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?
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$\begingroup$ Yes, that is the Taylor series at $a=0$, you can expand at any other point to get a different Taylor series. $\endgroup$– N. S.Nov 10, 2014 at 16:50
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$\begingroup$ What do you mean another form? Like a different Taylor Series expansion? $\endgroup$– graydadNov 10, 2014 at 16:50
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1$\begingroup$ There's a continued fraction, if you like. You have to be more specific about what you want... $\endgroup$– Karolis JuodelėNov 10, 2014 at 16:53
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1$\begingroup$ 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. $\endgroup$– user189855Nov 10, 2014 at 17:08
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1$\begingroup$ math.stackexchange.com/q/432771/52893 $\endgroup$– JohnDNov 10, 2014 at 17:38
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1 Answer
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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 $$\frac{\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}}{\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}} = x+\frac{x^3}{3}+\frac{2x^5}{15}+ \dots$$
