Question

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

Short answer

\tan^{-1}(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}.

Step-by-step solution

Understand the given function

The function given is \(\tan^{-1}(x)\), which is expressed as an integral: \(\tan^{-1}(x) = \int_{0}^{x} \frac{1}{1+u^2} \, du\). We'll find the Maclaurin series for \(\tan^{-1}(x)\).

Recall the geometric series expansion

The geometric series expansion for \(\frac{1}{1 - (-u^2)}\) is \(\frac{1}{1 - (-u^2)} = 1 + u^2 + u^4 + u^6 + \cdots\).

Substitute \(u^2\) in the series

Replace \(-u^2\) with \(u^2\) in the series to get: \(\frac{1}{1+u^2} = 1 - u^2 + u^4 - u^6 + \cdots\).

Integrate term by term

Integrate \(\frac{1}{1+u^2} = 1 - u^2 + u^4 - u^6 + \cdots\) term by term: \[ \int_{0}^{x}(1 - u^2 + u^4 - u^6 + \cdots) \, du = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \].

Write the Maclaurin series

The Maclaurin series for \( \tan^{-1}(x) \) is \(\tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}\).

Key concepts

Arctangent Function

The arctangent function, denoted as \(\tan^{-1}(x)\), is the inverse of the tangent function. This means that it undoes the action of the tangent. If y = tan(x), then x = tan^{-1}(y). The Maclaurin series for arctangent can be derived using integral representation. It relates to finding the integral of \(\frac{1}{1 + u^2}\) from 0 to x.
The function helps to understand arc lengths in geometric scenarios, and it's often used in trigonometry.
For small values of x, we can use the Maclaurin series to approximate \(\tan^{-1}(x)\).

Geometric Series

A geometric series is a series of the form \(\frac{1}{1 - r} = 1 + r + r^2 + r^3 + \cdots \) when \(|r| < 1\). For our case, \(\frac{1}{1 + u^2}\) can be rewritten using a geometric series by substituting \(-u^2\) for r. This gives us:

• \(\frac{1}{1 + u^2} = 1 - u^2 + u^4 - u^6 + \cdots \).

The geometric series helps simplify complex functions into an infinite sum of simpler terms that we can easily manage.

Term-by-Term Integration

When integrating a series term-by-term, each term of the series is integrated separately. This is allowed under certain conditions, ensuring the process remains valid. For the function \(\frac{1}{1 + u^2}\), we integrate each term of its geometric series expansion individually:
\(\frac{1}{1 + u^2} = 1 - u^2 + u^4 - u^6 + \cdots\)\[\int (1 - u^2 + u^4 - u^6 + \cdots) du = u - \frac{u^3}{3} + \frac{u^5}{5} - \frac{u^7}{7} + \cdots\]Bringing back the limits of integration from 0 to x, we get:\[\int_{0}^{x} (1 - u^2 + u^4 - u^6 + \cdots) du = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\]. This gives us the Maclaurin series of the integrated function.

Series Expansion

Series expansion involves expressing a function as an infinite sum of terms. These terms are computed from the function's derivatives at a single point.
For Maclaurin series, the point of expansion is 0. The series expansion of \(\tan^{-1}(x)\) was derived as:
\[\tan^{-1}(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}\]Here, each term differs by a pattern involving alternating signs and odd powers of x divided by their index.

• This expansion makes it easy to compute \(\tan^{-1}(x)\) for small values of x without using a calculator.
• This method is powerful in approximating functions analytically in calculus and analysis.

Integral Representation

Integral representation means expressing a function as an integral. This is useful because sometimes integrating a function term-by-term can simplify our calculations. For \(\tan^{-1}(x)\), its integral representation is:
\[\arctan(x) = \int_{0}^{x} \frac{du}{1 + u^2}\]

• This form helps in deriving the Maclaurin series for the arctangent function.
• We transformed the problem into integrating the series we derived from the geometric series representation.

By integrating the power series term-by-term, we found the Maclaurin series of the function. Integral representation is a powerful tool in mathematical analysis and helps us evaluate complex functions in simpler forms.