Question

Show that each of the following functions is of exponential order: (a) \(\sin \omega t, \omega\) a constant. (b) \(e^{\text {at }}, a\) a constant. (c) \(\ln (1+t)\). (d) \(t^{n}, n\) a positive integer.

Short answer

Question: Show that each of the following functions is of exponential order. a) \(\sin \omega t\) b) \(e^{\text {at }}\) c) \(\ln (1+t)\) d) \(t^n\) Answer: a) \(\sin \omega t\) is of exponential order with \(k = 0\) and \(M = 1\). b) \(e^{\text {at }}\) is of exponential order with \(k = |a|\) and \(M = 1\). c) \(\ln (1+t)\) is of exponential order with \(k = 1\) and \(M = 1\). d) \(t^n\) is of exponential order with \(k > n\).

Step-by-step solution

(a) Considering the function \(\sin \omega t\)

To show that \(\sin \omega t\) is of exponential order, we need to find positive constants \(k\) and \(M\) such that \(|\sin \omega t| \leq Me^{kt}\) for all \(t \geq 0\). Since the sine function is always bounded between -1 and 1, we have \(|\sin \omega t| \leq 1\) for all \(t \geq 0\). Hence, we can choose \(k = 0\) and \(M = 1\). This satisfies the condition, and therefore, \(\sin \omega t\) is of exponential order.

(b) Considering the function \(e^{\text {at }}\)

To show that \(e^{\text {at }}\) is of exponential order, we need to find positive constants \(k\) and \(M\) such that \(|e^{\text {at }}| \leq Me^{kt}\) for all \(t \geq 0\). We can choose \(k = |a|\) and \(M = 1\). Then, for all \(t \geq 0\), we have \(|e^{\text {at }}| \leq e^{|a|t} = 1 \cdot e^{|a|t} = Me^{kt}\). Hence, \(e^{\text {at }}\) is of exponential order.

(c) Considering the function \(\ln (1+t)\)

To show that \(\ln (1+t)\) is of exponential order, we need to find positive constants \(k\) and \(M\) such that \(|\ln (1+t)| \leq Me^{kt}\) for all \(t \geq 0\). Let's choose \(k = 1\) and \(M = 1\). The inequality \(|\ln (1+t)| \leq Me^{kt}\) becomes \(|\ln (1+t)| \leq e^t\). As t approaches 0, the function \(\ln(1+t)\) is smaller than \(e^t\). Since the exponential function \(e^t\) grows much faster than the logarithmic function \(\ln(1+t)\), the inequality holds for all \(t \geq 0\). Therefore, \(\ln (1+t)\) is of exponential order.

(d) Considering the function \(t^n\)

To show that \(t^n\) is of exponential order, we need to find positive constants \(k\) and \(M\) such that \(|t^n| \leq Me^{kt}\) for all \(t \geq 0\). Using the fact that \(\lim_{t\to\infty} \frac{t^n}{e^{kt}} = 0\) for any positive integers \(n\) and any \(k > 0\), we can choose \(k\) greater than \(n\). By the squeeze theorem, there exists an \(M\) such that \(|t^n| \leq Me^{kt}\) for all \(t \geq 0\). Hence, \(t^n\) is of exponential order.

Key concepts

Bounded Functions

A function is considered bounded if there is a real number that serves as an upper limit to its values over its domain. Specifically, for a function like \( \sin \omega t \), it is known to oscillate between \(-1\) and \(1\). This means that no matter what value \(t\) takes, \(|\sin \omega t|\) will always be less than or equal to 1.

• This behavior showcases the concept of boundedness: having upper and lower constraints.
• The importance of bounded functions in mathematics is their predictability and restraint within certain limits.
• Finding a bounding constant \(M\), like 1 in the sine function, allows us to control and compare functions against exponential growth conditions.

Understanding bounded functions helps in analyzing whether a function can be encompassed within exponential parameters.

Exponential Functions

Exponential functions have the general form of \( f(t) = ae^{bt} \), where \(a\) and \(b\) are constants. These functions are characterized by their rapid growth or decay, depending on the sign of \(b\).

• In the positive case, such as \(e^{at}\), the function grows exponentially, quickening its ascent as \(t\) increases.
• This form easily demonstrates an exponential order because it simplifies the inequality \(|f(t)| \leq Me^{kt}\) by matching forms on both sides.
• It's essential to recognize exponential functions as they often describe real-world growth processes, including populations and investments.

Exponential functions become useful benchmarks to determine if other types of functions can be bounded within exponential orders.

Logarithmic Functions

Logarithmic functions, like \( \ln(1+t) \), grow slower compared to exponential functions. They have the form \( g(t) = \ln(1+t) \) and reflect a gradual increase.

• These functions are the inverse of exponential functions; hence, their growth rate is considerably slower.
• The inequality \( |\ln(1+t)| \leq e^t \) holds because the exponential function \(e^t\) grows significantly faster than \(\ln(1+t)\), especially for larger \(t\).
• This relationship allows us to prove that logarithmic functions can be of exponential order by highlighting their slower growth rate.

Understanding the moderate pace of logarithmic growth, compared to exponential functions, plays a crucial role in categorizing function orders.

Polynomial Growth

Polynomial functions like \(t^n\), where \(n\) is a positive integer, are known for their growth rates which are moderate compared to exponential functions. The form \( h(t) = t^n \) demonstrates straightforward power law growth.

• A polynomial function's growth rate is lesser than that of an exponential one, where the latter eventually outpaces the former for large \(t\).
• The condition \(|t^n| \leq Me^{kt}\) can be satisfied by choosing a careful \(k\), because exponentials will eventually overpower the polynomial with adequate room for a bounding constant \(M\).
• This makes polynomial functions manageable within exponential limits over extended periods.

By recognizing the exponential dominance over polynomial functions as \(t\) increases, it becomes evident why polynomials can achieve exponential order under certain conditions.