Problem 3
Question
Find the Maclaurin polynomial of order 4 for \(f(x)\) and use it to approximate \(f(0.12) .\) $$ f(x)=\sin 2 x $$
Step-by-Step Solution
Verified Answer
The Maclaurin polynomial of order 4 is \(2x - \frac{4x^3}{3}\), and \(f(0.12) \approx 0.237696\).
1Step 1: Understand the Maclaurin Series
The Maclaurin series for a function \(f(x)\) is a Taylor series about \( x = 0 \). It is given by: \[ f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f^{(4)}(0)x^4}{4!} + \cdots \] For this problem, we need terms up to \(x^4\).
2Step 2: Differentiate the Function
Calculate the first four derivatives of \(f(x) = \sin(2x)\):\[ f'(x) = 2\cos(2x), \quad f''(x) = -4\sin(2x), \quad f'''(x) = -8\cos(2x), \quad f^{(4)}(x) = 16\sin(2x) \]
3Step 3: Evaluate the Derivatives at x = 0
Substitute \(x = 0\) into each derivative:\[ f(0) = \sin(0) = 0, \quad f'(0) = 2\cos(0) = 2, \quad f''(0) = -4\sin(0) = 0, \quad f'''(0) = -8\cos(0) = -8, \quad f^{(4)}(0) = 16\sin(0) = 0 \]
4Step 4: Construct the Maclaurin Polynomial
Using the derivatives evaluated at 0, construct the polynomial:\[ P_4(x) = 0 + 2\cdot x + 0\cdot \frac{x^2}{2} + (-8)\cdot \frac{x^3}{6} + 0\cdot \frac{x^4}{24} \]Simplify: \[ P_4(x) = 2x - \frac{4x^3}{3} \]
5Step 5: Approximate f(0.12)
Substitute \(x = 0.12\) into the polynomial:\[ P_4(0.12) = 2(0.12) - \frac{4(0.12)^3}{3} \]Calculate each term:\[ = 0.24 - \frac{4 \times 0.001728}{3} \]\[ = 0.24 - 0.002304 \]\[ = 0.237696 \]
Key Concepts
Taylor seriesPolynomial approximationTrigonometric functionsDifferentiation
Taylor series
The Taylor series is a mathematical tool that represents a function as an infinite sum of terms. Each term is derived from the derivatives of the function at a single point. For a function centered at zero, this series is known as the Maclaurin series.
This series allows us to approximate complex functions using simpler polynomial expressions. Taylor series can transform intricate function behaviors into a straightforward summation of polynomial terms:
This series allows us to approximate complex functions using simpler polynomial expressions. Taylor series can transform intricate function behaviors into a straightforward summation of polynomial terms:
- It is versatile and can approximate any smooth and continuous function.
- The convergence of the series depends on the function itself.
- This method calculates each term using derivatives of increasing order.
Polynomial approximation
Polynomial approximation involves using a polynomial to estimate a function. This approximation is beneficial, especially when dealing with functions that are difficult to analyze or compute directly.
Through polynomial approximation, complex functions, such as trigonometric or exponential ones, become more manageable. This is achieved by using truncated versions of Taylor or Maclaurin series, which can be calculated by considering only a finite number of terms from these series. Specifically for this exercise:
Through polynomial approximation, complex functions, such as trigonometric or exponential ones, become more manageable. This is achieved by using truncated versions of Taylor or Maclaurin series, which can be calculated by considering only a finite number of terms from these series. Specifically for this exercise:
- An order 4 Maclaurin polynomial provides a balance between simplicity and accuracy.
- Higher order polynomials yield better approximations over a wider range.
- For small input values, lower order terms significantly contribute to the approximation.
Trigonometric functions
Trigonometric functions, like sine and cosine, are fundamental in various fields of mathematics and science due to their periodic nature.
In this scenario, we have the function \(f(x) = \sin 2x\), which is a sine function with altered frequency. To apply polynomial approximations effectively:
In this scenario, we have the function \(f(x) = \sin 2x\), which is a sine function with altered frequency. To apply polynomial approximations effectively:
- Trigonometric functions often require polynomial representation for ease of calculation.
- Derivatives of these functions are cyclic and follow specific patterns.
- As derivatives of sine functions are linear combinations of sine and cosine, they simplify the process of generating polynomial terms.
Differentiation
Differentiation refers to the process of finding a derivative, signifying how a function changes at any given point.
In constructing a Maclaurin series, derivatives of the function at the point of interest are paramount. For the function \(f(x) = \sin 2x\), the process involves computing up to the fourth derivative:
In constructing a Maclaurin series, derivatives of the function at the point of interest are paramount. For the function \(f(x) = \sin 2x\), the process involves computing up to the fourth derivative:
- The first derivative, \(f'(x) = 2\cos(2x)\), indicates the rate of change of the sine function.
- Subsequent derivatives exhibit alternating patterns, reflecting the inherent properties of trigonometric functions.
- Evaluating these derivatives at \(x = 0\) simplifies calculations as many terms tend to zero.
Other exercises in this chapter
Problem 2
Indicate whether the given series converges or diverges. If it converges, find its sum. Hint: It may help you to write out the first few terms of the series $$
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An explicit formula for \(a_{n}\) is given. Write the first five terms of \(\left\\{a_{n}\right\\}\), determine whether the sequence converges or diverges, and,
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Find the terms through \(x^{5}\) in the Maclaurin series for \(f(x) .\) Hint: It may be easiest to use known Maclaurin series and then perform multiplications,
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find the power series representation for \(f(x)\) and specify the radius of convergence. Each is somehow related to a geometric series. $$ f(x)=\frac{1}{(1-x)^{
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