Equivalence of the order r

We shall write $f(t)\stackrel{r}{\sim} 0$ if f^(r)(t) is continuous (that is, $f^{(r)}\in {\rm C}$). If $f_{1}-f_{2}\stackrel{r}{\sim} 0$ then $f_{1} \stackrel{r}{\sim}f_{2}$. (It is not necessary that each of the functions f₁ and f₂ belongs to ${\rm C}^{r}$). We don't distinguish functions whose difference is a smooth function. Each continuous function is equivalent (of order r=0) to 0 and each polynomial P_n(t) is equivalent to 0 with the order $r=\infty$.

• Example 1: $H(t) \stackrel{r}{\not \sim}0$ (r=0,1,2,...) but $H^{(-1)}(t)=t_{+}\equiv tH(t)\in {\rm C}$. We may introduce the order r=-1 and write $H(t) \stackrel{-1}{\sim}0$.
• Example 2:  

  $$f_1(t) = e^{-\alpha t} \sin \omega_0 t \cdot H(t) \stackrel{0}{\sim} 0$$ (8)

   

  $$f_2(t) = t e^{-\alpha t} \cos \omega_0 t \cdot H(t) \stackrel{0}{\sim} 0$$ (9)

  but definitely $f_1\stackrel{1}{\not \sim}0$ and $f_2\stackrel{1}{\not \sim}0$.

Is it true that $f_{1}(t) \stackrel{0}{\sim} f_{2}(t)?$The answer is: yes. Is it true that $f_{1}(t) \stackrel{1}{\sim} f_{2}(t)?$The answer is: no, however $f_{1}(t) \stackrel{1}{\sim} \omega_{0}f_{2}(t)$.(It easily may be derived that $f_{1}(t) - \omega_{0} f_{2}(t)$ has a continuous derivative at the point t=0).