1. series expansion

Expand ln(1+sinx) in ascending powers of x up to and including the term in $x^{4}$

$ln(1+sinx) = ln [ 1+ (x-\frac{x^{3}}{3!}) +.....}]$

i know that the series expansion for sinx = $( x -\frac{x^{3}}{3!} )$

but dont know where to go from here. Any help appreciated

2. Re: series expansion

Hello, Tweety!

You were off to a good start . . .

$\text{Expand }\ln(1+\sin x)\,\text{ in ascending powers of }x\text{ up to and including the }x^4\text{ term.}$

We have: . $\ln(1 + u) \;=\;u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots$

And: . $\sin x \;=\;x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} - \cdots$

Hence: . $\ln(1 + \sin x) \;=\;\sin x - \frac{\sin^2\!x}{2} + \frac{\sin^3\!x}{3} - \frac{\sin^4\!x}{4} + \cdots$

Since we want terms up to $x^4$, we can use: . $\sin x \;\approx\;x - \frac{x^3}{3!}$

Substitute:
. $\ln(1 + \sin x) \;=\;\left(x-\tfrac{x^3}{6}\right) - \tfrac{1}{2}(x - \tfrac{x^3}{6})^2 + \tfrac{1}{3}(x - \tfrac{x^3}{6})^3 - \tfrac{1}{4}(x - \tfrac{x^3}{6})^4$

. . . $=\;(x-\tfrac{x^3}{6}) - \tfrac{1}{2}(x^2 - \tfrac{1}{3}x^4 + \tfrac{1}{36}x^6) + \tfrac{1}{3}(x^3 - \tfrac{1}{2}x^5 + \tfrac{1}{12}x^7 - \tfrac{1}{216}x^9)-\tfrac{1}{4}(x^4 - \tfrac{2}{3}x^6 + \cdots)$

. . . $=\;{\color{red}x - \tfrac{1}{6}x^3 -\tfrac{1}{2}x^2 + \tfrac{1}{6}x^4} - \tfrac{1}{72}^6 \;{\color{red}+\;\tfrac{1}{3}x^3} - \tfrac{1}{6}x^5 + \tfrac{1}{36}x^7 - \tfrac{1}{648}x^9 {\color{red}-\tfrac{1}{4}x^4} + \tfrac{1}{6}x^6 + \cdots$

. . . $\approx\;x - \tfrac{1}{2}x^2 + \tfrac{1}{6}x^3 - \tfrac{1}{12}x^4$