Chapter 11: Problem 8
Find Taylor series at \(x=0\) for the functions in Exercises \(7-18\) $$ x^{2} \sin x $$
Short Answer
Expert verified
The Taylor series for \(x^2 \sin x\) at \(x=0\) is \(x^3 - \frac{x^5}{6} + \frac{x^7}{120} + \cdots\).
Step by step solution
01
Identify the Function
The function given is \(x^2 \sin x\). We need to find its Taylor series expansion around \(x=0\).
02
Recall the Taylor Series Definition
The Taylor series of a function \(f(x)\) at \(x=0\) is given by \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\), where \(f^{(n)}(0)\) is the \(n\)-th derivative of \(f(x)\) evaluated at \(x=0\).
03
Find the Taylor Series of \(\sin x\)
Since \(\sin x\) is a well-known function, recall its Taylor series around 0: \(\sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\).
04
Multiply by \(x^2\)
To find the series for \(x^2 \sin x\), multiply each term in the \(\sin x\) series by \(x^2\):\[x^2 \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+3}}{(2n+1)!}\].
05
Write the Series Expansion
Write out the series expansion up to the first few terms:- First term: \(n=0\), \((-1)^0 \frac{x^3}{1!} = \frac{x^3}{1}\)- Second term: \(n=1\), \((-1)^1 \frac{x^5}{3!} = -\frac{x^5}{6}\)- Third term: \(n=2\), \((-1)^2 \frac{x^7}{5!} = \frac{x^7}{120}\)So, the series is \(x^3 - \frac{x^5}{6} + \frac{x^7}{120} + \cdots\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Maclaurin series
A Maclaurin series is a special type of Taylor series that specifically represents a function as an infinite sum of terms calculated from the function's derivatives at a single point, usually at 0. When discussing Maclaurin series, we are essentially discussing Taylor series centered at zero.
The general form of a Maclaurin series for a function \( f(x) \) is:
This formula provides a way to approximate complex functions with polynomials, which are easier to work with.
For example, the Maclaurin series for \( \sin x \) is:
The general form of a Maclaurin series for a function \( f(x) \) is:
- \( f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \)
This formula provides a way to approximate complex functions with polynomials, which are easier to work with.
For example, the Maclaurin series for \( \sin x \) is:
- \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \)
Power series
Power series provide a powerful language for expressing functions as infinite sums of powers of variables. They take the form:
If \( a = 0 \), the series is called a Maclaurin series, which we've explored earlier.
Power series allow us to work with functions through their polynomial representation, providing an approximate solution over a range.In practical applications, particularly in calculus and analysis, power series enable solving differential equations and integrating complex functions easily.
For the exercise function \( x^2 \sin x \), we used the power series expansion of \( \sin x \) and then multiplied through by \( x^2 \) to get our new series:
- \( \sum_{n=0}^{\infty} c_n (x-a)^n \)
If \( a = 0 \), the series is called a Maclaurin series, which we've explored earlier.
Power series allow us to work with functions through their polynomial representation, providing an approximate solution over a range.In practical applications, particularly in calculus and analysis, power series enable solving differential equations and integrating complex functions easily.
For the exercise function \( x^2 \sin x \), we used the power series expansion of \( \sin x \) and then multiplied through by \( x^2 \) to get our new series:
- \( x^2 \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+3}}{(2n+1)!} \)
Trigonometric series
Trigonometric series are power series that involve trigonometric functions like sine and cosine. They are very useful because they provide a way to express periodic functions in terms of simpler trigonometric functions.
A classic example is the sine function, which can be expressed using its well-known Maclaurin series:
In the exercise of finding the expansion for \( x^2 \sin x \), we used the already established trigonometric series for \( \sin x \) and systematically incorporated \( x^2 \).
By doing so, the function \( x^2 \sin x \) can be expressed in terms of its trigonometric components:
A classic example is the sine function, which can be expressed using its well-known Maclaurin series:
- \( \sin x = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!} \)
In the exercise of finding the expansion for \( x^2 \sin x \), we used the already established trigonometric series for \( \sin x \) and systematically incorporated \( x^2 \).
By doing so, the function \( x^2 \sin x \) can be expressed in terms of its trigonometric components:
- \( x^3 - \frac{x^5}{6} + \frac{x^7}{120} + \cdots \)