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Can i prove that

$$\sin(\pi) = 0 $$

by letting $$x= \pi$$ in its series expansion?? and somehow work something from there without just saying "the series is equal to $\sin(x)$ thus $\sin(\pi) = 0 $

Thanks in advance

Zubin Mukerjee
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Apple
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  • Do you know the Taylor series for $\sin(x)$? Here is a list of commonly used Taylor series – Zubin Mukerjee Mar 06 '19 at 05:27
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    What are your definitions of $\sin$ and $\pi$ here? – Travis Willse Mar 06 '19 at 05:29
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    What is your definition of $\pi$? If it’s only defined by its decimal expansion, the task would be hopeless. Some people might define $\pi$ to be the smallest positive root of the sine function, but that will not help you either. So I repeat: What is $\pi$? – Lubin Mar 06 '19 at 05:31
  • sin(\pi) is equal to 0 , i am wondering if this can be proven by working with just its taylor expansion. Say - i am given the taylor expansion of sin(x) (but i don't know its sin(x)) centered at 0 so Maclaurin. Now x is given the value of pi , and i am asked to prove that the sum is equal to 0. – Apple Mar 06 '19 at 05:32
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    I repeat: what is $\pi$? You need to specify some defining property of it. – Lubin Mar 06 '19 at 05:32
  • pi?? 3.14...... – Apple Mar 06 '19 at 05:33
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    @Lubin I'm not OP but maybe a good choice is $\pi$ is the ratio of a circle's circumference to its diameter – Zubin Mukerjee Mar 06 '19 at 05:34
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    No good to specify $\pi$ as a decimal unless you know all (infinitely many) digits of the expansion. (I gotta go to bed, sorry) – Lubin Mar 06 '19 at 05:35
  • This is an interesting Q & A to look at: https://math.stackexchange.com/questions/33732/prove-sin-pi-2-1-using-taylor-series – Christopher Marley Mar 06 '19 at 05:41
  • The answer from Qiaochu Yuan to the linked duplicate seems to cover this nicely. It derives the addition formula, which is the last link between $\sin \frac \pi 2 =1$ and $\sin \pi=0$ – Ross Millikan Mar 06 '19 at 05:57
  • I don't think it's an exact duplicate. The marked duplicate did not have a marked answer, and the answers I looked through didn't answer the question. I'm still interested in why $$\pi-\frac{\pi^3}{3!}+\frac{\pi^5}{5!}-\frac{\pi^7}{7!}...$$ would equal zero. – Christopher Marley Mar 06 '19 at 06:07
  • Qiaochu Yuan shows that $\sin \pi=0$. He uses the Taylor series to derive the differential equation, then uses the differential equation to show this. This proves that the series equals zero, but does not do it just by summing the series. He says he doesn't see a route other than through the differential equation. – Ross Millikan Mar 06 '19 at 06:10

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