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Magazine Chapter 9: Problem 25
Plot on the same axes the given function along with the Maclaurin polynomials of orders \(1,2,3\), and \(4 .\) $$ \sin e^{x} $$
Short Answer
Expert verified
Plot \( \sin(e^x) \) and its Maclaurin polynomials up to 4th order on a common graph.
Step by step solution
01
Understanding the Function
The function we need to plot is \( \sin(e^x) \). A Maclaurin series is a type of Taylor series expansion at \( x = 0 \). For \( \sin(e^x) \), we will expand its Maclaurin series up to the 4th order polynomial. Let's start by recalling the concept of Maclaurin series.
02
Derive Maclaurin Series
A Maclaurin polynomial for a function \( f(x) \) is given by:\[ f(x) \approx f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \]We need to calculate the derivatives of \( \sin(e^x) \) and evaluate them at \( x = 0 \).
03
Zero-order Polynomial
The zero-order Maclaurin polynomial is just the value of the function at \( x = 0 \).\[ f(0) = \sin(e^0) = \sin(1) \approx 0.8415 \]
04
First-order Polynomial
The first derivative of \( f(x) = \sin(e^x) \) is:\[ f'(x) = \cos(e^x) \cdot e^x \]Evaluating at \( x = 0 \):\[ f'(0) = \cos(e^0) \cdot e^0 = \cos(1) \cdot 1 \approx 0.5403 \]Thus, the first-order polynomial is:\[ P_1(x) = 0.8415 + 0.5403x \]
05
Second-order Polynomial
The second derivative is:\[ f''(x) = -\sin(e^x) \cdot e^{2x} + \cos(e^x) \cdot e^x \]Evaluating at \( x = 0 \):\[ f''(0) = -\sin(1) \cdot 1^2 + \cos(1) \cdot 1 \approx -0.8415 + 0.5403 = -0.3012 \]The second-order polynomial is:\[ P_2(x) = 0.8415 + 0.5403x - 0.1506x^2 \]
06
Third-order Polynomial
The third derivative will be:\[ f'''(x) = -\cos(e^x) \cdot e^{3x} + 2\sin(e^x) \cdot e^{2x} \]Evaluate at \( x = 0 \):\[ f'''(0) = -\cos(1) + 2\sin(1) \approx -0.5403 + 2 \cdot 0.8415 = 1.1427 \]The third-order polynomial is:\[ P_3(x) = 0.8415 + 0.5403x - 0.1506x^2 + \frac{1.1427}{6}x^3 \approx 0.8415 + 0.5403x - 0.1506x^2 + 0.1905x^3 \]
07
Fourth-order Polynomial
Calculate the fourth derivative:\[ f^{(4)}(x) = \sin(e^x) \cdot e^{4x} - 3\cos(e^x) \cdot e^{3x} + 3\sin(e^x) \cdot e^{2x} \]Evaluate at \( x = 0 \):\[ f^{(4)}(0) = \sin(1) - 3\cos(1) + 3\sin(1) \approx 4 \cdot 0.8415 - 3 \cdot 0.5403 = 1.6058 \]The fourth-order polynomial is:\[ P_4(x) = 0.8415 + 0.5403x - 0.1506x^2 + 0.1905x^3 + \frac{1.6058}{24}x^4 \approx 0.8415 + 0.5403x - 0.1506x^2 + 0.1905x^3 + 0.0669x^4 \]
08
Plot the Function and Polynomials
To visualize, plot \( \sin(e^x) \) along with its Maclaurin polynomials \( P_1(x) \), \( P_2(x) \), \( P_3(x) \), and \( P_4(x) \) on the same graph. Use a range for \( x \) such as \([-1, 1]\) to see how well the polynomials approximate the function near the origin.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Taylor series
The Taylor series is a powerful mathematical tool used to approximate functions by means of infinite sums of their derivatives at a specific point, typically a point where the function is easy to compute, like zero. In mathematical terms, the Taylor series of a function \( f(x) \) around a point \( a \) is given by:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \dots \]
When the point is specifically chosen as zero, the series is also known as a Maclaurin series.
When the point is specifically chosen as zero, the series is also known as a Maclaurin series.
- The Taylor series allows us to express functions analytically, leveraging the simplicity of polynomial expressions.
- For many practical purposes, only a finite number of terms is used, resulting in a polynomial that approximates the function.
polynomial approximation
Polynomial approximation is the process of estimating a function using polynomials. This technique is especially useful when dealing with functions that are complicated to compute directly. The key is to replace the function with a polynomial that closely imitates its behavior over a certain range. Maclaurin polynomials, derived from Taylor series centered at zero, are excellent for this purpose.
Why use polynomials?
Why use polynomials?
- Polynomials are simpler to handle as they involve only basic arithmetic operations.
- Computing powers and coefficients of polynomials is straightforward.
- They can be used to approximate functions near a point without computing complex derivatives for every operation.
sin function
The sine function, \( \sin(x) \), is a fundamental trigonometric function with numerous applications in mathematics, physics, and engineering. Its Maclaurin series, which offers a polynomial representation, is given as:\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \]
This series captures the oscillatory nature of the sine function around the origin.
This series captures the oscillatory nature of the sine function around the origin.
- The sine function is periodic with a period of \(2\pi\).
- Its value oscillates between \(-1\) and \(1\).
- When compounded with other functions, like the exponential function, the resulting expressions can be tackled using polynomial approximations.
exponential function
The exponential function \( e^x \) is a mathematical constant and function that is pivotal in describing growth processes. It is continuously increasing, with its Maclaurin series defined by:\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \]
This series results in a smooth and fast-growing function.
This series results in a smooth and fast-growing function.
- The function \( e^x \) has a growth rate proportional to its value, making it unique in calculus.
- When used in conjunction with trigonometric functions like sine, it yields complex behaviors and intricate plots.
- Polynomial approximations of \( \sin(e^x) \) combine both exponential and trigonometric features into simpler forms for easier computation and visualization.
