Problem 36 - Contemporary Precalculus

Question

The basic formula for compound interest \(T(x)=P(1+r)^{x}\) was discussed on page \(369 .\) Prove by induction that the formula is valid whenever \(x\) is a positive integer. [Note: \(P\) and \(r \text { are assumed to be constant. }]\)

Step-by-Step Solution

Answer

Question: Prove the validity of the compound interest formula for positive integer values of x by induction. Answer: By using mathematical induction, we have successfully shown that the compound interest formula is valid for all positive integer values of x. We first proved the base case for x = 1 and then assumed the formula to be valid for an arbitrary positive integer k. Following the inductive step, we showed that if the formula is valid for k, it is valid for k+1 as well.

Step 1: Base Case

When x = 1, the compound interest formula becomes: \(T(1) = P(1+r)^1\) By definition of compound interest, after 1 period, the investment will grow to P, the principal amount, with a rate of r, so: \(T(1) = P + Pr = P(1+r)\) This shows that the compound interest formula is valid when x = 1.

Step 2: Induction Hypothesis

Assume that the formula is valid for an arbitrary positive integer k: \(T(k) = P(1+r)^k\)

Step 3: Inductive Step

We must now show that if the formula is valid for k, it is also valid for k+1: \(T(k+1) = P(1+r)^{k+1}\) To do this, we will first find an expression for \(T(k+1) - T(k)\). This represents the increase in value from the k-th period to the (k+1)-th period. Recall that by the compound interest formula, the increase in the total amount from period k to period k+1 is just r times the total amount at period k: \(T(k+1) - T(k) = rT(k)\) Now, substitute the induction hypothesis, \(T(k) = P(1+r)^k\), into this equation: \(T(k+1) - P(1+r)^k = rP(1+r)^k\) Next, solve for \(T(k+1)\): \(T(k+1) = P(1+r)^k + rP(1+r)^k\)

Step 4: Simplification

Factor out the common term, \(P(1+r)^k\), from the expression for \(T(k+1)\) \(T(k+1) = P(1+r)^k(1 + r)\) Now, observe that \((1+r)\) in the parentheses can be written as \((1+r)^1\): \(T(k+1) = P(1+r)^k(1+r)^1\) Finally, apply the exponentiation rule: \((a^m)(a^n) = a^{m+n}\): \(T(k+1) = P(1+r)^{k+1}\) Since we have shown that if the formula is valid for k, it is also valid for k+1, we have completed the induction process and proven that the compound interest formula is valid for all positive integer values of x.

Key Concepts

Compound Interest

Compound interest is a powerful concept in finance that determines how money grows over time. Unlike simple interest, where interest is only paid on the principal, compound interest calculates interest on both the initial principal and any previously earned interest. This can significantly increase total earnings over a period, making it a preferred method of calculating returns for many investors.

• Formula: The basic formula for compound interest is given by \(T(x) = P(1+r)^x\), where:
  • \(T(x)\) is the total amount after \(x\) periods.
  • \(P\) is the principal or initial investment.
  • \(r\) is the interest rate per period.
  • \(x\) is the number of periods.
• Usage: Compound interest is crucial in personal finance, investing, and banking, indicating how wealth can grow by reinvesting earnings.

Understanding this formula helps in deciding the best investments and savings plans for maximizing potential returns.

Exponentiation Rule

The rule of exponentiation is key in understanding how powers work. It states that multiplying powers with the same base results in an addition of their exponents. This rule simplifies complex expressions and is crucial in mathematical calculations involving growth over time.

• For example, the rule states: \((a^m)(a^n) = a^{m+n}\).
• In our compound interest formula, when validating the inductive step, this rule helps show that \(T(k+1) = P(1+r)^{k+1}\).
• Exponentiation underlies many aspects of algebra and calculus and is necessary for handling exponential growth scenarios such as compound interest.

This rule is a basic yet essential tool for ensuring mathematical accuracy in calculations involving exponential terms.

Base Case

The base case is the starting point in a proof by induction. It is the foundation that ensures the property hold true for the initial step. To perform induction, we first verify that the statement is true for the smallest possible value, often \( x = 1 \).

• In the compound interest proof, the base case involves demonstrating that the formula holds for one compounding period: \(T(1) = P(1+r)^1 = P + Pr\).
• By establishing this, we ensure that our hypothesis begins from a point of truth, lending credibility to subsequent steps.

A strong base case ensures the principle to be proved isn't built on shaky ground, reassuring us of the validity of further steps in our proof.

Inductive Step

The inductive step is the heart of a proof by induction, establishing that if the formula works for one case, it will work for the next. This step generalizes from the base case and is crucial for proving mathematical statements involving sequences or iterations.

• Inductive Hypothesis: Assume the formula \( T(k) = P(1+r)^k \) is valid for a positive integer \( k \). This assumption is our stepping stone to extend the formula's validity forward.
• Goal: Show it holds for \( k+1 \), or \( T(k+1) = P(1+r)^{k+1} \).
• Procedure:
  • Calculate \( T(k+1) - T(k) \) to establish growth from period \( k \) to \( k+1 \).
  • Use the compound interest definition that each period grows by \( r \times T(k) \).
  • Simplify using the induction hypothesis, reinforcing \( T(k+1) = P(1+r)^{k+1} \).
• This step firmly ties the base case to every other possible instance, validating the compound interest formula for any positive integer \( x \).

Without a successful inductive step, the logical linkage from one case to the next would be broken, leaving the proof incomplete.