Question

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

Answer 1

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^((0))(x) = color(green)(f(x) = sin^2x)`
`color(green)(f'(x) = 2sinxcosx)`
`color(green)(f''(x)) = 2[sinx*-sinx + cosx*cosx] = 2(cos^2x - sin^2x)`
`= color(green)(2cos(2x))`
`color(green)(f'''(x)) = 2*-sin(2x)*2 = color(green)(-4sin(2x))`
`color(green)(f''''(x)) = -4cos(2x)*2 = color(green)(-8cos(2x))`
`color(green)(f'''''(x)) = 8sin(2x)*2 = color(green)(16sin(2x))`
`color(green)(f''''''(x)) = 16cos(2x)*2 = color(green)(32cos(2x))`

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 + 2/2 x^2 + 0 + (-8)/(24)x^4 + 0 + 32/720 x^6 ...`

`= color(blue)(x^2 - x^4/3 + 2/45 x^6 - ...)`