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Suppose \(p\) is a zero of multiplicity \(m\) of \(f\), where \(f^{(m)}\) is continuous on an open interval containing \(p\). Show that the following fixed- point method has \(g^{\prime}(p)=0\) : $$ g(x)=x-\frac{m f(x)}{f^{\prime}(x)} $$

Short Answer

Expert verified
By using the derivative and applying multiplcity at \(p\), we prove that \(g^{\prime}(p) = 0\).

Step by step solution

01

Derive \(g(x)\)

The derivative of \(g(x)\) using the quotient rule is : \( g^{\prime}(x) = 1 - \frac{mf^{\prime}(x)f^{\prime}(x) - mf(x)f^{\prime\prime}(x)}{(f^{\prime}(x))^2} = 1 - \frac{m(f(x)f^{\prime\prime}(x) + (f^{\prime}(x))^2)}{(f^{\prime}(x))^2} \)
02

Plug \(p\) into \(g^{\prime}(x)\)

Plug \(p\) into \(g^{\prime}(x)\) to get \(g^{\prime}(p)\). You will have: \( g^{\prime}(p) = 1-\frac{m(f(p)f^{\prime\prime}(p)+(f^{\prime}(p))^2)}{(f^{\prime}(p))^2} \). Since \(p\) is a zero of \(f\), this equation turns into \( g^{\prime}(p) = 1-\frac{m(0*f^{\prime\prime}(p)+(f^{\prime}(p))^2)}{(f^{\prime}(p))^2} = 1-\frac{m*(f^{\prime}(p))^2}{(f^{\prime}(p))^2} \).
03

Reduce the equation

Simplify the equation obtained in step 2 to get: \( g^{\prime}(p) = 1-m \)
04

Apply multiplicity

Since by definition, \(f\) has multiplicity \(m\) at \(p\), we know that \(f(p) = f^{\prime}(p) = ... = f^{(m-1)}(p) = 0\), and \(f^{(m)}(p) \neq 0\). Therefore by this, \(g^{\prime}(p) = 1-m = 0\).

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Key Concepts

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

Fixed-Point Method
The Fixed-Point Method is a numerical technique used to find solutions to equations where a function, denoted as \( g(x) \), can be iterated to converge to a solution. This technique is based on the concept of transforming an original function \( f(x) \) into a new form \( g(x) \) such that the solution of \( f(x) = 0 \) becomes the fixed point of \( g(x) \). Essentially, a point \( x \) is a fixed point of \( g \) if \( g(x) = x \).

The usefulness of this method lies in its simplicity: you start with an initial guess and repeatedly apply \( g(x) \) to obtain successive approximations that ideally converge to the actual solution. In our particular problem, the method is transformed into \( g(x) = x - \frac{mf(x)}{f'(x)} \), which accounts for the multiplicity of zeros. This adjustment helps ensure that the iterations converge more efficiently, especially when dealing with complex root structures.

To use this method effectively, one must ensure that \( g'(x) \) near the fixed point is less than 1 in magnitude, which guarantees that the iterations converge. If \( g'(p) = 0 \), as shown in the exercise, the fixed point has even better convergence properties.
Multiplicity of Zeros
Multiplicity of zeros refers to the number of times a particular root appears for an equation. For instance, if an equation \( f(x) = 0 \) has a root \( p \) and \( f(x) \) behaves like \( (x-p)^m \) near \( p \), then \( p \) has multiplicity \( m \).

In the scenario of this exercise, understanding the multiplicity is crucial because it impacts the derivative properties and influences methods like Newton's method or its adjusted form in the fixed-point iteration. A higher multiplicity implies that the root is also a critical point of the function, often leading to slower convergence if not properly adjusted. This is why the fixed-point method is modified in such cases: by multiplying the function \( f(x) \) by its multiplicity \( m \) and dividing by the derivative, \( f'(x) \), the convergence rate can be improved.

The trick lies in utilizing the fact that all derivatives up to \( f^{(m-1)}(p) \) vanish, as stated in the solution. This allows us to adjust \( g(x) \), providing more accurate solutions by "jumping" further with each iteration. Hence, understanding and incorporating multiplicity makes numerical methods more efficient and reduces the number of iterations needed.
Derivative Calculation
The derivative calculation is an essential tool in numerical analysis, providing insights into the behavior and properties of functions. In this exercise, the derivative plays a pivotal role in adjusting the fixed-point method to account for the multiplicity of zeros.

First, we calculate \( g'(x) \) using the quotient rule. The quotient rule is applied because \( g(x) \) involves a division of functions, namely, \( \frac{m f(x)}{f'(x)} \). When performed, the differentiation gets quite complex, involving products and chain rule applications. However, good practice in differentiation simplifies the process:
  • The derivative of \( g(x) \) has a major role in determining whether the function iterates towards convergence.
  • Setting \( g'(p) = 0 \) is the end goal, ensuring efficient convergence to the root.
This calculation reveals that, by letting \( f(p) = f'(p) = \dots = f^{(m-1)}(p) = 0 \), all terms relating to \( f^{(m)} \) naturally disappear, and \( g'(p) \) simplifies.

By reducing \( g'(p) = 1 - m \), the multiplicity adjustment underscores the effectiveness of adapting our iterative method to ensure convergence. Thus, computing derivatives accurately forms the backbone of controlling numerical method performances.

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Most popular questions from this chapter

A can in the shape of a right circular cylinder is to be constructed to contain \(1000 \mathrm{~cm}^{3}\). The circular top and bottom of the can must have a radius of \(0.25 \mathrm{~cm}\) more than the radius of the can so that the excess can be used to form a seal with the side. The sheet of material being formed into the side of the can must also be \(0.25 \mathrm{~cm}\) longer than the circumference of the can so that a seal can be formed. Find, to within \(10^{-4}\), the minimal amount of material needed to construct the can.

A drug administered to a patient produces a concentration in the blood stream given by \(c(t)=A t e^{-t / 3}\) milligrams per milliliter, \(t\) hours after \(A\) units have been injected. The maximum safe concentration is \(1 \mathrm{mg} / \mathrm{mL}\). a. What amount should be injected to reach this maximum safe concentration, and when does this maximum occur? b. An additional amount of this drug is to be administered to the patient after the concentration falls to \(0.25 \mathrm{mg} / \mathrm{mL}\). Determine, to the nearest minute, when this second injection should be given. c. Assume that the concentration from consecutive injections is additive and that \(75 \%\) of the amount originally injected is administered in the second injection. When is it time for the third injection?

Use the Bisection method to find solutions, accurate to within \(10^{-5}\) for the following problems. a. \(\quad 3 x-e^{x}=0\) for \(1 \leq x \leq 2\) b. \(\quad 2 x+3 \cos x-e^{x}=0 \quad\) for \(0 \leq x \leq 1\) c. \(\quad x^{2}-4 x+4-\ln x=0 \quad\) for \(1 \leq x \leq 2 \quad\) and \(\quad 2 \leq x \leq 4\) d. \(x+1-2 \sin \pi x=0 \quad\) for \(0 \leq x \leq 0.5 \quad\) and \(\quad 0.5 \leq x \leq 1\)

The following four methods are proposed to compute \(21^{1 / 3}\). Rank them in order, based on their apparent speed of convergence, assuming \(p_{0}=1\). a. \(\quad p_{n}=\frac{20 p_{n-1}+21 / p_{n-1}^{2}}{21}\) b. \(\quad p_{n}=p_{n-1}-\frac{p_{n-1}^{3}-21}{3 p_{n-1}^{2}}\) c. \(p_{n}=p_{n-1}-\frac{p_{n-1}^{4}-21 p_{n-1}}{p_{n-1}^{2}-21}\) d. \(\quad p_{n}=\left(\frac{21}{p_{n-1}}\right)^{1 / 2}\)

a. Sketch the graphs of \(y=e^{x}-2\) and \(y=\cos \left(e^{x}-2\right)\). b. Use the Bisection method to find an approximation to within \(10^{-5}\) to a value in \([0.5,1.5]\) with \(e^{x}-2=\cos \left(e^{x}-2\right)\)

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