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Chapter 1: Problem 4

Write a recursive Octave function that calculates \(a_{n}\) for any \(n \geq 0\) given $$ \begin{aligned} a_{0} &=100,000 \\ a_{n} &=1.05 a_{n-1}-1200, \quad n>0 . \end{aligned} $$

Short Answer

Expert verified
Define a recursive Octave function with a base case for \(n=0\) and use \(a_n = 1.05a_{n-1} - 1200\) otherwise.

Step by step solution

01

Understand the Problem

We need to write a recursive function in Octave that can compute the values of the sequence \(a_n\) for any non-negative integer \(n\). We are given that \(a_0 = 100,000\) and the formula \(a_n = 1.05 \times a_{n-1} - 1200\) for \(n > 0\). Our task is to implement this in Octave using recursion.
02

Base Case Identification

The base case for our recursive function will be when \(n = 0\) because we know \(a_0 = 100,000\). We need to return this value when the function is called with \(n = 0\).
03

Recursive Case Definition

For \(n > 0\), the function should call itself with the argument \(n-1\). We compute \(a_n\) using the formula \(a_{n} = 1.05 \times a_{n-1} - 1200\). This means our function should return \(1.05 \times \text{recursive function call result} - 1200\).
04

Write the Function Code

Let's write the Octave function using the defined base and recursive cases. The function will take an integer \(n\) as input and return \(a_n\).```octaveafunction = @(n) (n == 0) * 100000 + (n > 0) * (1.05 * afunction(n - 1) - 1200);```This defines an anonymous function in Octave where `afunction(n)` calculates \(a_n\). The base case checks if \(n == 0\) and returns 100000, otherwise, it executes the recursive case.

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Key Concepts

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

Sequence Calculation
Sequence calculation is a fundamental aspect in mathematical analysis and programming. In this specific exercise, we are working with a sequence defined by an initial term, followed by a recurring calculation.
Each term of the sequence is derived from the previous term using a specified formula.
  • The sequence begins with the base value for the first term, often labeled as \(a_0\).
  • Subsequent values are calculated using the recurrence relation \(a_n = 1.05 \times a_{n-1} - 1200\).
  • The goal is to determine \(a_n\) for any integer \(n \geq 0\).
When calculating such sequences, it鈥檚 vital to understand both the starting point of your sequence, \(a_0\), and the operation necessary to generate subsequent elements. By leveraging recursive programming, we can efficiently compute any term of the sequence when needed.
Base Case in Recursion
The base case in recursive programming is crucial because it serves as the stopping criteria for the recursion. Without it, recursion would continue indefinitely, leading to errors or infinite loops. In this problem, the base case is straightforward:
  • The base case is defined as \(a_0 = 100,000\), which is the initial term of the sequence.
This base case helps the recursive function recognize when to stop the recursion and return a value without any further recursive calls. By having a well-defined base case, the function can handle simple cases and provide a foundation upon which more complex calculations (like our sequence calculation) can be built.
Recursive Functions
Recursive functions simplify complex problems by breaking them down into smaller, more manageable tasks. In this case, recursion involves calling the same function within itself with altered input parameters until reaching the base case.
For this Octave function:
  • When \(n = 0\), the function directly returns 100,000.
  • For \(n > 0\), it uses the result of the function called with \(n-1\) to compute the current term \(a_n\).
  • The formula \(a_n = 1.05 \times a_{n-1} - 1200\) is applied, using the value of the previous sequence term.
Recursive functions are powerful tools in programming. They require careful planning of base and recursive cases, ensuring that each call moves closer to a base case.
Mathematical Modeling
Mathematical modeling involves creating abstract representations of real-world processes to understand and solve them more completely. In this exercise, the mathematical model is embodied by the sequence defined and calculated using recursion.
To effectively model a problem:
  • Start by understanding the initial conditions (\(a_0 = 100,000\)) and how each term relates to the previous one.
  • The mathematical model here uses the recurrence relation in recursion to reflect compounded interest minus regular deductions on a starting amount.
  • By implementing this model programmatically, we can simulate and predict future values based on these assumptions.
Mathematical modeling of sequences enables exploration and quantification of how changes in the initial terms and recurrence relations affect the overall system. In doing so, it aids in illustrating complex relationships in an understandable manner.

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Most popular questions from this chapter

Show that $$ (\sin h)(1-\cos h)=0+O\left(h^{3}\right) $$

Write a recursive Octave function that will calculate $$ \sum_{i=1}^{n} \frac{1}{i} $$

(The number in question 7 is an approximation of \(1 / \pi .\) Using Octave, find the absolute and relative errors in the approximation.

Find the rates of convergence of the following sequences as \(n \rightarrow \infty\) (a) \(\lim _{n \rightarrow \infty} \sin \frac{1}{n}=0\) (b) \(\lim _{n \rightarrow \infty} \sin \frac{1}{n^{2}}=0\) (c) \(\lim _{n \rightarrow \infty}\left(\sin \frac{1}{n}\right)^{2}=0\) (d) \(\lim _{n \rightarrow \infty}[\ln (n+1)-\ln (n)]=0\) For questions \(8-12,\) use the following definition for rate of convergence for a function. For a function \(f(h),\) we say \(\lim _{h \rightarrow a} f(h)=L\) with rate of convergence \(g(h)\) if \(|f(h)-L| \leq \lambda|g(h)|\) for some \(\lambda>0\) and all sufficiently small \(|h-a|\)

All of these equations are mathematically true. Nonetheless, floating point error causes some of them to be false according to Octave. Which ones? HINT: Use the boolean operator wo to check. For example, to check if \(\sin (0)=0,\) type \(\sin (0)==0\) into Octave. ans=1 means true (the two sides are equal according to Octave - no round-off error) and ans=0 means false (the two sides are not equal according to Octave round-off error). (a) \((2)(12)=9^{2}-4(9)-21\) (b) \(e^{3 \ln (2)}=8\) (c) \(\ln (10)=\ln (5)+\ln (2)\) (d) \(g\left(\frac{1+\sqrt{5}}{2}\right)=\frac{1+\sqrt{5}}{2}\) where \(g(x)=\sqrt[3]{x^{2}+x}\) (e) \(\lfloor 153465 / 3\rfloor=153465 / 3\) (f) \(3 \pi^{3}+7 \pi^{2}-2 \pi+8=((3 \pi+7) \pi-2) \pi+8\)
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