Question

Figure 7.1.4 suggests, but does not prove, that the function$f\left(t\right)=e^{t^2}$ is not of exponential order. How does the observation that$t^2>lnM+ct,$ for$M>0$ and t sufficiently large, show that $e^{t^2}>M{e}^{ct}$for any c?

Short answer

$e^{t^2}>M{e}^{ct}$for any c.

Step-by-step solution

Definition of Laplace Transform

Let f be a function defined for $t⩾0$ . Then the integral

$\mathcal{L}\left\{f\right(t\left)\right\}={\int }_0^x{e}^{-xt}f\left(t\right)dt$

is said to be the Laplace transform of f, provided that the integral converges.

Use Laplace Transform

Since, is an increasing function$,\left(e^y>e^t\right.\left.y>t\right)$.

So, $t^2>\ln M+ct$

Now,

$e^{t^2}>e^{\ln M+ct}\left(e^{a+b}=e^a{e}^b\right)$

$e^{t^2}>e^{\ln M}{e}^{ct}$

$e^{t^2}>M{e}^{ct}$

Hence, $e^{t^2}>M{e}^{ct}$for any$c$ .