# Is there another form of $\tan ( x )$?

The common form of $\tan(x)$ which depends on Taylor series is $$\tan(x)=x+\frac{x^3}{3}+\frac{2}{15}x^5+\cdots.$$ Is there an another form of this function depending on other method?

• Yes, that is the Taylor series at $a=0$, you can expand at any other point to get a different Taylor series. Nov 10, 2014 at 16:50
• What do you mean another form? Like a different Taylor Series expansion? Nov 10, 2014 at 16:50
• There's a continued fraction, if you like. You have to be more specific about what you want... Nov 10, 2014 at 16:53
• My teacher told me that there is another form of $tan(x)$ depends on fraction series but he didn't give me it because this is a homework.
– user189855
Nov 10, 2014 at 17:08
• math.stackexchange.com/q/432771/52893 Nov 10, 2014 at 17:38

As Simon S says, Taylor series centered at a point are unique. The only other different way I could think of writing tangent would be $$\tan(x) =\frac{\sin(x)}{\cos(x)} =\frac{\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}}{\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}}$$ but it is still true that $$\frac{\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}}{\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}} = x+\frac{x^3}{3}+\frac{2x^5}{15}+ \dots$$