Question

Find the Maclaurin series for the following functions. \(\arctan x=\int_{0}^{x} \frac{d u}{1+u^{2}}\)

Short answer

The Maclaurin series for \( \tan^{-1}(x) \) is \[ x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \cdots \].

Step-by-step solution

Understand the function

Given the function \(\frac{1}{1+u^2}\), which is the integrand in the provided integral. The goal is to find the Maclaurin series for the function \(\tan^{-1}(x) = \int_{0}^{x} \frac{du}{1+u^{2}}\).

Expand the integrand as a series

Expand \(\frac{1}{1+u^2}\) as a geometric series around 0: \( \frac{1}{1+u^2} = \sum_{n=0}^{\text{inf}} (-1)^n u^{2n}\).

Integrate each term of the series

Integrate term-by-term within 0 and x: \( \tan^{-1}(x) = \int_0^x \sum_{n=0}^{\text{inf}} (-1)^n u^{2n} \, du\) which simplifies to \( \tan^{-1}(x) = \sum_{n=0}^\text{inf} (-1)^n \int_0^x u^{2n} du\).

Compute the integrals

Evaluate the integral for each term: \( \int_0^x u^{2n} du = \frac{x^{2n+1}}{2n+1}\). So the series becomes \( \tan^{-1}(x) = \sum_{n=0}^{\text{inf}} \frac{(-1)^n x^{2n+1}}{2n+1} \).

Write the Maclaurin series

Combine the results into the Maclaurin series for \( \tan^{-1}(x) \): \[ \tan^{-1}(x) = x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \cdots \].

Key concepts

arctan function

The arctan function, also known as the inverse tangent function, is represented as \( \tan^{-1}(x) \). This function gives the angle whose tangent is the number x. It is an important function in trigonometry and calculus, especially when working with integrations and series expansions. One interesting fact about \(\tan^{-1}(x)\) is that it can be expressed as an integral: \[ \arctan(x) = \int_{0}^{x} \frac{du}{1+u^2} \]. This means that instead of evaluating the arctan function directly, we can understand and compute it through its integral representation.

geometric series

A geometric series is a series where each term is a constant multiple of the previous term. For example, in the series \(1 + x + x^2 + x^3 + \ldots\), each term is multiplied by x. To solve the given problem, we can represent \(\frac{1}{1+u^2} \) as a geometric series. Geometric series can be summed using the formula: \[ \frac{1}{1-r} = \sum_{n=0}^{\infty} r^n \], for \(|r| < 1\). In this exercise, by setting \(r = -u^2\), we convert \(\frac{1}{1+u^2} \) into a series: \[ \frac{1}{1+u^2} = \sum_{n=0}^{\infty} (-1)^n u^{2n} \]. This series expansion is an essential step towards finding the Maclaurin series for \(\tan^{-1}(x)\).

term-by-term integration

Term-by-term integration involves integrating each term of a series individually. For the function \(\arctan(x)\), we already expressed \(\frac{1}{1+u^2}\) as a series: \[ \frac{1}{1+u^2} = \sum_{n=0}^{\infty} (-1)^n u^{2n} \]. To find the Maclaurin series, integrate this series term by term: \[ \tan^{-1}(x) = \int_{0}^{x} \sum_{n=0}^{\infty} (-1)^n u^{2n} du \]. We can change the order of integration and summation because the series converges uniformly. Therefore, we get: \[ \tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \int_0^x u^{2n} du \]. By integrating each term, we evaluate \(\int_0^x u^{2n} du \) to get \(.\frac{x^{2n+1}}{2n+1} \).

power series expansion

A power series is an infinite series of the form \[ f(x) = \sum_{n=0}^{\infty} a_n x^n, \] where \(a_n\rm s\) are constants. The Maclaurin series is a special type of power series centered at \(x=0\). To find the power series expansion of \(\tan^{-1}(x)\), we start from its term-by-term integrated series and obtain: \[ \tan^{-1}(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \]. This series simplifies to the Maclaurin series for the arctan function: \[ \tan^{-1}(x) = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \]. Each term represents increasing powers of x, with alternating signs and decreasing coefficients.