/*! 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 4 Berechnen Sie das TAYLOR-Polynom... [FREE SOLUTION] | 91Ó°ÊÓ

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Berechnen Sie das TAYLOR-Polynom 2. Grades der Funktion \(f(x)=\sqrt{1+x^{2}}\) mit dem Entwicklungspunkt \(x_{0}=0\), und schätzen Sie die Genauigkeit der Approximation von \(f\) durch das TAYLOR-Polynom für \(x \in\left[0, \frac{1}{5}\right]\) ab.

Short Answer

Expert verified
The Taylor polynomial is \(T_2(x) = 1 + \frac{1}{2}x^2\). The approximation is accurate on \([0, \frac{1}{5}]\).

Step by step solution

01

Determine the function and derivatives

The function given is \( f(x) = \sqrt{1+x^2} \). First, find the derivatives needed for the Taylor polynomial. The first derivative is:\[ f'(x) = \frac{x}{\sqrt{1+x^2}}. \]The second derivative is:\[ f''(x) = \frac{1}{(1+x^2)^{3/2}}. \]
02

Evaluate the derivatives at \(x_0 = 0\)

Substitute \(x_0 = 0\) into the function and its derivatives. This gives:\[ f(0) = \sqrt{1+0^2} = 1, \]\[ f'(0) = \frac{0}{\sqrt{1+0^2}} = 0, \]\[ f''(0) = \frac{1}{(1+0^2)^{3/2}} = 1. \]
03

Construct the Taylor Polynomial of degree 2

The Taylor polynomial of degree 2 for \( f(x) \) at \( x_0 = 0 \) is:\[ T_2(x) = f(x_0) + f'(x_0)\cdot x + \frac{f''(x_0)}{2} x^2. \]Substituting the calculated values:\[ T_2(x) = 1 + 0 \cdot x + \frac{1}{2} x^2 = 1 + \frac{1}{2} x^2. \]
04

Estimate the approximation accuracy

We'll use the remainder term of the Taylor series to estimate the error. The error term for the second-degree polynomial is:\[ R_2(x) = \frac{f'''(\xi)}{6} x^3, \]where \( \xi \) is between \( 0 \) and \( x \). Calculate the third derivative:\[ f'''(x) = -\frac{3x}{(1+x^2)^{5/2}}. \]Over the interval \([0, \frac{1}{5}]\), the maximum value of \(|f'''(x)|\) can be estimated at \(|f'''(0)| = 0\). Thus, the remainder is small and the approximation is quite accurate for small \(x\).

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Key Concepts

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

Approximation Accuracy
When we work with Taylor polynomials, the goal is to approximate a function with a polynomial close to its behavior near a specific point. In this problem, we are approximating the function \( f(x) = \sqrt{1+x^2} \) at \( x_0 = 0 \) using a Taylor polynomial of degree 2. This polynomial is a simple form that uses the function's value and its first few derivatives evaluated at \( x_0 \) to mimic the function's shape.
The accuracy of this approximation depends greatly on how far \( x \) is from \( x_0 \) and the behavior of higher-order derivatives. For the interval \([0, \frac{1}{5}]\), a second-degree polynomial is suitable, especially since the third derivative and subsequent derivatives are relatively small, indicating that their influence on the polynomial's accuracy is negligible within this range.
  • The degree of the polynomial gives a balance between simplicity and accuracy.
  • The approximation generally gets better as the interval around \( x_0 \) gets smaller.
  • Checking higher-order derivatives helps in estimating how 'flat' the function is, which influences approximation.
Derivatives Calculation
A key part of forming a Taylor polynomial involves calculating derivatives. The function \( f(x) = \sqrt{1+x^2} \) requires derivatives to form its Taylor approximation. Each derivative gives insight into how the function behaves locally.
We start with the first derivative, \( f'(x) = \frac{x}{\sqrt{1+x^2}} \), which tells us about the slope of \( f \). The second derivative, \( f''(x) = \frac{1}{(1+x^2)^{3/2}} \), provides information about the curvature.
  • The first derivative at \( x_0 = 0 \) is \( 0 \), meaning the slope is flat at this point.
  • The second derivative at \( x_0 = 0 \) is \( 1 \), showing positive curvature, indicating the function is concave up at \( x_0 \).
This curvature information is crucial in shaping the Taylor polynomial, as it affects the coefficient of the \( x^2 \) term.
Maclaurin Series
The Taylor series centered at \( x = 0 \) is called the Maclaurin series. For \( f(x) = \sqrt{1+x^2} \), the Maclaurin series gives us a simplified polynomial that represents the function's behavior near zero.
In this case, we derive the second-degree Maclaurin polynomial, which is
\[ T_2(x) = 1 + \frac{1}{2} x^2. \]
This polynomial utilizes the first and second derivatives:
  • \( f(0) = 1 \) gives the constant term.
  • \( f'(0) = 0 \) results in no linear \( x \) term.
  • \( f''(0) = 1 \) contributes to the \( \frac{1}{2} x^2 \) term.
The Maclaurin series is particularly useful for computing approximate values of the function \( f \) over short intervals around \( x = 0 \), providing an easier computational method than working with non-polynomial functions.
Remainder Estimation
Even though a Taylor polynomial can approximate a function well, it is essential to know the possible error, especially when using the polynomial in place of the function over a range. The remainder term, \( R_2(x) \), quantifies this error for a second-degree Taylor polynomial:
\[ R_2(x) = \frac{f'''(\xi)}{6} x^3, \]
where \( \xi \) is a point between \( 0 \) and \( x \). Calculating \( f'''(x) = -\frac{3x}{(1+x^2)^{5/2}} \), we determine the possible error.
  • For small values of \( x \), such as within \([0, \frac{1}{5}]\), this error is minimal because \(|f'''(x)|\) remains small.
  • The evaluation of \( f'''(0) = 0 \) confirms that higher curvature impacts are nearly negligible in this region.
This understanding ensures confidence in the Taylor polynomial's use and provides a tangible estimate of its reliability.

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