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Some published papers and books give the impression that if you write down any infinite sequence of polynomials that follows a simple pattern, one will find that it's named after somebody and has an extensive literature. My question is whether the following is such a case: $$ f_n(x) = nx + \binom n 3 x^3 + \binom n 5 x^5 + \cdots $$ This has degree $n-1$ if $n$ is even and $n$ if $n$ is odd, so for each odd number there are two polynomials in the sequence with that degree.

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    $2f_n(x)=(1+x)^n-(1-x)^n$2012-12-23
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    "Some published papers and books?" I can't imagine why anybody would make this claim. *This* sequence is easy to come up with a closed form. Doubt it has a name.2012-12-23
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    Haste...... I stumbled across this while suffering another bout of fiddling with trigonometric identities and slapped it onto this board before I though further about it. So Colin McQuillan: why don't you post your comment as an answer.2012-12-23
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    This is known as the Hardy-McQuillan Stackexchange sequence.2012-12-23
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    OK, yet another tangent half-angle formula: If $\tan(\beta/2)=\left(\tan(\alpha/2)\right)^n$ then $\tan\beta$ $=\dfrac{(\sin\alpha)^n}{(1+\cos\alpha)^n-(1-\cos\alpha)^n}$.2012-12-23
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    . . . . .or maybe I'm missing a factor of $2$ on the right side.2012-12-23
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    If $\tan(\beta/2)=(\tan(\alpha/2))^n$, then $\tan\beta$ $=\dfrac{2(\sin(\alpha))^n}{(1+\cos\alpha)^n-(1-\cos\alpha)^n}$. The first version was missing a factor of $2$.2012-12-23
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    @ColinMcQuillan : Maybe you weren't "notified" of the comments above, but herewith, you are.2012-12-23

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Your formula has the following name: $$f_n(x)=\frac{(1+x)^n-(1-x)^n}2$$