Problem 102 - Calculus and its applications

Question

Find the function values that are approximations for e. Round to five decimal places. $$ \begin{aligned} &\text { For } f(t)=(1+t)^{1 / t}, \text { we have } e=\lim _{t \rightarrow 0} f(t) . \text { Find } f(1)\\\ &f(0.5), f(0.2), f(0.1), \text { and } f(0.001) \end{aligned} $$

Step-by-Step Solution

Answer

For the given \( t \) values: \( f(1) = 2, f(0.5) = 2.25, f(0.2) \approx 2.48832, f(0.1) \approx 2.59374, f(0.001) \approx 2.71692. \)

Step 1: Understanding the Function

The function given is \( f(t) = (1 + t)^{1/t} \), which is used to approximate the mathematical constant \( e \) as \( t \) approaches 0. We are tasked with finding this function's values for specific \( t \) values.

Step 2: Calculate \( f(1) \)

Replacing \( t \) with 1 in the function gives \( f(1) = (1+1)^{1/1} = 2^1 = 2. \)

Step 3: Calculate \( f(0.5) \)

For \( t = 0.5 \), the function becomes \( f(0.5) = (1+0.5)^{1/0.5} = 1.5^2 = 2.25. \)

Step 4: Calculate \( f(0.2) \)

Substitute \( t = 0.2 \) into the function: \( f(0.2) = (1+0.2)^{1/0.2} = 1.2^5. \) After calculating, \( f(0.2) \approx 2.48832. \)

Step 5: Calculate \( f(0.1) \)

Substitute \( t = 0.1 \) into the function: \( f(0.1) = (1+0.1)^{1/0.1} = 1.1^{10}. \) Compute to find \( f(0.1) \approx 2.59374. \)

Step 6: Calculate \( f(0.001) \)

For \( t = 0.001 \), the function is \( f(0.001) = (1+0.001)^{1/0.001} = 1.001^{1000}. \) After calculation, \( f(0.001) \approx 2.71692. \)

Key Concepts

Approximation Methods

Approximation methods in mathematics help us find values that are close enough to certain numbers or functions, especially when it is difficult or impossible to find an exact result. One common approach to approximation is through the use of limits. In this context, we approximate the mathematical constant \( e \), using the function \( f(t) = (1 + t)^{1/t} \). As \( t \) approaches zero, we observe how \( f(t) \) converges to \( e \). This method of using a sequence or a function that approximates a particular value is a powerful tool in calculus:

• By calculating \( f(t) \) for smaller values of \( t \), we refine our approximation of \( e \).
• Each calculation gives a closer approximation than the previous one as \( t \) becomes smaller.

This technique provides a way to approximate numbers or functions whenever we encounter complicated expressions in mathematics.

Mathematical Constant e

The mathematical constant \( e \) is an irrational number approximately equal to 2.71828. It is the base of the natural logarithm and is fundamental in the study of calculus. \( e \) has important properties that make it unique:

• \( e \) arises naturally in the process of limiting functions, particularly those involving growth and decay.
• An interesting property of \( e \) is its representation as a limit: \( e = \lim_{t \to 0} (1+t)^{1/t} \).

In our original problem, by substituting various values of \( t \) into the function \( f(t) = (1 + t)^{1/t} \), we can see how the values of the function get closer to \( e \) as \( t \) decreases. This relationship illustrates why \( e \) is often discovered through approximation techniques.

Exponential Functions

Exponential functions, characterized by a constant base raised to a variable exponent, are significant in mathematics due to their rapid growth. They are written in the form \( a^x \), where \( a \) is a positive constant. When the base is the constant \( e \), we get the natural exponential function, \( e^x \), widely used in calculus and real-world applications:

• Exponential functions model phenomena such as population growth, radioactive decay, and interest compounding.
• They exhibit constant relative growth, where the rate of change is proportional to the function's current value.

In our exercise, the function \( f(t) = (1+t)^{1/t} \) involves an exponent of the form \( 1/t \), demonstrating the concept of exponential growth as \( t \) decreases. This highlights the versatile nature of exponential functions in approximating constants and modeling various processes.