Problem 17 - Calculus

Question

Use Euler's method to estimate \(B(1),\) given that $$\frac{d B}{d t}=0.05 B$$ and \(B=1000\) when \(t=0 .\) Take: (a) \(\Delta t=1\) and 1 step (b) \(\Delta t=0.5\) and 2 steps (c) \(\Delta t=0.25\) and 4 steps (d) Suppose \(B\) is the balance in a bank account earning interest. Explain why the result of your calculation in part (a) is equivalent to compounding the interest once a year instead of continuously. (e) Interpret the result of your calculations in parts (b) and (c) in terms of compound interest.

Step-by-Step Solution

Answer

(a) 1050; (b) 1050.625; (c) 1051.45; (d) Equivalent to annual compounding; (e) Approaching continuous compounding.

Step 1: Euler's Method Explanation

Euler's method is a numerical technique used to estimate solutions of ordinary differential equations (ODEs). In this case, we use it to estimate the value of the function at a specific point based on initial conditions. Given the equation \(\frac{dB}{dt} = 0.05B\), we will estimate \(B(1)\) starting from \(B(0) = 1000\). The formula for Euler's method is \(B(t+\Delta t) = B(t) + \Delta t \cdot \frac{dB}{dt}\).

Step 2: Calculate for Δt = 1 (One Step)

For \(\Delta t = 1\), calculate \(B(1)\): 1. Determine \(\frac{dB}{dt}\) at \(t = 0\) using \(\frac{dB}{dt} = 0.05 \times 1000 = 50\).2. Update \(B\) using the Euler method: \(B(1) = 1000 + 1 \times 50 = 1050\).

Step 3: Calculate for Δt = 0.5 (Two Steps)

Subdivide the interval from \(t=0\) to \(t=1\) into two steps of \(\Delta t = 0.5\):1. First step (\(t = 0.5\)): - Calculate \(\frac{dB}{dt} = 0.05 \times 1000 = 50\). - Update \(B(0.5) = 1000 + 0.5 \times 50 = 1025\).2. Second step (\(t = 1\)): - Calculate \(\frac{dB}{dt} = 0.05 \times 1025 = 51.25\). - Update \(B(1) = 1025 + 0.5 \times 51.25 = 1050.625\).

Step 4: Calculate for Δt = 0.25 (Four Steps)

Subdivide the interval from \(t=0\) to \(t=1\) into four steps of \(\Delta t = 0.25\):1. First step (\(t = 0.25\)): - Calculate \(\frac{dB}{dt} = 0.05 \times 1000 = 50\). - Update \(B(0.25) = 1000 + 0.25 \times 50 = 1012.5\).2. Second step (\(t = 0.5\)): - Calculate \(\frac{dB}{dt} = 0.05 \times 1012.5 = 50.625\). - Update \(B(0.5) = 1012.5 + 0.25 \times 50.625 = 1025.15625\).3. Third step (\(t = 0.75\)): - Calculate \(\frac{dB}{dt} = 0.05 \times 1025.15625 = 51.2578125\). - Update \(B(0.75) = 1025.15625 + 0.25 \times 51.2578125 = 1038.470703125\).4. Fourth step (\(t = 1\)): - Calculate \(\frac{dB}{dt} = 0.05 \times 1038.470703125 = 51.92353515625\). - Update \(B(1) = 1038.470703125 + 0.25 \times 51.92353515625 = 1051.4515869140625\).

Step 5: Interpretation for Part (a)

The result from part (a) where \(B(1) = 1050\) can be considered as compounding interest once per year because we simply add the interest earned over a year (\(5\%\) of the initial balance). This discretized approximation mirrors annual compounding.

Step 6: Interpretation for Parts (b) and (c)

The gradual refinement of \(\Delta t\) from 1 to 0.5 and 0.25 increases the accuracy of interest compounding: - Part (b)'s result \(B(1) = 1050.625\) and part (c)'s result \(B(1) = 1051.4515869140625\) indicate a closer approximation to continuous compounding, as we apply the interest calculation more frequently within the time period.

Key Concepts

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that relate a function with its derivatives. They are used to describe dynamic systems that change over time, capturing the essence of many natural and physical processes. In the context of the given exercise, the ODE \( \frac{dB}{dt} = 0.05B \) describes how a balance \( B \) in a bank account grows at a constant relative rate of 5% per time unit.
This equation suggests that the growth rate of the balance is directly proportional to the current amount in the account, creating a positive feedback loop that results in exponential growth.

• They simplify complex phenomena into manageable mathematical representations.
• They often require initial conditions to provide complete solutions, as they define a family of possible functions.

ODEs are a crucial part of understanding and modeling how things change and evolve over time, whether in finance, physics, biology, or beyond.

Numerical Approximation

Numerical approximation involves finding an estimated rather than exact solution to mathematical problems. This approach is particularly useful when dealing with equations that can't be solved analytically. Euler's method, utilized in the exercise, is a straightforward numerical method used to solve ordinary differential equations. It is especially significant when the solutions involve functions that are difficult to express in closed form.

In Euler's method, you estimate the value of \( B(t) \) by initiating with an initial value and iteratively applying the formula:

• \( B(t+\Delta t) = B(t) + \Delta t \cdot \frac{dB}{dt} \).
• The smaller the \( \Delta t \), the more accurate the approximation becomes, approaching the continuous solution.

Euler's method is a foundational technique for introducing the concept of iterative solutions, paving the way to more complex and precise numerical methods used in simulations and computations.

Compound Interest

Compound interest refers to earning interest on both the original amount of money and the accumulated interest from previous periods. It's a powerful concept in finance that can significantly increase the growth of an investment or savings over time. In the given exercise, compound interest is analogized by the way interest is calculated more frequently as \( \Delta t \) decreases.

When \( \Delta t = 1 \), the interest was compounded annually, simply adding 5% once to get \( B(1) = 1050 \). As \( \Delta t \) becomes 0.5, and 0.25, we are, in essence, compounding more frequently, which mimics the idea of compounding semiannually and quarterly respectively.

• The results get closer to the continuous compounding limit of interest.
• Frequent compounding generally results in higher final amounts.

Understanding compound interest helps in realizing the effect of different compounding frequencies on savings and investments over time, demonstrating why it is especially favorable in financial growth contexts.

Initial Value Problem

An initial value problem involves solving a differential equation supplemented with a specified value at the start of the period. It provides a starting point or initial condition, allowing the calculation of specific solutions out of many possible solutions suggested by the ordinary differential equation. In the exercise, the initial value given is \( B(0) = 1000 \). This is the starting balance of the bank account when \( t = 0 \).

• The initial value provides context and a grounding axis for predictions and calculations.
• It's essential for the accuracy of numerical methods such as Euler's method, as these methods build incrementally from the initial condition.

The concept of initial value problems is fundamental in both theoretical studies and practical applications, from modeling financial scenarios to simulating dynamic systems in engineering and science.