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Is there some special name given to the two polynomials below: $${n \choose 1}x - {n \choose 3}x^3 + \ldots \label{} \tag{1}$$ and $$1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots \label{} \tag{2}$$.

gxyd
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  • How are these two series? $\binom {n}{k}=0$ if $k>n$. The first one is the finite sum $\sum_{i<n, i \text{ odd}} (-1)^i \binom{n}{i}x^i$, the other one is similar. – Lonidard Aug 08 '17 at 07:33
  • I don't have any intention for defining ${n \choose k}$ for $k > n$. Yes, I think you are right, they should be (better) called polynomials, because that I what my intention is as well. But is there any special names for these polynomials? Some special names like "Legendre polynomial" or "Chebyshev polynomial", similarly is there some special name for these polynomials then? – gxyd Aug 08 '17 at 09:59

2 Answers2

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HInt:try $$(1+ix)^n$$ wher $i^2=-1$ $$(1+ix)^n=\\\left(\begin{array}{c}n\\ 0\end{array}\right)1^{n}(ix)^{0}+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}(ix)^{1}+ \left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}(ix)^{2}+ \left(\begin{array}{c}n\\ 3\end{array}\right)1^{n-3}(ix)^{3}+ \left(\begin{array}{c}n\\ 4\end{array}\right)1^{n-4}(ix)^{4}+...=\\\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)(ix)+\left(\begin{array}{c}n\\ 20\end{array}\right)(ix)^{2}+\left(\begin{array}{c}n\\ 3\end{array}\right)(ix)^{3}+\left(\begin{array}{c}n\\ 4\end{array}\right)(ix)^{4}+...=\\( \left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 2\end{array}\right)x^2+\left(\begin{array}{c}n\\ 4\end{array}\right)x^4-...)+\\i( \left(\begin{array}{c}n\\ 1\end{array}\right)x-\left(\begin{array}{c}n\\ 3\end{array}\right)x^3+...)$$

Khosrotash
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  • That is good. This is from what I obtained it from, but I need to know if there is some special name given to these "polynomials"? – gxyd Aug 08 '17 at 10:08
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Define $$P_n(x)=1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots$$ $$Q_n(x)={n \choose 1}x - {n \choose 3}x^3 + \ldots$$

I have never heard of these polynomials before, but the polynomials $P_n$ bare some vague resemblances to Chebyschev polynomials $T_n(x)$.

Since $${n\choose 1}x - {n \choose 3}x^3 + \ldots=\Im(1+ix)^n,$$ $$1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots=\Re(1+ix)^n,$$ writing $(1+ix)$ in polar form yields $$ \begin{align} (1+ix)^n= & (\sqrt{1+x^2}e^{i\arctan(x)})^n \\ =&(1+x^2)^{\frac n2}e^{in\arctan(x)} \end{align}. $$ This shows that $$P_n(x)=(1+x^2)^{\frac n2}\cos\left(n \arctan(x)\right) \\ Q_n(x)=(1+x^2)^{\frac n2}\sin\left(n \arctan(x)\right).$$

Compare $$T_n(x)=\cos(n\arccos(x))$$ $$P_n(x)=(1+x^2)^\frac n2 \cos( n\arctan(x)).$$ Also, the family $P_n$ has the property that $$P_n(\tan(\theta))=(1+\tan^2(\theta))^{\frac n2}\cos(n\theta)=\frac{\cos(n\theta)}{\cos^{n}(\theta)},$$ mirroring $$T_n(\cos(\theta))=\cos(n\theta).$$ Through this relation, we can also express our polynomials as a function of Chebyschev polynomials $T_n$: $$P_n(\tan(\theta))=\frac{T_n(\cos(\theta))}{\cos^n(\theta)}.$$

Lonidard
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