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Chapter 10: Problem 7

Wenden Sie das Newton Verfahren an, um die Nullstelle \(x=0\) der beiden Funktionen $$ f(x)= \begin{cases}x^{4 / 3}, & x \geq 0 \\ -|x|^{4 / 3}, & x<0\end{cases} $$ und $$ g(x)= \begin{cases}\sqrt{x}, & x \geq 0 \\ -\sqrt{|x|}, & x<0\end{cases} $$ zu bestimmen. Falls das Verfahren konvergiert, geben Sie die Konvergenzordnung an und ein Intervall f眉r m枚gliche Startwerte.

Short Answer

Expert verified
Answer: The order of convergence for Newton's method when applied to the given functions f(x) and g(x) is quadratic. An appropriate interval for possible initial values could be \([-1, 1]\), as both functions are continuous and well-behaved around the root x = 0.

Step by step solution

01

Understand Newton's method

Newton's method is an iterative method for finding the roots of a real-valued function. If we have a function h(x), the Newton's method is defined as: \(x_{n+1} = x_n - \frac{h(x_n)}{h'(x_n)}\) where \(x_n\) denotes the approximation of the root at iteration n and \(h'(x_n)\) is the derivative of the function h(x) evaluated at \(x_n\). Now we will apply this method to the given functions f(x) and g(x).
02

Find the derivatives of f(x) and g(x)

Before applying the Newton's method, we need to compute the derivatives of f(x) and g(x). Let's do that: For f(x): If x 鈮� 0, \(f(x) = x^{4/3}\), and its derivative is \(f'(x) = \frac{4}{3} x^{1/3}\). If x < 0, \(f(x) = -|x|^{4/3}\), and its derivative is \(f'(x) = \frac{4}{3}x^{1/3}\). For g(x): If x 鈮� 0, \(g(x) = \sqrt{x}\), and its derivative is \(g'(x) = \frac{1}{2\sqrt{x}}\). If x < 0, \(g(x) = -\sqrt{|x|}\), and its derivative is \(g'(x) = \frac{-1}{2\sqrt{|x|}}\).
03

Apply Newton's method for f(x)

Using the derivatives we calculated in Step 2, we can now apply Newton's method to f(x). We have: For x 鈮� 0, \(x_{n+1} = x_n - \frac{x_n^{4/3}}{\frac{4}{3}x_n^{1/3}}\). For x < 0, \(x_{n+1} = x_n - \frac{-|x_n|^{4/3}}{\frac{4}{3}x_n^{1/3}}\).
04

Apply Newton's method for g(x)

Similarly, we will apply Newton's method to g(x). We have: For x 鈮� 0, \(x_{n+1} = x_n - \frac{\sqrt{x_n}}{\frac{1}{2\sqrt{x_n}}}\). For x < 0, \(x_{n+1} = x_n - \frac{-\sqrt{(|x_n|)}}{\frac{-1}{2\sqrt{|x_n|}}}\).
05

Determine the order of convergence and interval for initial values

Since Newton's method converges quadratically if the function is twice differentiable at the root, we can determine that both f(x) and g(x) exhibit quadratic convergence. In practice, however, the convergence can be slow and not guaranteed if the initial value is far from the root or the function has local extrema. The choice of an interval with possible initial values depends on how well-behaved the function is around the root and how fast the convergence is that you desire. For our given functions, an interval for possible initial values could be taken as \([-1, 1]\) since both functions are continuous and well-behaved around the root x = 0.

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Most popular questions from this chapter

Es wird eine H盲ngebr眉cke 眉ber eine \(30 \mathrm{~m}\) breiteBucht gebaut. Dabei ist die Form der Br眉cke durch die sogenannte Kettenlinie beschrieben, d.h., es gibt eine positive Konstante \(a>0\), sodass die Form durch den Graphen der Funktion \(f: \mathbb{R} \rightarrow \mathbb{R}\) mit $$ f(x)=h_{0}+a\left(\cosh \left(\frac{x-x_{0}}{a}\right)-1\right) $$ gegeben ist. Die Durchfahrtsh枚he f眉r Segelschiffe muss \(h_{0}=\) \(8 \mathrm{~m}\) betragen. Die Steilufer sind \(10 \mathrm{~m}\) und \(12 \mathrm{~m}\) 眉ber dem Wasserspiegel (siehe Abbildung). Bestimmen Sie mithilfe des Newton-Verfahrens den Parameter \(a>0\) und den Abstand \(x_{0}\) des Tiefpunkts zu einem der Ufer.

Berechnen Sie die Ableitungen der Funktionen \(f: D \rightarrow \mathbb{R}\) mit $$ \begin{aligned} &f_{1}(x)=\left(x+\frac{1}{x}\right)^{2}, \quad x \neq 0 \\ &f_{2}(x)=\cos \left(x^{2}\right) \cos ^{2} x, \quad x \in \mathbb{R} \\ &f_{3}(x)=\ln \left(\frac{\mathrm{e}^{x}-1}{\mathrm{e}^{x}}\right), \quad x \neq 0 \\ &f_{4}(x)=x^{\left(x^{x}\right)}, \quad x>0 \end{aligned} $$ auf dem jeweiligen Definitionsbereich der Funktion.

Zeigen Sie, dass der verallgemeinerte Mittelwertf眉r \(x \rightarrow 0\) gegen das geometrische Mittel positiver Zahlen \(a_{1}, \ldots a_{k} \in \mathbb{R}_{>0}\) konvergiert, d. h., es gilt: $$ \lim _{x \rightarrow 0}\left(\frac{1}{n} \sum_{j=1}^{n} a_{j}^{x}\right)^{\frac{1}{x}}=\sqrt[n]{\prod_{j=1}^{n} a_{j}} $$

\bullet\( Zeigen Sie, dass die Funktion \)f:[-1,2] \rightarrow \mathbb{R}\( mit \)f(x)=x^{4}\( konvex ist, (a) indem Sie nach Definition \)f(\lambda x+(1-\lambda) z) \leq \lambda f(x)+(1-\( \lambda) \)f(z)\( f眉r alle \)\lambda \in[0,1]\( pr眉fen, (b) mittels der Bedingung \)f^{\prime}(x)(y-x) \leq f(y)-f(x)$.

Zeigen Sie f眉r \(|x|<1\) die Taylorformel$$ \ln \frac{1-x}{1+x}=-2\left(x+\frac{x^{3}}{3}+\cdots+\frac{x^{2 n-1}}{2 n-1}\right)+R_{2 n}(x) $$ mit dem Restglied $$ R_{2 n}(x)=\frac{-x^{2 n+1}}{2 n+1}\left(\frac{1}{(1+t x)^{2 n+1}}+\frac{1}{(1-t x)^{2 n+1}}\right) $$ f眉r ein \(t \in(0,1)\). Approximieren Sie mithilfe des Taylorpolynoms vom Grad \(n=\) 2 den Wert \(\ln (2 / 3)\) und zeigen Sie, dass der Fehler kleiner als \(5 \cdot 10^{-4}\) ist.
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