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I currently looking at the polynomials you get from the series expansion of $$ \frac{1-x^2}{1-ax+2x^2}=1+a x +(a^2-3) x^2 +(a^3-5a) x^3 +\underbrace{(a^4-7a^2+6)}_{(a-1)(a+1)(a^2-6)} x^4+\dots $$ W|A helped here...

What I found is that from $x^{4}$ onwards, every sixth polynomial shares a factor of $(a^2-6)$. How to prove that?

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I will prove what you want kind of indirectly using Chebyshev polynomials of second kind. Actually once you formulate the question in terms of these polynomials this mysterious $a^2-6$ becomes quite meaningful. Let me start by The generating function of Chebyshev polynomial of second kind, $U_n$ (which you can take as the definition of these polynomials): $$ \frac{1}{1-2ax+x^2}=\sum_{n=0}^\infty U_n(a)x^n $$ So your function which I will call $I(x,a)$ is nothing but $$ I(x,a):=\frac{1-x^2}{1-ax +2x^2}\xrightarrow{t=\sqrt{2}x}\left[1-\frac{t^2}{2}\right]\sum_{n=0}^\infty U_n\left(\frac{a}{\sqrt{8}}\right)t^n$$ We are only interested in the value of this function when $a=\pm \sqrt{6}$. But then $\frac{a}{2\sqrt{2}}=\cos \pi/6$ or $=\cos(\pi-\pi/6)$ (You can already see where every sixth term pattern is coming from). Then your function becomes (I defined $X:=a/\sqrt{8}$) $$ I(X,t):=U_0(X) + tU_1(X)+ t^2\sum_{n=0}^\infty \left[U_{n+2}(X)-\frac{1}{2}U_n(X)\right]t^n $$ We are only interested in $n=6k+2$ and want to prove that $$ 2U_{4+6k}\left(\cos \frac{\pi}{6}\right)= U_{2+6k}\left(\cos \frac{\pi}{6}\right) $$ Now using yet another property of Chebyshev polynomial of second kind, in general if $z=\cos \psi$, then $$ U_m(z) = \frac{\sin((m+1)\psi)}{\sin \psi} $$ In our case then one can easily show with identity above that $ U_{4+6k}\left(\cos \frac{\pi}{6}\right)= (-1)^k $ and $ U_{2+6k}\left(\cos \frac{\pi}{6}\right)=2(-1)^k $. This show that $a=+\sqrt{6}$ is a root for every sixth polynomial after $x^4$ in the expansion of $I(x,a)$. Similarly you can check the $a=-\sqrt{6}$ is also a root.

These identities about Chebyshev polynomials are surprisingly easy to prove actually. For a fast reference about Chebyshev polynomials, look here, or wikipedia.

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  • $\begingroup$ Hi, thanks for pulling out Chebychev, a connection that was already assumed by Chris Godsil here... $\endgroup$
    – draks ...
    Nov 23 '15 at 22:16
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    $\begingroup$ @draks... I assume you've already worked through all these details, but in case not, I thought I would point out the following. In this answer, Chris Godsil defined the functions $p_n$, $n\ge0$, by the recurrence $$ p_{n+1}(a)=ap_n(a)-2p_{n-1}(a) $$ and the initial conditions $p_0(a)=1$, $p_1(a)=a$. $\endgroup$ Nov 25 '15 at 22:11
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    $\begingroup$ In comments to that answer, I showed that $$ p_n(a)=2^{n/2}U_n(a/2^{3/2})-2^{(n-2)/2}U_{n-2}(a/2^{3/2}). $$ In this answer, I showed that the generating function for the $p_n(a)$ is $$ \frac{1-x^2}{1-ax+2x^2}, $$ in other... $\endgroup$ Nov 25 '15 at 22:22
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    $\begingroup$ ...words, that $$ \begin{aligned} \frac{1-x^2}{1-ax+2x^2}&=\sum_{n=0}^\infty p_n(a)x^n\\ &=1+ax+\sum_{n=2}^\infty\left[2^{n/2}U_n(a/2^{3/2})-2^{(n-2)/2}U_{n-2}(a/2^{3/2})\right]x^n\\ &=1+ax+\sum_{n=2}^\infty \left[U_n(a/2^{3/2})-\frac{1}{2}U_{n-2}(a/2^{3/2})\right](\sqrt{2}x)^n \\ &=1+ax+2x^2\sum_{n=0}^\infty \left[U_{n+2}(a/2^{3/2})-\frac{1}{2}U_n(a/2^{3/2}) \right](\sqrt{2}x)^n. \end{aligned}$$ $\endgroup$ Nov 25 '15 at 22:26
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    $\begingroup$ If we define $t=\sqrt{2}x$ and $X=a/2^{3/2}$, then this last expression becomes $$ 1+2Xt+t^2\sum_{n=0}^\infty\left[U_{n+2}(X)-\frac{1}{2}U_n(X)\right]t^n. $$ Since $U_0(X)=1$ and $U_1(X)=2X$, this matches Hamed's expression. $\endgroup$ Nov 25 '15 at 22:27
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Consider $\displaystyle f(a,x)=\frac{1-x^2}{1-a x+2x^2}+{1\over2}=\frac{3-ax}{2(1-a x+2x^2)}$, which is just your function augmented by $1/2$. What you ask amounts at proving that the series expansions of both $f(\sqrt6, x)$ and $f(-\sqrt6, x)$ around $x=0$ do not contain powers of $x$ with exponent ${4+6k}$.

That is true because, by multiplying both numerator and denominator of $f(\pm\sqrt6, x)$ by $(1 \pm \sqrt6 x + 2 x^2)(1 + 2 x^2)$, one gets: $$ f(\pm\sqrt6, x) ={3/2 \pm \sqrt6 x + 3 x^2 \pm \sqrt6 x^3 \mp 2 \sqrt6 x^5\over 1+8x^6}\\ =(3/2 \pm \sqrt6 x + 3 x^2 \pm \sqrt6 x^3 \mp 2 \sqrt6 x^5) \sum_{k=0}^\infty(-8x^6)^k $$ and the polynomial factor does not contain $x^4$.

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  • $\begingroup$ Why do you add $1/2$? $\endgroup$
    – draks ...
    Nov 17 '15 at 12:08
  • $\begingroup$ On purely aesthetic grounds, to avoid the appearance of $-1/2$ in the subsequent formulae. $\endgroup$ Nov 17 '15 at 14:22
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    $\begingroup$ Clever approach! (+1) $\endgroup$
    – epi163sqrt
    Nov 17 '15 at 16:14
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Here is how you can answer a question like this in general, though I have not worked out the computation in this particular case. A polynomial is divisible by $a^2-6$ iff it vanishes when you plug in $a=\pm\sqrt{6}$. So your question really is: if $a=\pm\sqrt{6}$, does the coefficient of $x^{6n+4}$ in $f(x)=\frac{1-x^2}{1-ax+2x^2}$ vanish for each $n$?

To answer this, let's answer a much more general question. Suppose you have a power series $f(x)=\sum c_n x^n$ (convergent in some neighborhood of $x=0$) and you want to know whether all the coeffients $c_n$ vanish when $n\equiv k$ mod $m$, for some fixed $k$ and $m$. Let us write $f_j(x)=\sum c_{mn+j}x^{mn+j}$ for $j=0,1,\dots,m-1$; we want to know whether $f_k(x)$ is identically $0$. To answer this, let $\zeta$ be a primitive $m$th root of $1$, and consider the functions $g_j(x)=f(\zeta^jx)$, for $j=0,1,\dots,m-1$. Note that $g_j(x)=\sum_{\ell=0}^{m-1} \zeta^{j\ell} f_\ell(x)$. I claim that $mf_k(x)=\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)$. Indeed, we have $$\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)=\sum_j\sum_\ell \zeta^{-kj}\zeta^{j\ell}f_\ell(x)=\sum_\ell \left(\sum_{j=0}^{m-1} (\zeta^{\ell-k})^j\right) f_\ell(x).$$

Now $\zeta^{\ell-k}$ is an $m$th root of $1$, so the sum $\sum_{j=0}^{m-1} (\zeta^{\ell-k})^j$ vanishes unless $\zeta^{\ell-k}=1$, i.e. unless $\ell=k$, in which case the sum is $m$. So we get $\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)=mf_k(x)$.

Thus we can conclude that $f_k(x)=0$ iff $\sum_{j=0}^{m-1}\zeta^{-kj}g_j(x)=\sum_{j=0}^{m-1}\zeta^{-kj}f(\zeta^jx)=0$. This sum is one we can (in principle) compute directly from a formula for $f$. For instance, in your case, with $m=6$, $k=4$, and $f(x)=\frac{1-x^2}{1-ax+2x^2}$, you get the following sum: $$\frac{1-x^2}{1-ax+2x^2}+\zeta^2\frac{1-\zeta^2x^2}{1-a\zeta x+2\zeta^2x^2}+\zeta^4\frac{1-\zeta^4x^2}{1-a\zeta^2x+2\zeta^4x^2}+\frac{1-x^2}{1+ax+2x^2}+\zeta^2\frac{1-\zeta^2x^2}{1-a\zeta^4x+2\zeta^2x^2}+\zeta^4\frac{1-\zeta^4x^2}{1-a\zeta^5x+2\zeta^4x^2}.$$

Here $\zeta$ is a primitive $6$th root of $1$. While this is long and nasty, in principle it shouldn't be too hard to expand it all out and see if it vanishes identically if $a=\pm\sqrt{6}$.

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