Chapter 30: Problem 35
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:
- \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots \)
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:
- \( L(x) = f(a) + f'(a)(x-a) \)
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: \( f'''(x) = 27e^{3x} \)