Chapter 10: Problem 2
Find several terms in the power series expansion of \(e^{\sin x}\) by a method different from that in this section.
Short Answer
Expert verified
The series expansion of \( e^{\sin x} \) is approximately \( 1 + x + \frac{x^2}{2} - \frac{x^4}{4} \).
Step by step solution
01
Recognize the Exponential Power Series
The exponential function can be represented as a power series: \[ e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots \]For our problem, substitute \( y = \sin x \). Therefore:\[ e^{\sin x} = 1 + \sin x + \frac{(\sin x)^2}{2!} + \frac{(\sin x)^3}{3!} + \cdots \]
02
Expand Sin(x) Using Taylor Series
The sine function \( \sin x \) can be expanded using its Taylor series:\[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]We will use this series to find terms of \( e^{\sin x} \).
03
Plug the Taylor Series of Sin(x) into the Exponential Series
Substitute the Taylor series of \( \sin x \) into the series for \( e^{\sin x} \):\[ e^{\sin x} = 1 + (x - \frac{x^3}{3!} + \cdots) + \frac{(x - \frac{x^3}{3!} + \cdots)^2}{2!} + \cdots \]
04
Compute the First Few Terms
Calculate a few terms from the expansions:- First term: \( 1 \)- Second term: \( x \)- Third term coming from \( \frac{(\sin x)^2}{2!} \) involves squaring the sine series and then dividing by 2:\( \frac{(x - \frac{x^3}{6})^2}{2!} = \frac{x^2}{2} - \frac{x^4}{4} + \cdots \) Summing these gives the approximated expanded series.
05
Combine Results
Combine the results from the previous calculations to get the first few non-zero terms:\[ e^{\sin x} \approx 1 + x + \frac{x^2}{2} - \frac{x^4}{4} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential Function
The exponential function, denoted as \( e^y \), is a fundamental mathematical function that can be expanded into a power series. This is particularly helpful in calculus as it allows us to express functions in a form that is easier to manipulate mathematically. The exponential function has the remarkable property of being equal to its own derivative, and it can be represented as:
- \( e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \cdots \)
Taylor Series
A Taylor series is a powerful mathematical tool used to approximate functions by infinite sums of their derivatives evaluated at a point. It provides a polynomial approximation of a function around a particular point, commonly chosen to be zero, known as a Maclaurin series. The general form of a Taylor series expansion for a function \( f(x) \) around zero is:
- \( f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \cdots \)
- \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)
Sine Function
The sine function, \( \sin x \), is one of the fundamental trigonometric functions that appears frequently in mathematics, particularly in calculus and physics. The Taylor series expansion of \( \sin x \) gives:
- \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \)