# Was ist die Maclaurin-Reihe von f (x) = sin ^ 2 (x) ?

Die Maclaurin-Serie ist nur der Spezialfall für die Taylor-Serie 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 + ...

Also sollten wir nehmen n Ableitungen, bis wir ein Muster finden.

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))

Ich denke, das ist ungefähr so ​​weit, wie wir gehen müssen. Mal sehen, was wir bekommen!

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 - ...)