On the properties of function periodicity If one of the following conditions holds for any X in the field of F (x), then f (x) is a periodic function, and the period T = 2A (a is not equal to 0) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; f(x+a)=f(x)+1\f(x)-1 f(x+a)=1-f(x)\1+f(x) f(x+a)=f(x-a) How can I prove this

On the properties of function periodicity If one of the following conditions holds for any X in the field of F (x), then f (x) is a periodic function, and the period T = 2A (a is not equal to 0) & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; f(x+a)=f(x)+1\f(x)-1 f(x+a)=1-f(x)\1+f(x) f(x+a)=f(x-a) How can I prove this

2、f(x+a+a)=-f(x+a)=-[-f(x)]=f(x);
3、f(x+a+a)=-1/f(x+a)=-1/[-1/f(x)]=f(x)
4、f[-(x-a)+a]=-f[-(x-a)]=-[-f(-x)]=f(-x);

Symmetry and periodicity of functions
They all seem to have a property in the form of F (a + b) = f (c + D)

F (A-X) = f (B + x) f (x) with respect to x = (a + b) / 2 symmetry
f(x-a)=f(x+b) T=a+b

What is function periodicity
What's the use
Main questions

The concept of function periodicity. Teaching process designer: last lesson, we learned how to use the sine line in the unit circle as the image of sine function. Today we will use the image of sine function to study an important property of trigonometric function. Please observe the image of y = SiNx, X ∈ R: (teacher put the picture on the top left of the blackboard...)

The operation of periodic function
Let the minimum positive periods of two functions f (x) and G (x) be T1 and T2 respectively, and T1 and T2 have a common multiple of integers. Ask if f (x) + G (x) and f (x) g (x) are also periodic functions, and if so, what are their minimum positive periods? Please give a detailed proof
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Let the minimum positive periods of two functions f (x) and G (x) be T1 and T2 respectively, and T1 and T2 have a common multiple of integers. Ask if f (x) + G (x) and f (x) g (x) are also periodic functions, and if so, what are their minimum positive periods? Please give a detailed proof
f(x+T1)=f(x);
g(x+T2)=g(x);
So:
f(x+n T1)=f(x);
g(x+n T2)=g(x);
If T3 = K1, T1 = K2, T2; K1, K2 are integers,
Then,
f(x+T3)=f(x);
g(x+T3)=g(x);
therefore
f(x+T3)+g(x+T3)=f(x)g(x),
f(x+T3)g(x+T3)=f(x)g(x),
So T3 is the period of F (x) + G (x), f (x) g (x)
It is difficult to say the minimum period. There are many cases
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The calculation of function period
Let y = 2cosx / 2 - 3sinx / 3
What is the period of Y?
There is a formula and calculation process to write out

The first is to find the period of sin4x
It is Π / 2
Because the absolute value is added
The negative part of the image becomes positive
So the cycle has doubled
It's Π / 4

The calculation of periodic function
f(x)=2sin(π-x)cosx
Finding the minimum positive period of F (x)
Finding the maximum and minimum of F (x) in the region [- π / 6, π / 2]

∵f(x)=2sin(π-x)cosx=2sinxcosx=sin2x
(1)∴T=2π/2=π
(2)∵x∈[-π/6,π/2]
The image shows that f (x) max = f (π / 4) = 1, f (x) min = f (- π / 6) = - (√ 3) / 2

Prove that the function is a periodic function
Given f (x + 1) = - f (x), find f (x) as a periodic function, and find its period

F (x + 1 + 1) = - f (x + 1) = f (x), the period is 2

Prove periodic function. Thank you
Given f (x) = - f (2-x) f (x) = f (4-x), it is proved that f (x) is a periodic function

From topic to stem
f(x)=
f(4-x)=-f(2-x)
Let 2-x = t get
f(t+2)=-f(t)
So f (T + 4) = - f (T + 2) = f (T)
So f (x) is a function of period 4

It is proved that the original function of periodic odd function is periodic function, and the two functions have the same period

First of all, the integral of periodic odd function in one period is 0, which can be proved! Construct f (x) = ∫ x, 0f (T) DT, that is, the original function of F (x), where x and 0 are their upper and lower limits respectively. Then f (x, t) = ∫ x T, 0f (T) DT let u = T-T, then f (x + T) = ∫ x, one TF (U, t) DU + T = ∫ x, 0f (U) DU + 0, one TF (U) Du and ∫ 0, One TF (U) Du = 0, so f (x + T) = f (x). As for the simple self proof why the odd function integral is 0 in a period, let's see!

Let f (x) satisfy f (x1) + F (x2) = 2F [(x1 + x2) / 2] * f [(x1-x2) / 2], and f (PI / 2) = 0, X belongs to R. it is proved that f (x) is a periodic function
The process of proof

It is proved that if X1 = x + π, X2 = x, then
f(x+π)+f(x)=2f((2x+π)/2)f(π/2)=0
f(x)=-f(x+π)
So f (x + π) = - f (x + 2 π)
So f (x) = - f (x + π) = - (- f (x + 2 π)) = f (x + 2 π)
So f (x) is a periodic function with a period of 2 π