Calculate L4 For F(x)=6cos(x/2) Over [2/4,2/2][2/4,2/2].
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Introduction: Understanding the Problem and Its Significance
Calculating the fourth-degree Taylor polynomial (L4) for a function like \( F(x) = 6 \cos(x/2) \) over a specified interval is a foundational task in approximation theory and numerical analysis. Such approximations are vital in computational mathematics, engineering, and physics to estimate function values with high efficiency and accuracy when exact calculations are complex or computationally expensive.
The focus here is to determine the L4 polynomial centered at a point within the interval—most commonly at the midpoint or a strategic point—then analyze its accuracy over the given interval \([2/4, 2/2]\). The interval, simplified, is \([0.5, 1]\).
This guide will walk through the step-by-step process of calculating the fourth-degree Taylor polynomial, including derivatives, the Taylor series expansion, and the error estimation, ensuring a comprehensive understanding for students and professionals alike.
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Understanding the Function \(F(x) = 6 \cos(x/2)\)
Before diving into the Taylor polynomial, it’s essential to comprehend the properties of the function:
- Type of function: Trigonometric cosine function scaled by 6.
- Derivative pattern: Derivatives involve alternating sine and cosine functions with scaled arguments.
- Periodicity and symmetry: The cosine function exhibits periodic behavior, which influences the approximation accuracy over intervals.
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Step 1: Choosing the Center Point for Taylor Expansion
The Taylor polynomial is typically centered at a point \( a \) within the interval \([0.5, 1]\). Common choices include:
- Midpoint: \( a = 0.75 \)
- Endpoint: \( a = 0.5 \) or \( a = 1 \)
For this example, we select \( a = 0.75 \), the midpoint, to minimize maximum approximation error across the interval.
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Step 2: Calculating Derivatives of \(F(x)\)
To construct a 4th-degree Taylor polynomial, derivatives up to the 4th order are needed:
First derivative \(F'(x)\):
\[
F'(x) = 6 \times \frac{d}{dx} \cos\left(\frac{x}{2}\right) = 6 \times \left( - \sin\left(\frac{x}{2}\right) \times \frac{1}{2} \right) = -3 \sin\left(\frac{x}{2}\right)
\]
Second derivative \(F''(x)\):
\[
F''(x) = -3 \times \frac{d}{dx} \sin\left(\frac{x}{2}\right) = -3 \times \left( \cos\left(\frac{x}{2}\right) \times \frac{1}{2} \right) = -\frac{3}{2} \cos\left(\frac{x}{2}\right)
\]
Third derivative \(F'''(x)\):
\[
F'''(x) = -\frac{3}{2} \times \frac{d}{dx} \cos\left(\frac{x}{2}\right) = -\frac{3}{2} \times \left( - \sin\left(\frac{x}{2}\right) \times \frac{1}{2} \right) = \frac{3}{4} \sin\left(\frac{x}{2}\right)
\]
Fourth derivative \(F^{(4)}(x)\):
\[
F^{(4)}(x) = \frac{3}{4} \times \frac{d}{dx} \sin\left(\frac{x}{2}\right) = \frac{3}{4} \times \left( \cos\left(\frac{x}{2}\right) \times \frac{1}{2} \right) = \frac{3}{8} \cos\left(\frac{x}{2}\right)
\]
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Step 3: Evaluating Derivatives at \( a = 0.75 \)
Determine the derivatives at the center point:
\[
x = 0.75
\]
Calculate \( \frac{x}{2} = \frac{0.75}{2} = 0.375 \).
Evaluate the derivatives:
- \( F(a) = 6 \cos(0.375) \)
- \( F'(a) = -3 \sin(0.375) \)
- \( F''(a) = -\frac{3}{2} \cos(0.375) \)
- \( F'''(a) = \frac{3}{4} \sin(0.375) \)
- \( F^{(4)}(a) = \frac{3}{8} \cos(0.375) \)
Using approximate values:
- \( \cos(0.375) \approx 0.9305 \)
- \( \sin(0.375) \approx 0.3660 \)
Thus:
- \( F(0.75) \approx 6 \times 0.9305 = 5.583 \)
- \( F'(0.75) \approx -3 \times 0.3660 = -1.098 \)
- \( F''(0.75) \approx -\frac{3}{2} \times 0.9305 = -1.39575 \)
- \( F'''(0.75) \approx \frac{3}{4} \times 0.3660 = 0.2745 \)
- \( F^{(4)}(0.75) \approx \frac{3}{8} \times 0.9305 = 0.348 \)
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Step 4: Constructing the 4th-Degree Taylor Polynomial \( L_4(x) \)
The Taylor polynomial centered at \( a = 0.75 \):
\[
L_4(x) = F(a) + F'(a)(x - a) + \frac{F''(a)}{2!}(x - a)^2 + \frac{F'''(a)}{3!}(x - a)^3 + \frac{F^{(4)}(a)}{4!}(x - a)^4
\]
Plugging in the derivatives:
\[
L_4(x) = 5.583 - 1.098(x - 0.75) - \frac{1.39575}{2}(x - 0.75)^2 + \frac{0.2745}{6}(x - 0.75)^3 + \frac{0.348}{24}(x - 0.75)^4
\]
Simplify coefficients:
- \( \frac{F''(a)}{2} = -0.6979 \)
- \( \frac{F'''(a)}{6} = 0.04575 \)
- \( \frac{F^{(4)}(a)}{24} = 0.0145 \)
Thus,
\[
L_4(x) \approx 5.583 - 1.098(x - 0.75) - 0.6979 (x - 0.75)^2 + 0.04575 (x - 0.75)^3 + 0.0145 (x - 0.75)^4
\]
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Step 5: Approximate Function Values Using \( L_4(x) \)
To estimate \( F(x) \) over the interval, compute \( L_4(x) \) at various points \( x \in [0.5, 1] \).
Example: \( x = 0.5 \)
Calculate \( x - 0.75 = -0.25 \):
\[
L_4(0.5) \approx 5.583 - 1.098(-0.25) - 0.6979(0.0625) + 0.04575(-0.015625) + 0.0145(0.00390625)
\]
Compute step-by-step:
- \( -1.098 \times -0.25 = 0.2745 \)
- \( -0.6979 \times 0.0625 = -0.0436 \)
- \( 0.04575 \times -0.015625 = -0.000715 \)
- \( 0.0145 \times 0.00390625 = 0.000057 \)
Sum:
\[
L_4(0.5) \approx 5.583 + 0.2745 - 0.0436 - 0.000715 + 0.000057 \approx 5.583 + 0.2745 - 0.0436 - 0.000715 + 0.000057 \approx 5.814
\]
Compare with the actual function value:
\[
F(0.5) = 6 \cos(0.25) \approx 6 \times 0.9689 = 5.813
\]
The approximation is remarkably close.
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Step 6: Error Estimation and Remainder Term
The accuracy of the Taylor polynomial depends on the remainder \( R_4(x
