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Chapter 1: Problem 56

If you invest \(x\) dollars at 4\(\%\) interest compounded annually,then the amount \(A(x)\) of the investment after one year is \(A(x)=1.04 x .\) Find \(A \circ A, A\) o \(A\) o \(A,\) and \(A \circ A \circ A \circ A .\) What do these compositions represent? Find a formula for the com- position of \(n\) copies of \(A .\)

Short Answer

Expert verified
Each composition represents one additional year of investment compounding, and the formula is \(A^{(n)}(x) = (1.04)^n x\).

Step by step solution

01

Understanding the Problem

We are given the function \(A(x) = 1.04x\), which represents the amount after one year given an initial investment \(x\) dollars at 4% compounded annually. We need to evaluate the compositions \(A \circ A\), \(A \circ A \circ A\), \(A \circ A \circ A \circ A\), and eventually derive a general formula for \(n\) compositions.
02

Evaluating the Composition \( A \circ A \)

For the function \(A \circ A\), we substitute \(A(x)\) into itself: \[ A(A(x)) = A(1.04x) = 1.04(1.04x) = (1.04)^2 x.\] This represents the amount after two years with compounding.
03

Evaluating the Composition \( A \circ A \circ A \)

Next, find \(A(A(A(x)))\): \[ A(A(A(x))) = A((1.04)^2 x) = 1.04((1.04)^2 x) = (1.04)^3 x.\] This represents the amount after three years with compounding.
04

Evaluating the Composition \( A \circ A \circ A \circ A \)

Now, we compute \(A(A(A(A(x))))\): \[ A(A(A(A(x)))) = A((1.04)^3 x) = 1.04((1.04)^3 x) = (1.04)^4 x.\] This signifies the amount after four years with compounding.
05

Finding the General Formula for \( A^{(n)}(x) \)

From the previous results, we observe that each time we compose further, the exponent of \(1.04\) increases by 1. Thus, for \(n\) compositions, the formula is:\[ A^{(n)}(x) = (1.04)^n x.\] This reflects the amount after \(n\) years with compounding.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Composition
Function composition is a fundamental concept in mathematics where we apply one function to the result of another function. In this exercise, function composition is used to calculate the future value of an investment over multiple years based on compound interest.

When we compose the function, like \( A \circ A \) or \( A(A(x)) \), it means we apply the function \( A \) to itself. This is a repeated process, indicative of adding more layers of complexity or compounding in the context of investments.
  • \( A(x) = 1.04x \) represents the amount after one year.
  • \( A \circ A \) indicates two years with compounding, leading to \( (1.04)^2 x \).
  • \( A \circ A \circ A \) or \( A^{(3)}(x) \) equates to three years, yielding \( (1.04)^3 x \).
  • General formula \( A^{(n)}(x) \) is \( (1.04)^n x \), denoting \( n \) years of growth.
Function composition vividly underscores how repetitive applications of the same operation can evolve using mathematical modeling. It elegantly portrays exponential growth over time.
Exponential Growth
Exponential growth occurs when a quantity increases over time at a consistent rate, and it's vividly illustrated in the problem of compounded interest.

Every time we apply the function \( A(x) \), we multiply the initial amount by the same factor, 1.04 (reflecting 4% growth per year). This consistent percentage increase is key to exponential characteristics.

  • With exponential growth, small annual percentage increases can lead to significant total gains over long periods.
  • After each composition, you notice the exponent on 1.04 growing, corresponding directly to the number of years.
Exponential growth is apparent as each subsequent function application results in the investment growing more significantly, explaining why long-term investments with compound interest can lead to substantial growth.
Mathematical Modeling
Mathematical modeling is a powerful tool that transforms real-world phenomena into mathematical expressions. In this problem, we are modeling how an investment grows over time, specifically through the mechanism of compound interest.

Using the function \( A(x) = 1.04x \), we create a model that shows how the value of an investment changes annually. As we compose the function repeatedly, this model becomes more complex yet accurately represents this real-world scenario.

  • The function \( A(x) \) depicts a year's worth of economic growth.
  • Through composition, the model extends to multiple years, illustrating the cumulative effect.
  • The model captures the principle of accumulated growth effectively through the pattern \( (1.04)^n x \).
This exercise exemplifies how mathematical modeling helps us explore and plan financial scenarios, providing clarity and insights into how investments may perform over time.

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Most popular questions from this chapter

Data points \((x, y)\) are given. (a) Draw a scatter plot of the data points. (b) Make semilog and log-log plots of the data. (c) Is a linear, power, or exponential function appropriate for modeling these data? (d) Find an appropriate model for the data and then graph the model together with a scatter plot of the data. $$\begin{array}{|c|c|c|c|c|c|c|}\hline x & {2} & {4} & {6} & {8} & {10} & {12} \\\ \hline y & {0.08} & {0.12} & {0.18} & {0.26} & {0.35} & {0.53} \\\ \hline\end{array}$$ \(\begin{array}{|c|c|c|c|c|c|c|}\hline x & {0.5} & {1.0} & {1.5} & {2.0} & {2.5} & {3.0} \\ \hline y & {4.10} & {3.71} & {3.39} & {3.2} & {2.78} & {2.53} \\ \hline\end{array}\)

Us population The table gives the population of theUnited States, in millions, for the years \(1900-2010 .\) Use a calculator with exponential regression capability to model the US population since \(1900 .\) Use the model to estimate the population in 1925 and to predict the population in the year \(2020 .\) $$\begin{array}{|c|c|}\hline \text { Year } & {\text { Population }} \\\ \hline 1900 & {76} \\ \hline 1910 & {92} \\ \hline 1920 & {106} \\ \hline 1940 & {131} \\ {1950} & {150} \\ \hline\end{array}$$$$ \begin{array}{|c|c|}\hline \text { Year } & {\text { Population }} \\ \hline 1960 & {179} \\ {1970} & {203} \\ \hline 1990 & {220} \\ {2000} & {281} \\\ {2010} & {310} \\ \hline\end{array}$$

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