(0)
(0-1)
若(0)成立,则 f ( x + n × T ) = f ( x ) f(x+n\times{T})=f(x) f(x+n×T)=f(x), n ∈ Z n\in\mathbb{Z} n∈Z(1)
(2)
为周期
设 g ( x ) g(x) g(x)的周期为 T ′ T' T′,根据周期函数的定义, g ( x + T ′ ) g(x+T') g(x+T′)= g ( x ) g(x) g(x)(3)
从验证的角度说明(2)式为 g ( x ) g(x) g(x)的周期
记 u ( x ) = a x + b u(x)=ax+b u(x)=ax+b ; g ( x ) = f ( u ( x ) ) g(x)=f(u(x)) g(x)=f(u(x))= f ( a x + b ) f(ax+b) f(ax+b), g ( x ) g(x) g(x)的周期设为 T ′ T' T′
则 g ( x + T ′ ) = g ( x ) g(x+T')=g(x) g(x+T′)=g(x),即 f ( a ( x + T ′ ) + b ) f(a(x+T')+b) f(a(x+T′)+b)= f ( a x + b ) f(ax+b) f(ax+b)
f ( a ( x + T ′ ) + b ) f(a(x+T')+b) f(a(x+T′)+b)= f ( a x + a T ′ + b ) f(ax+aT'+b) f(ax+aT′+b)= f ( a x + b + a T ′ ) f(ax+b+aT') f(ax+b+aT′),由式(0), f ( a x + b + a T ′ ) = f ( a x + b ) f(ax+b+aT')=f(ax+b) f(ax+b+aT′)=f(ax+b)
再由式(0),(1),若 T T T是 f ( x ) f(x) f(x)的最小正周期,则 a T ′ aT' aT′是 T T T的整数倍, a T ′ = n T aT'=nT aT′=nT, k ∈ Z k\in\mathbb{Z} k∈Z,从而 T ′ = n T a T'=\frac{nT}{a} T′=anT,取最小正周期,即 ∣ T n a ∣ |\frac{Tn}{a}| ∣aTn∣取最小值,则 ∣ n ∣ = 1 |n|=1 ∣n∣=1,所以 T ′ = T ∣ a ∣ T'=\frac{T}{|a|} T′=∣a∣T
Note:上述结论在三角函数上用的很多,例如
设 f ( x + T ) = f ( x ) f(x+T)=f(x) f(x+T)=f(x), h ( x ) = 1 f ( x ) h(x)=\frac{1}{f(x)} h(x)=f(x)1,则 f ( x ) , h ( x ) f(x),h(x) f(x),h(x)都是以T为周期的周期函数
例如, sin x , 1 sin x = csc x \sin{x},\frac{1}{\sin{x}}=\csc{x} sinx,sinx1=cscx都是周期为 2 π 2\pi 2π的函数
f ( x + T ) = f ( x ) f(x+T)=f(x) f(x+T)=f(x), g ( x + T ) = g ( x ) g(x+T)=g(x) g(x+T)=g(x),则 h ( x ) = f ( x ) ± g ( x ) h(x)=f(x)\pm{g(x)} h(x)=f(x)±g(x)满足 h ( x + T ) = h ( x ) h(x+T)=h(x) h(x+T)=h(x)
f ( x + T 1 ) = f ( x ) f(x+T_1)=f(x) f(x+T1)=f(x), g ( x + T 2 ) = g ( x ) g(x+T_2)=g(x) g(x+T2)=g(x)则 h ( x ) = f ( x ) ± g ( x ) h(x)=f(x)\pm{g(x)} h(x)=f(x)±g(x)满足 h ( x + T ′ ) = h ( x ) h(x+T')=h(x) h(x+T′)=h(x),其中 T ′ = l c m ( T 1 , T 2 ) T'=lcm(T_1,T_2) T′=lcm(T1,T2)
周期分别为 T 1 , T 2 T_1,T_2 T1,T2的两个周期函数之和函数或差函数是周期为 T ′ T' T′( T 1 , T 2 T_1,T_2 T1,T2的最小公倍数)
明白这一点很重要,在傅里叶级数中,会用到这一点
更一般的,记具有不同周期的周期函数 t i ( x ) t_i(x) ti(x)的和函数 s ( x ) = ∑ i = 1 n t i ( x ) s(x)=\sum\limits_{i=1}^{n}t_i(x) s(x)=i=1∑nti(x)
f ( x + t 0 ) = f ( x ) f(x+t_0)=f(x) f(x+t0)=f(x); g ( x + t 1 ) = g ( x ) g(x+t_1)=g(x) g(x+t1)=g(x)
记 h ( x ) = f ( x ) + g ( x ) h(x)=f(x)+g(x) h(x)=f(x)+g(x); l c m ( t 0 , t 1 ) = k 0 t 0 = k 1 t 1 lcm(t_0,t_1)=k_0t_0=k_1t_1 lcm(t0,t1)=k0t0=k1t1
(1)
f ( x + T ) = f ( x ) f(x+T)=f(x) f(x+T)=f(x),则 y ( x ) = f ( x ) + c y(x)=f(x)+c y(x)=f(x)+c满足 y ( x + T ) = y ( x ) y(x+T)=y(x) y(x+T)=y(x)
y ( x ) = f ( x ) + c y(x)=f(x)+c y(x)=f(x)+c; y ( x + t ) = f ( x + t ) + c = f ( x ) + c y(x+t)=f(x+t)+c=f(x)+c y(x+t)=f(x+t)+c=f(x)+c
可见, y ( x + t ) = y ( x ) y(x+t)=y(x) y(x+t)=y(x),
一般地,周期函数 f ( x ) f(x) f(x)加上一个常数得到的新函数的周期和 f ( x ) f(x) f(x)的周期一致
f ( x + T ) = f ( x ) f(x+T)=f(x) f(x+T)=f(x),则 y ( x ) = k f ( x ) y(x)=kf(x) y(x)=kf(x),( k k k为常数),满足 y ( x + T ) = y ( x ) y(x+T)=y(x) y(x+T)=y(x)
f ( x + T ) = f ( x ) f(x+T)=f(x) f(x+T)=f(x), y ( x ) = k f ( x ) + c y(x)=kf(x)+c y(x)=kf(x)+c满足 y ( x + T ) = y ( x ) y(x+T)=y(x) y(x+T)=y(x)
sin 2 x \sin^{2}x sin2x= 1 2 ( 1 − cos 2 x ) \frac{1}{2}(1-\cos{2x}) 21(1−cos2x),其周期为 cos 2 x \cos{2x} cos2x的周期,即 π \pi π
类似的, sin 4 x \sin^4{x} sin4x= 1 4 ( 1 − cos 2 x ) 2 \frac{1}{4}(1-\cos{2x})^2 41(1−cos2x)2= 1 4 ( 1 − 2 cos 2 x + cos 2 2 x ) \frac{1}{4}(1-2\cos{2x}+\cos^2{2x}) 41(1−2cos2x+cos22x)
sin 2 x \sqrt{\sin^2{x}} sin2x= ∣ sin x ∣ |\sin{x}| ∣sinx∣周期为 π \pi π
设 s ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n c o s ( n x ) + b n s i n ( n x ) ) 对于 cos n x , sin n x , n = 1 , 2 , ⋯ 这些函数的周期分别是 { 2 π n } = 2 π 1 , 2 π 2 , 2 π 3 , . . . 事实上 , 2 π ( 2 π n ) = n , 而 n = 1 , 2 , 3 , . . . s ( x ) 的各个周期函数的周期的最小公倍数为 2 π 因此 : s ( x + 2 π ) = s ( x ) 设s(x)=\frac{a_0}{2}+\sum\limits_{n=1}^{\infin}(a_ncos(nx)+b_nsin(nx)) \\对于\cos{nx},\sin{nx},n=1,2,\cdots \\这些函数的周期分别是\{\frac{2\pi}{n}\}=\frac{2\pi}{1},\frac{2\pi}{2},\frac{2\pi}{3},... \\事实上,\frac{2\pi}{(\frac{2\pi}{n})}=n,而n=1,2,3,... \\ s(x)的各个周期函数的周期的最小公倍数为2\pi \\因此:s(x+2\pi)=s(x) 设s(x)=2a0+n=1∑∞(ancos(nx)+bnsin(nx))对于cosnx,sinnx,n=1,2,⋯这些函数的周期分别是{n2π}=12π,22π,32π,...事实上,(n2π)2π=n,而n=1,2,3,...s(x)的各个周期函数的周期的最小公倍数为2π因此:s(x+2π)=s(x)