that associates the argument x with the smallest integer not exceeding the given one.

Solution. Often the function f (x) is denoted by (x) - the fractional part of the number x, i.e.

F(x) \u003d (x) \u003d x - E (x).

Let f(х) be a periodic function, i.e. there exists a number T >0 such that x - E(x) = x + T - E(x + T). Let's write this equation

(x) + E(x) - E(x) = (x + T) + E(x + T) - E(x + T),

(x) + (x + T) - true for any x from the domain D f, provided that T ≠ 0 and T є Z. The smallest positive of them is T = 1, i.e. T = 1 such that

X + T - E (x + T) \u003d x - E (x),

Moreover, (х ± Тk) є D f , where k є Z.

Answer: This function is periodic.

№ 7

Is the function f(x) = Sin x periodic? 2 .

Solution. Let's say f(x) = Sin x 2 periodic function. Then, by definition of a periodic function, there is a number T ≠ 0 such that: Sin x 2 \u003d Sin (x + T) 2 for any x є D f.

Sin x 2 \u003d Sin (x + T) 2 \u003d 0,

2 Cos x 2 + (x + T) 2 / 2 Sin x 2 - (x + T) 2 / 2 \u003d 0, then

Cos x 2 + (x + T) 2 / 2 = 0 or Sin x 2 - (x + T) 2 / 2 = 0.

Consider the first equation:

Cos x 2 + (x + T) 2 / 2 \u003d 0,

X 2 + (x + T) 2 / 2 \u003d π (1 + 2 k) / 2 (k є Z),

T \u003d √ π (1 + 2 k) - x 2 - x. (one)

Consider the second equation:

Sin x 2 - (x + T) 2 / 2 \u003d 0,

X + T \u003d √- 2πk + x 2,

T \u003d √x 2 - 2πk - x. (2)

It can be seen from expressions (1) and (2) that the found values ​​of T depend on x, i.e. there is no T>0 such that

Sin x 2 \u003d Sin (x + T) 2

For any x from the domain of this function. f(x) is not periodic.

Answer: no

№ 8

Investigate the periodicity of the function f(x) = Cos 2 x.

Solution. Let's represent f(x) by the double angle cosine formula

F(x) = 1/2 + 1/2 Cos 2x.

Let f 1 (x) = ½, then T 1 - it can be any real number; f 2 (x) \u003d ½ Cos 2x is a periodic function, because product of two periodic functions having a common period T 2 = pi. Then the smallest positive period of this function

T \u003d LCM (T 1, T 2) \u003d π.

So, the function f(x) = Cos 2 x – π – is periodic.

Answer: π is periodic.

№ 9

Can the domain of a periodic function be:

A) the half-line [a, ∞),

B) cut?

Solution. No, because

A) by definition of a periodic function, if х є D f, then x ± ω also

Must belong to the scope of the function. Let x = a, then

X 1 \u003d (a - ω) є [a, ∞);

B) let x = 1, then x 1 \u003d (1 + T) є.

№ 10

Can a periodic function be:

A) strictly monotonous;

B) even;

B) not even?

Solution. a) Let f(x) be a periodic function, i.e. there exists T≠0 such that for any x from the domain of the functions D f whats

(x ± T) є D f and f (x ± T) \u003d f (x).

Fix any x 0 º D f , because f(x) is periodic, then (x 0 + T) є D f and f (x 0) \u003d f (x 0 + T).

Assume that f(x) is strictly monotone and on the entire domain of definition D f , for example, increases. Then by definition of an increasing function for any x 1 and x 2 from the domain D f from inequality x 1 2 it follows that f(x 1 ) 2 ). In particular, from the condition x 0 0 + T, it follows that

F(x 0 ) 0 +T), which contradicts the condition.

This means that a periodic function cannot be strictly monotonic.

b) Yes, a periodic function can be even. Let's take a few examples.

F (x) \u003d Cos x, Cos x \u003d Cos (-x), T \u003d 2π, f (x) is an even periodic function.

0 if x is a rational number;

D(x) =

1 if x is an irrational number.

D(x) = D(-x), the domain of the function D(x) is symmetric.

The Direchlet function D(x) is an even periodic function.

f(x) = (x),

f (-x) \u003d -x - E (-x) \u003d (-x) ≠ (x).

This function is not even.

c) A periodic function can be odd.

f (x) \u003d Sin x, f (-x) \u003d Sin (-x) \u003d - Sin \u003d - f (x)

f(x) is an odd periodic function.

f (x) - Sin x Cos x, f (-x) \u003d Sin (-x) Cos (-x) \u003d - Sin x Cos x \u003d - f (x),

f(x) is odd and periodic.

f(x) = ℓ Sin x, f(-x) = ℓ Sin(- x) = ℓ -Sin x ≠ - f(x),

f(x) is not odd.

f(х) = tg x is an odd periodic function.

Answer: no; Yes; Yes.

№ 11

How many zeros can a periodic function have on:

one) ; 2) on the entire real axis, if the period of the function is equal to T?

Solution: 1. a) On the segment [a, b], a periodic function may not have zeros, for example, f(x) = C, C≠0; f (x) \u003d Cos x + 2.

b) On the segment [a, b], a periodic function can have an infinite number of zeros, for example, the Direchlet function

0 if x is a rational number,

D(x) =

1 if x is an irrational number.

c) On the segment [a, b], a periodic function can have a finite number of zeros. Let's find this number.

Let T be the period of the function. Denote

X 0 = (min x є(a,b), such that f(х) = 0).

Then the number of zeros on the segment [a, b]: N = 1 + E (in x 0 /T).

Example 1. x є [-2, 7π / 2], f (x) \u003d Cos 2 х is a periodic function with period Т = π; X 0 = -π/2; then the number of zeros of the function f(x) on the given segment

N \u003d 1 + E (7π / 2 - (-π / 2) / 2) \u003d 1 + E (8π / 2π) \u003d 5.

Example 2. f (x) \u003d x - E (x), x є [-2; 8.5]. f(х) – periodic function, Т + 1,

x 0 = -2. Then the number of zeros of the function f(x) on the given segment

N \u003d 1 + E (8.5 - (-2) / 1) \u003d 1 + E (10.5 / 1) \u003d 1 + 10 \u003d 11.

Example 3. f (x) \u003d Cos x, x є [-3π; π], T 0 \u003d 2π, x 0 \u003d - 5π / 2.

Then the number of zeros of this function on a given segment

N \u003d 1 + E (π - (-5π / 2) / 2π) \u003d 1 + E (7π / 2π) \u003d 1 + 3 \u003d 4.

2. a) An infinite number of zeros, because X 0 є D f and f(х 0 ) = 0, then for all numbers

X 0 + Tk, where k є Z, f (x 0 ± Tk) = f (x 0 ) =0, and points of the form x 0 ± Tk is an infinite set;

b) not have zeros; if f(х) is periodic and for any

х є D f function f(x) >0 or f(x)

F(x) \u003d Sin x +3.6; f(x) = C, C ≠ 0;

F(x) \u003d Sin x - 8 + Cos x;

F(x) = Sin x Cos x + 5.

№ 12

Can the sum of non-periodic functions be periodic?

Solution. Yes maybe. For example:

1. f1 (х) = х is non-periodic, f 2 (x) \u003d E (x) - non-periodic

F (x) \u003d f 1 (x) - f 2 (x) \u003d x - E (x) - periodic.

1. f1 (x) \u003d x - non-periodic, f (x) \u003d Sin x + x - non-periodic

F (x) \u003d f 2 (x) - f 1 (x) = Sin x - periodic.

Answer: yes.

№ 13

The function f(x) and φ(x) are periodic with periods T 1 and T 2 respectively. Is their product always a periodic function?

Solution. No, only if T 1 and T 2 - comparable. For example,

F(x) \u003d Sin x Sin πx, T 1 \u003d 2π, T 2 \u003d 2; then T 1 /T 2 = 2π/2 = π is an irrational number, so f(х) is not periodic.

f (x) \u003d (x) Cos x \u003d (x - E (x)) Cos x. Let f 1 (x) \u003d x - E (x), T 1 \u003d 1;

f 2 (x) \u003d Cos (x), T 2 \u003d 2π. T 2 /T 1 = 2π/1 = 2π, so f(x) is not periodic.

Answer: No.

Tasks for independent solution

Which of the functions are periodic, find the period?

1. f (x) \u003d Sin 2x, 10. f (x) \u003d Sin x / 2 + tg x,

2. f (x) \u003d Cos x / 2, 11. f (x) \u003d Sin 3x + Cos 4x,

3. f (x) \u003d tg 3x, 12. f (x) \u003d Sin 2 x+1,

4. f(x) = Cos (1 - 2x), 13. f(x) = tg x + ctg√2x,

5. f (x) \u003d Sin x Cos x, 14. f (x) \u003d Sin πx + Cos x,

6. f (x) \u003d ctg x / 3, 15. f (x) \u003d x 2 - E (x 2),

7. f (x) \u003d Sin (3x - π / 4), 16. f (x) \u003d (x - E (x)) 2 ,

8. f (x) \u003d Sin 4 x + Cos 4 x, 17. f (x) \u003d 2 x - E (x),

9. f(x) = Sin 2 x, 18. f(x) = x – n + 1 if n ≤ x≤ n + 1, n = 0, 1, 2…

№ 14

Let f(x) - T be a periodic function. Which of the functions are periodic (find T)?

1. φ(x) = f(x + λ) is periodic, because "shift" along the Ox axis does not affect ω; its period ω = T.
2. φ(х) = а f(х + λ) + в is a periodic function with period ω = Т.
3. φ(x) = f(kx) is a periodic function with period ω = T/k.
4. φ(x) \u003d f (ax + b) - a periodic function with a period ω \u003d T / a.
5. φ(x) = f(√x) is not periodic, because its domain of definition Dφ = (x/x ≥ 0), while the domain of definition of a periodic function cannot be a semiaxis.
6. φ(x) = (f(x) + 1/(f(x) - 1) is a periodic function, because

φ (x + T) \u003d f (x + T) + 1 / f (x + T) - 1 \u003d φ (x), ω \u003d T.

1. φ (x) \u003d a f 2 (x) + in f (x) + c.

Let φ 1 (x) = a f 2 (x) - periodic, ω 1 = t/2;

φ 2 (х) = in f(х) – periodic, ω 2=T/T=T;

φ 3 (х) = с – periodic, ω 3 - any number;

then ω = LCM(Т/2; Т) = Т, φ(х) is periodic.

Otherwise, because the domain of definition of this function is the entire number line, then the set of values ​​of the function f - E f є D φ , so the function

φ(х) is periodic and ω = Т.

1. φ(х) = √φ(х), f(х) ≥ 0.

φ(х) is periodic with period ω = Т, because for any x, the function f(x) takes the values ​​f(x) ≥ 0, i.e. its set of values ​​E f є D φ , where

Dφ is the domain of definition of the function φ(z) = √z.

№ 15

Is the function f(x) = x 2 periodic?

Solution. Consider x ≥ 0, then for f(x) there is an inverse function √x, which means that on this interval f(x) is a monotonic function, then it cannot be periodic (see No. 10).

№ 16

Given a polynomial P(x) = a 0 + a 1 x + a 2 x + ... a n x.

Is P(x) a periodic function?

Solution. 1. If the identity is constant, then P(x) is a periodic function, i.e. if a i = 0, where i ≥ 1.

2. Let P(x) ≠ c, where c is some constant. Let P(x) be a periodic function, and let P(x) have real roots, then since P(x) is a periodic function, then there must be an infinite number of them. And according to the fundamental theorem of algebra, their number k is such that k ≤ n. So P(x) is not a periodic function.

3. Let P(x) be a polynomial that is identically nonzero and has no real roots. Let's say P(x) is a periodic function. We introduce the polynomial q(x) = a 0 , q(х) is a periodic function. Consider the difference P(x) - q(x) = a 1 x 2 + ... + a n x n.

Because there is a periodic function on the left side of the equality, then the function on the right side is also periodic, moreover, it has at least one real root, x \u003d 0. If the function is periodic, then there must be an infinite number of zeros. We got a contradiction.

P(x) is not a periodic function.

№ 17

The function f(t) – T is periodic. Is the function f to (t), where

k є Z, a periodic function, how are their periods related?

Solution. The proof will be carried out by the method of mathematical function. Let

f 1 = f(t), then f 2 = f 2 (t) = f(t) f(t),

F 3 \u003d f 3 (t) \u003d f (t) f 2 is a periodic function according to the property of item 4.

………………………………………………………………………….

Let f k-1 = f k-1 (t) is a periodic function and its period T k-1 commensurate with the period T. We multiply both parts of the last equality by f(t), we get f k-1 f(t) = f(t) f k-1 (t),

F to = f to (t) is a periodic function by property item 4. ω ≤ Т.

№ 18

Let f(x) be an arbitrary function defined on . Is the function f((x)) periodic?

A n e t: yes, because the set of values ​​of the function (x) belongs to the domain of definition of the function f(x), then by property item 3 f((x)) is a periodic function, its period ω = T = 1.

№ 19

F(x) is an arbitrary function defined on [-1; 1], is the function f(sinx) periodic?

Answer: yes, its period is ω = T = 2π (the proof is similar to #18).

Studying the phenomena of nature, solving technical problems, we are faced with periodic processes that can be described by functions of a special kind.

A function y = f(x) with domain D is called periodic if there exists at least one number T > 0 such that the following two conditions are satisfied:

1) the points x + T, x − T belong to the domain D for any x ∈ D;

2) for each x from D we have the relation

f(x) = f(x + T) = f(x − T).

The number T is called the period of the function f(x). In other words, a periodic function is a function whose values ​​are repeated after a certain interval. For example, the function y = sin x is periodic (Fig. 1) with a period of 2π.

Note that if the number T is the period of the function f(x), then the number 2T will also be its period, as well as 3T, and 4T, etc., i.e., a periodic function has infinitely many different periods. If among them there is the smallest (not equal to zero), then all other periods of the function are multiples of this number. Note that not every periodic function has such a smallest positive period; for example, the function f(x)=1 has no such period. It is also important to keep in mind that, for example, the sum of two periodic functions having the same smallest positive period T 0 does not necessarily have the same positive period. So, the sum of functions f(x) = sin x and g(x) = −sin x does not have the smallest positive period at all, and the sum of functions f(x) = sin x + sin 2x and g(x) = −sin x, whose least periods are 2π has the smallest positive period equal to π.

If the ratio of the periods of two functions f(x) and g(x) is a rational number, then the sum and product of these functions will also be periodic functions. If the ratio of the periods of the everywhere defined and continuous functions f and g is an irrational number, then the functions f + g and fg will already be non-periodic functions. So, for example, the functions cos x sin √2 x and cosj √2 x + sin x are non-periodic, although the functions sin x and cos x are periodic with a period of 2π, the functions sin √2 x and cos √2 x are periodic with a period of √2 π .

Note that if f(x) is a periodic function with period T, then the complex function (if, of course, it makes sense) F(f(x)) is also a periodic function, and the number T will serve as its period. For example, the functions y \u003d sin 2 x, y \u003d √ (cos x) (Fig. 2.3) are periodic functions (here: F 1 (z) \u003d z 2 and F 2 (z) \u003d √z). However, one should not think that if the function f(x) has the smallest positive period T 0 , then the function F(f(x)) will have the same smallest positive period; for example, the function y \u003d sin 2 x has the smallest positive period, which is 2 times less than the function f (x) \u003d sin x (Fig. 2).

It is easy to show that if the function f is periodic with period T, is defined and differentiable at every point of the real line, then the function f "(x) (derivative) is also a periodic function with period T, but the antiderivative function F (x) (see Integral calculus) for f(x) will be a periodic function only if

F(T) − F(0) = T o ∫ f(x) dx = 0.