Chapter 10: Problem 12
Use power series operations to find the Taylor series at \(x=0\) for the functions. $$x^{2} \sin x$$
Short Answer
Expert verified
The Taylor series for \( x^2 \sin x \) is \( x^3 - \frac{1}{6}x^5 + \frac{1}{120}x^7 + \cdots \).
Step by step solution
01
Recall the Taylor Series for Sine Function
The Taylor series for the sine function around zero is given by the formula: \( \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \). This series expansion will be useful in finding the Taylor series for \( x^2 \sin x \).
02
Multiply the Sine Series by \(x^2\)
The function we are interested in is \( x^2 \sin x \). We multiply each term in the series expansion for \( \sin x \) by \( x^2 \). This gives us: \( x^2 \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+3} \).
03
Write Terms of the New Series
Start substituting the first few values of \( n \) into the series to develop the Taylor series:- For \( n = 0 \): \( \frac{(-1)^0}{1!} x^{3} = x^3 \)- For \( n = 1 \): \( \frac{(-1)^1}{3!} x^{5} = -\frac{1}{6} x^5 \)- For \( n = 2 \): \( \frac{(-1)^2}{5!} x^{7} = \frac{1}{120} x^7 \)Continuing this pattern, the terms start to emerge.
04
Construct the Taylor Series
Based on the terms from Step 3, the Taylor series up to the first three non-zero terms for \( x^2 \sin x \) is: \[ x^2 \sin x = x^3 - \frac{1}{6} x^5 + \frac{1}{120} x^7 + \cdots \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series is a special type of infinite series that is expressed in the form:
\[ \sum_{n=0}^{\infty} a_{n} (x-c)^n \]
Here, \( a_n \) are coefficients, \( x \) is a variable, and \( c \) is a constant called the center of the series.
\[ \sum_{n=0}^{\infty} a_{n} (x-c)^n \]
Here, \( a_n \) are coefficients, \( x \) is a variable, and \( c \) is a constant called the center of the series.
- If \( c = 0 \), the power series is centered at zero, which is also known as a Maclaurin series.
- This structure allows us to represent complex functions through simpler polynomial forms.
- The series converges within a certain radius of convergence, which depends on the specific function and the center \( c \).
Sine Function
The sine function, denoted as \( \sin x \), is one of the fundamental trigonometric functions. Its values are derived from the ratios of sides in a right triangle or from unit circle coordinates. The sine function has a distinctive wave-like pattern that repeats every \( 2\pi \) radians.
- Its Maclaurin series (a special case of the Taylor series at \( x = 0 \)) is given by the alternating infinite series:
\[ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} \] - This series is characterized by only odd powers of \( x \) because the sine function's derivatives at zero yield alternating positives and negatives.
- The series perfectly replicates the \( \sin x \) curve when plotted, offering a way to calculate sine values using basic arithmetic operations without a calculator.
Series Expansion
Series expansion involves expressing a function as the sum of terms of a series. This approach provides a polynomial approximation to more complex functions, making calculations easier.
- The Taylor series is a type of series expansion centered at a point \( c \), allowing one to approximate a function using its derivatives at that point. A Taylor series at \( x = 0 \) is called a Maclaurin series, like the sine series.
- Expanding functions transforms them into a form with powers of \( x \) and easily computable coefficients. For example, multiplying a series by \( x^2 \) as seen in \( x^2 \sin x \) means adjusting the powers of \( x \) without changing the alternating pattern of coefficients.
- Recognizing patterns, such as alternating signs or incremental changes in powers and coefficients, simplifies creating a series expansion of any function.