How do you find the Maclaurin Series for #x^2 - sinx^2#?

Question

Emily Ammerman · Accepted Answer

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

First we must find the series for #sin(x)#

let# f(x) = sin(x) # #f(0) = sin(0) = 0# #f'(0) = cos(0) = 1# #f''(0) = -sin(0) = 0# #f'''(0) = -cos(0) = -1#

We can now utilize the Macuarin series;

#f(x) = f(0) + f'(0)x + (f''(0)x^2)/(2!) + (f'''(0)x^3)/(3!) + ...#

Hence #sin(x) = x - x^3/(3!) + x^5/(5!) - ...#

Hence for #sin(x^2)# we replace each #x# by #x^2# in the series for #sin(x)#

#sin(x^2) = (x^2) - (x^2)^3/(3!) + (x^2)^5/(5!) - ...#

# = x^2 - x^6/(3!) + x^10/(5!) - ....#

Hence #x^2 - sin(x^2) => #

#x^2 - (x^2-x^6/(3!) +x^10/(5!) - ... ) #

#=> x^6/(3!) - x^10/(5!) + ...#

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

Cameron Alexander · Answer

undefined To find the Maclaurin series for $x^2 - \sin(x^2)$, we first need to express $\sin(x^2)$ as a Maclaurin series and then subtract it from the Maclaurin series of $x^2$. The Maclaurin series for $\sin(x^2)$ is given by:

$$ \sin(x^2) = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \ldots $$

Subtracting this from the Maclaurin series of $x^2$ yields the Maclaurin series for $x^2 - \sin(x^2)$:

$$ x^2 - \sin(x^2) = x^2 - (x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \frac{x^{14}}{7!} + \ldots) $$
$$ = \frac{x^6}{3!} - \frac{x^{10}}{5!} + \frac{x^{14}}{7!} - \ldots $$

So, the Maclaurin series for $x^2 - \sin(x^2)$ is $ \frac{x^6}{3!} - \frac{x^{10}}{5!} + \frac{x^{14}}{7!} - \ldots $.