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Let $P_n(x)$ be Legendre polynomials: $$\frac{1}{\sqrt{1-2tx+t^2}}=\sum\limits_{n=0}^{\infty}P_n(x)t^n.$$ Usual arguments from the theory of formal groups allow to prove that for any $n$ $$P_n(x)=Q_n(P_1(x),P_2(x),P_4(x), \ldots,P_{2^k}(x),\ldots),$$ where $Q_n$ is a polynomial with integer coefficients, depending on $P_{2^k}(x)$ such that $2^k\le n$. (Over $\mathbb Q$ every $P_n$ is a polynomial of $P_1$ and $P_2$.) For example $$P_1=x,\quad P_2=\frac{1}{2}(3x^2-1),\quad P_3=\frac{1}{2}(5x^3-3x)=P_1(3P_2-2P_1^2),$$ $$P_4=\frac{1}{8}(35x^4-30x^2+3),\quad P_5=\frac{x}{8}(63x^4-70x^2+15)=P_1(5P_4+4P_1^2(P_1^2-5P_2)).$$ Here we have binary expasion in leading terms: $$P_3=3P_2P_1+\ldots,\quad P_5=5P_4P_1+\ldots$$ Probably this property (existence of $Q_n$) is known. Can you give any references concerning this fact?

For any $n$ the polynomial $Q_n$ is unique if (thanks to Victor Kleptsyn) we consider symbols $P_1$, $P_2$, $\ldots$ as coefficients of the formal group law. I don't know whether it is interesting to study properties of $Q_n$ but I'll be greatful for any additional information.

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  • $\begingroup$ Would you be more precise on the form of $Q$? How does $Q$ depend on $n$? Which indices $2^j$ are to be considered? $\endgroup$
    – Pietro Majer
    Commented Feb 4, 2015 at 13:23
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    $\begingroup$ I think that this is true for any sequence of polynomials $(T_n)$, $\deg T_n=n$. In fact a more general statement seems to be true namely $x^n= Q_n(T_1(x),T_2(x),\dotsc, T_{2^k}(x),\dotsc)$, where $Q_n(t_1, t_2,\dotsc, t_{2k},\dotsc)$ is a polynomial. $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 4, 2015 at 13:33
  • $\begingroup$ That's why I'm asking. In that generality it's just binary expansion of $n$ and linear independence of monomials. Maybe the OP looks for an explicit $Q_n$? $\endgroup$
    – Pietro Majer
    Commented Feb 4, 2015 at 15:35
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    $\begingroup$ @LiviuNicolaescu: in the question it is mentioned that $Q$ should be a polynomial with \emph{integer}, not \emph{rational} coefficients. For instance, $P_4$ has a leading coefficient $1/8$, and I would say that you cannot represent it in such a way: given that the leading coefficient of $P_2$ is only $1/2$. Alexey: I join Pietro Majer's question -- are you interested in any such $Q$, or should it have some other properties? $\endgroup$
    – Victor Kleptsyn
    Commented Feb 4, 2015 at 19:24
  • $\begingroup$ @Pietro Majer: yes, $Q$ depend on $n$ $\endgroup$
    – Alexey Ustinov
    Commented Feb 5, 2015 at 1:51

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