Exponential Order: Definition and Examples

Exponential Order

A function \( f \) is called an exponential order function if there is a constant \( a \) and positive constants \( t_0 \) and \( M \) such that:

\[ |f(t)| \leq M e^{at} \] for all \( t \gt t_0 \) where \( f(t) \) is defined.

In simple terms, if a function \( f \) follows this rule for some constant \( a \), we say that \( f \) is of exponential order.

This inequality sets an upper bound on the function’s growth, thereby enabling effective analysis in various mathematical and engineering applications.

Definition and Explanation

The criterion for a function to be of exponential order is both simple and powerful. Essentially, it means that the absolute value of \( f(t) \) is controlled by an exponential function for sufficiently large \( t \). In other words, no matter how complex \( f(t) \) may appear, its long-term behavior is confined within the limits dictated by the exponential function \( M e^{at} \). This controlled growth is vital when solving differential equations and analyzing system stability.

Examples-1

\( f(t)=c \) is of exponential order where \( c \) is a constant.

Since \( f(t)=c \), we first take the absolute value: \[ |f(t)| = |c|. \]

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( b \) such that \[ |f(t)| \le M e^{bt} \quad \text{for all } t > t_0. \]

Choose \( M = |c| + 1 \) (any value greater than \( |c| \)), \( b = 0 \), and any \( t_0 > 0 \). Then, for all \( t > t_0 \): \[ \begin{align} |f(t)| &= |c| \nonumber\\ &\leq |c| + 1 \nonumber\\ &\leq (|c| + 1)e^{0\cdot t} \nonumber\\ &= Me^{0\cdot t}\nonumber \end{align} \]

Since the condition holds for some constants \( M > 0 \), \( t_0 > 0 \), and \( b = 0 \), we conclude that \( f(t) = c \) is of exponential order.

Examples-2

\( f(t)=e^{at} \) is of exponential order where \(a\) is a constant.

Since \( f(t)=e^{at} \), we first take the absolute value: \[ |f(t)| = |e^{at}| = e^{at}. \]

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( b \) such that \[ |f(t)| \le M e^{bt} \quad \text{for all } t > t_0. \]

Case 1: \( a \ge 0 \)
Choose \( M = 1 \), \( b = a \), and \( t_0 = 0 \). Then, for all \( t \ge 0 \): \[ |f(t)| = e^{at} = 1\cdot e^{at} = M e^{bt}. \]

Case 2: \( a \lt 0 \)
In this case, \( e^{at} \) is a decreasing function and satisfies \( e^{at} \le 1 \) for \( t \ge 0 \). Choose \( M = 1 \), \( b = 0 \), and \( t_0 = 0 \). Then, for all \( t \ge 0 \): \[ |f(t)| = e^{at} \le 1 = 1\cdot e^{0\cdot t} = M e^{bt}. \]

Since in both cases the condition \[ |f(t)| \le M e^{bt} \quad \text{for all } t > t_0 \] is satisfied, we conclude that \( f(t)=e^{at} \) is of exponential order.

Examples-3

\( f(t)=t^n \) is of exponential order where \( n > 0 \).

Since \( f(t)=t^n \), we take the absolute value: \[ |f(t)| = t^n \quad \text{for } t \ge 0. \]

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and \( a > 0 \) such that \[ t^n \le M e^{at} \quad \text{for all } t > t_0. \]

Let \( a > 0 \) be any positive constant. Note that \[ \lim_{t \to \infty} \frac{t^n}{e^{at}} = 0, \] which implies that for sufficiently large \( t \), the exponential \( e^{at} \) dominates the polynomial \( t^n \). In other words, there exists some \( t_0 > 0 \) such that for all \( t > t_0 \), \[ t^n \lt e^{at}. \]

Now, choose \( M \) to be any constant satisfying \[ M \ge \max \left\{ \sup_{0 \le t \le t_0} \frac{t^n}{e^{at}},\, 1 \right\}. \] Then for all \( t > t_0 \): \[ t^n \le e^{at} \le M e^{at}. \]

Since the condition \[ |f(t)| \le M e^{at} \] holds for some constants \( M > 0 \), \( t_0 > 0 \), and \( a > 0 \), we conclude that \( f(t)=t^n \) is of exponential order.

Examples-4

\( f(t)=\sin at \) is of exponential order where \( a \) is a constant.

Since \( f(t)=\sin at \), we first take the absolute value: \[ |f(t)| = |\sin at|. \]

We know that for all \( t \), the sine function is bounded: \[ |\sin at| \leq 1. \]

Choose \( M = 1 \), \( b = 0 \), and \( t_0 = 0 \). Then, for all \( t \ge 0 \): \[ |f(t)| \leq 1 = Me^{0 \cdot t}. \]

Since the condition holds for some constants \( M > 0 \), \( t_0 \ge 0 \), and \( b = 0 \), we conclude that \( f(t)=\sin at \) is of exponential order.

Examples-5

\( f(t)=\cos at \) is of exponential order where \( a \) is a constant.

Since \( f(t)=\cos at \), we first consider the absolute value: \[ |f(t)| = |\cos at| \leq 1 \quad \text{for all } t. \]

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( b \) such that \[ |f(t)| \le M e^{bt} \quad \text{for all } t > t_0. \]

Choose \( M = 1 \) (any value \( \geq 1 \) will work), \( b = 0 \), and any \( t_0 \ge 0 \). Then, for all \( t \ge t_0 \): \[ |f(t)| \leq 1 = 1\cdot e^{0\cdot t} = M e^{0\cdot t}. \]

Since the condition holds for some constants \( M > 0 \), \( t_0 \ge 0 \), and \( b = 0 \), we conclude that \( f(t)=\cos at \) is of exponential order.

Examples-6

\( f(t)=t^{n}e^{at} \) is of exponential order, where \( n > 0 \) and \( a \) is a constant.

Since \( f(t)=t^{n}e^{at} \), we first take the absolute value: \[ |f(t)| = |t^{n}e^{at}| = t^{n}\,|e^{at}|. \]

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( \alpha \) such that \[ t^{n}e^{at} \le M e^{\alpha t} \quad \text{for all } t > t_0. \]

Case 1: \( a \ge 0 \)
Choose any \( \delta > 0 \) and set \( \alpha = a + \delta \). Then, for large \( t \), the exponential \( e^{\delta t} \) eventually dominates the polynomial \( t^{n} \) because \[ \lim_{t \to \infty} \frac{t^n}{e^{\delta t}} = 0. \] Thus, there exists some \( t_0 > 0 \) such that for all \( t > t_0 \): \[ t^{n} \lt e^{\delta t}. \] Multiplying both sides by \( e^{at} \) gives \[ t^{n}e^{at} \lt e^{at}e^{\delta t} = e^{(a+\delta)t} = e^{\alpha t}. \] Now, choose \( M \) to be any constant satisfying \[ M \ge \max\left\{ \sup_{0 \le t \le t_0} \frac{t^{n}e^{at}}{e^{\alpha t}},\, 1 \right\}. \] Then, for all \( t > t_0 \): \[ t^{n}e^{at} \le e^{\alpha t} \le M e^{\alpha t}. \]

Case 2: \( a \lt 0 \)
Here, \( e^{at} \) decays as \( t \) increases, and the product \( t^{n}e^{at} \) attains a maximum at some finite \( t \). Define \[ M = \max_{t \ge 0} \{ t^{n}e^{at} \}, \] and choose \( \alpha = 0 \) and \( t_0 = 0 \). Then for all \( t \ge 0 \): \[ t^{n}e^{at} \le M = M e^{0\cdot t}. \]

In both cases, we have found constants \( M > 0 \), \( t_0 \ge 0 \), and a constant (either \( \alpha > 0 \) or \( \alpha = 0 \)) such that \[ |f(t)| = t^{n}e^{at} \le M e^{\alpha t} \quad \text{for all } t > t_0. \] Therefore, we conclude that \( f(t)=t^{n}e^{at} \) is of exponential order.

Examples-7

\( f(t)=t^{n}\sin(at) \) where \( n > 0 \) and \( a \) is a constant is of exponential order.

Since \( f(t)=t^{n}\sin(at) \), we first take the absolute value: \[ |f(t)| = |t^{n}\sin(at)| \le t^{n}, \] because \( |\sin(at)| \le 1 \) for all \( t \).

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( b \) such that \[ |f(t)| \le M e^{bt} \quad \text{for all } t > t_0. \]

Notice that \( t^n \) is a polynomial and grows slower than any exponential function \( e^{bt} \) when \( b > 0 \). In fact, for any fixed \( b > 0 \), \[ \lim_{t \to \infty}\frac{t^n}{e^{bt}} = 0. \] This implies that there exists some \( t_0 \ge 0 \) such that for all \( t > t_0 \), \[ t^n \le e^{bt}. \]

Thus, if we choose \( M \ge 1 \) (for example, \( M = 1 \)) and any \( b > 0 \) along with the corresponding \( t_0 \), then for all \( t > t_0 \) we have: \[ |f(t)| \le t^n \le M e^{bt}. \]

Since the condition holds for some constants \( M > 0 \), \( t_0 \ge 0 \), and \( b > 0 \), we conclude that \( f(t)=t^{n}\sin(at) \) is of exponential order.

Examples-8

\( f(t)=t^{n}\cos at \) where \(n \gt 0\) and \(a\) is a constant.

Since \( f(t)=t^{n}\cos(at) \), we first take the absolute value: \[ |f(t)| = |t^{n}\cos(at)| \le t^{n}, \] because \( |\cos(at)| \le 1 \) for all \( t \).

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( b \) such that \[ |f(t)| \le M e^{bt} \quad \text{for all } t \gt t_0. \]

Notice that \( t^n \) is a polynomial and grows slower than any exponential function \( e^{bt} \) when \( b > 0 \). In fact, for any fixed \( b > 0 \), \[ \lim_{t \to \infty}\frac{t^n}{e^{bt}} = 0. \] This implies that there exists some \( t_0 \ge 0 \) such that for all \( t > t_0 \), \[ t^n \le e^{bt}. \]

Thus, if we choose \( M \ge 1 \) (for example, \( M = 1 \)) and any \( b > 0 \) along with the corresponding \( t_0 \), then for all \( t > t_0 \) we have: \[ |f(t)| \le t^n \le M e^{bt}. \]

Since the condition holds for some constants \( M > 0 \), \( t_0 \ge 0 \), and \( b > 0 \), we conclude that \( f(t)=t^{n}\cos(at) \) is of exponential order.

Examples-9

\( f(t)=e^{bt}\sin at \) is of exponential order, where \(a\) and \(b\) are constants.

Since \( f(t)=e^{bt}\sin at \), we first take the absolute value: \[ |f(t)| = |e^{bt}\sin at| \le e^{bt} \, |\sin at| \le e^{bt}, \] because \( |\sin at| \le 1 \) for all \( t \).

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( \alpha \) such that \[ |f(t)| \le M e^{\alpha t} \quad \text{for all } t > t_0. \]

Case 1: \( b \ge 0 \)
Choose \( M = 1 \), \( \alpha = b \), and \( t_0 = 0 \). Then for all \( t \ge 0 \): \[ |f(t)| \le e^{bt} = 1\cdot e^{bt} = M e^{\alpha t}. \]

Case 2: \( b \lt 0 \)
In this case, since \( e^{bt} \) is a decreasing function and satisfies \( e^{bt} \le 1 \) for \( t \ge 0 \), choose \( M = 1 \), \( \alpha = 0 \), and \( t_0 = 0 \). Then for all \( t \ge 0 \): \[ |f(t)| \le e^{bt} \le 1 = 1\cdot e^{0\cdot t} = M e^{\alpha t}. \]

Since in both cases the condition \[ |f(t)| \le M e^{\alpha t} \quad \text{for all } t > t_0 \] is satisfied, we conclude that \( f(t)=e^{bt}\sin at \) is of exponential order.

Examples-10

\( f(t)=e^{bt}\cos at \) is of exponential order, where \(a\) and \(b\) are constants.

Since \( f(t)=e^{bt}\cos at \), we first take the absolute value: \[ |f(t)| = |e^{bt}\cos at| \le e^{bt} \, |\cos at| \le e^{bt}, \] because \( |\cos at| \le 1 \) for all \( t \).

We want to show that there exist constants \( M > 0 \), \( t_0 \ge 0 \), and a constant \( \alpha \) such that \[ |f(t)| \le M e^{\alpha t} \quad \text{for all } t > t_0. \]

Case 1: \( b \ge 0 \)
Choose \( M = 1 \), \( \alpha = b \), and \( t_0 = 0 \). Then for all \( t \ge 0 \): \[ |f(t)| \le e^{bt} = 1\cdot e^{bt} = M e^{\alpha t}. \]

Case 2: \( b \lt 0 \)
In this case, since \( e^{bt} \) is a decreasing function and satisfies \( e^{bt} \le 1 \) for \( t \ge 0 \), choose \( M = 1 \), \( \alpha = 0 \), and \( t_0 = 0 \). Then for all \( t \ge 0 \): \[ |f(t)| \le e^{bt} \le 1 = 1\cdot e^{0\cdot t} = M e^{\alpha t}. \]

Since in both cases the condition \[ |f(t)| \le M e^{\alpha t} \quad \text{for all } t > t_0 \] is satisfied, we conclude that \( f(t)=e^{bt}\cos at \) is of exponential order.

Examples-11

\( f(t)=\sinh at \) is of exponential order, where \( a \) is a constant.

Since \( f(t)=\sinh at \), recall that \[ \sinh at = \frac{e^{at} – e^{-at}}{2}. \] Taking the absolute value gives \[ |f(t)| = |\sinh at| \le \frac{e^{at} + e^{-at}}{2}. \]

Notice that for any real \( a \) and for all \( t \ge 0 \), \[ e^{at} \le e^{|a|t} \quad \text{and} \quad e^{-at} \le e^{|a|t}. \] Therefore, \[ |f(t)| \le \frac{e^{|a|t} + e^{|a|t}}{2} = e^{|a|t}. \]

Choose \( M = 1 \), \( \alpha = |a| \), and \( t_0 = 0 \). Then, for all \( t \ge 0 \), \[ |f(t)| \le e^{|a|t} = M e^{\alpha t}. \]

Since we have found constants \( M > 0 \), \( t_0 \ge 0 \), and \( \alpha > 0 \) such that \[ |f(t)| \le M e^{\alpha t}, \] we conclude that \( f(t)=\sinh at \) is of exponential order.

Examples-12

\( f(t)=\cosh at \) is of exponential order, where \( a \) is a constant.

Since \( f(t)=\cosh at \), recall that \[ \cosh at = \frac{e^{at}+e^{-at}}{2}. \]

Taking the absolute value (noting that \(\cosh at \ge 0\) for all \(t\)), we have: \[ |f(t)| = \cosh at = \frac{e^{at}+e^{-at}}{2}. \]

For any real \(a\) and all \(t \ge 0\), observe that \[ e^{at} \le e^{|a|t} \quad \text{and} \quad e^{-at} \le e^{|a|t}. \] Hence, \[ |f(t)| \le \frac{e^{|a|t}+e^{|a|t}}{2} = e^{|a|t}. \]

Choose \( M = 1 \), \( \alpha = |a| \), and \( t_0 = 0 \). Then, for all \( t \ge 0 \): \[ |f(t)| \le e^{|a|t} = M e^{\alpha t}. \]

Since the condition \[ |f(t)| \le M e^{\alpha t} \] holds for some constants \( M > 0 \), \( t_0 \ge 0 \), and \( \alpha > 0 \), we conclude that \( f(t)=\cosh at \) is of exponential order.