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I'm learning about taylor series and I tried coming up with one (a Maclaurin) for $\sin x$. I came up with the following: $$\sin x = \sum_{n=1}^\infty (n\bmod 2) (-1)^?\frac{x^n}{n!}$$ This way, all the even $n$ terms become $0$ and excluded. However, I can't figure out which power to put on the $(-1)$ term such that the odd $n$ terms alternate between being positive and negative. I thought about using the Fibonacci sequence where it turns out that starting from $F(1)$, every third number is odd, and starting from $F(2)$ every third number is even. This would work if the pattern existed for every fourth number, rather than third. Is there any way to fill in that question mark? Thanks.

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$?=\frac{n-1}{2}$ works.

Now, if you denote $k=\frac{n-1}{2}$, your formula becomes a nice expression in $k$, which is the usual one.

Now, if you want an expression which is integer also for even $n$, you could use $\frac{n(n-1)}{2}$ which is the same as $1+2+..+n-1$.

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  • $\begingroup$ Thank you! I thought about this for way too long... haha $\endgroup$ – Archie Gertsman Mar 17 '17 at 0:58
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    $\begingroup$ @ArchieGertsman If you want a general method, usually the sum of the powers of nth roots of unity generate $n$-periodic sequences. So expressions of the form $C_1+C_2(-1)^n+C_ni^n+C_4(-i)^n$ should also work for the power. In this case I think that $\frac{1}{2} (1+(-1)^{n+1}+i^{n+1} +(-i)^{n+1})$ would also work instead of $?$... $\endgroup$ – N. S. Mar 17 '17 at 1:06

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