1
$\begingroup$

I know that the maclaurin series of $\tan(x)$ is $\tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+...$, then shouldn't be $\tan(x+x^2)=(x+x^2)+\frac{1}{3}(x+x^2)^3+\frac{2}{15}(x+x^2)^5+...$?

Mathematica actually gives me $\tan(x+x^2)=x+x^2+\frac{x^3}{3}+o(x^4)$ to order 3.

$\endgroup$
  • $\begingroup$ We need to ignore terms containing $$x^4$$ $\endgroup$ – lab bhattacharjee Jul 2 at 15:43
1
$\begingroup$

Expanding the monomial terms gives Mathematica's result – and you can stop at $(x+x^2)^3$: $$(x+x^2)+\frac13(x+x^2)^3=x+x^2+\frac13(x^3\color{lightgrey}{+3x^4+3x^5+x^6})=x+x^2+\frac{x^3}3$$

$\endgroup$
2
$\begingroup$

Alternatively: $$\begin{align}y&=\tan{(x+x^2)} \Rightarrow \color{red}{y(0)=0}\\ \arctan y&=x+x^2\\ \frac1{1+y^2}\cdot y'&=1+2x\\ y'&=(1+\color{red}y^2)(1+2x) \Rightarrow \color{blue}{y'(0)=1}\\ y''&=2\color{red}yy'(1+2x)+2(1+\color{red}y^2)\Rightarrow \color{green}{y''(0)=2}\\ y'''&=2\color{blue}{y'}^2(1+2x)+2\color{red}yy''(1+2x)+8\color{red}yy' \Rightarrow \color{brown}{y'''(0)=2}\\ y&=\tan(x+x^2)=\color{red}{y(0)}+\frac{\color{blue}{y'(0)}}{1!}x+\frac{\color{green}{y''(0)}}{2!}x^2+\frac{\color{brown}{y'''(0)}}{3!}x^3+O(x^4)=\\ &=x+x^2+\frac{x^3}{3}+O(x^4).\end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.