How do you find the Maclaurin Series for #sin(x^2)/x#?

Question

Charles Anctil · Accepted Answer

#sin(x^2)/x = sum_(n=0)^(∞) (-1)^(n) (x^(4n+1))/((2n+1)!)#

#= x/(1!) - x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!) +...#

In order to find the Maclaurin series of #(sin(x^2))/(x)#, we first need to take note of some prominent, already-derived Maclaurin series.

We know that the Maclaurin series of #sin(x)# is the following:

#sin(x) = x-(x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + ... = sum_(n=0)^(∞) (-1)^(n) (x^(2n+1))/((2n+1)!)#

Given this known Maclaurin series, we can now see that

#sin(x^2) = sum_(n=0)^(∞) (-1)^(n) ((x^2)^(2n+1))/((2n+1)!) = sum_(n=0)^(∞) (-1)^(n) (x^(4n+2))/((2n+1)!) #

Note that we still have another factor we have to consider, namely the division by #x#, which we can easily incorporate into our series.

Since we now have the series

#sin(x^2)/x = 1/x * sum_(n=0)^(∞) (-1)^(n) (x^(4n+2))/((2n+1)!)#

All we need to do is to divide by an #x#, giving us

#sum_(n=0)^(∞) (-1)^(n) (x^(4n+1))/((2n+1)!) = x/(1!) - x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!) + ...#

We can check the first #4# terms of this Maclaurin series, as shown in the graph below.

Graph of #sin(x^2)/x#
graph{sin(x^2)/x [-10, 10, -5, 5]}

Graph of #x/(1!) -x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!)#
graph{x-x^(5)/(3!)+x^(9)/(5!)-x^(13)/(7!) [-10, 10, -5, 5]}

Overlapping both graphs produces the following:

Emily Ammerman · Answer

undefined To find the Maclaurin series for $ \frac{\sin(x^2)}{x} $, follow these steps:

1. Find the Maclaurin series for $ \sin(x^2) $.
2. Divide each term of the resulting series by $ x $.
3. Simplify the series to obtain the Maclaurin series for $ \frac{\sin(x^2)}{x} $.

Let's go through these steps:

1. The Maclaurin series for $ \sin(x) $ is:
$$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$

Substitute $ x^2 $ for $ x $ to get the Maclaurin series for $ \sin(x^2) $:
$$ \sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \cdots $$

2. Divide each term by $ x $:
$$ \frac{\sin(x^2)}{x} = x - \frac{x^5}{3!} + \frac{x^9}{5!} - \frac{x^{13}}{7!} + \cdots $$

3. Simplify the series:
$$ \frac{\sin(x^2)}{x} = x - \frac{x^5}{6} + \frac{x^9}{120} - \frac{x^{13}}{5040} + \cdots $$

Therefore, the Maclaurin series for $ \frac{\sin(x^2)}{x} $ is:
$$ \frac{\sin(x^2)}{x} = x - \frac{x^5}{6} + \frac{x^9}{120} - \frac{x^{13}}{5040} + \cdots $$