expand-ln-1-sin-x-right-up-to-the-term-in-x-3-

Question Number 90018 by Rio Michael last updated on 20/Apr/20

expand, \ln (1 + \sin x) right up to the term in x^{3}

Commented by mathmax by abdo last updated on 21/Apr/20

s i n x = x - \frac{x^{3}}{6} + o (x^{3}) \Rightarrow l n (1 + s i n x) = l n (1 + x - \frac{x^{3}}{6} + o (x^{3}))

{l n}^{^{'}} (1 + u) = \frac{1}{1 + u} = 1 - u + u^{2} - u^{3} + o (u^{3}) \Rightarrow

l n (1 + u) = u - \frac{u^{2}}{2} + \frac{u^{3}}{3} + o (u^{3}) \Rightarrow

l n (1 + x - \frac{x^{3}}{6}) = x - \frac{x^{3}}{6} - \frac{1}{2} {(x - \frac{x^{3}}{6})}^{2} + \frac{{(x - \frac{x^{3}}{6})}^{3}}{3} + o (x^{3})

= x - \frac{x^{3}}{6} - \frac{1}{2} (x^{2} - \frac{1}{3} x^{4} + \frac{x^{6}}{36}) + \frac{1}{3} x^{3} {(1 - \frac{x^{2}}{6})}^{3} + o (x^{3})

= x - \frac{x^{3}}{6} - \frac{1}{2} x^{2} + \frac{1}{6} x^{4} - \frac{x^{6}}{72} + \frac{1}{3} x^{3} (1 - 3 \frac{x^{2}}{6} + \dots) + o (x^{3})

= x - \frac{x^{3}}{6} - \frac{1}{2} x^{2} + \frac{1}{3} x^{3} + o (x^{3})

= x - \frac{1}{2} x^{2} + \frac{1}{6} x^{3} + o (x^{3})

Commented by Rio Michael last updated on 21/Apr/20

thanks sir

Answered by mr W last updated on 21/Apr/20

y = \ln (1 + \sin x) \Rightarrow y (0) = 0

y^{(1)} = \frac{\cos x}{1 + \sin x} \Rightarrow y^{(1)} (0) = 1

y^{(2)} = - \frac{\sin x}{1 + \sin x} - \frac{\cos^{2} x}{{(1 + \sin x)}^{2}} \Rightarrow y^{(2)} (0) = - 1

y^{(3)} = - \frac{\cos x}{1 + \sin x} + \frac{3 \sin x \cos x}{{(1 + \sin x)}^{2}} + \frac{{2 \cos}^{3} x}{{(1 + \sin x)}^{3}} \Rightarrow y^{(3)} (0) = 1

\Rightarrow \ln (1 + \sin x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{6} + o (x^{3})

Commented by Rio Michael last updated on 21/Apr/20

thanks sir