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Solve the given problems. If \(f(x)=e^{3 x},\) compare the Maclaurin expansion with the linearization for \(a=0\)

Short Answer

Expert verified
The linearization and Maclaurin series up to the first degree term for \(e^{3x}\) at \(a=0\) are both \(1 + 3x\).

Step by step solution

01

Understand Maclaurin Series

The Maclaurin series is a special case of the Taylor series expansion about the point 0. For a function \( f(x) \), its Maclaurin series is given by \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots \). Let us apply this to \( f(x) = e^{3x} \).
02

Calculate Derivatives

Calculate the derivatives of \( f(x) = e^{3x} \):1. \( f'(x) = 3e^{3x} \)2. \( f''(x) = 9e^{3x} \)3. \( f'''(x) = 27e^{3x} \) and so on.
03

Evaluate at \(x=0\)

Substitute \( x=0 \) into each of the derivatives:1. \( f(0) = e^{0} = 1 \)2. \( f'(0) = 3e^{0} = 3 \)3. \( f''(0) = 9e^{0} = 9 \)4. \( f'''(0) = 27e^{0} = 27 \)These give the coefficients for the Maclaurin series.
04

Write Maclaurin Series

Using the coefficients, the Maclaurin series for \( f(x) = e^{3x} \) up to the quadratic term is: \[ f(x) \approx 1 + 3x + \frac{9x^2}{2!} = 1 + 3x + \frac{9}{2}x^2 \]
05

Understand Linearization

The linearization of a function around a point \( a \) is given by \( L(x) = f(a) + f'(a)(x-a) \) which simplifies to \( L(x) = f(0) + f'(0)x \) at \( a=0 \).
06

Calculate Linearization for \(e^{3x}\) at \(a=0\)

Using the values from step 3, the linearization becomes:\[ L(x) = 1 + 3x \]
07

Compare Results

Both the Maclaurin series up to the first degree term and the linearization provide: \( 1 + 3x \). This shows that up to the linear term, the Maclaurin series and linearization agree.

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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 tool in calculus used to approximate complex functions using polynomials. It generalizes the idea of representing a function as an infinite sum of terms calculated from the values of its derivatives at a single point. For any smooth function \( f(x) \) around a point \( a \), the Taylor series is:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots \)
The more terms included, the better the approximation becomes, especially near the point \( a \). A special case of the Taylor series is the Maclaurin series, where \( a = 0 \). Such series expansions are essential in mathematical analysis and are particularly useful because they simplify calculations and predictions in physics and engineering.In our example \( f(x) = e^{3x} \), the Maclaurin series is a specific case where \( a = 0 \). Differentiating repeatedly, and using those values in the series expansion formula, helps in creating a polynomial that closely mimics \( e^{3x} \) around zero.
Linearization
Linearization is a method of approximating functions using linear equations. This involves reducing a complicated function to a simple linear form, making it easier to handle for small changes around a point. The linearization of a function \( f(x) \) at a given point \( a \) is given by:
  • \( L(x) = f(a) + f'(a)(x-a) \)
This is essentially the equation of the tangent line at \( a \), which gives a straight-line approximation to the function close to \( a \). The simplicity of this approximation is why linearization is widely used in physics and engineering for analyzing systems that experience minor deviations from equilibrium.In our specific case for \( f(x) = e^{3x} \), when we compute the linearization at \( a=0 \), we find that \( L(x) = 1 + 3x \). This approximation accurately represents the function for small values of \( x \), with greater deviations as \( x \) moves away from zero.
Derivative Calculations
Calculating derivatives is a fundamental part of understanding calculus and applies directly to both Taylor series and linearization. A derivative represents the rate of change of a function's output relative to changes in its input. For functional approximation techniques such as Taylor series and linearization, derivatives provide the necessary coefficients for the polynomial expressions.To construct a Maclaurin series or find a function's linearization, you need to determine successive derivatives of the function. For our example function \( f(x) = e^{3x} \), the derivative calculations are as follows:
  • First derivative: \( f'(x) = 3e^{3x} \)
  • Second derivative: \( f''(x) = 9e^{3x} \)
  • Third derivative: \( f'''(x) = 27e^{3x} \)
Evaluating these derivatives at \( x = 0 \) gives us the coefficients: \( f(0) = 1 \), \( f'(0) = 3 \), \( f''(0) = 9 \), and so on. These coefficients are then used in the Maclaurin series expansions to approximate the function. Understanding how to derive these coefficients is crucial for anyone working with calculus-based models.

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