Problem 56
Question
Use the remainder term to estimate the absolute error in approximating the following quantities with the nth-order Taylor polynomial centered at \(0 .\) Estimates are not unique. $$\cos 0.45 ; n=3$$
Step-by-Step Solution
Verified Answer
The approximate error in estimating the cosine of 0.45 using a 3rd order Taylor polynomial centered at 0 is approximately 0.000222.
1Step 1: Identify the Taylor polynomial for the cosine function
The Taylor polynomial for the cosine function centered at \(0\) is given by:
$$ T_n(x) = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \frac{(-1)^k x^{2k}}{(2k)!} $$
For this problem, we have \(n=3\), thus, the \(3^{rd}\) order Taylor polynomial for the cosine function is:
$$ T_3(x) = \sum_{k=0}^{\lfloor\frac{3}{2}\rfloor} \frac{(-1)^k x^{2k}}{(2k)!} = 1 - \frac{x^2}{2} $$
2Step 2: Find the \((n+1)^{th}\) derivative of the cosine function
In this case, the \((n+1)^{th}\) derivative corresponds to the \(4^{th}\) derivative of the cosine function. The derivatives of the cosine function follow a cyclical pattern:
1. Derivative: \(-\sin x\)
2. Derivative: \(-\cos x\)
3. Derivative: \(\sin x\)
4. Derivative: \(\cos x\)
Thus, the \(4^{th}\) derivative of the cosine function is again the cosine function:
$$ f^{(4)}(x) = \cos x $$
3Step 3: Identify the interval and find the maximum value for the \(4^{th}\) derivative
Since we are using the Taylor polynomial centered at \(0\) to approximate the cosine function at \(0.45\), we need to find the maximum value for the \(4^{th}\) derivative in the interval \([0, 0.45]\). We know that the cosine function varies between \(-1\) and \(1\). Therefore, the maximum value for the \(4^{th}\) derivative in the given interval is:
$$ M = |\cos x|_{max} = 1 $$
4Step 4: Apply Lagrange's form of the remainder to estimate the error
The remainder term, \(R_n(x)\), in Lagrange's form can be written as:
$$ R_n(x) = \frac{M |x - 0|^4}{(n+1)!} $$
Calculate the absolute value of the remainder term for estimating the error using \(n=3\) and \(x=0.45\).
$$ |R_3(0.45)| = \frac{1 \cdot |0.45|^4}{(3+1)!} = \frac{|0.45|^4}{4!} \approx 0.000222 $$
By analyzing the question and solving it step-by-step, we found that the absolute error in approximating \(\cos(0.45)\) using a \(3^{rd}\) order Taylor polynomial centered at \(0\) is approximately \(0.000222\).
Key Concepts
Absolute ErrorTaylor SeriesCosine Function
Absolute Error
When approximating functions using polynomials, it's important to evaluate how accurate that approximation truly is. **Absolute error** is a metric used to estimate this accuracy. It measures the difference between the actual value and the approximated value given by a mathematical method like a Taylor polynomial.
The formula for calculating absolute error given the Taylor polynomial's remainder term, also known as Lagrange's remainder, is:
This approach gives us a good estimate for how much our approximation might differ from reality. Estimating the absolute error helps students understand the limits of mathematical models and the precision needed for various calculations.
The formula for calculating absolute error given the Taylor polynomial's remainder term, also known as Lagrange's remainder, is:
- \( R_n(x) = \frac{M |x|^{n+1}}{(n+1)!} \)
This approach gives us a good estimate for how much our approximation might differ from reality. Estimating the absolute error helps students understand the limits of mathematical models and the precision needed for various calculations.
Taylor Series
A Taylor series is a powerful mathematical tool that expresses a function as an infinite sum of terms. Each of these terms is calculated using the function's derivatives at a particular point, typically around 0 or another convenient point.
The Taylor polynomial is a finite version of this series, offering an approximation when calculating or computing functions, like the cosine function, where exact values are challenging to derive by hand.
To form the Taylor polynomial for the cosine function around 0, for instance, we use the Taylor series formula:
Understanding Taylor series and how to construct Taylor polynomials is instrumental in various science and engineering fields, enabling approximations where direct calculations would be complex or impossible.
The Taylor polynomial is a finite version of this series, offering an approximation when calculating or computing functions, like the cosine function, where exact values are challenging to derive by hand.
To form the Taylor polynomial for the cosine function around 0, for instance, we use the Taylor series formula:
- \[ T_n(x) = \sum_{k=0}^{\lfloor\frac{n}{2}\rfloor} \frac{(-1)^k x^{2k}}{(2k)!} \]
Understanding Taylor series and how to construct Taylor polynomials is instrumental in various science and engineering fields, enabling approximations where direct calculations would be complex or impossible.
Cosine Function
The **cosine function** is a fundamental concept in trigonometry, associated with angles and periodic phenomena. It's denoted as \( \cos(x) \) and forms the basis for cyclical patterns often seen in nature and engineering.
When approximating the cosine function, such as \( \cos(0.45) \), Taylor polynomials come into play. These approximations leverage the cyclical nature of the cosine function's derivatives, which follow this sequence:
Knowing the behavior and properties of the cosine function aids in understanding its representation through Taylor series and evaluating the approximation accuracy using polynomials.
When approximating the cosine function, such as \( \cos(0.45) \), Taylor polynomials come into play. These approximations leverage the cyclical nature of the cosine function's derivatives, which follow this sequence:
- 1st Derivative: \(-\sin x\)
- 2nd Derivative: \(-\cos x\)
- 3rd Derivative: \(\sin x\)
- 4th Derivative: \(\cos x\)
Knowing the behavior and properties of the cosine function aids in understanding its representation through Taylor series and evaluating the approximation accuracy using polynomials.
Other exercises in this chapter
Problem 56
Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the followi
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Write the following power series in summation (sigma) notation. $$1+\frac{x}{2}+\frac{x^{2}}{4}+\frac{x^{3}}{6}+\cdots$$
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Identify the functions represented by the following power series. $$\sum_{k=0}^{\infty}(-1)^{k} \frac{x^{2 k}}{4^{k}}$$
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Find the remainder in the Taylor series centered at the point a for the following functions. Then show that \(\lim _{n \rightarrow \infty} R_{n}(x)=0\) for all
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