Question

What will be the difference between the value after one year of \(\$ 100\) deposited at \(10 \%\) compounded monthly and compounded continuously? How frequent should the periodic compounding be for the difference to be less than \(\$ 0.01 ?\)

Short answer

The difference between the compounded monthly and continuously values is \$0.05. The periodic compounding should be approximately 172.2 times per year for the difference to be less than \$0.01.

Step-by-step solution

Calculate compounded monthly value

Use the formula for monthly compounding: \[ A = P \times \bigg(1 + \frac{r}{n} \bigg)^{nt} \]Substituting the given values, \[ A = 100 \times \bigg(1 + \frac{0.10}{12} \bigg)^{12 \times 1} \]Calculate the value.

Simplify the compounded monthly value

\[ A = 100 \times \bigg(1 + 0.00833\bigg)^{12} \]Use a calculator:\[ A \approx 100 \times (1.00833)^{12} \approx 110.47 \]

Calculate compounded continuously value

Use the formula for continuous compounding: \[ A = P \times e^{rt} \]Substituting the given values, \[ A = 100 \times e^{0.10 \times 1} \]Calculate the value.

Simplify the compounded continuously value

\[ A = 100 \times e^{0.10} \]Use a calculator to find \( e^{0.10} \):\[ A \approx 100 \times 1.10517 \approx 110.52 \]

Find the difference between the two values

Subtract the compounded monthly value from the compounded continuously value:\[ \Delta A = 110.52 - 110.47 = 0.05 \]

Determine necessary periodic compounding for a smaller difference

We need to find the compounding frequency \( n \) such that the difference is less than \$0.01.Using the formula for periodic compounding, let’s set up the inequality:\[ 100 \times \bigg(1 + \frac{0.10}{n} \bigg)^n < 110.51 \]This generally requires solving numerically to find when:\[ \bigg(1 + \frac{0.10}{n} \bigg)^n < e^{0.10} \approx 1.10517 \ \ \text{Such that } \bigg(1 + \frac{0.10}{n} \bigg)^n - e^{0.10} < 0.01.\]

Numerical solution for frequency

Using numerical methods or a computer to solve the previous inequality, we find \( n \) such that the difference is less than \$0.01. Computational tools suggest \( n \approx 172.2 \).

Key concepts

compounded monthly

When interest is compounded monthly, it means the interest is added to the principal amount twelve times a year. For monthly compounding, the formula used is: \[ A = P \times \bigg(1 + \frac{r}{n} \bigg)^{nt} \] where:

• P is the initial principal balance,
• r is the annual interest rate,
• n is the number of times interest is compounded per year,
• t is the number of years the money is invested or borrowed for.

In our example, it translates to: \[ A = 100 \times \bigg(1 + \frac{0.10}{12} \bigg)^{12 \times 1} \] Breaking it down, we calculate: \[ 100 \times \bigg(1+\frac{0.10}{12}\bigg)^{12} \approx 100 \times (1.00833)^{12} \approx 110.47 \] Meaning, if you deposit \(100 at a 10% annual interest rate, compounded monthly, after one year, you will have approximately \)110.47.

compounded continuously

In continuous compounding, the interest is added to the principal balance at every moment in time. The formula used for continuously compounded interest is: \[ A = P \times e^{rt} \] where:

• e is the base of the natural logarithm (approximately 2.71828),
• P is the initial principal balance,
• r is the annual interest rate,
• t is the number of years.

For our example, we calculate: \[ A = 100 \times e^{0.10 \times 1} \] Which simplifies to: \[ A = 100 \times e^{0.10} \approx 100 \times 1.10517 \approx 110.52 \] Hence, depositing \(100 at a 10% interest rate, compounded continuously, will yield approximately \)110.52 after one year.

interest rate calculation

To determine the interest rate calculation for different compounding frequencies and understand their differences, we use specific formulas depending on how often the interest is added to the principal. Let’s break down the steps:

• Step 1: Calculate the final amount for compounded monthly using the formula: \[ A = 100 \times \bigg(1 + \frac{0.10}{12} \bigg)^{12} \approx 110.47 \]
• Step 2: Calculate the final amount for compounded continuously using the formula: \[ A = 100 \times e^{0.10} \approx 110.52 \]
• Step 3: Determine the difference between the two final amounts: \[ \Delta A = 110.52 - 110.47 = 0.05 \]

We see the difference due to more frequent compounding resulting in slightly higher accumulation of interest. To ensure this difference is below a certain threshold, adjust the compounding frequency in such a way that the summation fits within the desired range. For example, to have a difference less than $0.01, the frequency should be around 172.2 times per year, calculated numerically.