Question

Use the binomial series to find the Maclaurin series for the function. $$ f(x)=\sqrt{1+x^{2}} $$

Short answer

The Maclaurin series for the function \(f(x)=\sqrt{1+x^{2}}\) is \(f(x) = 1 + \sum_{n=1}^{\infty} (-1)^n \frac{(2n)!}{4^n (n!)^2} x^{2n}\)

Step-by-step solution

Use Binomial Theorem

Apply the binomial series theorem for any \(|x|<1\), \((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^{2} + \frac{n(n-1)(n-2)}{3!}x^{3} + \dots\). When there is a square root or any fractional exponent, the formula \((1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^{2} + \frac{n(n-1)(n-2)}{3!}x^{3} + \dots\) where \(n = \frac{1}{2}\) is used. In this case, \(f(x)=\sqrt{1+x^{2}}\) can be written as \(f(x)=(1+x^{2})^{1/2}\).

Expand the Function using the Binomial Theorem

Using the binomial theorem to expand \(f(x)=(1+x^{2})^{1/2}\) results in \(f(x) = 1 + \frac{1}{2}x^2 - \frac{1}{8}x^4 + \frac{1}{16}x^6 - \dots\). This is a series centered at 0, also known as a Maclaurin series.

General Form of the Maclaurin Series

In general form, the Maclaurin series can be written as \(f(x) = 1 + \sum_{n=1}^{\infty} (-1)^n \frac{(2n)!}{4^n (n!)^2} x^{2n}\). It helps not only to identify the pattern but also to state the Maclaurin series in a more compact way.