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Chapter 1: Problem 10

The number \(e\) can be defined by \(e=\sum_{n=0}^{\infty}(1 / n !)\), where \(n !=n(n-1) \cdots 2 \cdot 1\) for \(n \neq 0\) and \(0 !=1\). Compute the absolute error and relative error in the following approximations of \(e\) : a. \(\sum_{n=0}^{5} \frac{1}{n !}\) b. \(\sum_{n=0}^{10} \frac{1}{n !}\)

Short Answer

Expert verified
The absolute and relative errors quantify the accuracy of the approximation of \(e\) when using a finite sum. Both errors will decrease as the number of terms in the series increases, which means the approximation gets more accurate.

Step by step solution

01

Calculate the Approximations

First, calculate the approximations of \(e\) by summing the series \(e=\sum_{n=0}^{\infty}(1 / n!)\) up to n=5 and n=10 respectively.
02

Calculate the Absolute Error

Calculate the absolute error using the formula: Absolute error = |True value - Approximate value|. The true value in this case is the constant \(e\), and the approximate values are what we computed in step 1.
03

Calculate the Relative Error

Next, calculate the relative error, given by the formula: Relative error = (Absolute error / True value) * 100. Use the values found in the previous steps.
04

Summarize your findings

Report the absolute and relative error for the approximations for both cases.

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Key Concepts

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

Error Analysis
Error Analysis is crucial in numerical analysis, especially when working with mathematical series approximations. It helps us understand how close an approximation is to the true value. In this problem, we are approximating the mathematical constant \(e\) using a series.

In any calculation or approximation, two main types of errors usually occur: absolute error and relative error.
  • Absolute Error: This measures the difference between the true value and the approximate value. It is calculated using the formula: \(\text{Absolute error} = |\text{True value} - \text{Approximate value}|\). This provides a direct look at how much the approximation is off by.
  • Relative Error: This measures the size of the absolute error in relation to the true value, usually represented as a percentage. The formula is \(\text{Relative error} = (\text{Absolute error} / \text{True value}) \times 100\). This helps to understand how significant the error is compared to the size of the true value itself.
Understanding these errors is important because it helps in assessing the quality of an approximation and deciding whether it is accurate enough for practical purposes.
Factorials in Mathematics
Factorials are a mathematical operation used often in the calculation of series expansions, such as the computation of \(e\). A factorial, denoted by \(n!\), involves multiplying a series of descending natural numbers.

For instance, \(n!\) for \(n = 5\) is calculated as \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials grow rapidly with increasing \(n\), which is why they are often used in mathematical series to express terms that decrease in influence as \(n\) becomes larger.
  • Basic Properties: By definition, \(0! = 1\), which is an essential base case for mathematical formulas. This property is used extensively in combinatorial calculations and certain series representations.
  • Applications: Factorials are essential in permutations, combinations, and series approximations like the exponential function. Their rapid growth helps in balancing series terms, making certain approximations more manageable.
Factorials form the backbone of many series expansions and are an indispensable tool in both theoretical and applied mathematics.
Mathematical Series Approximations
In mathematics and numerical analysis, series approximations help us compute complex numbers like \(e\) accurately without direct methods. Series approximations break down complex problems by summing up simpler terms.

The constant \(e\) can be approximated by the series \(e = \sum_{n=0}^{\infty}(1 / n!)\), which sums the reciprocals of factorials.
  • Finite Series: As seen in the exercise, considering terms up to a finite \(n\) provides an approximate value of \(e\). For instance, summing from \(n=0\) to \(5\) offers an initial approximation, while extending it to \(n=10\) delivers a more refined estimation.
  • Convergence: The series for \(e\) converges quickly because the factorial in the denominator grows rapidly, making each subsequent term smaller. This property is what makes it feasible to use series approximations for accurate calculations.
These methods are critical in computational mathematics, providing solutions where direct computations are impractical or impossible.

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Most popular questions from this chapter

The number \(e\) is defined by \(e=\sum_{n=0}^{\infty}(1 / n !)\), where \(n !=n(n-1) \cdots 2 \cdot 1\) for \(n \neq 0\) and \(0 !=1\). Use four-digit chopping arithmetic to compute the following approximations to \(e\), and determine the absolute and relative errors. a. \(\quad e \approx \sum_{n=0}^{5} \frac{1}{n !}\) b. \(\quad e \approx \sum_{j=0}^{5} \frac{1}{(5-j) !}\) c. \(\quad e \approx \sum_{n=0}^{10} \frac{1}{n !}\) d. \(\quad e \approx \sum_{j=0}^{10} \frac{1}{(10-j) !}\)

The Taylor polynomial of degree \(n\) for \(f(x)=e^{x}\) is \(\sum_{i=0}^{n}\left(x^{i} / i !\right)\). Use the Taylor polynomial of degree nine and three-digit chopping arithmetic to find an approximation to \(e^{-5}\) by each of the following methods. a. \(\quad e^{-5} \approx \sum_{i=0}^{9} \frac{(-5)^{i}}{i !}=\sum_{i=0}^{9} \frac{(-1)^{i} 5^{i}}{i !}\) b. \(\quad e^{-5}=\frac{1}{e^{5}} \approx \frac{1}{\sum_{i=0}^{9} \frac{5^{i}}{i !}}\) c. An approximate value of \(e^{-5}\) correct to three digits is \(6.74 \times 10^{-3}\). Which formula, (a) or (b), gives the most accuracy, and why?

The Fibonacci sequence also satisfies the equation $$F_{n} \equiv \tilde{F}_{n}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$$ a. Write a Maple procedure to calculate \(F_{100}\). b. Use Maple with the default value of Digits followed by evalf to calculate \(\tilde{F}_{100}\). c. Why is the result from part (a) more accurate than the result from part (b)? d. Why is the result from part (b) obtained more rapidly than the result from part (a)? e. What results when you use the command simplify instead of evalf to compute \(\tilde{F}_{100}\) ?

Let \(f \in C[a, b]\), and let \(p\) be in the open interval \((a, b)\). a. Suppose \(f(p) \neq 0\). Show that a \(\delta>0\) exists with \(f(x) \neq 0\), for all \(x\) in \([p-\delta, p+\delta]\), with [p \(-\delta, p+\delta]\) a subset of \([a, b]\). b. Suppose \(f(p)=0\) and \(k>0\) is given. Show that a \(\delta>0\) exists with \(|f(x)| \leq k\), for all \(x\) in \([p-\delta, p+\delta]\), with \([p-\delta, p+\delta]\) a subset of \([a, b]\).

a. Show that the polynomial nesting technique described in Example 6 can also be applied to the evaluation of $$ f(x)=1.01 e^{4 x}-4.62 e^{3 x}-3.11 e^{2 x}+12.2 e^{x}-1.99 $$ b. Use three-digit rounding arithmetic, the assumption that \(e^{1.53}=4.62\), and the fact that \(e^{n x}=\left(e^{x}\right)^{n}\) to evaluate \(f(1.53)\) as given in part (a). c. Redo the calculation in part (b) by first nesting the calculations. d. Compare the approximations in parts (b) and (c) to the true three-digit result \(f(1.53)=-7.61\).
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