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Chapter 3: Problem 7

Start from \(x_{0}=6\) and \(x_{0}=2\). Compute \(x_{1}, x_{2}, \ldots\) to test convergence: $$ x_{n+1}=\frac{1}{2} x_{n}-1 $$

Short Answer

Expert verified
Both sequences converge, with \(x_0 = 6\) converging starting from \(x_3\) and \(x_0 = 2\) steadily decreasing.

Step by step solution

01

Apply Recurrence Formula for Initial Values

We start with the initial values given, which are \(x_0 = 6\) and \(x_0 = 2\). We apply the recurrence formula \(x_{n+1} = \frac{1}{2} x_n - 1\) to compute the next term \(x_1\).
02

Compute Next Term for \(x_0 = 6\)

Substitute \(x_0 = 6\) into the formula \(x_{n+1} = \frac{1}{2} x_n - 1\): \[ x_1 = \frac{1}{2} \cdot 6 - 1 = 3 - 1 = 2 \]So, \(x_1 = 2\).
03

Compute Next Term for \(x_0 = 2\)

Substitute \(x_0 = 2\) into the same formula:\[ x_1 = \frac{1}{2} \cdot 2 - 1 = 1 - 1 = 0 \]So, \(x_1 = 0\).
04

Continue Applying Recurrence for Further Terms, \(x_0 = 6\) Sequence

Now apply the formula to find \(x_2\):\[ x_2 = \frac{1}{2} \cdot 2 - 1 = 1 - 1 = 0 \]Continue for \(x_3\):\[ x_3 = \frac{1}{2} \cdot 0 - 1 = 0 - 1 = -1 \]
05

Continue Applying Recurrence for Further Terms, \(x_0 = 2\) Sequence

For the sequence starting with \(x_0 = 2\), find \(x_2\):\[ x_2 = \frac{1}{2} \cdot 0 - 1 = 0 - 1 = -1 \]And \(x_3\):\[ x_3 = \frac{1}{2} \cdot (-1) - 1 = -0.5 - 1 = -1.5 \]
06

Analyze Convergence

Observe that the sequence for \(x_0 = 6\) approached -1 starting from \(x_3\). Similarly, the sequence from \(x_0 = 2\) is also approaching a closer negative value and continues decreasing. Both sequences converge towards negative values.

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Key Concepts

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

Convergence Testing
Convergence testing helps us determine if a sequence approaches a specific value as it is extended. It is essential to analyze whether terms within the sequence become closer to a particular number or if they continue to fluctuate without settling down. In our exercise, convergence testing was achieved by continuing the sequence calculation and observing the trend.
  • The starting term in both sequences have different initial conditions, creating diverse paths of convergence.
  • As we apply the recurrence relation repeatedly, the sequences seem to approach a certain value indicating convergence.
  • For example, in the sequence starting from 6, the terms quickly shift towards -1 by the third term and stabilize, suggesting a convergence towards this number.
Convergence testing is a vital tool in understanding sequence behavior, especially when dealing with infinite or large number sequences. It ensures that we know if or when a sequence will settle into a predictable and stable value.
Sequence Computation
Sequence computation involves generating subsequent numbers in a sequence using a specific formula or process. This exercise uses a recurrence relation to compute each next term based on the previous one.
  • Starting from an initial value, here given as 6 or 2, the recurrence formula \(x_{n+1} = \frac{1}{2}x_n - 1\) is applied.
  • This pattern reveals subsequent terms, which are calculated step-by-step until a specific part of the sequence is obtained.
  • The computation shows how changing the initial value affects the resulting sequence pattern.
This methodical approach is crucial in various mathematical contexts, including calculus and computer algorithms, where sequences need to be calculated precisely and efficiently based on an established rule.
Initial Conditions
Initial conditions are the starting values given in a sequence problem that significantly impact the resulting behavior of the sequence. In our exercise, these conditions were \(x_0 = 6\) and \(x_0 = 2\).
  • The choice of initial conditions can lead to entirely different sequences, even with an identical recurrence relation as seen here.
  • For \(x_0 = 6\), the sequence diminished towards -1, different compared to the sequence starting from \(x_0 = 2\).
  • These initial conditions provide the initial step needed to initiate calculations and direct how the sequence develops over time.
Understanding the role of initial conditions helps in predicting the long-term behavior of sequences and is critical in modeling real-world phenomena where initial circumstances can change outcomes significantly.

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Most popular questions from this chapter

Decide between circle-ellipse-parabola-hyperbola, based on the \(X Y\) equation with \(X=x-1\) and \(Y=y+3\). (a) \(x^{2}-2 x+y^{2}+6 y=6\) (b) \(x^{2}-2 x-y^{2}-6 y=6\) (c) \(x^{2}-2 x+2 y^{2}+12 y=6\) (d) \(x^{2}-2 x-y=6\).

Rescale \(y=\sin x\) so \(X\) is in degrees, not radians, and \(Y\) changes from meters to centimeters.

On the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\), solve for \(y\) when \(x=c=\sqrt{a^{2}-b^{2}}\). This height above the focus will be valuable in proving Kepler's third law.

Evaluate the limits in 13-22 by l'H么pital's Rule. \(\lim _{x \rightarrow 1} \frac{x-1}{\sin x}\)

Problems \(15-20\) are about parabolas, \(21-34\) are about ellipses, \(35-41\) are about hyperbolas. The parabola \(y=9-x^{2}\) opens with vertex at Centering by \(Y=y-9\) yields \(Y=-x^{2}\)
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