1
$\begingroup$

I've been looking around and I see no formulas given in any of the sources I've been able to find for the infinite product representing $\tan\left(x\right)$. Is it simply the ratio of the infinite products for $\sin\left(x\right)$ and $\cos\left(x\right)$? In which case, it should look like $$ \frac{x\cdot\prod\limits_{n=1}^{\infty}\left(1-\frac{x^2}{n^2\pi^2}\right)}{\prod\limits_{n=1}^{\infty}\left(1-\frac{4x^2}{\pi^2\left(2n-1\right)^2}\right)}=\frac{\sin\left(x\right)}{\cos\left(x\right)}=\tan\left(x\right)? $$ Forgive me if there is an obvious solution to this problem that I haven't seen as of yet. I've taken up to differential equations and multivariable calc so far in my major.

Brandon

$\endgroup$
  • $\begingroup$ It's fine, I remember the numerator and I imagine the denominator works. Be aware that the evident factoring(s) lead to products that do not converge. As long as you keep the squares in place all is good. $\endgroup$ – Will Jagy Sep 28 '14 at 4:08
  • $\begingroup$ A bunch of $\pi$'s got lost, of course. $\endgroup$ – user138530 Sep 28 '14 at 4:55
  • $\begingroup$ Woops, sorry! I'll add them... $\endgroup$ – jm324354 Sep 28 '14 at 5:58
  • $\begingroup$ Now it is okay. $\endgroup$ – Lucian Sep 28 '14 at 15:30
0
$\begingroup$
  • The upper product should start at $n=1$, otherwise you have division by $0$.
  • The x before the infinite product in the numerator should be changed to $\pi x$.
  • The arguments of the sine, cosine, and tangent functions should be $\pi x$ rather than x. Unless you want to imply that $~\sin(0)=\sin(\pm1)=\sin(\pm2)=\sin(\pm3)=\ldots=0$, as opposed to $\sin(0)=\sin(\pm\pi)=\sin(\pm2\pi)=\sin(\pm3\pi)=\ldots=0$.
$\endgroup$
  • $\begingroup$ Yeah someone commented above and said that the pi's were missing. I just copied it wrong from the formula sheet and haven't fixed it yet. $\endgroup$ – jm324354 Sep 28 '14 at 15:11
  • $\begingroup$ Fixed it now, at least according to Wolfram. $\endgroup$ – jm324354 Sep 28 '14 at 15:15
  • $\begingroup$ I didn't know you could do step three...neat, thanks. $\endgroup$ – jm324354 Sep 28 '14 at 22:36

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.