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Chapter 4: Problem 1

For the sequence \(a_{0}, a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{0}=1\) and that for each \(n \in \mathbb{N} \cup\\{0\\}, a_{n+1}=(n+1) a_{n}\). Use mathematical induction to prove that for each \(n \in \mathbb{N} \cup\\{0\\}, a_{n}=n !\)

Short Answer

Expert verified
For the base case, we check \(n=0\). We have \(a_{0}=1\) and \(0!=1\), thus \(a_{0}=0!\) is true. Assuming the proposition holds for some \(n=k\), that is, \(a_{k}=k!\), we need to prove that \(a_{k+1}=(k+1)!\) using the induction hypothesis and the given relationship \(a_{n+1}=(n+1)a_{n}\). Substituting the induction hypothesis \(a_{k}=k!\) into \(a_{k+1}=(k+1)a_{k}\) gives \(a_{k+1}=(k+1)(k!)\). Since \((k+1)!=(k+1)k!\), we find the last two equations equivalent, thus \(a_{k+1}=(k+1)!\). Hence, by the principle of mathematical induction, the statement \(a_{n}=n!\) holds for all \(n \in \mathbb{N} \cup\\{0\\}\).

Step by step solution

01

Base Case

Let's check the base case for \(n=0\): We have \(a_{0}=1\) and \(0!=1\). Therefore, the statement \(a_{0}=0!\) is true.
02

Induction Hypothesis

Assume that the proposition holds for some \(n=k\), that is, \(a_{k}=k!\).
03

Induction Step

Our task is to prove that \(a_{k+1}=(k+1)!\) using the induction hypothesis and the given relationship \(a_{n+1}=(n+1)a_{n}\). Using the given relationship for \(n=k\), we have: \[a_{k+1}=(k+1)a_{k}\] Now, let's substitute the induction hypothesis \(a_{k}=k!\) into the above equation: \[a_{k+1}=(k+1)(k!)\] Recall that \((k+1)!=(k+1)k!\). Comparing the last two equations, we find that they are the same: \[a_{k+1}=(k+1)! \]
04

Conclusion

By the principle of mathematical induction, the statement \(a_{n}=n!\) holds for all \(n \in \mathbb{N} \cup\\{0\\}\).

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Key Concepts

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

Mathematical Reasoning
At its core, mathematical reasoning involves the process of reaching a valid conclusion through logical steps. In your subject of interest, such as the verification of sequences, this form of reasoning is crucial. A sequence, represented as a list of numbers following a specific pattern, requires careful analysis to deduce properties or formulas that apply to all of its members.

For instance, in the exercise given, the student must discern a pattern from a recursive relationship鈥攅ach term is defined based on the previous one. The ultimate goal is to show that this sequence matches that of the factorial sequence for all natural numbers, including zero. Mathematical reasoning steps in to provide a structured approach to navigate from known information (such as the starting term of the sequence and the recursive formula) to an overarching conclusion about the entire sequence.
Proof Writing
Proof writing is the art of persuasively explaining why a particular mathematical statement is true. In our exercise, the proof method of choice is mathematical induction, which checks a base case and then assumes the statement for a general case 'k' to prove it for 'k+1'.

To write a clear and cogent proof, begin with the statement to be proven and follow with a methodology that supports this statement. A proof by induction usually has the following structure: the verification of the base case, the formulation of the induction hypothesis, and the induction step to extend the assumption to the next case. Always conclude your proof by articulating that the principle of induction has been satisfied, thereby reinforcing the truth of the statement for all relevant cases.
Induction Hypothesis
The induction hypothesis is a pivotal step within proof writing using mathematical induction. After establishing the truth of the base case, you assume, without proving, that the property holds for an arbitrary term 'k'. This assumption is the induction hypothesis. In the provided exercise, this step translates to accepting that the k-th term of the sequence, denoted as 'a_k', equals 'k!'.

The strength of your hypothesis isn鈥檛 in its immediate truth, but in its strategic use to bridge from 'k' to 'k+1'. Remember that this hypothesis is not proven yet鈥攊t is postulated to help us step forward. It is the logical leap that makes the domino effect possible, hoping that if the statement holds for 'k', then it must also hold as we increment to 'k+1'.
Sequence Notation
Sequence notation is a concise way to describe the elements of a sequence using subscripts. In our example, the notation 'a_n' represents the n-th term of the sequence. The zero subscript in 'a_0=1' indicates the starting point or the base case when working with mathematical induction.

Understanding this notation is essential as it helps visualize the progression of terms in a sequence. It is also invaluable when using recursive definitions, where each term is defined in terms of previous terms. For instance, the recursive formula 'a_{n+1}=(n+1)a_n' tells you how to calculate the next term from the current term. By mastering sequence notation, you significantly enhance your ability to navigate through problems involving sequences and series.

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Most popular questions from this chapter

One of the most interesting results in trigonometry is De Moivre's Theorem, which relates the complex number \(i\) to the trigonometric functions. Recall that the number \(i\) is a complex number whose square is \(-1,\) that is, \(i^{2}=-1 .\) One version of the theorem can be stated as follows: If \(x\) is a real number, then for each nonnegative integer \(n,\) $$ [\cos x+i(\sin x)]^{n}=\cos (n x)+i(\sin (n x)) $$ This theorem is named after Abraham de Moivre \((1667-1754),\) a French mathematician. (a) The proof of De Moivre's Theorem requires the use of the trigonometric identities for the sine and cosine of the sum of two angles. Use the Internet or a book to find identities for \(\sin (\alpha+\beta)\) and \(\cos (\alpha+\beta)\). (b) To get a sense of how things work, expand \([\cos x+i(\sin x)]^{2}\) and write the result in the form \(a+b i\). Then use the identities from part (1) to prove that \([\cos x+i(\sin x)]^{2}=\cos (2 x)+i(\sin (2 x))\) (c) Use mathematical induction to prove De Moivre's Theorem.

Most of the work done in constructing a proof by induction is usually in proving the inductive step. This was certainly the case in Proposition 4.2 . However, the basis step is an essential part of the proof. Without it, the proof is incomplete. To see this, let \(P(n)\) be $$ 1+2+\cdots+n=\frac{n^{2}+n+1}{2} $$ (a) Let \(k \in \mathbb{N}\). Complete the following proof that if \(P(k)\) is true, then \(P(k+1)\) is true. Let \(k \in \mathbb{N}\). Assume that \(P(k)\) is true. That is, assume that $$ 1+2+\cdots+k=\frac{k^{2}+k+1}{2} $$ The goal is to prove that \(P(k+1)\) is true. That is, we need to prove that $$ 1+2+\cdots+k+(k+1)=\frac{(k+1)^{2}+(k+1)+1}{2} $$ To do this, we add \((k+1)\) to both sides of equation (1). This gives $$ \begin{aligned} 1+2+\cdots+k+(k+1) &=\frac{k^{2}+k+1}{2}+(k+1) \\ &=\cdots \end{aligned} $$ (b) Is \(P(1)\) true? Is \(P(2)\) true? What about \(P(3)\) and \(P(4) ?\) Explain how this shows that the basis step is an essential part of a proof by induction.

For which natural numbers \(n\) is \(n^{2}<2^{n} ?\) Justify your conclusion.

Use mathematical induction to prove each of the following: (a) For each natural number \(n, 3\) divides \(\left(4^{n}-1\right)\). (b) For each natural number \(n, 6\) divides \(\left(n^{3}-n\right)\).

The Lucas numbers are a sequence of natural numbers \(L_{1}, L_{2}, L_{3}, \ldots, L_{n}, \ldots\) which are defined recursively as follows: \- \(L_{1}=1\) and \(L_{2}=3,\) and \- For each natural number \(n, L_{n+2}=L_{n+1}+L_{n}\). List the first 10 Lucas numbers and the first ten Fibonacci numbers and then prove each of the following propositions. The Second Principle of Mathematical Induction may be needed to prove some of these propositions. (a) For each natural number \(n, L_{n}=2 f_{n+1}-f_{n}\) (b) For each \(n \in \mathbb{N}\) with \(n \geq 2,5 f_{n}=L_{n-1}+L_{n+1}\). (c) For each \(n \in \mathbb{N}\) with \(n \geq 3, L_{n}=f_{n+2}-f_{n-2}\).
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