Special substitutions reveal scaling 常数.
译文:Setting m=1 m=1m=1 译文: gives f(kt2)=f(t)2 f(kt^2)=f(t)^2f(kt2)=f(t)2 译文:, while setting n=1 n=1n=1 译文: gives f(f(t))=k2t f(f(t))=k^2t f(f(t))=k2t. 应用ing these identities 到 t t t 且 kt kt kt 译文: yields f(kt)=kf(t) f(kt)=kf(t)f(kt)=kf(t).