# What is the Maclaurin series of f(x) = sin^2(x)?

Nov 2, 2015

The Maclaurin series is just the special case for the Taylor series centered around $a = 0$.

sum_(n=1)^N (f^((n))(0))/(n!)x^n

= (f(0))/(0!)x^0 + (f'(0))/(1!)x^1 + (f''(0))/(2!)x^2 + (f'''(0))/(3!)x^3 + ...

So, we should take $n$ derivatives until we find a pattern.

${f}^{\left(0\right)} \left(x\right) = \textcolor{g r e e n}{f \left(x\right) = {\sin}^{2} x}$
$\textcolor{g r e e n}{f ' \left(x\right) = 2 \sin x \cos x}$
$\textcolor{g r e e n}{f ' ' \left(x\right)} = 2 \left[\sin x \cdot - \sin x + \cos x \cdot \cos x\right] = 2 \left({\cos}^{2} x - {\sin}^{2} x\right)$
$= \textcolor{g r e e n}{2 \cos \left(2 x\right)}$
$\textcolor{g r e e n}{f ' ' ' \left(x\right)} = 2 \cdot - \sin \left(2 x\right) \cdot 2 = \textcolor{g r e e n}{- 4 \sin \left(2 x\right)}$
$\textcolor{g r e e n}{f ' ' ' ' \left(x\right)} = - 4 \cos \left(2 x\right) \cdot 2 = \textcolor{g r e e n}{- 8 \cos \left(2 x\right)}$
$\textcolor{g r e e n}{f ' ' ' ' ' \left(x\right)} = 8 \sin \left(2 x\right) \cdot 2 = \textcolor{g r e e n}{16 \sin \left(2 x\right)}$
$\textcolor{g r e e n}{f ' ' ' ' ' ' \left(x\right)} = 16 \cos \left(2 x\right) \cdot 2 = \textcolor{g r e e n}{32 \cos \left(2 x\right)}$

I think that's about as far as we may need to go. Let's see what we get!

sum_(n=1)^6 (f^((n))(0))/(n!)x^n

= (sin^2(0))/(0!)x^0 + (2sin(0)cos(0))/(1!)x^1 + (2cos(2*0))/(2!)x^2 + (-4sin(2*0))/(3!)x^3 + (-8cos(2*0))/(4!)x^4 + (16sin(2*0))/(5!)x^5 + (32cos(2*0))/(6!)x^6 + ...

$= 0 + 0 + \frac{2}{2} {x}^{2} + 0 + \frac{- 8}{24} {x}^{4} + 0 + \frac{32}{720} {x}^{6} \ldots$

$= \textcolor{b l u e}{{x}^{2} - {x}^{4} / 3 + \frac{2}{45} {x}^{6} - \ldots}$