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Chapter 12: Problem 35

Estimate to within 0.01 by using series. $$\int_{0}^{1} \arctan x^{2} d x$$

Short Answer

Expert verified
The estimated value of the integral \(\int_{0}^{1} \arctan x^{2} dx\) to within 0.01 would be obtained by summing the terms of series \(\sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(2n+1)}\) until the absolute value of a term becomes less than 0.01.

Step by step solution

01

Expressing arctan(x^2) as a power series

We know that the power series for \(\arctan(x)\) is \(\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}\) for \(|x|<1\). Substituting \(x^{2}\) in place of \(x\), we get the power series for \(\arctan(x^{2})\) as \(\sum_{n=0}^{\infty} \frac{(-1)^n (x^{2})^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1}\).
02

Integrating term by term

The next step is to integrate the power series term by term from 0 to 1. The result is \(\int_{0}^{1} \arctan(x^{2}) dx = \int_{0}^{1} \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1} x^{4n+2} dx = \sum_{n=0}^{\infty} \frac{(-1)^n}{(4n+3)(2n+1)}\).
03

Estimating to within 0.01

We need to calculate terms of the series until the absolute value of the (n+1)th term is less than 0.01. The absolute error in the integral is less than the first neglected term (Weierstrass M-test).

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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 representation of a function as an infinite sum of terms, each being a product of a coefficient and a power of a variable. It is generally written in the form \[ \sum_{n=0}^{\infty} a_n x^n \] where \( a_n \) represents the coefficient for the \( n \)-th term, and \( x \) is the variable.

Power series are particularly useful because they allow for functions to be written in a form that is easy to manipulate, especially for calculation purposes. These series have a radius of convergence, within which they converge to the function they represent. For the power series of the arctangent function (\
Term by Term Integration
Term by term integration is a technique that allows us to integrate a power series by integrating each term individually. This technique is handy when dealing with an infinite series representation of a function. Mathematically, if we have a power series \[ \sum_{n=0}^{\infty} a_n x^n \], we can integrate it term by term as follows: \[ \int \sum_{n=0}^{\infty} a_n x^n dx = \sum_{n=0}^{\infty} \int a_n x^n dx \].

The result is another power series where each new coefficient is the original coefficient multiplied by the reciprocal of one plus the exponent on the variable, reflecting the rule for integrating power functions. This technique can be used as long as the series converges on the interval of integration and the function represented by the series is continuous on that interval.

Applying this to our exercise, we can integrate the series representation of the arctangent function term by term to calculate the integral of \( \arctan(x^2) \), given by: \[ \int_{0}^{1} \arctan(x^2) dx \].

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Most popular questions from this chapter

Expand \(f(x)\) in powers of \(x\) $$f(x)=x \ln \left(1+x^{3}\right)$$

Let \(\sum_{i=0}^{\infty} a_{z}\) be a convergent series and let \(R_{o}=\sum_{j=n+1}^{\infty} a_{k} .\) Prove that \(R_{n} \rightarrow 0\) as \(n \rightarrow \infty\). Note that if \(s_{n}\) is the nth partial sum of the series, then \(\sum_{k=9}^{\infty} a_{k}=s_{n}+R_{m} ; R_{n}\) is called the remainder.

Show that if \(\sum a_{2}\) is absolutely convergent and \(\left|b_{k}\right| \leq\left|a_{k}\right|\) for at \(J k,\) then \(\sum b_{k}\) is absolutely convergent.

(a) Show that if the series \(\sum a_{k}\) converges and the series \(\sum b_{k}\) diverges, then the series \(\sum\left(a_{k}+b_{k}\right)\) diverges. (b) Give examples to show that if \(\sum a_{k}\) and \(\sum b_{k}\) both diverge, then each of the series \(\sum\left(a_{k}+b_{k}\right) \quad\) and \(\sum\left(a_{k}-b_{\varepsilon}\right)\) may converge or may diverge.

Let \(\sum_{i=0}^{\infty} a_{k} x^{k}\) be a power series with radius of convergence \(r, r\) possibly infinite. (a) Giventhat \(\left|a_{k}\right|^{1 / k} \rightarrow \rho\) show that, if \(\rho: j\), then \(r=1 / \rho\) and, if \(\rho=0,\) then \(r=\infty\) (b) Given that \(\left|a_{k+1} / a_{k}\right| \rightarrow \lambda\) show that, if \(\lambda \neq 0\), then \(r=1 / \lambda\) and if \(\lambda=0,\) then \(r\) \(\infty\)
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