Problem 1 Man bestimme die Taylor-Reihe de... [FREE SOLUTION] (2024)

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Chapter 22: Problem 1

Man bestimme die Taylor-Reihe der Funktion $x \longmapsto x^{a}$ mitEntwicklungspunkt $a \in \mathbb{R}_{+}^{*}$

Short Answer

Expert verified

The Taylor series for the function $f(x)=x^{a}$ around the point $a$ is \[f(x) = a^{a} + \sum_{n=1}^{\infty} \frac{(a)_n (x-a)^n}{n!}\] where (a)_n denotes the falling factorial of $a$.

Step by step solution

Root Selection

We first need to decide which root we will use to expand the power series around. Since the problem specifically mentions $a \in \mathbb{R}_{+}^{*}$, that means the root will be the point $a$.

Taylor Series Expansion

The general definition of the Taylor series for a function $f(x)$ about the point $a$ is: \[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^{2}}{2} + \frac{f'''(a)(x-a)^3}{3!} + \ldots\] We now need to apply this formula to our function $f(x) = x^{a}$ and use our development point $a$.

Calculating Derivatives

The derivative of $f(x) = x^{a}$ is $f'(x) = a * x^{a-1}$, the second derivative is $f''(x) = a * (a-1) * x^{a-2}$, and, generally, the nth derivative is given by $f^{(n)}(x) = a * (a-1) * \ldots * (a-n+1) * x^{a-n}$.

Substituting the Derivatives

Substituting these derivatives into the expansion, we find: \[f(x) = a^{a} + a * a^{a-1} * (x-a) + \frac{a * (a-1) * a^{a-2} * (x-a)^{2}}{2} + \frac{a * (a-1) * (a-2) * a^{a-3} * (x-a)^3}{3!} + \ldots\]

Simplifying the Expansion

Each term can be simplified by recognising that the numerator of the fraction in each term forms a falling factorial, while the denominator is a rising factorial. Thus, the series simplifies to: \[f(x) = a^{a} + \sum_{n=1}^{\infty} \frac{(a)_n (x-a)^n}{n!}\] where $(a)_n$ denotes the falling factorial of $a$.

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Most popular questions from this chapter

Aufgabe $22 \mathbf{G}^{*}$. Man beweise die Funktionalgleichung des Arcus-Tangens: Für $x, y \in \mathbb{R}$ mit $|\arctan x+\arctan y|<\frac{\pi}{2}$gilt $$ \arctan x+\arctan y=\arctan \frac{x+y}{1-x y} $$ Man folgere hieraus die,,Machinsche Formel" $$ \frac{\pi}{4}=4 \arctan \frac{1}{5}-\arctan \frac{1}{239} $$ und die Reihenentwicklung $$ \frac{\pi}{4}=\frac{4}{5} \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1}\left(\frac{1}{5}\right)^{2 k}-\frac{1}{239} \sum_{k=0}^{\infty}\frac{(-1)^{k}}{2 k+1}\left(\frac{1}{239}\right)^{2 k} $$ Welche Glieder muss man berücksichtigen, um $\pi$ mit einer Genauigkeit von$10^{-12}$ zu berechnen?. Sei $p$ eine natürliche Zahl mit $1 \leq p \leq n+1$. Man beweise für dasRestglied $R_{n+1}$ der Taylorschen Formel (An. 1, $\S 22$, Satz 1): Es gibtein \xi zwischen $a$ und $x$, so dass $$ R_{n+1}(x)=\frac{f^{(n+1)}(\xi)}{p \cdot n !}(x-\xi)^{n+1-p}(x-a)^{p} $$ (Dies ist das sogenannte Schlömilchsche Restglied.)Man bestimme die Taylor-Reihen der Funktionen sin und cos mit einem beliebigenEntwicklungspunkt $a \in \mathbb{R}$.Man berechne den Anfang der Taylor-Reihe der Funktion $f:]-2,2[\rightarrow\mathbb{R}$ $$ f(x):=\frac{\sin x}{2+x} $$ mit Entwicklungspunkt 0 bis einschlieBlich des Gliedes 5 . Ordnung. Aufgabe 22 D. Durch Integration der Taylor-Reihe der Ableitung von $\arcsin:]-1,1[\longrightarrow \mathbb{R}$Aufgabe $22 \mathrm{I}^{*}$. Man zeige: Die Taylor-Entwicklung der Funktion$\frac{1}{\cos x}$ um den Nullpunkt hat die Gestalt $$ \frac{1}{\cos x}=\sum_{k=0}^{\infty} \frac{E_{2 k}}{(2 k) !} x^{2 k} $$ mit positiven ganzen Zahlen $E_{2 k}$. Man berechne $E_{0}, E_{2}, E_{4}, \ldots, E_{10}$

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Problem 1 Man bestimme die Taylor-Reihe de... [FREE SOLUTION] (2024)

Short Answer

Step by step solution

Root Selection

Taylor Series Expansion

Calculating Derivatives

Substituting the Derivatives

Simplifying the Expansion

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Discrete Mathematics

Calculus

Geometry

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

Privacy Overview

References