Problem 63 - Algebra and Trigonometry

Question

Compound Interest \(\quad\) A savings account earns 5\(\%\) interest compounded annually. If you invest \(x\) dollars in such an account, then the amount \(A(x)\) of the investment after one year is the initial investment plus \(5 \% ;\) that is, \(A(x)=x+0.05 x=1.05 x .\) Find $$\begin{array}{l}{A \circ A} \\ {A \circ A \circ A} \\ {A \circ A \circ A \circ A}\end{array}$$ What do these compositions represent? Find a formula for what you get when you compose \(n\) copies of \(A .\)

Step-by-Step Solution

Answer

The compositions represent annual compound interest. Formula: \(A_n(x) = (1.05)^n x\).

Step 1: Understanding the Composition

Composing functions involves applying one function to the result of another. Here, the function \(A(x) = 1.05x\) is applied multiple times. We need to find expressions for \(A \circ A\), \(A \circ A \circ A\), and \(A \circ A \circ A \circ A\).

Step 2: Finding A∘A

First, apply \(A\) to \(x\) to get \(A(x) = 1.05x\). Now apply \(A\) again to the result: \(A(A(x)) = A(1.05x) = 1.05 \cdot 1.05x = (1.05)^2 x\). Thus, \(A \circ A = (1.05)^2 x\).

Step 3: Finding A∘A∘A

Next, apply \(A\) to the result \((1.05)^2 x\). We have \(A((1.05)^2 x) = 1.05 \times (1.05)^2 x = (1.05)^3 x\). Thus, \(A \circ A \circ A = (1.05)^3 x\).

Step 4: Finding A∘A∘A∘A

Apply \(A\) once more to the result \((1.05)^3 x\) to get: \(A((1.05)^3 x) = 1.05 \times (1.05)^3 x = (1.05)^4 x\). Thus, \(A \circ A \circ A \circ A = (1.05)^4 x\).

Step 5: General Formula for n Compositions

When composing \(A\) with itself \(n\) times, we apply \(A(x) = 1.05x\) repeatedly, resulting in \(A_n(x) = (1.05)^n x\). This composition represents the compound interest growth compounded annually over \(n\) years.

Key Concepts

Composition of Functions

When we talk about the composition of functions, we are applying one function to the results of another function. It's like a process where you input a value, transform it with the first function, and then take that result and transform it again with the second function.
For example, let's consider the function in our problem: \( A(x) = 1.05x \). In the context of compound interest, composing this function with itself represents applying the function multiple times.

• First application: \( A(x) = 1.05x \)
• Second application: \( A(A(x)) = 1.05(1.05x) = (1.05)^2x \)
• Third application: \( A(A(A(x))) = 1.05((1.05)^2x) = (1.05)^3x \)
• Fourth application, which completes our problem context: \( A(A(A(A(x)))) = (1.05)^4x \)

Each additional composition models another year of compound interest growth.

Exponential Growth

Exponential growth is a process where quantities increase rapidly at a consistent rate over time. Think of it like a snowball rolling down a hill, gathering more snow, and growing larger more quickly the longer it rolls.
In our context of compound interest, the equation \( A(x) = 1.05x \) reflects exponential growth. Each year, the amount of money in the account grows by 5%.

• Year 1 growth: \( A(x) = 1.05x \)
• Year 2 growth: \( (1.05)^2x \) – 5% more on both the initial amount and the 5% already earned
• Year 3 growth: \( (1.05)^3x \) – this extra interest multiplication shows the power of exponential growth

As you see, the function grows more significantly each year, as it includes interest on both the initial principal and the previously earned interest.

Annual Compounding

In finance, annual compounding refers to the process of calculating interest on the initial principal and also on the accumulated interest from previous periods within a year. It's a common method used by banks to help your savings grow.
Unlike simple interest, where you only earn interest on the original amount invested, annual compounding allows you to earn interest on both your initial deposit and the interest your deposit has already earned.
This is what makes compound interest such a powerful tool for growing wealth over time.
In our exercise, the investment \( x \) grows annually by 5%, increasing not just on the initial investment but also on each year's growth:

• After one year: \( 1.05x \)
• After two years: \( (1.05)^2x \)
• Continuing in this pattern for \( n \) years results in \( (1.05)^nx \)

This formula shows the total amount in the account after \( n \) years of annual compounding, reflecting a thorough application of both the composition of functions and exponential growth principles.