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Chapter 2: Problem 23

Find a recurrence relation for the balance \(B(k)\) owed at the end of \(k\) months on a loan of \(\$ 5000\) at a rate of 7\(\%\) if a payment of \(\$ 100\) is made each month. [Hint: Express \(B(k)\) in terms of \(B(k-1) ;\) the monthly interest is \((0.07 / 12) B(k-1) . ]\)

Short Answer

Expert verified
The recurrence relation is \[ B(k) = B(k-1) \bigg(1 + \frac{0.07}{12}\bigg) - 100 \].

Step by step solution

01

Identify initial balance

The initial balance on the loan is given as $5000. This is the amount owed at the beginning, so we can write this as: \[ B(0) = 5000 \]
02

Determine the monthly interest

The annual interest rate is 7%. To find the monthly interest rate, divide the annual rate by 12: \[ \text{Monthly interest rate} = \frac{0.07}{12} \]
03

Express the interest for the current month

For any given month \(k\), the interest on the balance owed from the previous month \(B(k-1)\) can be expressed as: \[ \text{Monthly interest} = \frac{0.07}{12} B(k-1) \]
04

Construct the balance update equation

The new balance owed at the end of month \(k\), \(B(k)\), is the previous balance \(B(k-1)\) plus the interest accrued minus the payment made: \[ B(k) = B(k-1) + \frac{0.07}{12} B(k-1) - 100 \]
05

Simplify the recurrence relation

Combine like terms in the equation from the previous step to simplify it: \[ B(k) = B(k-1) \bigg(1 + \frac{0.07}{12}\bigg) - 100 \]

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Key Concepts

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

recurrence relations in finance
Recurrence relations are a fundamental concept in finance that help model the evolution of quantities over time. These mathematical formulas express the current value of a variable based on previous values, incorporating elements like interest rates and periodic payments. By recursively defining the loan balance or investment value, you can predict future amounts without multiple complex calculations. In this exercise, the idea is to find the balance owed on a loan over time using a recurrence relation. It's particularly useful for understanding how regular payments and interest impact financial outcomes.
monthly interest calculation
Monthly interest calculation is a key step in understanding how a loan grows over time. The annual interest rate given in this exercise is 7%. To get the monthly interest rate, we divide the annual rate by 12, since there are 12 months in a year. So:
  • Annual interest rate: 7%
  • Monthly interest rate: \[ \frac{0.07}{12} \]
This monthly rate is then applied to the balance owed from the previous month to find the new interest amount. This calculation helps to see how debt grows with each passing month.
loan balance computation
Loan balance computation involves updating the amount owed by accounting for new interest and payments. Here鈥檚 the step-by-step breakdown:
  • Initial loan amount: \(5000 (\( B(0) \))
  • Monthly interest: \[ \frac{0.07}{12} B(k-1) \]
  • Monthly payment: \)100
Combining these: \[ B(k) = B(k-1) + \left( \frac{0.07}{12} \right) B(k-1) - 100 \] This formula shows that each month, the new balance is the previous balance plus the monthly interest on that balance, minus the monthly payment. By repeatedly applying this formula, you can track the balance over any number of months.
discrete mathematics application
Discrete mathematics often involves solving problems where variables change over distinct, separate intervals - like months in our loan problem. Recurrence relations, a part of discrete mathematics, help in modeling these changes. In this case, we use: \[ B(k) = B(k-1) \left( 1 + \frac{0.07}{12} \right) - 100 \] This equation relies on the previous month's balance (a discrete step) to compute the current month's balance. Discrete mathematics provides tools and methods, such as this recurrence relation, to solve real-world financial problems systematically and accurately. It鈥檚 a powerful approach to understanding how quantities evolve over specified intervals.

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Most popular questions from this chapter

Let \(g(x)=\lfloor x\rfloor .\) Find a) \(g^{-1}(\\{0\\})\) b) \(g^{-1}(\\{-1,0,1\\})\) c) \(g^{-1}(\\{x | 0 < x < 1\\})\)

Show that if \(A, B,\) and \(C\) are sets, then $$\begin{aligned}|A \cup B \cup C| &=|A|+|B|+|C|-|A \cap B| \\ &-|A \cap C|-|B \cap C|+|A \cap B \cap C| \end{aligned}$$ (This is a special case of the inclusion-exclusion principle, which will be studied in Chapter \(8 . )\)

Find a matrix \(\mathbf{A}\) such that $$ \left[\begin{array}{lll}{1} & {3} & {2} \\ {2} & {1} & {1} \\ {4} & {0} & {3}\end{array}\right] \mathbf{A}=\left[\begin{array}{rrr}{7} & {1} & {3} \\\ {1} & {0} & {3} \\ {-1} & {-3} & {7}\end{array}\right] $$

$$ \mathbf{A}=\left[\begin{array}{llll}{1} & {1} & {1} & {3} \\ {2} & {0} & {4} & {6} \\ {1} & {1} & {3} & {7}\end{array}\right] $$ a) What size is \(A ?\) b) What is the third column of A? c) What is the second row of A? d) What is the element of A in the (3,2) th position? e) What is A'?

In this exercise, we prove the Schr枚der-Bernstein theorem. Suppose that \(A\) and \(B\) are sets where \(|A| \leq|B|\) and \(|B| \leq|A| .\) This means that there are injections \(f : A \rightarrow B\) and \(g : B \rightarrow A\) . To prove the theorem, we must show that there is a bijection \(h : A \rightarrow B,\) implying that \(|A|=|B|\) To build \(h,\) we construct the chain of an element \(a \in A .\) This chain contains the elements \(a, f(a), g(f(a)), f(g(f(a))), g(f(g(f(a)))), \ldots\) It also may contain more elements that precede \(a,\) extending the chain backwards. So, if there is a \(b \in B\) with \(g(b)=a\) , then \(b\) will be the term of the chain just before \(a .\) Because \(g\) may not be a surjection, there may not be any such \(b,\) so that \(a\) is the first element of the chain. If such a \(b\) exists, because \(g\) is an injection, it is the unique element of \(B\) mapped by \(g\) to \(a ;\) we denote it by \(g^{-1}(a) .\) (Note that this defines \(g^{-1}\) as a partial function from \(B\) to \(A .\) . We extend the chain backwards as long as possible in the same way, adding \(f^{-1}\left(g^{-1}(a)\right), g^{-1}\left(f^{-1}\left(g^{-1}\left(g^{-1}(a)\right)\right), \ldots \text { To construct }\right.\) the proof, complete these five parts. a) Show that every element in \(A\) or in \(B\) belongs to exactly one chain. b) Show that there are four types of chains: chains thaform a loop, that is, carrying them forward from every element in the chain will eventually return to this element (type 1\()\) , chains that go backwards without stopping (type 2\()\) , chains that go backwards and end nin the set \(A\) (type \(3 ),\) and chains that go backwards and end in the set \(B\) (type 4\()\) . c) We now define a function \(h : A \rightarrow B .\) We set \(h(a)=\) \(f(a)\) when \(a\) belongs to a chain of type \(1,2,\) or \(3 .\) Show that we can define \(h(a)\) when \(a\) is in a chain of type \(4,\) by taking \(h(a)=g^{-1}(a)\) . In parts \((\mathrm{d})\) and (e), we show that this function is a bijection from \(A\) to \(B,\) proving the theorem. d) Show that \(h\) is one-to-one. (You can consider chains of types \(1,2,\) and 3 together, but chains of type 4 separately.) e) Show that \(h\) is onto. (You need to consider chains of types \(1,2,\) and 3 together, but chains of type 4 separately.)
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