I have found, by accident, an identity that relates a sum of Chebyshev polynomials of the first kind to a Chebyshev polynomial of the second kind. It goes as follows:


Given an integer partition of $n$, let $g\_a$ be the number of times $a$ appears in said partition.${}^1$ Then the following identity holds for all $n \in\mathbb{N}$:


$$
U\_n(x) = \sum\_{\substack{n\_i>0\\ \sum\_i n\_i = n}} \frac{1}{\prod\_{a\in \{n\_i\}} g\_a!} \prod\_{i=1} \frac{2}{n\_i} T\_{n\_i}(x)\,.
$$


The sum is over all integer partitions of $n$, the product is on all $n\_i$'s in the partition, with repetitions.


I have a very roundabout way to prove this identity (I'll skip the details). The left hand side is obtained by contracting two symmetric traceless tensors of $SO(4)$. That is, letting $|x|=|y|=1$ and $x,y\in \mathbb{R}^4$ then
$$
(x^{i\_1}\cdots x^{i\_n} - \mathrm{traces}) (y\_{i\_1}\cdots y\_{i\_n} - \mathrm{traces}) \propto U\_n(x\cdot y)\,.
$$
The right hand side instead comes from the same contraction but in spinor notation. Namely we let
$$
\mathrm{x} = \left(\begin{matrix}x\_3-x\_4 & x\_1 - i x\_2 \\ x\_1 + i x\_2 & -x\_3-x\_4\end{matrix}\right)\,,\quad \bar{\mathrm{x}} = \epsilon\, \mathrm{x}\, \epsilon^T\,,
$$
and $\mathrm{y}$ in a similar way ($\epsilon$ is the Levi Civita tensor). Then we introduce two dimensional spinors $\eta,\tilde{\eta}$ let $\partial\_{\eta^\alpha}\eta^\beta = \delta\_\alpha^\beta$ (similar for $\tilde{\eta}$) and finally
$$
(\partial\_\eta \mathrm{x} \partial\_{\tilde{\eta}})^n (\eta \mathrm{y}\tilde{\eta})^n \sim \sum\_{\substack{n\_i>0\\ \sum\_i n\_i = n}} \# \prod\_{i} \mathrm{tr}\,((\mathrm{x}\bar{\mathrm{y}})^{n\_i})\,.
$$
The sum over partitions comes from a combinatoric argument. Then it's a simple exercise to show that $\mathrm{tr}\,((\mathrm{x}\bar{\mathrm{y}})^n) \propto T\_n(x\cdot y)$.


My questions are


1. Is this identity known already?
2. If not, could you come up with some more direct argument to prove it?




---


$\;{}^1$ For example, $(1,1,1,2,2,3)$ is an integer partition of $n=10$ with $g\_1 =3,\, g\_2=2,\,g\_3=1$.



Curious identity between the two kinds of Chebyshev polynomials
Here is how to prove it with more standard methods. First of all, let me
restate your identity:



> 
> **Definition.** Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. A
>  *partition* shall mean [an integer
>  partition](https://en.wikipedia.org/wiki/Partition_(number_theory)), i.e., a
>  weakly decreasing finite list of positive integers. If $\lambda$ is a partition
>  and $i$ is a positive integer, then $m\_{i}\left( \lambda\right) $ shall mean
>  the number of times that $i$ appears as entry of $\lambda$. (For example,
>  $m\_{3}\left( \left( 4,3,3,1\right) \right) =2$ and $m\_{2}\left( \left(
> 4,3,3,1\right) \right) =0$.) The *size* $\left\vert \lambda\right\vert $ of
>  a partition is defined to be the sum of all entries of $\lambda$. If
>  $n\in\mathbb{N}$, then a *partition of $n$* means a partition of size $n$. We
>  write "$\lambda\vdash n$" for "$\lambda$ is a partition of $n$".
> 
> 
> **Definition.** We let $T\_n\left(x\right)$ denote [the Chebyshev polynomials of the first kind](https://en.wikipedia.org/wiki/Chebyshev_polynomials), which can be defined (e.g.) by the recurrence $T\_0\left(x\right) = 1$ and $T\_1\left(x\right) = x$ and $T\_{n+1}\left(x\right) = 2x T\_n\left(x\right) - T\_{n-1}\left(x\right)$. We let $U\_n\left(x\right)$ denote [the Chebyshev polynomials of the second kind](https://en.wikipedia.org/wiki/Chebyshev_polynomials), which can be defined (e.g.) by the recurrence $U\_0\left(x\right) = 1$ and $U\_1\left(x\right) = 2x$ and $U\_{n+1}\left(x\right) = 2x U\_n\left(x\right) - U\_{n-1}\left(x\right)$.
> 
> 
> **Theorem 1.** For any $n\in\mathbb{N}$, we have
>  \begin{equation}
> U\_{n}\left( x\right) =\sum\_{\lambda=\left( \lambda\_{1},\lambda\_{2}
> ,\ldots,\lambda\_{k}\right) \vdash n}\left( \prod\_{i=1}^{\infty}\dfrac
> {1}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left( \lambda\right) !}\right)
> \cdot\prod\_{j=1}^{k}\left( 2T\_{\lambda\_{j}}\left( x\right) \right) .
> \end{equation}
> 
> 
> 


To prove this, I will use two well-known generating-function identities for
Chebyshev polynomials, both of which appear on [the
Wikipedia](https://en.wikipedia.org/wiki/Chebyshev_polynomials#Definition):
\begin{equation}
\sum\_{n=0}^{\infty}T\_{n}\left( x\right) t^{n}=\dfrac{1-tx}{1-2tx+t^{2}}
\label{darij1.eq.T-gen}
\tag{1}
\end{equation}
and
\begin{equation}
\sum\_{n=0}^{\infty}U\_{n}\left( x\right) t^{n}=\dfrac{1}{1-2tx+t^{2}
}.
\label{darij1.eq.U-gen}
\tag{2}
\end{equation}
These are identities in the ring $\left( \mathbb{Q}\left[ x\right] \right)
\left[ \left[ t\right] \right] $ of formal power series in the variable
$t$ over the polynomial ring $\mathbb{Q}\left[ x\right] $. Both identities can easily be derived from the above recurrent definitions of $T\_n\left(x\right)$ and $U\_n\left(x\right)$.


Now, subtracting the equality $\underbrace{T\_{0}\left( x\right) }
\_{=1}\underbrace{t^{0}}\_{=1}=1$ from the identity \eqref{darij1.eq.T-gen}, we obtain
\begin{equation}
\sum\_{n=1}^{\infty}T\_{n}\left( x\right) t^{n}=\dfrac{1-tx}{1-2tx+t^{2}
}-1=t\cdot\dfrac{x-t}{1-2tx+t^{2}}.
\end{equation}
Dividing both sides of this by $t$, we obtain
\begin{equation}
\sum\_{n=1}^{\infty}T\_{n}\left( x\right) t^{n-1} = \dfrac{x-t}{1-2tx+t^{2}}.
\end{equation}
Integrating both sides of this equality over $t$, we find
\begin{align}
\sum\_{n=1}^{\infty}T\_{n}\left( x\right) \dfrac{t^{n}}{n} & =\int\dfrac
{x-t}{1-2tx+t^{2}}dt\nonumber\\
& =\dfrac{1}{2}\log\dfrac{1}{1-2tx+t^{2}}
\label{darij1.eq.T-ge2}
\tag{3}
\end{align}
(as you can easily check by differentiation). (Note that this identity also
appears on the Wikipedia, under the guise of $\sum\_{n=1}^{\infty}T\_{n}\left(
x\right) \dfrac{t^{n}}{n}=\log\dfrac{1}{\sqrt{1-2tx+t^{2}}}$, apparently
because someone finds square roots simpler than division by $2$.)


Multiplying both sides of the equality \eqref{darij1.eq.T-ge2}
by $2$, we obtain
\begin{equation}
2\sum\_{n=1}^{\infty}T\_{n}\left( x\right) \dfrac{t^{n}}{n}=\log\dfrac
{1}{1-2tx+t^{2}}.
\end{equation}
Hence,
\begin{equation}
\log\dfrac{1}{1-2tx+t^{2}}=2\sum\_{n=1}^{\infty}T\_{n}\left( x\right)
\dfrac{t^{n}}{n}=\sum\_{n=1}^{\infty}2T\_{n}\left( x\right) \dfrac{t^{n}}{n},
\end{equation}
so that
\begin{equation}
\dfrac{1}{1-2tx+t^{2}}=\exp\left( \sum\_{n=1}^{\infty}2T\_{n}\left( x\right)
\dfrac{t^{n}}{n}\right) .
\end{equation}
Hence, \eqref{darij1.eq.U-gen} becomes
\begin{equation}
\sum\_{n=0}^{\infty}U\_{n}\left( x\right) t^{n}=\dfrac{1}{1-2tx+t^{2}}
=\exp\left( \sum\_{n=1}^{\infty}2T\_{n}\left( x\right) \dfrac{t^{n}}
{n}\right) .
\label{darij1.eq.T-ge3}
\tag{4}
\end{equation}


Now, we recall one of the staple formulas of algebraic combinatorics (probably
in EC or Wilf or similar sources):



> 
> **Proposition 2.** Let $R$ be a commutative $\mathbb{Q}$-algebra (for example,
>  $\mathbb{Q}$ or $\mathbb{Q}\left[ x\right] $). Let $b\_{1},b\_{2},b\_{3}
> ,\ldots\in R$ and $c\_{0},c\_{1},c\_{2},\ldots\in R$ be such that
>  \begin{equation}
> \sum\_{n=0}^{\infty}c\_{n}t^{n}=\exp\left( \sum\_{n=1}^{\infty}b\_{n}\dfrac
> {t^{n}}{n}\right)
> \end{equation}
>  in the ring $R\left[ \left[ t\right] \right] $ of formal power series.
>  Then,
>  \begin{equation}
> c\_{n}=\sum\_{\lambda=\left( \lambda\_{1},\lambda\_{2},\ldots,\lambda\_{k}\right)
> \vdash n}\left( \prod\_{i=1}^{\infty}\dfrac{1}{i^{m\_{i}\left( \lambda\right)
> }m\_{i}\left( \lambda\right) !}\right) \cdot\prod\_{j=1}^{k}b\_{\lambda\_{j}}
> \end{equation}
>  for each $n\in\mathbb{N}$.
> 
> 
> 


*Proof of Proposition 2.* An infinite sequence $\left( k\_{1},k\_{2}
,k\_{3},\ldots\right) \in\mathbb{N}^{\infty}$ of nonnegative integers will be
called a *weak composition* if all but finitely many $i\geq1$ satisfy
$k\_{i}=0$. There is a bijection
\begin{align}
\left\{ \text{partitions}\right\} & \rightarrow\left\{ \text{weak
compositions}\right\} ,\nonumber\\
\lambda & \mapsto\left( m\_{1}\left( \lambda\right) ,m\_{2}\left(
\lambda\right) ,m\_{3}\left( \lambda\right) ,\ldots\right)
\label{darij1.pf.p2.1}
\tag{5}
\end{align}
(since any partition $\lambda$ is uniquely determined by the numbers
$m\_{1}\left( \lambda\right) ,m\_{2}\left( \lambda\right) ,m\_{3}\left(
\lambda\right) ,\ldots$ which record how often each positive integer appears
in $\lambda$). We notice that any partition $\lambda$ satisfies
\begin{equation}
1m\_{1}\left( \lambda\right) +2m\_{2}\left( \lambda\right) +3m\_{3}\left(
\lambda\right) +\cdots=\left\vert \lambda\right\vert
\label{darij1.pf.p2.2}
\tag{6}
\end{equation}
(because $\left\vert \lambda\right\vert $ is the sum of all entries of
$\lambda$, while $1m\_{1}\left( \lambda\right) +2m\_{2}\left( \lambda\right)
+3m\_{3}\left( \lambda\right) +\cdots$ is what becomes of this sum after
equal addends are bunched together). Moreover, any partition $\lambda=\left(
\lambda\_{1},\lambda\_{2},\ldots,\lambda\_{k}\right) $ satisfies
\begin{equation}
\prod\_{i=1}^{\infty}b\_{i}^{m\_{i}\left( \lambda\right) }=\prod\_{j=1}
^{k}b\_{\lambda\_{j}}
\label{darij1.pf.p2.3}
\tag{7}
\end{equation}
(for a similar reason: the product $\prod\_{i=1}^{\infty}b\_{i}^{m\_{i}\left(
\lambda\right) }$ is what will become of the product $\prod\_{j=1}
^{k}b\_{\lambda\_{j}}$ if you bunch factors corresponding to equal entries of
$\lambda$ together).


We have the following product rule (i.e., analogue of the distributivity law)
for infinite products of infinite sums: If $\left( a\_{i,k}\right)
\_{i\geq1\text{ and }k\geq0}$ is a family of elements of $R\left[ \left[
t\right] \right] $ satisfying $a\_{i,0}=1$ for each $i\geq1$, then
\begin{equation}
\prod\_{i=1}^{\infty}\sum\_{k=0}^{\infty}a\_{i,k}=\sum\_{\substack{\left(
k\_{1},k\_{2},k\_{3},\ldots\right) \text{ is a}\\\text{weak composition}}
}\prod\_{i=1}^{\infty}a\_{i,k\_{i}},
\label{darij1.pf.p2.prodrule}
\tag{8}
\end{equation}
provided that everything formally converges (i.e., for each given
$N\in\mathbb{N}$, all but finitely many pairs $\left( i,k\right) \in\left\{
1,2,3,\ldots\right\} ^{2}$ satisfy
$t^N \mid a\_{i,k}$ in $R\left[\left[t\right]\right]$).


We have
\begin{align}
\sum\_{n=0}^{\infty}c\_{n}t^{n} & =\exp\left( \sum\_{n=1}^{\infty}b\_{n}
\dfrac{t^{n}}{n}\right) =\prod\_{n=1}^{\infty}\underbrace{\exp\left(
b\_{n}\dfrac{t^{n}}{n}\right) }\_{\substack{=\sum\_{k=0}^{\infty}\dfrac{1}
{k!}\left( b\_{n}\dfrac{t^{n}}{n}\right) ^{k}\\\text{(since }\exp
z=\sum\_{k=0}^{\infty}\dfrac{1}{k!}z^{k}\text{)}}}\nonumber\\
& \qquad \left(\text{since $\exp\left(\sum\_{i\in I} a\_i\right) = \prod\_{i\in I} \exp a\_i$ for any family $\left(a\_i\right)\_{i\in I}$}\right)
\nonumber\\
& =\prod\_{n=1}^{\infty}\sum\_{k=0}^{\infty}\dfrac{1}{k!}\left( b\_{n}
\dfrac{t^{n}}{n}\right) ^{k}=\prod\_{i=1}^{\infty}\sum\_{k=0}^{\infty}\dfrac
{1}{k!}\underbrace{\left( b\_{i}\dfrac{t^{i}}{i}\right) ^{k}}\_{=\dfrac
{b\_{i}^{k}t^{ik}}{i^{k}}}\nonumber\\
& \qquad\left( \text{here, we have renamed the index }n\text{ as }i\right)
\nonumber\\
& =\prod\_{i=1}^{\infty}\sum\_{k=0}^{\infty}\dfrac{1}{k!}\cdot\dfrac{b\_{i}
^{k}t^{ik}}{i^{k}}=\prod\_{i=1}^{\infty}\sum\_{k=0}^{\infty}\dfrac{b\_{i}
^{k}t^{ik}}{i^{k}k!}\nonumber\\
& =\sum\_{\substack{\left( k\_{1},k\_{2},k\_{3},\ldots\right) \text{ is
a}\\\text{weak composition}}}\prod\_{i=1}^{\infty}\dfrac{b\_{i}^{k\_{i}}
t^{ik\_{i}}}{i^{k\_{i}}k\_{i}!}\nonumber\\
& \qquad\left( \text{by the product rule \eqref{darij1.pf.p2.prodrule},
applied to }a\_{i,k}=\dfrac{b\_{i}^{k}t^{ik}}{i^{k}k!}\right) \nonumber\\
& =\sum\_{\lambda\text{ is a partition}}\underbrace{\prod\_{i=1}^{\infty}
\dfrac{b\_{i}^{m\_{i}\left( \lambda\right) }t^{im\_{i}\left( \lambda\right)
}}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left( \lambda\right) !}}\_{=\left(
\prod\_{i=1}^{\infty}\dfrac{1}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left(
\lambda\right) !}\right) \left( \prod\_{i=1}^{\infty}b\_{i}^{m\_{i}\left(
\lambda\right) }\right) \left( \prod\_{i=1}^{\infty}t^{im\_{i}\left(
\lambda\right) }\right) }\nonumber\\
& \qquad\left(
\begin{array}
[c]{c}
\text{here, we have substituted }\left( m\_{1}\left( \lambda\right)
,m\_{2}\left( \lambda\right) ,m\_{3}\left( \lambda\right) ,\ldots\right) \\
\text{for }\left( k\_{1},k\_{2},k\_{3},\ldots\right) \text{ in the sum, due to
the bijection \eqref{darij1.pf.p2.1}}
\end{array}
\right) \nonumber\\
& =\sum\_{\lambda\text{ is a partition}}\left( \prod\_{i=1}^{\infty}\dfrac
{1}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left( \lambda\right) !}\right)
\left( \prod\_{i=1}^{\infty}b\_{i}^{m\_{i}\left( \lambda\right) }\right)
\underbrace{\left( \prod\_{i=1}^{\infty}t^{im\_{i}\left( \lambda\right)
}\right) }\_{\substack{=t^{1m\_{1}\left( \lambda\right) +2m\_{2}\left(
\lambda\right) +3m\_{3}\left( \lambda\right) +\cdots}\\=t^{\left\vert
\lambda\right\vert }\\\text{(by \eqref{darij1.pf.p2.2})}}}\nonumber\\
& =\sum\_{\lambda\text{ is a partition}}\left( \prod\_{i=1}^{\infty}\dfrac
{1}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left( \lambda\right) !}\right)
\left( \prod\_{i=1}^{\infty}b\_{i}^{m\_{i}\left( \lambda\right) }\right)
t^{\left\vert \lambda\right\vert }.
\label{darij1.pf.p2.6}
\tag{9}
\end{align}


Now, let $n\in\mathbb{N}$. Comparing coefficients of $t^{n}$ on both sides of
the equality \eqref{darij1.pf.p2.6}, we obtain
\begin{align\*}
c\_{n} & =\underbrace{\sum\_{\substack{\lambda\text{ is a partition;}
\\\left\vert \lambda\right\vert =n}}}\_{\substack{=\sum\_{\lambda\vdash
n}\\\text{(since the partitions of }n\\\text{are precisely the partitions
}\lambda\\\text{with }\left\vert \lambda\right\vert =n\text{)}}}\left(
\prod\_{i=1}^{\infty}\dfrac{1}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left(
\lambda\right) !}\right) \prod\_{i=1}^{\infty}b\_{i}^{m\_{i}\left(
\lambda\right) }\\
& =\sum\_{\lambda\vdash n}\left( \prod\_{i=1}^{\infty}\dfrac{1}{i^{m\_{i}\left(
\lambda\right) }m\_{i}\left( \lambda\right) !}\right) \prod\_{i=1}^{\infty
}b\_{i}^{m\_{i}\left( \lambda\right) }\\
& =\sum\_{\lambda=\left( \lambda\_{1},\lambda\_{2},\ldots,\lambda\_{k}\right)
\vdash n}\left( \prod\_{i=1}^{\infty}\dfrac{1}{i^{m\_{i}\left( \lambda\right)
}m\_{i}\left( \lambda\right) !}\right) \underbrace{\left( \prod
\_{i=1}^{\infty}b\_{i}^{m\_{i}\left( \lambda\right) }\right) }
\_{\substack{=\prod\_{j=1}^{k}b\_{\lambda\_{j}}\\\text{(by
\eqref{darij1.pf.p2.3})}}}\\
& =\sum\_{\lambda=\left( \lambda\_{1},\lambda\_{2},\ldots,\lambda\_{k}\right)
\vdash n}\left( \prod\_{i=1}^{\infty}\dfrac{1}{i^{m\_{i}\left( \lambda\right)
}m\_{i}\left( \lambda\right) !}\right) \prod\_{j=1}^{k}b\_{\lambda\_{j}}.
\end{align\*}
This proves Proposition 2. $\blacksquare$


*Proof of Theorem 1.* Recall the identity \eqref{darij1.eq.T-ge3}. Thus,
Proposition 2 (applied to $R=\mathbb{Q}\left[ x\right] $ and $c\_{n}
=U\_{n}\left( x\right) $ and $b\_{n}=2T\_{n}\left( x\right) $) yields that
\begin{equation}
U\_{n}\left( x\right) =\sum\_{\lambda=\left( \lambda\_{1},\lambda\_{2}
,\ldots,\lambda\_{k}\right) \vdash n}\left( \prod\_{i=1}^{\infty}\dfrac
{1}{i^{m\_{i}\left( \lambda\right) }m\_{i}\left( \lambda\right) !}\right)
\cdot\prod\_{j=1}^{k}\left( 2T\_{\lambda\_{j}}\left( x\right) \right)
\end{equation}
for each $n\in\mathbb{N}$. This proves Theorem 1. $\blacksquare$


