Chapter 11: Problem 26
\(23-26\) Evaluate the indefinite integral as a power series. What is the radius of convergence? $$ \int \tan ^{-1}\left(x^{2}\right) d x $$
Short Answer
Expert verified
The radius of convergence is 1.
Step by step solution
01
Express the Function as a Power Series
To evaluate the indefinite integral \( \int \tan^{-1}(x^2) \, dx \) as a power series, first express \( \tan^{-1}(x) \) as a power series. We have:\[\tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}, \quad |x| \leq 1.\]For \( \tan^{-1}(x^2) \), substitute \( x^2 \) for \( x \):\[\tan^{-1}(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1}.\]
02
Integrate the Power Series Term by Term
Now integrate the power series term by term:\[\int \tan^{-1}(x^2) \, dx = \int \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1} \right) \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int x^{4n+2} \, dx.\]Integrating each term: \[\int x^{4n+2} \, dx = \frac{x^{4n+3}}{4n+3} + C.\]So the integrated series is:\[\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \cdot \frac{x^{4n+3}}{4n+3} + C.\]
03
Determine the Radius of Convergence
To find the radius of convergence of the power series, we need to look at the original series we used for \( \tan^{-1}(x^2) \), which has the convergence condition based on the geometric series\[\left| x^2 \right| \leq 1 \Rightarrow |x|^2 \leq 1 \Rightarrow |x| \leq 1.\]Thus, the radius of convergence is \( R = 1 \).
04
Write the Final Expression for the Integral
The integral expressed as a power series is:\[\int \tan^{-1}(x^2) \, dx = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{(2n+1)(4n+3)} + C,\]where \( C \) is the constant of integration.
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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 essentially an infinite sum of terms involving powers of a particular variable. In this context, a power series can be used to represent complicated functions in a simpler form. For example, consider the function \( an^{-1}(x)\). We can express it as a power series:
The power series representation lets us explore different functions by substituting variables. For instance, in the case of \( an^{-1}(x^2)\), we substitute \(x^2\) for \(x\), leading to a new series representation. Using power series expansions helps us turn complex integrals into manageable algebraic sums.
- \( an^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \)
The power series representation lets us explore different functions by substituting variables. For instance, in the case of \( an^{-1}(x^2)\), we substitute \(x^2\) for \(x\), leading to a new series representation. Using power series expansions helps us turn complex integrals into manageable algebraic sums.
Radius of Convergence
The radius of convergence is a key concept that tells us the range of values for which a power series converges to a function. When working with series like that of \( an^{-1}(x)\), we need to determine where the series will accurately represent the function.To identify the radius of convergence for a power series, we typically look at the original series and apply certain convergence tests like the ratio test. In the case of \(\tan^{-1}(x^2)\), its power series shares the same convergence condition as its original form:
- \( |x| \leq 1 \)
Term-by-Term Integration
Term-by-term integration is a technique that allows us to integrate each term of a power series individually. This can dramatically simplify the process of finding the integral of a complex function.When dealing with the series for \( an^{-1}(x^2)\), we can integrate the series one term at a time. If \(\sum_{n=0}^{\infty} a_n x^n \) is the power series representation of a function, its integral is obtained as:
- \( \sum_{n=0}^{\infty} \int a_n x^n \, dx = \sum_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1} + C \)