Prove that tan($Z$) = $\frac{e_1t - e_3t^3 + e_5t^5 - \cdots}{1 - e_2t^2 + e_4t^4 - \cdots}$

Question

Let $e_r$ and $p_r$ denote the $r$-th elementary symmetric function and power sum, respectively. Let $t$ be a formal variable and define $$Z := p_1t-p_3t^3/3+ p_5t^5/5- \cdots $$ Prove that the following equality of formal power series holds in $\Lambda[[t]]$:

$$tan(Z) = \frac{e_1t-e_3t^3+e_5t^5-\cdots} {1-e_2t^2+e_4t^4-\cdots}$$

$\textbf{My work:}$

We can show that $\exp(p_1t-p_2t^2/2+p_3t^3/3-\cdots) = \sum_{i=1}^\infty e_it^i \hspace{7mm} (1)$.

Note that $$tan(Z) = \frac{sin(Z)}{cos(Z)} = \frac{\frac{exp(iZ) - exp(-iZ)}{2i}}{\frac{exp(iZ) + exp(-iZ)}{2}} = \frac{\frac{exp(2iZ) - 1}{2i}}{\frac{exp(2iZ) + 1}{2}} = -i\frac{exp(2iZ) - 1}{exp(2iZ) + 1}$$

Then I have tried to find a way to use identity (1), but haven't found it so far since $2iZ$ is not equal to $p_1t-p_2t^2/2+p_3t^3/3-\cdots$

Any hints will be appreciated as I've bee working on this for a long time.

Answer 1

Starting from the identity

$$ \sum_{k=0}^{\infty} e_k t^k = \exp\left( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} p_k t^k \right), $$

we apply the substitutions $t \mapsto it$ and $t \mapsto -it$, respectively. Then

$$ A + iB = \exp(Y + iZ) \qquad\text{and}\qquad A - iB = \exp(Y - iZ), $$

where

\begin{align*} A &= \sum_{k=0}^{\infty} (-1)^k e_{2k} t^{2k}, & B &= \sum_{k=0}^{\infty} (-1)^k e_{2k+1} t^{2k+1}, \\ Y &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{2k} p_{2k} t^{2k}, & Z &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2k+1} p_{2k+1} t^{2k+1}. \end{align*}

From this, we obtain

$$ A = e^Y \cos Z, \qquad B = e^Y \sin Z $$

and hence the desired conclusion follows from

$$ \tan Z = \frac{\sin Z}{\cos Z} = \frac{B}{A}. $$

Answer 2

From a formal point of view, it is not too difficult.

$$Z=\sum _{i=0}^\infty (-1)^i\,\frac{ p_{2 i+1}}{2 i+1}\,\, t^{2 i+1}\tag 1$$ $$\tan(Z)=\sum _{i=0}^\infty a_i\, t^{2i+1}\tag 2$$ where the first coefficients are $$\left( \begin{array}{cc} 0 & p_1 \\ 1 & \frac{1}{3} \left( p_1^3-p_3\right) \\ 2 & \frac{1}{15} \left(2 p_1^5-5 p_1^2\,p_3+3 p_5\right) \\ 3 & \frac{1}{315} \left(17 p_1^7-70 p_1^4\,p_3+63 p_1^2\, p_5+35 p_1\,p_3^2-45 p_7\right) \\ \end{array} \right)$$ where you recognize the front coefficient $$(-1)^n \,\frac{4^n \left(1-4^n\right) B_{2 n}}{(2 n)!}$$ as in the expansion of $\tan(x)$.

Now, since $(2)$ does not contain a conatant term, transform the series as its corresponding $[2m+1,2m]$ Padé approximant (this is your rhs).

$$\frac{\sum_{i=0}^m b_i\,t^{2i+1}}{1+\sum_{i=1}^m c_i\,t^{2i} } $$

For example $$b_0=a_0$$ $$b_1=\frac{ a_1^2\,a_3-a_1\,a_2^2-a_0\,a_1\, a_4+a_0\, a_2\,a_3}{a_1\, a_3-a_2^2}$$

$$c_1=\frac{a_1\, a_4-a_2\, a_3}{a_2^2-a_1\, a_3}$$ $$c_2=\frac{a_2\, a_4-a_3 ^2 }{a_2^2-a_1\, a_3}$$