Prove the period of a function Let a > 0, if f (x) + F (x + a) + F (x + 2a) + F (x + 3a) + F (x + 4a) = f (x) f (x + a) f (x + 2a) f (x + 3a) f (x + 4a), then the period is t = 5A Prove the proposition Your proof after quotient is f (x + 5a) = f (x + a), indicating that the period is 4A instead of 5A

f(x)+f(x+a)+f(x+2a)+f(x+3a)+f(x+4a)=f(x)f(x+a)f(x+2a)f(x+3a)f(x+4a)
Let x = x + a
f(x+a)+f(x+2a)+f(x+3a)+f(x+4a)+f(x+5a)=f(x+a)f(x+2a)f(x+3a)f(x+4a)f(x+5a)
The difference between the two methods is as follows
f(x+5a)-f(x)=【f(x+5a)-f(x)】【f(x+a)f(x+2a)f(x+3a)f(x+4a)】
arrangement
【f(x+5a)-f(x)】【f(x+a)f(x+2a)f(x+3a)f(x+4a)-1】=0
If f (x + 5a) - f (x) = 0, then f (x + 5a) = f (x) is proved
otherwise
f(x+a)f(x+2a)f(x+3a)f(x+4a)=1
Let x = x + a
f(x+2a)f(x+3a)f(x+4a)f(x+5a)=1
The comparison of the two formulas is as follows:
If f (x + 5a) / F (x) = 1, then f (x + 5a) = f (x) is proved
Let x = x + a get f (x + a) + F (x + 2a) + F (x + 3a) + F (x + 4a) + F (x + 5a) = f (x + a) f (x + 2a) f (x + 3a) f (x + 4a) f (x + 5a) by subtracting the formula in the question
f(x+5a)-f(x)=[f(x+5a)-f(x)]f(x+a)f(x+2a)f(x+3a)f(x+4a)
Proof of the proposition f (x + 5a) - f (x) = 0
Or F (x + a) f (x + 2a) f (x + 3a) f (x + 4a) = 1, let x = x + a have f (x + 2a) f (x + 3a) f (x + 4a) f (x + 5a) = 1, then f (x + 5a) = f (x) proposition is proved
If the length of a rectangle is reduced by 5cm and the width is increased by 2cm, it becomes a square, and the areas of the two figures are equal. If the length of the rectangle is xcm and the width is YCM, then the equations ()
A. x−5＝y+2xy＝(x−5)(y+2)B. x−5＝y+22(x−5)＝5yC. x−5＝y+2xy＝(x+5)(y+2)D. x+5＝y−2xy＝(x−5)(y+2)
From the meaning of the question, we get: X − 5 = y + 2XY = (x − 5) (y + 2)
A proof of periodic function
If the function f (x) defined on R is symmetric with respect to x = a or x = B (b > A), it is proved that f (x) is a periodic function and 2b-2a is one of its periods
Because the image of F (x) is symmetric with respect to the line x = B and x = a
So f (x) = f (2a-x) f (x) = f (2b-x)
f(2a-x)=f(2b-x)
Let 2a-x = t, then x = 2a-t
The original formula becomes f (T) = f (2b-2a + T) = f (T + (2b-2a))
Because of the arbitrariness of T, f (x) is a periodic function and T = 2b-2a
222222222 (22 16:53:54)
If the inequality system x2m-1 has no solution, then the value range of M is
This inequality is incomplete
Where are your inequalities?
Where is the group?? I see you. Don't run
How to prove the periodicity of the following function
How to prove that the period of F (x + 2) - f (x + 1) = f (x) is 6
f(x+1)=f(x+2)+f(x)
f(x+2)=f(x+3)+f(x+1)
Add the two formulas to get
f(x)=-f(x+3)
f(x+3)=-f(x+6)
The two formulas are subtracted to get the
f(x)=f(x+6)
Typhoon is a kind of natural disaster. It takes the typhoon center as the center and forms a cyclone storm within tens of kilometers, which has strong destructive power. According to meteorological observation, there is a typhoon center at B, 220km south of a coastal city. The maximum wind force in the center is 12. Every 20km away from the typhoon center, the wind force will weaken by one. The typhoon center is moving northward at a speed of 15km / h If the wind force of the city reaches or exceeds level 4, the city will be affected by the typhoon. (1) will the city be affected by the typhoon? Why? (hint: ad ⊥ BC over a is better than D) (2) if the typhoon affects the city, how long will the typhoon affect the city? (3) What is the maximum wind force of the city affected by typhoon?
(1) The city will be affected by the typhoon. The reason is as follows: as shown in the figure, ad ⊥ BC is used as ad ⊥ BC through A. in RT △ abd, ∵ abd = 30 °, ab = 220, ∵ ad = 12ab = 110, ∵ the city is affected by the typhoon when the wind force reaches or exceeds level 4, ∵ the radius of the affected area is 20 × (12-4) =
Proof of periodicity of Dirichlet function
D (x) = 1, X is a rational number; D (x) = 0, X is an irrational number. Therefore, for any rational number a, D (x + a) = D (x), that is, rational numbers are periodic
(2014. The first mock exam in Hohhot) {an}, known for any positive integer n, a1+a2+a3+... +An = 2N-1, then A12 + A22 + A32 + +An2 & nbsp; equals ()
A. （2n-1）2B. 13(2n−1)C. 13(4n−1)D. 4n-1
∵a1+a2+a3+… +an=2n-1… ①∴a1+a2+a3+… +an-1=2n-1-1… ② The results show that an = 2N-1, an2 = 22n-2, an2} is an equal ratio sequence with 1 as the first term and 4 as the common ratio, A12 + A22 + A32 + +An2 = 1 − 4n1 − 4 = 13 (4N − 1), so C
How to prove several conclusions about periodicity of periodic function
1. F (x + a) = f (x + b) (a ≠ b) period
2. The period of F (x + a) = - f (x) (a ≠ 0)
3. The period of F (x + a) = 1 / F (x) (a ≠ 0, f (x) ≠ 0)
Although I know that these cycles are lb-al 2a and 2A respectively, when I look at others' proofs, the first one is to let x + a = y, and the second one is to let x + a = X. then I get f (x + 2a) = f (x). Why is the first one replaced by Y, and after the second one is replaced, I have to add the original x, so the first one is not needed
Your problem is to turn it into an explicit definition of cycle
2. One more minus sign. How can I get rid of this minus sign,
F (x + 2a) = - f (x + a) = - [- f (x)] = f (x), so t = 2A
3. The position is incorrect. F (x) runs to the denominator,
f(x+2a)=1/[f(x+a)]=1/[1/f(x)]=f(x)
The condition of 2 and 3 is to give a law of F, which is not the source law of the definition of periodic function. How can it be transformed into a standard definition? It should be transformed into a standard definition according to the given form;
My father is three times as old as my son this year. My son is 24 years older than my father. How old are my father and my son?
Set your son x years old
From the meaning of the title, we get 3x-x = 24
The solution is x = 12
　　　　3x＝36
A: my son is 12 years old and my father is 36 years old
Arithmetic method
24 / 2 = 12 years old
12 * 3 = 36 years old
A: my son is 12 years old and my father is 36 years old

RELATED INFORMATIONS