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Chapter 18: Problem 1

Calculate the Bernoulli numbers \(B_{8}, B_{10}\), and \(B_{12} .\) The sequence of Bernoulli numbers is usually completed by setting \(B_{0}=1, B_{1}=-\frac{1}{2}\), and \(B_{k}=0\) for \(k\) odd and greater than 1 .

Short Answer

Expert verified
Question: Determine the Bernoulli numbers \(B_8\), \(B_{10}\), and \(B_{12}\). Answer: \(B_8 = \frac{1}{30}\), \(B_{10} = -\frac{5}{66}\), \(B_{12} = \frac{691}{2730}\).

Step by step solution

01

Understand the Bernoulli number recursive formula

The recursive formula for calculating Bernoulli numbers is given by: \( B_n = \displaystyle-\frac{1}{n+1}\sum_{k=0}^{n-1} \binom{n+1}{k} B_k\) This formula allows us to calculate Bernoulli numbers based on the previous ones. In our case, we are given initial values for \(B_0=1, B_1 = -\frac{1}{2}\), and \(B_k = 0\) for all odd k > 1, which will serve as the starting point for our calculations.
02

Calculate B_2

We will begin by calculating \(B_2\) using the recursive formula: \(B_2 = -\frac{1}{3}\sum_{k=0}^{1} \binom{3}{k} B_k = -\frac{1}{3}\left(\binom{3}{0} B_0 + \binom{3}{1} B_1\right)\) Now, substitute the given values for \(B_0\) and \(B_1\): \(B_2 = -\frac{1}{3}\left(\binom{3}{0} \cdot 1 + \binom{3}{1} \cdot-\frac{1}{2}\right) = -\frac{1}{3}\left(1 - \frac{3}{2}\right) = \frac{1}{6}\)
03

Calculate B_4

Next, calculate \(B_4\) using the recursive formula: \(B_4 = -\frac{1}{5}\sum_{k=0}^{3} \binom{5}{k} B_k = -\frac{1}{5}\left(\binom{5}{0} B_0 + \binom{5}{1} B_1 + \binom{5}{2} B_2 + \binom{5}{3} B_3\right)\) Using the values we have for \(B_0\), \(B_1\), \(B_2\) and \(B_k=0\) for odd k > 1: \(B_4 = -\frac{1}{5}\left(1 - \frac{5}{2} + 10\cdot\frac{1}{6}\right) = -\frac{1}{5}\left(1 - \frac{5}{2} + \frac{5}{3}\right) = -\frac{1}{30}\)
04

Calculate B_6

Now, calculate \(B_6\) using the recursive formula: \(B_6 = -\frac{1}{7}\sum_{k=0}^{5} \binom{7}{k} B_k = -\frac{1}{7}\left(\binom{7}{0} B_0 + \binom{7}{1} B_1 + \binom{7}{2} B_2 + \binom{7}{3} B_3 + \binom{7}{4} B_4\right)\) Using the values we have for \(B_0\), \(B_1\), \(B_2\), \(B_4\) and \(B_k=0\) for odd k > 1: \(B_6 = -\frac{1}{7}\left(1 - \frac{7}{2} + 21\cdot\frac{1}{6} -1\cdot-\frac{1}{30}\right) = -\frac{1}{7}\left(1 - \frac{7}{2} + \frac{7}{2} + \frac{1}{30}\right) = \frac{1}{42}\)
05

Calculate B_8, B_10 and B_12

We have all the previous Bernoulli numbers, and we can now easily calculate \(B_8\), \(B_{10}\), and \(B_{12}\). For \(B_8\): \(B_8 = -\frac{1}{9}\sum_{k=0}^{7} \binom{9}{k} B_k = -\frac{1}{9}\left(\binom{9}{0} B_0 + \binom{9}{1} B_1 + \binom{9}{2} B_2 + \binom{9}{3} B_3 + \binom{9}{4} B_4 + \binom{9}{5} B_5 + \binom{9}{6} B_6\right)\) Substitute the known Bernoulli number values and calculate \(B_8\): \(B_8 = -\frac{1}{9}\left(1 -\frac{9}{2} + 36\cdot\frac{1}{6} + 84\cdot-\frac{1}{30} + 126\cdot\frac{1}{42}\right) = \frac{1}{30}\) For \(B_{10}\) and \(B_{12}\), we follow the same process and get: \(B_{10} = -\frac{5}{66}\) \(B_{12} = \frac{691}{2730}\)
06

Final results

The Bernoulli numbers in their simplest reduced form are: \(B_8 = \frac{1}{30}\) \(B_{10} = -\frac{5}{66}\) \(B_{12} = \frac{691}{2730}\)

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Key Concepts

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

Understanding Mathematical Sequences
Mathematical sequences are ordered lists of numbers following specific patterns or rules. They play a significant role in various fields within mathematics, including algebra, calculus, and number theory. Sequences like the ones involving Bernoulli numbers can be either finite or infinite and may be governed by explicit formulas or by recursive relationships.

A sequence can be thought of as a function whose domain is the set of natural numbers, and its range is the set of values that form the sequence. For example, Bernoulli numbers form a sequence where each number is related to the previous ones through a recursive formula. Understanding the nature of these sequences is essential in tackling problems that require calculating specific terms like the ones shown in the exercise for finding the Bernoulli numbers.
Recursive Sequences and Their Significance
Recursive sequences use previous terms to define the next ones. They start with one or several initial terms, and each subsequent term is then determined by applying a fixed rule to the preceding terms. This property is evident in Bernoulli numbers where each number depends on the ones before it.The recursive formula provided in the exercise solution for Bernoulli numbers is a typical example of how these numbers are generated. To calculate a specific number in the sequence, the formula applies a sum over the product of binomial coefficients and previous Bernoulli numbers. By iterating this process, the entire sequence can be built up step by step. Recursive sequences are prevalent in combinatorial contexts and solving them often requires both analytical skills and sometimes also the use of programming to automate calculations for larger terms.
Combinatorics in the Context of Bernoulli Numbers
Combinatorics is the field of mathematics that studies combinations, permutations, and the counting principles used to solve problems concerning discrete structures. It is closely related to sequences and recursion because many combinatorial problems involve counting sequences and employ recursive reasoning.In the case of Bernoulli numbers, we see combinatorial principles at work in the recursive formula through the presence of the binomial coefficient \( \binom{n+1}{k} \), which counts the number of ways to choose a subset of size k from a larger set of n+1 elements. This coefficient is a fundamental concept in combinatorics and appears in various formulas across mathematics. The exercise requires the understanding of these coefficients to compute Bernoulli numbers accurately, showcasing how combinatorics intertwines with number theory and recursive sequences.

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Most popular questions from this chapter

Imagine an urn with two balls, each of which may be either white or black. One of these balls is drawn and is put back before a new one is drawn. Suppose that in the first two draws white balls have been drawn. What is the probability of drawing a white ball on the third draw?

Show that if an event of unknown probability happens \(n\) times in succession, the odds are \(2^{n+1}-1\) to 1 for more than an even chance of its happening again.

De Moivre's result developing the normal curve implies that the probability \(P_{\epsilon}\) of an observed result lying between \(p-\epsilon\) and \(p+\epsilon\) in \(n\) trials is given by $$ P_{\epsilon}=\frac{1}{\sqrt{2 \pi n p(1-p)}} \int_{-n \epsilon}^{n \epsilon} e^{-\frac{t^{2}}{2 n p(1-p)}} d t $$ Change variables by setting \(u=t / \sqrt{n p(1-p)}\) and use symmetry to show that this integral may be rewritten as $$ P_{\epsilon}=\frac{2}{\sqrt{2 \pi}} \int_{0}^{\frac{\sqrt{n} \epsilon}{\sqrt{p(1-p)}}} e^{-\frac{1}{2} u^{2}} d u $$ Calculate this integral for Bernoulli's example, using \(p=\) \(.6, \epsilon=.02\), and \(n=6498\), and show that in this case \(P_{\epsilon}=\) \(0.999\), a value giving moral certainty. (Use a graphing utility.) Find a value for \(n\) that gives \(P_{\epsilon}=0.99\).

Use Bernoulli's formula to show that if greater certainty is wanted in the problem of Exercise 7 , say, \(c=10,000\), then the number of trials necessary is \(N=31,258\).

The so-called St. Petersburg Paradox was a topic of debate among those mathematicians involved in probability theory in the eighteenth century. The paradox involves the following game between two players. Player \(A\) flips a coin until a tail appears. If it appears on his first flip, player \(B\) pays him 1 ruble. If it appears on the second flip, \(B\) pays 2 rubles, on the third, 4 rubles, \(\ldots\), on the \(n\)th flip, \(2^{n-1}\) rubles. What amount should \(A\) be willing to pay \(B\) for the privilege of playing? Show first that \(A\) 's expectation, namely, the sum of the probabilities for each possible outcome of the game multiplied by the payoff for each outcome, is $$ \sum_{i=0}^{\infty} \frac{1}{2^{i}} 2^{i-1} $$ and then that this sum is infinite. Next, play the game 10 times and calculate the average payoff. What would you be willing to pay to play? Why does the concept of expectation seem to break down in this instance?
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