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Chapter 1: Problem 6

Compute the absolute error and relative error in approximations of \(p\) by \(\bar{p}\). (a) \(p=\sqrt{2}, \bar{p}=1.414\) (b) \(p=10^{\pi}, \tilde{p}=1400\) (c) \(p=9 !, \tilde{p}=\sqrt{18 \pi}(9 / e)^{9}\)

Short Answer

Expert verified
(a) Absolute error: 0.000213562, Relative error: 0.000151; (b) Absolute error: 14.544, Relative error: 0.0105; (c) Absolute error: 239877, Relative error: 0.661.

Step by step solution

01

Understanding Absolute Error

The absolute error is defined as the absolute difference between the exact value \(p\) and the approximation \(\bar{p}\). Mathematically, it is expressed as \(|p - \bar{p}|\).
02

Calculate Absolute Error for (a)

For \(p = \sqrt{2}\) and \(\bar{p} = 1.414\), the absolute error is:\[|\sqrt{2} - 1.414| \approx |1.414213562 - 1.414| = 0.000213562\]
03

Understanding Relative Error

The relative error is the absolute error divided by the exact value \(p\). It is given by \(\frac{|p - \bar{p}|}{|p|}\).
04

Calculate Relative Error for (a)

Given our previously found absolute error of 0.000213562, the relative error is:\[\frac{0.000213562}{\sqrt{2}} \approx \frac{0.000213562}{1.414213562} \approx 0.000151\]
05

Calculate Absolute Error for (b)

For \(p = 10^{\pi}\), calculate its value and find the absolute error using \(\tilde{p} = 1400\):\[10^{\pi} \approx 1385.455731\]\[|10^{\pi} - 1400| = |1385.455731 - 1400| = 14.544269\]
06

Calculate Relative Error for (b)

The relative error is given by:\[\frac{14.544269}{10^{\pi}} = \frac{14.544269}{1385.455731} \approx 0.0105\]
07

Calculate Factorial for (c)

Calculate \(9!\) since it's not straightforward. \[9! = 362880\]
08

Calculate Approximate Value for (c)

Calculate \(\tilde{p} = \sqrt{18\pi}(9/e)^9\) to find its numerical value:\[\tilde{p} \approx \sqrt{56.5487} \times 8103.085 \approx 602756.709\]
09

Calculate Absolute Error for (c)

With \(p = 362880\) and \(\tilde{p} = 602756.709\), calculate the absolute error:\[|9! - \tilde{p}| = |362880 - 602756.709| = 239876.709\]
10

Calculate Relative Error for (c)

The relative error for \(c\) is:\[\frac{239876.709}{362880} \approx 0.661]\]

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Key Concepts

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

Absolute Error
In numerical analysis, the concept of absolute error provides insight into how far our approximation is from the true value. This measure is crucial in understanding the precision of an estimate. Absolute error is calculated using the formula:
  • \(|p - \bar{p}|\)
Here, \(p\) represents the exact value while \(\bar{p}\) is our approximation. This formula yields a non-negative value, as it uses absolute values.
In the context of our exercise, consider example (a) where you approximate \(\sqrt{2}\) using the value \(1.414\). The absolute error computed here is:
  • \(|\sqrt{2} - 1.414| \approx 0.000213562\)
Even slight deviations can be noted, highlighting why understanding absolute error is key to ensuring accurate numerical approximations.
Relative Error
Relative error offers another layer of understanding by comparing the size of the absolute error to the exact value, \(p\). This is important because it conveys how significant the absolute error is in the context of the value being approximated. The formula is:
  • \(\frac{|p - \bar{p}|}{|p|}\)
This ratio helps us understand whether an error is large or small relative to the true size of the number. A small relative error suggests a high level of accuracy in the approximation. For instance, the relative error for example (a) is:
  • \(\frac{0.000213562}{\sqrt{2}} \approx 0.000151\)
This demonstrates that the approximation is quite close, as the relative error is very small, affirming that the degree of deviation is minimal in comparison with \(\sqrt{2}\).
Factorial Calculation
Factorial calculation, indicated by the symbol \(!\), is essential in many areas of mathematics, including combinatorics and numerical analysis. Calculating the factorial of a number \(n\) involves multiplying all whole numbers from 1 to \(n\). For example, \(9!\), which is part of our exercise, results in:
  • \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880\)
Understanding factorials is crucial, as it often underpins many approximations and calculations such as series expansions. In the given exercise, calculating \(9!\) was a step toward evaluating the approximation \(\tilde{p} = \sqrt{18\pi}(9/e)^9\), a more complex mathematical expression that needs its simplified factorial base to find accurate error measurements.

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Most popular questions from this chapter

The Tower of Hanoi, part 1. The Tower of Hanoi is a game played with a number of different sized disks stacked on a pole in decreasing size, the largest on the bottom and the smallest on top. There are two other poles, initially with no disks on them. The goal is to move the entire stack of disks to one of the initially empty poles following two rules. You are allowed to move only one disk at a time from one pole to another. You may never place a disk upon a smaller one. [3] (a) Starting with a stack of three disks, what is the minimum number of moves it takes to complete the game? Answer this question with a number. (b) Starting with a stack of four disks, what is the minimum number of moves it takes to complete the game? i. Answer this question recursively. ii. Answer this question with a number based on your recursive answer.

Some linearly convergent sequences and their limits are given. Find the (fastest) rate of convergence of the form \(O\left(\frac{1}{n^{p}}\right)\) or \(O\left(\frac{1}{a^{n}}\right)\) for each. If this is not possible, suggest a reasonable rate of convergence. (a) \(6, \frac{6}{7}, \frac{6}{49}, \frac{6}{343}, \frac{6}{2401}, \ldots \rightarrow 0\) (b) \(\left\langle\frac{11 n-2}{n+3}\right\rangle \rightarrow 11\) (c) \(\left\langle\frac{\sin n}{\sqrt{n}}\right\rangle \rightarrow 0\), (d) \(\left\langle\frac{4}{10^{n}+35 n+9}\right\rangle \rightarrow 0^{[\text {[s] }}\) (e) \(\left\langle\frac{1}{n}+\frac{1}{n^{2}}\right\rangle \rightarrow 0\) (f) \(\left\langle\frac{4}{10^{n}-35 n-9}\right\rangle \rightarrow 0^{[s]}\) (g) \(\left\langle\frac{2 n}{\sqrt{n^{2}+3 n}}\right\rangle \rightarrow 2^{[A]}\) (h) \(\left\langle\frac{5^{n}-2}{5^{n}+3}\right\rangle \rightarrow 1\) (i) \(\langle\sqrt{n+47}-\sqrt{n}\rangle \rightarrow 0^{[A]}\) (j) \(\left\langle\frac{n^{2}}{3 n^{2}+1}\right\rangle \rightarrow \frac{1}{3}\) (k) \(\left\langle\frac{\pi}{e^{n}-\pi^{n}}\right\rangle \rightarrow 0\) (1) \(\left\langle\frac{n^{2}}{2^{n}}\right\rangle \rightarrow 0^{[\mathbb{S}]}\) (m) \(\left\langle\frac{7+\cos (5 n)}{n^{3}+1}\right\rangle \rightarrow 0\) (n) \(\left\langle\frac{8 n^{2}}{3 n^{2}+12}+\frac{n}{3 n+10}\right\rangle \rightarrow 3\) (o) \(\left\langle\frac{2 n^{2}+3 n}{1-n^{2}}\right\rangle \rightarrow-2^{[A]}\) (p) \(\left\langle\frac{3 n^{5}-5 n}{1-n^{5}}\right\rangle \rightarrow-3\)

Write a for loop that outputs the sequence of numbers. (a) 7,8,9,10,11,12,13,14,15 (b) 20,19,18,17,16,15,14,13 (c) 12,12.333,12.667,13,13.333,13.667,14 (d) 1,9,25,49,81,121,169,225,289,361,441 (e) 1, .5, .25, .125, .0625, .03125, .015625

Let the sequences \(\left(b_{n}\right)\) and \(\left\langle c_{n}\right)\) be defined as follows. $$ \begin{array}{l} b_{0}=\frac{1}{3} ; \quad b_{n+1}=4 b_{n}-1, \quad n \geq 0 \\ c_{0}=\frac{1}{10} ; \quad c_{n+1}=4 c_{n}\left(1-c_{n}\right), \quad n \geq 0 \end{array} $$ (a) Write a function that uses a for loop to calculate the \(n^{\text {th }}\) term of \(\left\langle b_{n}\right\rangle\). Your function should have one argument, \(n\). (b) Write a function that uses a for loop to calculate the \(n^{\text {th }}\) term of \(\left\langle c_{n}\right\rangle .\) Your function should have one argument, \(n\). (c) Write a program that calls these functions to calculate \(b_{30}\) and \(c_{30}\). How accurate are these calculations? HINT \(b_{30}=\frac{1}{3}\) and \(c_{30}=.32034\) accurate to 5 decimal places. (d) Can you think of a way to make these calculations more dependable (more accurate)?

Find \(\xi(x)\) as guaranteed by Taylor's theorem in the following situation. (a) \(f(x)=\cos (x), x_{0}=0, n=3, x=\pi\). (b) \(f(x)=e^{x}, x_{0}=0, n=3, x=\ln 4\). (c) \(f(x)-\ln (x), x_{0}-1, n-4, x-2\).
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