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Magazine Chapter 3: Problem 18
What is the order of convergence of the iteration $$ x_{n+1}=\frac{x_{n}\left(x_{n}^{2}+3 a\right)}{3 x_{n}^{2}+a} $$ as it converges to the fixed point \(\alpha=\sqrt{a}\) ?
Short Answer
Expert verified
Order of convergence is 2
Step by step solution
01
Define the Iteration and Fixed Point
The given iteration is \[ x_{n+1} = \frac{x_{n}\left(x_{n}^{2}+3 a\right)}{3 x_{n}^{2}+a} \]We want to determine the order of convergence as it approaches the fixed point \(\alpha = \sqrt{a}\).
02
Set Up the Linearization
Linearize the iteration function around the fixed point \(\alpha = \sqrt{a}\). Let \(e_n = x_n - \alpha\) and expand \(x_{n+1}\) in a Taylor series around \(e_n\).
03
Taylor Expansion
Use the Taylor series to expand the right-hand side of the iteration around \(e_n = x_n - \sqrt{a}\). Let鈥檚 denote the function by \(f(x) = \frac{x\left(x^{2}+3 a\right)}{3 x^{2}+a}\). We need the derivatives at \(x = \sqrt{a}\).
04
Calculate Derivatives
Find the first derivative of \(f(x)\) at \(x = \sqrt{a}\). This requires calculating \(f'(x)\). Evaluate \(f'(x)\) at the fixed point \(\sqrt{a}\).
05
Determine Leading Error Term
From the Taylor expansion, determine the leading error term. If \(f'(\sqrt{a}) = 0\), determine the next non-zero term in the series to find the convergence order.
06
Analyze Convergence Order
If \(x_{n+1} - \sqrt{a}\) behaves like \(C(x_n - \sqrt{a})^k\), then the order of convergence is \(k\). Evaluate to determine the value of \(k\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
fixed point
A fixed point in mathematical terms refers to a value that remains fixed under a given function. In the context of an iteration, this means that for some function \(f\), the fixed point \(\alpha\) satisfies \(f(\alpha) = \alpha\). In our exercise, the iteration is given by:
\[x_{n+1} = \frac{x_{n} \left(x_{n}^{2} + 3a\right)}{3x_{n}^{2} + a}\]
We want to see how this sequence of \(x_n\) converges to the fixed point \(\alpha = \sqrt{a}\). This means if we plug \(\sqrt{a}\) back into the function, we should get \(\sqrt{a}\) again.
\[x_{n+1} = \frac{x_{n} \left(x_{n}^{2} + 3a\right)}{3x_{n}^{2} + a}\]
We want to see how this sequence of \(x_n\) converges to the fixed point \(\alpha = \sqrt{a}\). This means if we plug \(\sqrt{a}\) back into the function, we should get \(\sqrt{a}\) again.
iteration
Iteration is the process of repeatedly applying a function to its own output. To determine how quickly our iteration approaches the fixed point, we need to analyze the behavior of a sequence created by this iteration. Starting with an initial guess, we calculate subsequent values:
For our specific iteration, we calculate:
\[x_{n+1} = \frac{x_{n} \left(x_{n}^{2} + 3a\right)}{3x_{n}^{2} + a}\]
until \(x_n\) get sufficiently close to \(\sqrt{a}\).
- Start with an initial value \(x_0\).
- Apply the function to get \(x_1\).
- Repeat to get \(x_2\), then \(x_3\), and so on.
For our specific iteration, we calculate:
\[x_{n+1} = \frac{x_{n} \left(x_{n}^{2} + 3a\right)}{3x_{n}^{2} + a}\]
until \(x_n\) get sufficiently close to \(\sqrt{a}\).
Taylor series
The Taylor series helps us approximate complex functions using polynomials. To understand the convergence of our iteration, we expand the iteration function using a Taylor series around the fixed point \(\alpha = \sqrt{a}\). For the function:
\[f(x) = \frac{x \left(x^{2} + 3a\right)}{3x^{2} + a}\]
we express it as a series expansion:
\[f(x) = f(\alpha) + f'(\alpha) (x - \alpha) + \frac{f''(\alpha)}{2!} (x - \alpha)^2 + \dots\]
In our case, \(f(\alpha)\) equals \(\sqrt{a}\), and we need to find the derivatives of \(f\) at \(\alpha = \sqrt{a}\) to understand how the errors evolve.
\[f(x) = \frac{x \left(x^{2} + 3a\right)}{3x^{2} + a}\]
we express it as a series expansion:
\[f(x) = f(\alpha) + f'(\alpha) (x - \alpha) + \frac{f''(\alpha)}{2!} (x - \alpha)^2 + \dots\]
In our case, \(f(\alpha)\) equals \(\sqrt{a}\), and we need to find the derivatives of \(f\) at \(\alpha = \sqrt{a}\) to understand how the errors evolve.
linearization
Linearization simplifies the process of analyzing convergence by considering the linear part of the Taylor series. In our case, we let \(e_n = x_n - \alpha\).
We want to see how \(e_n\) behaves as \(n\) increases. This requires finding the first derivative, \(f'(\alpha)\):
\[f'(x) = \frac{d}{dx} \left( \frac{x (x^2 + 3a)}{3x^2 + a} \right)\]
and evaluate it at \(x = \sqrt{a}\). By doing this, we can see how changes in \(x_n\) affect changes in \(x_{n+1}\). If \(f'(\alpha) = 0\), then higher-order terms dominate.
We want to see how \(e_n\) behaves as \(n\) increases. This requires finding the first derivative, \(f'(\alpha)\):
\[f'(x) = \frac{d}{dx} \left( \frac{x (x^2 + 3a)}{3x^2 + a} \right)\]
and evaluate it at \(x = \sqrt{a}\). By doing this, we can see how changes in \(x_n\) affect changes in \(x_{n+1}\). If \(f'(\alpha) = 0\), then higher-order terms dominate.
error analysis
Error analysis involves examining how the error \(e_n = x_n - \alpha\) changes with each iteration. Specifically, after obtaining the Taylor series expansion and derivatives:
\[e_{n+1} \approx f'(\alpha) e_n\]
If \(f'(\alpha) = 0\), we need to consider the higher-order terms from the Taylor series such as \(f''(\alpha)\). The order of convergence \(k\) tells us how quickly \(e_n\) reduces. If \(e_{n+1} \sim C e_n^2\), we see quadratic convergence (\(k=2\)); if \(e_{n+1} \sim C e_n^3\), we see cubic convergence (\(k=3\)).
For our iteration, after evaluating the derivatives, we determine how the error behaves, leading to the conclusion about the order of convergence.
\[e_{n+1} \approx f'(\alpha) e_n\]
If \(f'(\alpha) = 0\), we need to consider the higher-order terms from the Taylor series such as \(f''(\alpha)\). The order of convergence \(k\) tells us how quickly \(e_n\) reduces. If \(e_{n+1} \sim C e_n^2\), we see quadratic convergence (\(k=2\)); if \(e_{n+1} \sim C e_n^3\), we see cubic convergence (\(k=3\)).
For our iteration, after evaluating the derivatives, we determine how the error behaves, leading to the conclusion about the order of convergence.
