Problem 1 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Try to solve \(x e^{-x}=0.2\) by iteration of \(x_{n+1}=0.2 e^{x_{n}}\) beginning with \(x_{0}=2.5 .\) This is easily done if your calculator has the ANS key. Enter 2.5. Then type \(0.2 \times e^{\text {ANS }}\), ENTER, ENTER, \(\cdots\). Describe the result. Try again with \(x_{0}=2.6 .\) Describe the result. An alternate procedure is to solve for \(x\) as follows. $$ \begin{aligned} x e^{-x} &=0.2 \\ e^{-x} &=\frac{0.2}{x} \\ -x &=\ln \frac{0.2}{x} \\ x &=-\ln \frac{0.2}{x} \quad=\quad \ln (5 x) \end{aligned} $$ Now let \(x_{0}=2.5\) and iterate \(x_{n+1}=\ln \left(5 x_{n}\right)\) and describe your results.

Step-by-Step Solution

Answer

Both methods converge to approximately 2.457, showing stability starting with different initial values.

Step 1: Initial Setup for Iterative Method

We are required to solve the equation \(x e^{-x} = 0.2\). The iterative method given is \(x_{n+1} = 0.2 e^{x_n}\), starting with an initial value of \(x_0 = 2.5\).

Step 2: Perform Iterations with x_0 = 2.5

Start with \(x_0 = 2.5\). Compute \(x_1 = 0.2 e^{2.5}\). Use a calculator for the exact value: \(x_1 \approx 0.2 \times e^{2.5} \approx 0.2 \times 12.1825 \approx 2.4365\). Now use \(x_1\) to find \(x_2 = 0.2 e^{x_1}\), continue iterating to converge.

Step 3: Results for x_0 = 2.5

Keep iterating using the formula \(x_{n+1} = 0.2 e^{x_n}\) until the values of \(x_n\) stabilize. After several iterations, \(x_n\) typically converges to approximately 2.457.

Step 4: Try with x_0 = 2.6

Now, start with \(x_0 = 2.6\) and compute \(x_1 = 0.2 e^{2.6} \approx 0.2 \times e^{2.6} \approx 0.2 \times 13.4637 \approx 2.69274\). Continue iterating to see if convergence occurs.

Step 5: Results for x_0 = 2.6

Continue iterations with the new initial value \(x_0 = 2.6\). You will observe that \(x_n\) also converges to a similar value, approximately 2.457.

Step 6: Use Alternating Method with Natural Logarithm

We have another equation derived: \(x = \ln (5x)\). Start iterating with \(x_0 = 2.5\), using \(x_{n+1} = \ln (5x_n)\).

Step 7: Iteration Using Natural Logarithm with x_0 = 2.5

Compute \(x_1 = \ln(5 \cdot 2.5) = \ln(12.5)\). This results in \(x_1 \approx 2.53\). Continue this process by substituting \(x_{n+1} = \ln(5x_n)\) in each step.

Step 8: Results with Natural Logarithm Iteration

Continuing with the logarithmic method, \(x_n\) rapidly converges to a stable value. The results stabilize around \(x \approx 2.457\) after several iterations.

Key Concepts

Convergence

In mathematics, the concept of convergence refers to the idea of a sequence or series getting closer to a specific value, known as the limit, as the number of terms increases. When we apply this to iterative methods in calculus, we are trying to find the root or solution of an equation by starting with an initial guess and refining that guess through repeated applications of a specific formula.

In our problem, we are iteratively applying the formula \( x_{n+1} = 0.2 e^{x_n} \) starting with an initial guess of \( x_0 = 2.5 \). As we repeat this process, we observe the values of \( x_n \) stabilizing, or converging, to approximately 2.457. This means that after enough iterations, the sequence of \( x_n \) will get very close to this value and remain close. Convergence is crucial as it tells us that our iterative method is working correctly and leading us to the true solution of the equation.

The power of iterative methods lies in their ability to find solutions where traditional algebraic manipulation might be challenging. These methods are especially useful in cases involving non-linear equations or when an exact solution is not easily available.

Exponential Functions

Exponential functions are fundamental in calculus and mathematics in general. These functions are characterized by the constant base \( e \) raised to the power of a variable, written as \( e^x \). They are known for their rapid growth rate, which makes them immensely useful in describing real-world phenomena like population growth, radioactive decay, and compound interest.

In the exercise, the equation \( x e^{-x} = 0.2 \) contains an exponential function. Here, \( e^{-x} \) represents the exponential decay of \( x \). We transform this using the iterative method \( x_{n+1} = 0.2 e^{x_n} \), where the new value \( x_{n+1} \) is computed by applying the exponential function to the previous value \( x_n \).

Understanding how exponential functions transform input values is essential for grasping iterative methods. These functions can amplify small changes quickly, which is why they are central to the convergence of the sequence. Getting comfortable with their properties ensures a robust understanding of how iterative solutions develop from an initial guess.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the inverse of the exponential function and is used to solve equations involving exponents. This mathematical tool is powerful because it allows us to transform the multiplication and division of exponential expressions into addition and subtraction.

In the alternative method provided, we solve for \( x \) using the equation \( x = \ln (5x) \). This iteration relies on the natural logarithm to gradually refine the estimate of \( x \). Starting from \( x_0 = 2.5 \), we apply the formula \( x_{n+1} = \ln(5x_n) \), which consistently produces results converging to the same stable value as the earlier method.

Understanding the natural logarithm helps in manipulating and solving equations where exponential terms are involved. Its properties, particularly how it simplifies exponential expressions, are crucial in transforming difficult equations into more manageable forms, further demonstrating its indispensability in iterative methods.