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Chapter 1: Problem 1

Write down \(x_{1}, x_{2}\) and \(x_{3}\) for the sequences defined by (a) \(x_{n}=\frac{n^{2}}{n+2}\) (b) \(x_{n+1}=x_{n}+4, x_{0}=2\) (c) \(x_{n+1}=\frac{-x_{n}}{4}, x_{0}=256\)

Short Answer

Expert verified
(a) \(x_1=\frac{1}{3}\), \(x_2=1\), \(x_3=\frac{9}{5}\); (b) \(x_1=6\), \(x_2=10\), \(x_3=14\); (c) \(x_1=-64\), \(x_2=16\), \(x_3=-4\).

Step by step solution

01

Determine initial terms for sequence (a)

The sequence is given by the formula \(x_{n} = \frac{n^2}{n+2}\). We need to calculate \(x_1\), \(x_2\), and \(x_3\).- For \(x_1\): \[x_1 = \frac{1^2}{1+2} = \frac{1}{3}\]- For \(x_2\):\[x_2 = \frac{2^2}{2+2} = \frac{4}{4} = 1\]- For \(x_3\):\[x_3 = \frac{3^2}{3+2} = \frac{9}{5}\]
02

Determine initial terms for sequence (b)

The sequence is defined recursively by \(x_{n+1} = x_{n} + 4\) with initial term \(x_0=2\). We need to find \(x_1\), \(x_2\), and \(x_3\).- \(x_1 = x_0 + 4 = 2 + 4 = 6\)- \(x_2 = x_1 + 4 = 6 + 4 = 10\)- \(x_3 = x_2 + 4 = 10 + 4 = 14\)
03

Determine initial terms for sequence (c)

This sequence is also recursive, defined by \(x_{n+1} = \frac{-x_{n}}{4}\) starting with \(x_0 = 256\). Calculate \(x_1\), \(x_2\), and \(x_3\).- \(x_1 = \frac{-256}{4} = -64\)- \(x_2 = \frac{-(-64)}{4} = 16\)- \(x_3 = \frac{-16}{4} = -4\)

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Key Concepts

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

Recursive Sequences
Recursive sequences are fascinating in mathematics as they define each term based on previous terms. Unlike formulas that directly calculate a term based on its position, recursive sequences build terms step-by-step, which can be both intriguing and challenging for learners. An example of a recursive sequence is: - If you start with an initial value, denoted as \( x_0 \). - Then, each next term is found through a specific rule, such as \( x_{n+1} = x_n + 4 \). In this rule, you add 4 to the previous term to get the next term.
  • A sequence might begin with an initial term like \( x_0 = 2 \).
  • Using the rule \( x_{n+1} = x_n + 4 \), the sequence develops into \( 2, 6, 10, 14,... \).
This step-by-step approach encourages a deep understanding of how sequences are put together.
Initial Terms
Initial terms are integral starting points in sequences that determine how the rest of the sequence unfolds. Without an initial term, a recursive sequence cannot be initiated, as the sequence heavily relies on this first value to generate subsequent terms. Consider your starting point like planting a seed:
  • The initial term \( x_0 \) acts as the seed.
  • From there, the rule of the sequence takes over to grow the rest of the terms.
If you have \( x_0 = 256 \) with the recursive rule \( x_{n+1} = \frac{-x_n}{4} \), the sequence is like a path that you travel from that starting point. As you move from \( x_0 = 256 \) to \( x_1, x_2, x_3 \), each term forms from the previous one, demonstrating the power of initial terms in shaping the sequence.
Sequence Formulae
Sequence formulae act like blueprints or maps for sequences. They allow mathematicians to jump straight to any term in the sequence without having to calculate all the preceding terms. Sequence formulae can express sequences in closed form, bypassing the step-by-step method of recursive sequences.Consider the formula \( x_n = \frac{n^2}{n+2} \):
  • This closed-formula directly calculates specific terms, \( x_1, x_2, x_3 \), and so on, without needing to know \( x_0 \).
  • For instance, to find \( x_3 \), simply substitute \( n \) with 3: \( x_3 = \frac{3^2}{3+2} = \frac{9}{5} \).
Closed-form expressions in sequence formulae provide efficient tools for getting specific sequence terms quickly, particularly essential in complex problems where computing each term recursively would be cumbersome.

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Most popular questions from this chapter

An isosceles trapezium has non-parallel sides of length \(20 \mathrm{~cm}\) and the shorter parallel side is \(30 \mathrm{~cm}\), as illustrated in Figure \(1.8 .\) The perpendicular distance between the parallel sides is \(h \mathrm{~cm}\). Show that the area of the trapezium is \(h\left(30+\sqrt{\left.\left(400-h^{2}\right)\right)} \mathrm{cm}^{2}\right.\)

Find the values of \(A\) and \(B\) such that (a) \(\frac{1}{(x+1)(x-2)} \equiv \frac{A}{x+1}+\frac{B}{x-2}\) (b) \(3 x+2=A(x-1)+B(x-2)\) (c) \(\frac{5 x+1}{\sqrt{\left(x^{2}+x+1\right)}}=\frac{A(2 x+1)+B}{\sqrt{\left(x^{2}+x+1\right)}}\)

The arithmetic-geometric inequality $$ \frac{x+y}{2} \geqslant \sqrt{x y} $$ mplies $$ \left(\frac{x+y}{2}\right)^{2} \geqslant x y $$ Use the substitution \(x=\frac{1}{2}(a+b), y=\frac{1}{2}(c+d)\), where \(a, b, c\) and \(d>0\), to show that $$ \left(\frac{a+b}{2}\right)\left(\frac{c+d}{2}\right) \leqslant\left(\frac{a+b+c+d}{4}\right)^{2} $$ and hence that $$ \left(\frac{a+b}{2}\right)^{2}\left(\frac{c+d}{2}\right)^{2} \leqslant\left(\frac{a+b+c+d}{4}\right)^{4} $$ By applying the arithmetic-geometric inequality to the first two terms of this inequality, deduce. that $$ a b c d \leqslant\left(\frac{a+b+c+d}{4}\right)^{4} $$ and hence $$ \frac{a+b+c+d}{4} \geqslant \sqrt[4] a b c d $$

Rearrange the following quadratic expressions by completing the square. (a) \(x^{2}+x-12\) (b) \(3-2 x+x^{2}\) (c) \((x-1)^{2}-(2 x-3)^{2}\) (d) \(1+4 x-x^{2}\)

Newton's recurrence formula for determining the root of a certain equation is $$ x_{n+1}=\frac{x_{n}^{2}-1}{2 x_{n}-3} $$Taking \(x_{0}=3\) as your initial approximation, obtain the root correct to \(4 \mathrm{sf}\). By setting \(x_{n+1}=x_{n}=\alpha\) show that the fixed points of the iteration are given by the equation \(\alpha^{2}-3 \alpha+1=0\)
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