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Problem 468

# (a) Prove that $$\mathrm{f}(\mathrm{t})=\mathrm{t}^{\mathrm{n}}, \mathrm{n}>0$$, is of exponential order $$\alpha$$ on $$[0, \infty]$$ for all $$\alpha>0$$. (b) Prove that $$\mathrm{f}(\mathrm{t})=\sin \mathrm{kt}$$ is of exponential order $$\alpha$$ on $$[0, \infty]$$ for all $$\alpha>0$$

Expert verified
We proved that the functions f(t) = t^n, n>0, and f(t) = sin(kt) are of exponential order α on [0, ∞] for all α>0. For f(t) = t^n, we used L'Hôpital's Rule to show that the limit t^n/e^(αt) goes to 0 as t approaches infinity. For f(t) = sin(kt), we showed that |sin(kt)| ≤ e^(αt) holds true for all α>0 since e^(αt) is non-negative and |sin(kt)| is always less than or equal to 1.
See the step by step solution

## Step 1: Define the function

We are given the function f(t) = t^n, where n>0.

## Step 2: Show that f(t) is of exponential order α

To show that f(t) is of exponential order α, we need to find a positive constant M and α>0 such that |f(t)| ≤ M * e^(αt) for all t≥0. Note that as t tends to infinity, t^n tends to infinity as well, which means that there exists no M such that the inequality holds for all t. However, for every α>0, we can take a constant M>0 that satisfies: lim_{t->∞} t^n/(M * e^(αt))=0 From the L'Hôpital's Rule, we can apply it n times on the limit: lim_{t->∞} t^n / e^(αt) = 0 Since the limit as t goes to infinity is 0, we can conclude that the function f(t) = t^n is of exponential order α for all α>0. (b) Prove that f(t) = sin(kt) is of exponential order α on [0, ∞] for all α>0

## Step 1: Define the function

We are given the function f(t) = sin(kt).

## Step 2: Show that f(t) is of exponential order α

To show that f(t) = sin(kt) is of exponential order α, we need to find a positive constant M and α>0 such that |f(t)| ≤ M * e^(αt) for all t≥0. Since the maximum value of sin(kt) is 1, we can take M=1. Now, we need to find α>0 such that |sin(kt)| ≤ e^(αt) for all t≥0. The inequality always holds because e^(αt) is always non-negative, and |sin(kt)| is always less than or equal to 1. Hence, the function f(t) = sin(kt) is of exponential order α on [0, ∞] for all α>0.

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