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Magazine Chapter 11: Problem 21
Find the Taylor series of \(x e^{x^{2}}\) at \(x=0\).
Short Answer
Expert verified
The Taylor series of \( x e^{x^2} \) at \( x = 0 \) is \( x + x^3 \).
Step by step solution
01
- Understand the Function and Taylor Series
The goal is to find the Taylor series of the function \( x e^{x^2} \) at \( x = 0 \). A Taylor series of a function \( f(x) \) at \( x = 0 \) is given by: \[ f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \frac{f^{(4)}(0)}{4!}x^4 + \text{...} \] We need to find the derivatives of \( f(x) = x e^{x^2} \) at \( x = 0 \).
02
- Calculate the Zeroth Derivative
The zeroth derivative is simply the function evaluated at \( x = 0 \.\) \[ f(0) = 0 e^{0^2} = 0 \] So, the constant term in the Taylor series is 0.
03
- Calculate the First Derivative
Use the product rule to differentiate \( f(x) = x e^{x^2} \.\) \[ f'(x) = \frac{d}{dx}\big(x e^{x^2}\big) = e^{x^2} + 2x^2 e^{x^2} \] Evaluating at \( x = 0 \), \[ f'(0) = e^{0^2} + 2(0)^2 e^{0^2} = 1 \] This gives us the coefficient for the \( x \) term.
04
- Calculate the Second Derivative
Differentiate \( f'(x) \). \[ f''(x) = \frac{d}{dx}\big(e^{x^2} + 2x^2 e^{x^2}\big) = 2x e^{x^2} + 4x e^{x^2} + 4x^3 e^{x^2} \] Simplify this, \[ f''(x) = 6x e^{x^2} + 4x^3 e^{x^2} \] Evaluating at \( x = 0 \), \[ f''(0) = 6(0) e^{0^2} + 4(0)^3 e^{0^2} = 0 \] Thus, the coefficient for the \( x^2 \) term is 0.
05
- Calculate the Third Derivative
Differentiate \( f''(x) \). \[ f'''(x) = \frac{d}{dx}\big(6x e^{x^2} + 4x^3 e^{x^2}\big) = 6 e^{x^2} + 12x^2 e^{x^2} + 12x^2 e^{x^2} + 8x^3 e^{x^2} + 12x^4 e^{x^2} \] Simplify, \[ f'''(x) = 6 e^{x^2} + 24x^2 e^{x^2} + 8x^3 e^{x^2} + 12x^4 e^{x^2} \] Evaluating at \( x = 0 \), \[ f'''(0) = 6 e^{0^2} = 6 \] So, the coefficient for the \( x^3 \) term is \ \frac{6}{3!} = 1 \.
06
- Form the Taylor Series
Combine the terms found: \[ f(x) = 0 + x + 0 x^2 + \frac{6}{3!}x^3 + \text{...} \] Simplifying, \[ f(x) = x + x^3 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Derivatives
Derivatives are mathematical tools used to determine the rate at which a function is changing at any given point. When dealing with Taylor series, derivatives play a key role because the coefficients of the series are built from them.
Here鈥檚 how:
Here鈥檚 how:
- The first derivative (\( f'(x) \)) tells us the rate of change of the function.
- The second derivative (\( f''(x) \)) gives us information about the curvature of the function.
- Higher-order derivatives provide even more detailed information about how the function behaves.
Maclaurin series
The Maclaurin series is a special case of the Taylor series, centered at 0. It鈥檚 essentially a way to approximate a function as an infinite sum of terms calculated from the values of the function鈥檚 derivatives at 0. The general form is:
\[\begin{equation} f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \text{...} \end{equation}\]
This means:
\[\begin{equation} f(x) = \frac{f(0)}{0!} + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \text{...} \end{equation}\]
This means:
- The constant term (the zeroth term) is the value of the function at 0.
- The coefficient of the x term is the function鈥檚 first derivative at 0, divided by 1!.
- The coefficient of the x2 term is the function鈥檚 second derivative at 0, divided by 2!.
- This pattern continues for all higher-degree terms.
Exponential function
The exponential function (\( e^x \)) is ubiquitous in mathematics and its derivative is remarkably simple because it鈥檚 the same as the function itself. This makes it a common subject in Taylor series problems. For example:
\[\begin{equation} e^{x^2} \text{ has derivatives that follow the pattern of multiplying by 2x each time.} \end{equation}\]
Hence, understanding the fundamental property that the exponent remains an exponent even after differentiation is key. The exponential function plays a significant role in the problem:
\[\begin{equation} e^{x^2} \text{ has derivatives that follow the pattern of multiplying by 2x each time.} \end{equation}\]
Hence, understanding the fundamental property that the exponent remains an exponent even after differentiation is key. The exponential function plays a significant role in the problem:
- \[\begin{equation} f(x) = x e^{x^2} \end{equation}\] incorporates both the linear x term and the exponential function.
- Calculating the derivatives of this, particularly because of the product rule, brings out the combined effect of these functions.
