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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?

Aditya Hase
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  • Yes, that is the Taylor series at $a=0$, you can expand at any other point to get a different Taylor series. – N. S. Nov 10 '14 at 16:50
  • What do you mean another form? Like a different Taylor Series expansion? – graydad Nov 10 '14 at 16:50
  • I mean that the form not depend on Taylor series –  Nov 10 '14 at 16:51
  • Depends on what you are asking. In the world of Taylor series, not really. But you should look into Fourier Series – jameselmore Nov 10 '14 at 16:51
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    There's a continued fraction, if you like. You have to be more specific about what you want... – Karolis Juodelė Nov 10 '14 at 16:53
  • Is there another power series expansion of $\tan$ centered at $0$? No, power series expansions of a function about a given point are unique. To prove this compare their derivatives, which must be equal. – Simon S Nov 10 '14 at 16:54
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    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. –  Nov 10 '14 at 17:08
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    http://math.stackexchange.com/q/432771/52893 – JohnD Nov 10 '14 at 17:38
  • not this, I need the fraction series –  Nov 10 '14 at 17:51
  • My teacher told me that there is another form of $\tan(x)$ depends on fraction series** - Could he have meant $~\displaystyle\sum_{n=-\infty}^\infty\frac1{x-n} ~=~ \frac1x+\sum_{n=1}^\infty\frac{2x}{x^2-n^2} ~=~ \pi\cdot\cot(\pi x)$ ? – Lucian Nov 11 '14 at 04:53
  • yes, this is a good answer. Can you give me the source of this form? –  Nov 11 '14 at 23:20
  • The proof of this formula consists in differentiating the natural logarithm of Euler's infinite product formula for the sine function. See Basel problem for more information. – Lucian Nov 12 '14 at 23:21
  • Are you aware that $$\tan(x) = \frac{\sin x}{\cos x}$$? – Xetrov May 04 '17 at 15:36

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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 $$\frac{\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}}{\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}} = x+\frac{x^3}{3}+\frac{2x^5}{15}+ \dots$$

graydad
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