|
I am solving a problem that involves expanding $\sin(x-x^2)$. Since $$\sin(x)=x-x^3/6+x^5/120-...$$ i try to substitute $x$ to $x-x^2$ to arrive at expansion of $x-x^2-x^3/6+x^4/2-x^5/2+x^6/6+...$ Now that would be good enough, but if I look up at this expansion on WolframAlpha I see that it is supposed to look like $x-x^2-x^3/6+x^4/2-59x^5/120+x^6/8$ So does that mean that I cannot make that substitution? Where is the mistake? |
||||
|
|
It works out OK, you obviously just had a mistake in your substitution: $$\sin(x-x^2) \sim (x-x^2) -\frac{1}{6} \left(x-x^2\right)^3 + \frac{1}{120} \left(x-x^2\right)^5 = $$ $$x - x^2 -\frac{x^3}{6} +\frac{x^4}{2} -\frac{59 x^5}{120} +\frac{x^6}{8}...$$ |
|||
|
|
|
$\sin (a-b) = \sin a \; \cos b - \cos a \; \sin b$ , expand each term as usual and collect the coefficients of $x$ have same power. I guess this is more complicated than substituting $x-x^2$ for $u$ in expansion of $\sin u$. Just another method. |
|||
|
|
|
$$\sin(x-x^2)=\sum_{k=0}^\infty\frac{(-1)^n}{(2n+1)!}(x-x^2)^{2n+1}=\sum_{k=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}(1-x)^{2n+1}=$$ $$=x-x^2-\frac{(x-x^2)^3}{6}+\frac{(x-x^2)^5}{120}-\ldots=x-x^2-\frac{x^3-3x^4+3x^5-x^6}{6}+\ldots=$$ $$x-x^2-\frac{1}{6}x^3+\frac{1}{2}x^4+\ldots$$ |
|||
|
|
