Features of constructing a graph of periodic functions
The graph of a periodic function is usually first plotted on the interval [ x 0 ; x 0 + T). Perform parallel transfer of graph points to the entire definition area.
Examples of periodic functions and their graphs.
Examples of periodic functions are trigonometric functions. Let's look at the main ones.
Function F(x) =sin(x)
a) Domain of definition: D (sin x) = R .
b) Set of values: E (sin x) = [– 1 , 1] .
c) Even, odd: the function is odd.
d) Periodicity: a periodic function with a main period.
e) Zeros of the function: sin x = 0 for , n Z.
f) Intervals of constant sign of the function:
g) Intervals of monotonicity: the function increases as ;
the function decreases as ,
h) Extrema of the function:
; .
The graph of the function y= sin x is shown in the figure.
Function F(x) = cos(x)
a) Domain of definition.
b) Multiple values: E (cos x) = [ – 1 , 1 ] .
c) Even, odd: the function is even.
G ) Periodicity: the function is periodic with a main period.
d) Zeros of the function: at .
e) Intervals of constancy of sign:
g) Intervals of monotony:
the function increases as ;
the function decreases as
h) Extremes:
Graph of a function y=cos x shown in the figure.
Function F(x) = tan(x)
a) Scope of definition:
b) Set of values: E()
c) Even, odd. The function is odd.
d) Frequency. Periodic function with main period
e) Zeros of the function: tan x = 0 for x = n, n Z.
f) Intervals of constancy of signs:
g) Intervals of monotonicity: the function increases on each interval that entirely belongs to its domain of definition.
h) Extremes: no.
Graph of a function y= tg x shown in the figure.
Function F(x) = cot(x)
a) Domain of definition: D (ctg x) = R\ ( n(n Z) ).
b) Multiple values: E (ctg x) = R .
c) Even, odd is an odd function.
d) Periodicity: periodic function with main period T = .
e) Zeros of the function: cot x = 0 at x = /2 + n, n Z.
f) Intervals of constancy of signs;
g) Intervals of monotonicity: the function decreases on each interval that entirely belongs to its domain of definition.
h) Extremes: no.
The graph of the function y = ctg x is shown in the figure.
Interesting graphs are obtained using superposition - the formation of complex functions based on trigonometric periodic functions.
Graph of a periodic function
II. Applications of periodic functions. Periodic fluctuations.
Oscillations.
Oscillations are processes that differ in varying degrees of repeatability. Oscillations are processes that repeat at regular intervals (however, not all repeating processes are oscillations). Depending on the physical nature of the repeating process, vibrations are distinguished between mechanical, electromagnetic, electromechanical, etc. During mechanical vibrations, the positions and coordinates of bodies periodically change. For electrical - voltage and current. Depending on the nature of the impact on the oscillating system, free oscillations, forced oscillations, self-oscillations and parametric oscillations are distinguished.
Repetitive processes continuously occur inside any living organism, for example: heart contractions, lung function; we shiver when we are cold; we hear and speak thanks to the vibrations of the eardrums and vocal cords; When we walk, our legs make oscillatory movements. The atoms of which we are made vibrate. The world we live in is prone to fluctuations.
Periodic fluctuations.
Periodic are called such oscillations in which all the characteristics of movement are repeated after a certain period of time.
For periodic oscillations, the following characteristics are used:
period of oscillation T, equal to the time during which one complete oscillation occurs;
oscillation frequencyν, equal to the number of oscillations performed in one second (ν = 1/T);
Parametric oscillations are carried out when the parameters of an oscillating system are periodically changed (a person swinging on a swing periodically raises and lowers his center of gravity, thereby changing the parameters of the system). Under certain conditions, the system becomes unstable - a random deviation from the equilibrium position leads to the emergence and increase of oscillations. This phenomenon is called parametric excitation of oscillations (i.e., oscillations are excited by changing the parameters of the system), and the oscillations themselves are called parametric. Despite their different physical nature, vibrations are characterized by the same patterns, which are studied by general methods. An important kinematic characteristic is the shape of the vibrations. It is determined by the type of the time function that describes the change in one or another physical quantity during oscillations. The most important are those oscillations in which the fluctuating quantity changes over time according to the law of sine or cosine. They are called harmonic. This type of oscillation is especially important for the following reasons. Firstly, vibrations in nature and technology often have a character very close to harmonic. Secondly, periodic processes of a different form (with a different time dependence) can be represented as an imposition, or superposition, of harmonic oscillations.
Appendix No. 7
Municipal educational institution
Secondary school No. 3
Teacher
Korotkova
Asya Edikovna
Kurganinsk
2008
CONTENT
Introduction……………………………………………………………… 2-3
Periodic functions and their properties……………. 4-6
Problems…………………………………………………………………… 7-14
Introduction
Let us note that periodicity problems in educational and methodological literature have a difficult fate. This is explained by a strange tradition of allowing certain negligence in determining periodic functions, which 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 will be 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.
Many of the newer alternative school textbooks face similar problems.
In the textbook by A.N. Kolmogorov the following definition is given: “Speaking about the periodicity of a 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, also contains points obtained from x by parallel translation along axis Ox (to the right and left) at a distance T. The function f is called periodic with period T ≠ 0, if for any of the domain of definition the values of this function at points x, x – T, x + T are equal, i.e. f (x + T) = f (x) = f (x – T).” Further in the textbook it is written: “Since sine and cosine are defined on the entire number line and Sin (x + 2π) = Sin x,
Cos (x + 2π) = Cos x for any x, sine and cosine are the period of a function with a period of 2π.”
In this example, for some reason, the required condition in the definition is not checked:
Sin (x – 2π) = Sin x. What's the matter? The fact is that this condition in the definition is superfluous. Indeed, if T > 0 is the period of the function f(x), then T will also be the period of this function.
I would like to give one more definition from M.I. Bashmakov’s textbook “Algebra and beginnings of analysis grades 10-11.” “The function y = f(x) is called periodic if there is a number T ≠ 0 such that the equality
f (x + T) = f (x) holds identically for all values of x.”
The above definition says nothing about the domain of a function, although it means x in the domain of definition, not any real x. By this definition, the function y = Sin (√x) can be periodic 2 , defined only for x ≥ 0, which is incorrect.
In the Unified State Exam there are tasks for periodicity. In one scientific periodical magazine, as a training session for section C of the Unified State Examination, a solution to the problem was given: “is the function y (x) = Sin 2 (2+x) – 2 Sin 2 Sin x Cos (2+x) periodic?”
The solution shows that y (x – π) = y (x) in the answer there is 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 complex trigonometric education to solve this problem? After all, here you can focus on the concept of periodicity, as the key one in the condition of the problem.
Solution.
f 1 (x) = Sin x – periodic function with period Т = 2π
f 2 (x) = Cos x is a periodic function with period T = 2π, then 2π is the period for functions f 3 (x) = Sin (2 + x) and f 4 (x) = Cos (2 + x), (this follows from the definition of periodicity)
f 5 (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 on 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 = period of the function f (t), then the number kω, where k = ±1, ±2, ±3, ... are also periods of the function f(t).
EXAMPLE f (t) = Sin t. 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 – period of the function f (t) = Sin t.
2. If the function f(t) – ω is a periodic function, then the functions f (аt), where а є R, and f (t + с), where с is an arbitrary constant, are also periodic.
Let's 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/a.
Example 1. Find the period of the function y = Sin t/2.
Example 2. Find the period of the function y = Sin (t + π/3).
Let f(t) = Sin t; y 0 = Sin (t 0 + π/3).
Then the function f(t) = Sin t will take the same value 0 at 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 = Sin (t + π/3) is accepted π/3 time units earlier than the function f(t) “shifted” to the left by π/3. Obviously, the period of the function will not change due to this, i.e. T y = 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 definition 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 Examine the function for periodicity: F(x) = ℓ sinx.
Domain of this function D f coincides with the set of real numbers R. f (x) = Sin x.
The set of values for this function is [-1; 1]. 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 – 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
C 1 f 1 (t) + C 2 f 2 (t) and f 1 (t) f 2 (t) are periodic (C 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 ω 1 and ω 2 (NOC).
Consider the function C 1 f 1 (t) + C 2 f 2 (t).
Indeed, ω = LCM (ω 1 , ω 2 ) - 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 – period of function f 1 (t) f 2 (t), because
f 1 (t + ω) f 2 (t + ω =f 1 (t +ω 1 q) f 2 (t =ω 2 p) = f 1 (t) f 2 (t).
Definition: Let f 1 (t) and f (t) are periodic functions with periods ω, respectively 1 and ω 2 , then two periods are said to be commensurate ifω 1 /ω 2 = r is a rational number.
3) If 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) = C, where C is an arbitrary constant. This function is periodic. Its period is any rational number, which means it does not have the smallest positive period.
5) The statement is also true for a larger number of 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 – the smallest positive period.
f 2 (x) = Cos x, T 2 = 2π.
Ratio T 1 / T 2 = 2π/2π = 1 – rational number, i.e. periods of functions f 1 (x) and f 2 (x) are commensurate. This means that 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 = Cos x - Cos (x + T),
2 Cos 2х+ π/2 · Sin Т/2 = 2 Sin 2х+Т/2 · Sin Т/2,
Sin T/2 (Cos T+2x/2 - Sin T+2x/2) =0,
√2 Sin Т/2 Sin (π/4 – Т+2х/2) = 0, therefore,
Sin Т/2 = 0, then Т = 2πk.
Because (х ± 2πk) є D f , where f(x) = Sin x + Cos x,
f(x + t) = f(x), then the function f(x) is periodic with the smallest positive period 2π.
Example 2. Is the function f(x) = Cos 2x · Sin x periodic, what is its period?
Solution. Let f 1 (x) = Cos 2x, then T 1 = 2π: 2 = 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 = NOC
(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 opposite statement is not true. That is, if all periods are commensurable, then it does not follow 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 paragraph 2.
Example 1. Find the period of the function y = Sin (2x – 5).
Solution. y = Sin (2x – 5) = Sin (2(x – 5/2)).
The graph of the function y is obtained from the graph of the function Sin x, first by “compressing” by two times, 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 obtain the period of this function using the property of step 6:
Т = 2π/2 = 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, Т = ω
Example 1. f(t) = Sin t, Т = 2πk. Its derivative f"(t) = Cos t
F"(t) = Cos t, Т = 2πk, k є Z.
Example 2. f(t) = Cos t, Т = 2πk. Its derivative
F"(t) = - Sin t, T = 2πk, k є Z.
Example 3. f(t) =tg t, its period T = πk.
F"(t) = 1/ Cos 2 t is also periodic by property of step 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 definition of a periodic function. Let us 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 = Sin πt - Sin π (t + T),
2 Cos 2t + Т/2 Sin Т/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 , in which the left side of the last equality becomes zero.
Then we have: 1) Cos 2t 0 +T/2 Sin T/2 = 0. Let’s resolve relative to T.
Sin Т/2 = 0 at Т = 2 πk, where k є Z.
2) Cos 2πt 0 + πt 0 /2 Sin πT/2 = 0. Let’s resolve relative to T.
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 a rational number. That is, our assumption that the function f(t) is periodic was incorrect.
Secondly, the solution is much simpler if you use the above properties of periodic functions:
Let f 1 (t) = Sin t, T 1 = 2 π; f 2 (t) = Sin πt, T 2 - 2π/π = 2. Then, T 1 / T 2 = 2π/2 = π is an irrational number, i.e. periods T 1, T 2 are not commensurate, which means 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 = 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) – periodic; T = 2π/5.
Answer: 2π/5.
№ 4
Is the function F(x) = arccos x + arcsin x periodic?
Solution. Let's consider this function
F(x) = arccos x + arcsin x = π - arcsin x + arcsin x = π,
those. F(x) is a periodic function (see property of paragraph 5, example 1.).
Answer: yes.
№ 5
Is the function periodic?
F(x) = Sin 2x + Cos 4x + 5?
solution. Let f 1 (x) = Sin 2x, then T 1 = π;
F 2 (x) = Cos 4x, then T 2 = 2π/4 = π/2;
F 3 (x) = 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 T = LCM (π, π/2) = π. That is, f(x) is periodic with period T = π.
Answer: yes.
№ 6
Is the function f(x) = x – E(x) periodic, where E(x) is a function that assigns the argument x to 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) = (x) = x – E(x).
Let f(x) be a periodic function, i.e. there is a number T > 0 such that x – E(x) = x + T – E(x + T). Let's write down this equality
(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) = x – E(x),
Moreover, (x ± Tk) 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 assume that f(x) = Sin x 2 periodic function. Then, by the definition of a periodic function, there is a number T ≠ 0 such that: Sin x 2 = Sin (x + T) 2 for any x є D f.
Sin x 2 = Sin (x + T) 2 = 0,
2 Cos x 2 + (x+T) 2 /2 Sin x 2 -(x+T) 2 /2 = 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 = 0,
X 2 + (x+T) 2 /2 = π(1+2 k)/2 (k є Z),
Т = √ π(1+2 k) – x 2 – x. (1)
Consider the second equation:
Sin x 2 -(x+T) 2 /2 = 0,
X + T = √- 2πk + x 2,
T = √x 2 - 2πk – x. (2)
From expressions (1) and (2) it is clear that the found values of T depend on x, i.e. there is no T>0 such that
Sin x 2 = Sin (x+T) 2
For any x from the domain of definition of this function. f(x) is not periodic.
Answer: no
№ 8
Examine the function f(x) = Cos for periodicity 2 x.
Solution. Let us represent f(x) using 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) = ½ Cos 2x is a periodic function, because product of two periodic functions having a common period T 2 = π. Then the smallest positive period of this function
T = LOC (T 1, T 2) =π.
So, function f(x) = Cos 2 x – π – periodic.
Answer: π is periodic.
№ 9
Can the domain of a periodic function be:
A) half-line [a, ∞),
B) segment?
Solution. No, because
A) by definition of a periodic function, if x є D f, then x ± ω too
Must belong to the domain of the function. Let x = a, then
X 1 = (a – ω) є [a, ∞);
B) let x = 1, then x 1 = (1 + T) є .
№ 10
Can a periodic function be:
A) strictly monotonous;
B) even;
C) not even?
Solution. a) Let f(x) be a periodic function, i.e. there exists Т≠0 such that for any x from the domain of definition of the functions D f why
(x ±T) є D f and f (x±T) = f(x).
Let's fix any x 0 є Df , because f(x) is periodic, then (x 0 +T) є D f and f(x 0) = f(x 0 +T).
Let us assume that f(x) is strictly monotone and throughout 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 of definition 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 give a few examples.
F(x) = Cos x, Cos x = Cos (-x), T = 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 definition of the function D(x) is symmetrical.
The Direchlet function D(x) is an even periodic function.
f(x) = (x),
f(-x) = -x – E(-x) = (-x) ≠ (x).
This function is not even.
c) A periodic function may be odd.
f(x) = Sin x, f(-x) = Sin (-x) = - Sin = - f(x)
f(x) is an odd periodic function.
f(x) – Sin x Cos x, f(-x) = Sin (-x) Cos (-x) = - Sin x Cos x = - f(x) ,
f(x) – odd and periodic.
f(x) = ℓ Sin x, f(-x) = ℓ Sin(- x) = ℓ -Sin x ≠ - f(x),
f(x) is not odd.
f(x) = tan x – odd periodic function.
Answer: no; Yes; Yes.
№ 11
How many zeros can a periodic function have on:
1) ; 2) on the entire numerical 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) = Cos x + 2.
b) On the interval [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 interval [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. Let's denote
X 0 = (min x є(a,b), such that f(x) = 0).
Then the number of zeros on the segment [a, b]: N = 1 + E (c-x 0 /T).
Example 1. x є [-2, 7π/2], f(x) = Cos 2 x – periodic function with period T = π; X 0 = -π/2; then the number of zeros of the function f(x) on a given interval
N = 1 + E (7π/2 – (-π/2)/2) = 1 + E (8π/2π) = 5.
Example 2. f(x) = x – E(x), x є [-2; 8.5]. f(x) – periodic function, T + 1,
x 0 = -2. Then the number of zeros of the function f(x) on a given interval
N = 1 + E (8.5 – (-2)/1) = 1 + E (10.5/1) = 1 + 10 = 11.
Example 3. f(x) = Cos x, x є [-3π; π], T 0 = 2π, x 0 = - 5π/2.
Then the number of zeros of this function on a given interval
N = 1 + E (π – (-5π/2)/2π) = 1 + E (7π/2π) = 1 + 3 = 4.
2. a) An infinite number of zeros, because X 0 є D f and f(x 0 ) = 0, then for all numbers
Х 0 +Тk, where k є Z, f(х 0 ± Тk) = f(х 0 ) =0, and points of the form x 0 ± Tk is an infinite set;
b) have no zeros; if f(x) is periodic and for any
x є D f function f(x) >0 or f(x)
F(x) = Sin x +3.6; f(x) = C, C ≠ 0;
F(x) = 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:
- f 1 (x) = x – non-periodic, f 2 (x) = E(x) – non-periodic
F(x) = f 1 (x) – f 2 (x) = x – E(x) – periodic.
- f 1 (x) = x – non-periodic, f(x) = Sin x + x – non-periodic
F(x) = 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 when T 1 and T 2 – are commensurate. For example,
F(x) = Sin x Sin πx, T 1 = 2π, T 2 = 2; then T 1 / T 2 = 2π/2 = π is an irrational number, which means f(x) is not periodic.
f(x) = (x) Cos x = (x – E(x)) Cos x. Let f 1 (x) = x – E(x), T 1 = 1;
f 2 (x) = Cos (x), T 2 = 2π. T 2 /T 1 = 2π/1 = 2π, which means f(x) is not periodic.
Answer: No.
Problems to solve independently
Which of the functions are periodic, find the period?
1. f(x) = Sin 2x, 10. f(x) = Sin x/2 + tan x,
2. f(x) = Cos x/2, 11. f(x) = Sin 3x + Cos 4x,
3. f(x) = tan 3x, 12. f(x) = Sin 2 x+1,
4. f(x) = Cos (1 – 2x), 13. f(x) = tan x + ctg√2x,
5. f(x) = Sin x Cos x, 14. f(x) = Sin πx + Cos x,
6. f(x) = ctg x/3, 15. f(x) = x 2 – E(x 2),
7. f(x) = Sin (3x – π/4), 16. f(x) = (x – E(x)) 2 ,
8. f(x) = Sin 4 x + Cos 4 x, 17. f(x) = 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)?
- φ(x) = f(x + λ) – periodic, because the “shift” along the Ox axis does not affect ω; its period ω = T.
- φ(x) = a f(x + λ) + в – periodic function with period ω = T.
- φ(х) = f(kh) – periodic function with period ω = Т/k.
- φ(x) = f(ax + b) is a periodic function with period ω = T/a.
- φ(x) = f(√x) is not periodic, because its domain of definition Dφ = (x/x ≥ 0), and a periodic function cannot have a domain defined by a semi-axis.
- φ(x) = (f(x) + 1/(f(x) – 1) is a periodic function, because
φ(x +T) = f(x+T) + 1/f(x +T) – 1 = φ(x), ω = T.
- φ(x) = a f 2 (x) + in f(x) + c.
Let φ 1 (x) = a f 2 (x) – periodic, ω 1 = t/2;
φ 2 (x) = in f(x) – periodic, ω 2 = T/T = T;
φ 3 (x) = с – periodic, ω 3 – any number;
then ω = LCM(T/2; T) = T, φ(x) 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 , which means the function
φ(x) is periodic and ω = T.
- φ(x) = √φ(x), f(x) ≥ 0.
φ(x) – periodic with period ω = T, 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 – 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 monotone 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 equal to a constant, then P(x) is a periodic function, i.e. if a i = 0, where i ≥ 1.
2. Let P(x) ≠ с, where с is some constant. Let's say P(x) is 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. This means that P(x) is not a periodic function.
3. Let P(x) be an identically nonzero polynomial and it has no real roots. Let's say P(x) is a periodic function. Let us introduce the polynomial q(x) = a 0 , q(x) is a periodic function. Consider the difference P(x) - q(x) = a 1 x 2 + … +a n x n.
Because On the left side of the equality there is a periodic function, then the function on the right side is also periodic, and it has at least one real root, x = 0. Because 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
Given a function f(t) – T – periodic. Is the function f to (t), where
k є Z, a periodic function, how are their periods related?
Solution. We will carry out the proof using the mathematical function method. Let
f 1 = f(t), then f 2 = f 2 (t) = f(t) f(t),
F 3 = f 3 (t) = f (t) f 2 is a periodic function according to the property of step 4.
………………………………………………………………………….
Let f k-1 = f k-1 (t) – periodic function and its period T k-1 comparable with the period T. Multiplying both sides of the last equality by f(t), we obtain f k-1 f(t) = f(t) f k-1 (t),
F k = f k (t) is a periodic function according to the property of step 4. ω ≤ T.
№ 18
Let f(x) be an arbitrary function defined on . Is the function f((x)) periodic?
Answer: yes, because the set of values of the function (x) belongs to the domain of definition of the function f(x), then by property of 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 = Т = 2π (proof similar to No. 18).
HARMONIC ANALYSIS
Introduction.
Modern development of technology places increased demands on the mathematical training of engineers. As a result of the formulation and study of a number of specific problems in mechanics and physics, the theory of trigonometric series arose. Fourier series play a vital role in all areas of technology based on the theory of oscillations and the theory of spectral analysis. For example, in data transmission systems to describe signals, the practical use of spectral representations invariably leads to the need for experimental implementation of the Fourier decomposition. The role of trigonometric series in electrical engineering is especially great in the study of periodic non-sinusoidal currents: the amplitude spectrum of a function is found using the Fourier series in complex form. The Fourier integral is used to represent non-periodic processes.
Trigonometric series find important applications in numerous branches of mathematics and provide particularly convenient methods for solving difficult problems in mathematical physics, for example, the problem of the vibration of a string and the problem of the propagation of heat in a rod.
Periodic functions.
Many problems in science and technology involve periodic functions that reflect cyclic processes.
Definition 1. Periodic phenomena are those that repeat in the same sequence and in the same form at certain intervals.
Example. In spectral analysis - spectra.
Definition 2. Function at = f(x) is called periodic with a period T, If f(x + T) = f(x) in front of everyone X And x + T from the domain of the function.
In the figure, the period of the depicted function T = 2.
Definition 3. The smallest positive period of a function is called the fundamental period.
Where we have to deal with periodic phenomena, trigonometric functions are almost always encountered.
Function period 
is equal to , the period of the functions 
equal to .
Period of trigonometric functions with argument ( Oh) is found by the formula:
.
Example. Find the basic period of functions 1) 
.
Solution. 1) . 2) .
Lemma. If f(x) has a period T, then the integral of this function, taken within limits differing by T, does not depend on the choice of the lower limit of integration, i.e. = .
The main period is difficult periodic function at = f(x) (consisting of the sum of periodic functions) is the least common multiple of the periods of the component functions.
That is, if f(x) = f 1 (x) + f 2 (x), T 1 – function period f 1 (x), T 2 – period of the function f 2 (x), then the smallest positive period T must satisfy the condition:
T = nT 1 + kT 2 where(*) –
In normal school tasks prove periodicity of one or another function is usually not difficult: so, to make sure that the function $y=sin\frac34 x+sin\frac27 x$ is periodic, it is enough to simply note that the product $T=4\times7\times 2\pi$ is its period: if we add the number T to x, then this product will “eat up” both denominators and under the sine sign only the integer multiples of $2\pi$ will be superfluous, which will be “eaten up” by the sine itself.
But proof of non-periodicity of one or another function directly by definition may not be simple at all. Thus, to prove the non-periodicity of the function $y=\sin x^2$ considered above, you can write out the equality $sin(x+T)^2=\sin x^2$, but do not solve this trigonometric equation out of habit, but guess and substitute it in it x=0, after which the following will happen almost automatically: $\sin T^2=0$, $T^2=k\pi$, where k is some integer greater than 0, i.e. $T=\sqrt (k\pi)$, and if we now guess to substitute $x=\sqrt (\pi)$ into it, it turns out that $\sin(\sqrt(\pi)+\sqrt(k\ pi))=0$, whence $\sqrt(\pi)+\sqrt(k\pi)=n\pi$, $1+\sqrt(k)=n\sqrt(\pi)$, $1+k+ 2\sqrt(k)=n^2\pi$, $2\sqrt(k)=n^2\pi-1-k=n^2\pi=m$, $4k=n^4(\pi) ^2+2mn^2x+m^2$, and thus the number p is the root of the equation $n^4x^2+2mn^2\pi+m^2-4k=0$, i.e. is algebraic, which is not true: $\pi$ is, as we know, transcendental, i.e. is not the root of any algebraic equation with integer coefficients. However, in the future we will receive a much simpler proof of this statement - but with the help of mathematical analysis.
When proving the non-periodicity of functions, an elementary logical trick often helps: if all periodic functions have some property, but a given function does not have it, then it naturally is not periodic. Thus, a periodic function takes on any value infinitely many times, and therefore, for example, the function $y=\frac(3x^2-5x+7)(4x^3-x+2)$ is not periodic, since the value is 7 it only accepts at two points. Often, to prove non-periodicity, it is convenient to use the features of its domain of definition, and to find the desired property of periodic functions sometimes you have to show some imagination.
Let us also note that very often when asked what a non-periodic function is, one hears an answer in the style that we talked about in connection with even and odd functions, is when $f(x+T)\neq f(x)$, which, of course, is unacceptable.
And the correct answer depends on the specific definition of a periodic function, and, based on the definition given above, we can, of course, say that a function is non-periodic if it does not have a single period, but this will be a “bad” definition that does not give direction evidence of non-periodicity. And if we decipher it further, describing what the sentence “the function f does not have a single period” means, or, what is the same, “no number $T \neq 0$ is a period of the function f”, then we get that the function f is not periodic if and only if for every $T \neq 0$ there is a number $x\in D(f)$ such that either at least one of the numbers $x+T$ and $x-T$ is not belongs to D(f), or $f(x+T)\neq f(x)$.
You can say it another way: “There is a number $x\in D(f)$ such that the equality $f(x+T) = f(x)$ does not hold” - this equality may not hold for two reasons: or it doesn't make sense, i.e. one of its parts is undefined, or - otherwise, be incorrect. For interest, we add that the language effect that we talked about above also manifests itself here: for the equality “not to be true” and “to be false” are not the same thing - equality may not yet have meaning.
A detailed elucidation of the causes and consequences of this linguistic effect is in fact the subject not of mathematics, but of the theory of language, linguistics, or more precisely, its special section: semantics - the science of meaning, where, however, these questions are very complex and do not have an unambiguous solution. And mathematics, including school mathematics, is forced to put up with these difficulties and overcome linguistic “troubles” - while and because it uses, along with symbolic, natural language.
Goal: summarize and systematize students’ knowledge on the topic “Periodicity of Functions”; develop skills in applying the properties of a periodic function, finding the smallest positive period of a function, constructing graphs of periodic functions; promote interest in studying mathematics; cultivate observation and accuracy.
Equipment: computer, multimedia projector, task cards, slides, clocks, tables of ornaments, elements of folk crafts
“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. Report the topic and objectives of the lesson.
II. Checking homework.
We check homework using samples and discuss the most difficult points.
III. Generalization and systematization of knowledge.
1. Oral frontal work.
Theory issues.
1) Form a definition of the period of the function
2) Name 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) Using a circle, 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 an angle of 540º is one of the periods of the function y= cos(2x)
3. Prove that an 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 come across the words PERIOD, PERIODICITY?
Student answers: A period in music is a structure in which a more or less complete musical thought is presented. A geological period is part of an era and is divided into epochs with a period from 35 to 90 million years.
Half-life of a radioactive substance. Periodic fraction. Periodicals are printed publications that appear within strictly defined deadlines. Mendeleev's periodic system.
6. The figures show parts of the graphs of periodic functions. Determine 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 encountered the construction of repeating elements?
Student answer: Elements of ornaments, folk art.
IV. Collective problem solving.
(Solving problems on slides.)
Let's consider one of the ways to study a function for periodicity.
This method avoids the difficulties associated with proving that a particular period is the smallest, and also eliminates the need to touch upon questions about arithmetic operations on periodic functions and 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: 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's put x=-0.25 we get
(T)=0<=>T=n, n € Z
We have obtained that all periods of the function in question (if they exist) are among the integers. Let's choose the smallest positive number among these numbers. This 1 . Let's check whether it will actually be 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.
Problem 2. Show that the function f(x)=cos 2 (x) is periodic and find its main period.
Problem 3. Find the main period of the function
f(x)=sin(1.5x)+5cos(0.75x)
Let us assume the T-period of the function, then for any X the ratio is valid
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)
| sin(1.5T)+5cos(0.75T)=5 – sin(1.5T)+5cos(0.75T)=5 |
Adding it up, we get:
10cos(0.75T)=10
2π n, n € Z
Let us choose the smallest positive number from all the “suspicious” numbers for the period 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)
This means that this is the main period of the function f.
Problem 4. Let’s check whether the function f(x)=sin(x) is periodic
Let T be the period of the function f. Then for any x
sin|x+Т|=sin|x|
If x=0, then sin|Т|=sin0, sin|Т|=0 Т=π n, n € Z.
Let's assume. That for some n the number π n is the period
the function under consideration π n>0. Then sin|π n+x|=sin|x|
This implies that n must be both an even and an odd number, but this is impossible. Therefore, this function is not periodic.
Task 5. Check if the function is periodic
f(x)= 
Let T be the period of f, then
, hence sinT=0, Т=π n, n € Z. Let us assume that for some n the number π n is indeed the period of this function. Then the number 2π n will be the period
Since the numerators are equal, their denominators are also equal, therefore
This means that the function f is not periodic.
Work in groups.
Tasks for group 1.
Tasks for group 2.
Check if the function f is periodic and find its fundamental period (if it exists).
f(x)=cos(2x)+2sin(2x)
Tasks for group 3.
At the end of their work, the groups present their solutions.
VI. Summing up the lesson.
Reflection.
The teacher gives students cards with drawings and asks them to paint part of the first drawing in accordance with the extent to which they think they have mastered the methods of studying a function for periodicity, and in part of the second drawing - in accordance with their contribution to the work in the lesson.
VII. Homework
1). Check if the function f is periodic and find its fundamental 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/
- Mordkovich A.G. Algebra and beginnings of analysis with in-depth study.
- Mathematics. Preparation for the Unified State Exam. Ed. Lysenko F.F., Kulabukhova S.Yu.
- Sheremeteva T.G. , Tarasova E.A. Algebra and beginning analysis for grades 10-11.