/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 Man entwickle \(f(x, y)=x^{y}\) ... [FREE SOLUTION] | 91影视

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Chapter 24: Problem 9

Man entwickle \(f(x, y)=x^{y}\) an der Stelle \(\tilde{x}=\) \((1,1)^{\mathrm{T}}\) in ein Taylorpolynom bis zu Termen zweiter Ordnung und berechne damit n盲herungsweise \(\sqrt[10]{(1.05)^{9}}\).

Short Answer

Expert verified
A) Compute the partial derivatives of $f(x, y)$ up to second-order terms, evaluate them at the point (1,1), write the Taylor polynomial, and approximate the given expression. B) Compute the gradient for all points and find the optimal point for approximation. C) Calculate the polynomial with third-order derivatives and find the approximation using a test point. D) Apply the chain rule and find the approximation by calculating the limit of the function.

Step by step solution

01

Compute the partial derivatives

First, we need to compute the partial derivatives of \(f(x, y)\) up to the second order. The first-order partial derivatives are: $$ \frac{\partial f}{\partial x} = y x^{y-1} \\ \frac{\partial f}{\partial y} = x^y \ln x $$ The second-order partial derivatives are: $$ \frac{\partial^2 f}{\partial x^2} = y(y-1) x^{y-2} \\ \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} = y x^{y-1} \ln x \\ \frac{\partial^2 f}{\partial y^2} = x^y (\ln x)^2 $$
02

Evaluate the partial derivatives at the given point

Now, we need to evaluate the computed partial derivatives of \(f(x, y)\) at \((\tilde{x},\tilde{y}) = (1,1)\). Plugging in the values, we get: $$ \frac{\partial f}{\partial x}(1,1) = 1 \\ \frac{\partial f}{\partial y}(1,1) = 0 \\ \frac{\partial^2 f}{\partial x^2}(1,1) = 0 \\ \frac{\partial^2 f}{\partial x \partial y}(1,1) = 0 \\ \frac{\partial^2 f}{\partial y^2}(1,1) = 0 $$
03

Write the Taylor polynomial up to second-order terms

Now we can write the Taylor polynomial for \(f(x,y)\) up to second-order terms: $$ T_2(x,y) = f(1,1) + \frac{\partial f}{\partial x}(1,1)(x-1) + \frac{\partial f}{\partial y}(1,1)(y-1) + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(1,1)(x-1)^2 + \frac{\partial^2 f}{\partial x \partial y}(1,1)(x-1)(y-1) + \frac{1}{2}\frac{\partial^2 f}{\partial y^2}(1,1)(y-1)^2 $$ Plugging in the values obtained in step 2: $$ T_2(x,y) = 1 +(x-1) $$
04

Approximate the given expression

Now, we can use the Taylor polynomial to approximate \(\sqrt[10]{(1.05)^9}\). Since \(f(x, y) = x^y\), we want to find \(T_2(1.05, 0.9)\). Plugging in these values, we get: $$ T_2(1.05,0.9) = 1 +(1.05 - 1) = 1.05 $$ So, the approximation of \(\sqrt[10]{(1.05)^9}\) using the second-order Taylor polynomial is 1.05.

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Most popular questions from this chapter

Gegeben ist die Funktion \(f: \mathbb{R}^{3} \rightarrow \mathbb{R}\) $$ f(x, y, z)=\mathrm{e}^{\cos ^{2}\left(x y^{3} z\right)}-\sqrt{\mathrm{e}} $$ Man begr眉nde, warum sich \(f(x, y, z)=0\) in einer Umgebung von \(P=\left(x_{0}, y_{0}, z_{0}\right)=\left(\pi, 1, \frac{1}{4}\right)\) lokal nach \(z\) aufl枚sen l盲sst, und berechne dort die partiellen Ableitungen \(z_{x}\left(x_{0}, y_{0}\right)\) und \(z_{y}\left(x_{0}, y_{0}\right)\)

Transformieren Sie den Ausdruck $$ W=\frac{1}{\sqrt{x^{2}+y^{2}}}\left(x \frac{\partial U}{\partial x}+y \frac{\partial U}{\partial y}\right) $$ auf Polarkoordinaten. (Hinweis: Setzen Sie dazu \(u(r, \varphi)=\) \(U(r \cos \varphi, r \sin \varphi) .)\)

Man betrachte die Schar aller Strecken von \((0, t)^{\mathrm{T}}\) nach \((1-t, 0)^{\mathrm{T}}\) mit \(t \in[0,1]\) und bestimme die Einh眉llende dieser Strecken.

Bestimmen Sie die Werte und Fehler der folgenden Gr枚脽en: \- Zylindervolumen \(V\), $$ \begin{aligned} &V=r^{2} \pi h \\ &r=(10.0 \pm 0.1) \mathrm{cm}, \quad h=(50.0 \pm 0.1) \mathrm{cm} \end{aligned} $$ \- Beschleunigung \(a\), $$ \begin{aligned} &s=\frac{1}{2} a t^{2} \\ &s=(100.0 \pm 0.5) \mathrm{m}, \quad t=(3.86 \pm 0.01) \mathrm{s} \end{aligned} $$ \- Widerstand \(R_{12}\) bei Parallelschaltung, $$ \begin{aligned} &\frac{1}{R_{12}}=\frac{1}{R_{1}}+\frac{1}{R_{2}} \\ &R_{1}=(100 \pm 5) \Omega, \quad R_{2}=(50 \pm 5) \Omega \end{aligned} $$

Untersuchen Sie die beiden Funktionen \(f\) und \(g\), $$ \mathbb{R}^{2} \rightarrow \mathbb{R} $$ $$ f(x)= \begin{cases}\frac{x_{1} x_{2}^{3}}{\left(x_{1}+x_{2}\right)^{2}} & \text { f眉r } x \neq 0 \\ 0 & \text { f眉r } x=0\end{cases} $$ $$ g(x)= \begin{cases}\frac{x_{1}^{3}, x_{2}^{2}}{\left(x_{1}^{2}+x_{2}^{2}\right)^{2}} & \text { f眉r } x \neq 0 \\\ 0 & \text { f眉r } x=0\end{cases} $$ auf Stetigkeit im Ursprung,
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