A math problem about the cycle of a function from high school or above

There is a formula that y=cos(x a) then the minimum positive period of this function is 2 *a.

Then y1=cos(x 2) his period is 4 and y2=2sin(x 3) is 6

Then there is the synthesis of the 2 periods of the function. Let's take a simple example. There were two people running.

One person runs one lap in 4 minutes, and one person runs one lap in 6 minutes, so when can these two people run in the same position as they first started? Obviously, it must be two people running an integer lap at the same time, which should be easy to understand. Then the first person is in ......minutes to run back to the starting point, the second person is in ......minutes to run back to the starting point, apparently at ......Minutes later, two people ran to the starting point at the same time.

Therefore, the answer to this question is 12

The minimum period of cos(x 2) is 2 (1 2)=4 The minimum period of sin(x 3) is 2 (1 3)=6 Let the minimum period of the function of this problem be , according to the definition of period.

y(x)=y(x+θ)

i.e. cos(x 2)+2sin(x 3)=cos[(x+) 2]+2sin[(x+) 3].

cos(x/2)+2sin(x/3)=cos(x/2+θ/2)+2sin(x/3+θ/3)③

According to the definition of sine and cosine periods.

cos(x/2+2π)=cos(x/2)

sin(x/3+2π)=sin(x/3)

The conditions for the formula to be established are:

2 is an integer multiple of 2; 3 is an integer multiple of 2.

is an integer multiple of 4; is an integer multiple of 6.

So is an integer multiple of 12.

The minimum positive period is 12

For the function y=f(x), if there is a constant t that is not zero, such that f(x+t)=f(x) is true when x takes the beginning of every value in the defined domain, then the function y=f(x) is called the periodic function, and the non-zero constant t is called the period of the function.

Periodic function properties:

1) If t(≠0) is the period of f(x), then -t is also the period of f(x).

2) If t(≠0) is the period of f(x), then nt(n is any non-zero integer) is also the period of f(x).

3) If t1 and t2 are both periods of f(x), then t1 t2 is also periods of f(x).

4) If f(x) has a minimum positive period t*, then any positive period t of f(x) must be a positive integer multiple of t*.

5) t* is the minimum positive highlight pure width period of f(x), and t1 and t2 are the two periods of f(x) respectively, then (q is the set of rational numbers).

6) If t1 and t2 are two periods of f(x) and are irrational numbers, then f(x) does not have a minimum positive period.

7) The domain m of the periodic function f(x) must be an unbounded set of both parties.

1: Proof 4 is a period of f(x), which is equivalent to f(x)=f(x+4) for all x r

f(x)=-f(x+2)

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

f(x)=f(x=4)

Proven. Variant: Similarly, for all x r,f(x+2)=-1 f(x), for all x r,f(x)≠0

f(x+4)=-1 f(x+2)=f(x). 2: Proof: f(x) is an even function, so there is f(x)=f(-x) and f(x) with 2 as the period, so there is f(x)=f(x-2) f(

Because f(x+2)=-f(x), which is obtained by replacing x with x+2, f(x+4)=-f(x+2)=-(-f(x))=f(x), 4 is a period of f(x).

f(x+2)= 1 f(x)You'd better change the letter, it's easy for beginners to understand f(t+2)=1 f(t),f(t)*f(t+2)=1, take t=x and substitute it to get f(x)*f(x+2)=1

Take t=x+2 and substitute it to obtain f(x+2)*f(x+4)=1, and according to the method of equal substitution, f(x)=f(x+4).

It shows that the function takes 4 as the period, that is, the independent variable is added to 4, and the value of the function is unchanged: f(3) = f(7) = f(11) = f(15)...=f(3+4*502)=f(2011)

The period is t

Then if two numbers differ by integer multiples of the period, their function values are equal, because f(3)=f(3+4)=f(7)=f(7+4)=f(11)=......And so on.

All the way up to =f(2011).

Understood this way.

f(3)=f(7)=f(11)=f(15)..=f(4*502+3)

Add four to each.

f(x+2)=1 f(x)so f(x+4)=1 f(x+2)=1 1 f(x) so f(x+4)=f(x) so the period t=4 f(2011) is 502 4s and 3 so f(2011)=f(3).

f(x-4)=-f(x), then f(x-4-4)=-f(x-4), f(x-8)=f(x).

If x is subtracted by 4, a negative sign should be added before the original function.

Generally, you can guess the existence of a cycle through the question, and then make an argument.

Knowing that f(x) is a function defined on r, and satisfies f(2 x) f(x) f(x 1), when x ( 6,0 , f(x) and the function g(x) lg(x 2 2) 1 have the same increase or decrease, compare the magnitude of f( 49) and f(88).

f(2＋x)＋f(x)＝f(x＋1)

Let x=t, there is.

f(2＋t)＋f(t)＝f(t＋1)--1

Let x=t-1, yes.

f(1＋t)＋f(t-1)＝f(t)--2

1 (2, remove the same term.

We have f(2+t)+f(t-1)=0

F(3+t)+f(t)=0 can be pushed

F(6+t)+f(3+t)=0 can be pushed

f(3+t)+f(t)=f(6+t)+f(3+t).

It follows that f(t) = f(t+6), and it is clear that f(x) is a function of period 6.

So f(-49) = f(-1).

f(88)=f(-2)

g(x) is monotonically decreasing at (-6,0).

f(x) has the same monotonicity as g(x) at (-6,0).

So f(-2)> f(-1).

That is, f( 49) When I see (-6,0), I immediately guess that the period is 6, or 3, or a multiple of 6, and then I do a conversion to find that the period is 6

For periodic problems like this, as soon as I get the relationship between functions, I will replace them and find the direct relationship between x and x+n, so you are much closer to the period.

Generally by the subject suspects the existence of a cycle, which is then proven.

The function f(x) is a known function defined in r, satisfying f(2 + x) of +(x) = (x + 1), when x(6,0] the function f(x) has the same variability as the function g(x) = lg(x -2 2)-1, comparing the size of f(-49) and f(88) > tons >

2 + x）+ f（x）= f（x +1）

f（2 + t）+ f（t ）= f（t +1）--

order x = t-1，1 + t）+ f（t-1）= f（t） -2

1(2, elimination.

2 + t）+ f（t-1）= 0

It is possible to start a similar project (3 + t) + f(t) = 0

Issue (6 + t) + (3 + t) = 0

It can be deduced that f(3 + t) + f(t) = f(6 + t) + (3 + t).

It can be drawn. f(t) = f(t +6) and f(x) is obviously a function f(-49) = f(-1).

f（88）= f（-2）

g(x) is monotonically decreasing at (-6,0).

The monotonicity of the functions f(x) and g(x).

6,0），f（-2）> f（-1）

When I see (-6,0), I immediately suspect that 6 cycles, or 3, or multiples of 6, and then do the conversion process to get 6 cycles.

The topic is like this loop, and as soon as I get my hands on the relationship between the functions I will replace the x+n to get the direct relationship, then, you and your cycle, much closer.

f(-1)=1

f(0)=0

f(1)=f(0)-f(-1)=-1

f(2)=f(1)-f(0)=-1

f(3)=0

f(4)=1

f(5)=1

f(6)=0

f(7)=-1=f(1)

f(8)=-1=f(2)

The kind of common approach you want doesn't exist! You can only observe and try to generalize the rules.

In fact, this problem has to be solved by a common method, which is to solve the general formula of f(n) according to the second-order linear difference equation, and then find it. However, the general formula for the elimination of the f(n) is like a general term like the Fibonacci sequence, which is inconvenient for this problem.

This problem has a special bridge compass, so the periodicity in it is not a common feature of this type of problem, in other words, this question if.

The coefficients between f(-1), f(0), and f(n)f(n-1)f(n-2) are not special cherry blossom values, so there is no periodicity.