According to this (equation 21 and 26) how to show that \begin{gather} \ln(1+\Pi_1t+\Pi_2t^2+\Pi_3t^3+...) =\Pi_1t+\frac{1}{2}(-\Pi^2_1+2\Pi_2)t^2+\frac{1}{3}(\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3)t^3+.. .=\sum_{k=1}^{\infty}\frac{s_k}{k}t^k \end{gather} $s_1=\Pi_1 \\ s_2=-\Pi^2_1+2\Pi_2 \\ s_3=\Pi^3_1-3\Pi_1\Pi_2+3\Pi_3 $
And, I've been curious about how did they derive this equation $(-1)^{p-1}s_p = $ \begin{vmatrix} \Pi_1 & 1 & 0 & 0 &\dots & 0 \\ 2\Pi_2 & \Pi_1 & 1& 0 &\ddots &0 \\ 3\Pi_3 & \Pi_2 & \Pi_1 & 1 &\ddots & 0 \\ 4\Pi_4 & \Pi_3 &\Pi_2 &\Pi_1 &\ddots&0 \\ \vdots & \vdots & \vdots & \vdots & \ddots &1 \\ p\Pi_{p} & p\Pi_{p-1} & p\Pi_{p-2} & p\Pi_{p-3} & \dots & \Pi_1 \end{vmatrix} Is there any elementary way to show? Thanks in advance
Given a generating function $$ f(x) := 1+a_1t+a_2t^2+a_3t^3+\dots \tag{1} $$ whose logarithm is $$ \ln(f(x)) = \sum_{k=1}^{\infty}\frac{s_k}{k}t^k, \tag{2} $$ differentiate both sides to get $$ \frac{d}{dx} \ln(f(x)) = f'(x)/f(x) = \sum_{k=1}^{\infty}s_k t^{k-1}. \tag{3} $$ Multiply both sides by $\,f(x)\,$ to get $$ f'(x) = \sum_{k=1}^\infty k\,a_k\,t^{k-1} = \left(1 + \sum_{k=1}^\infty a_k\,t^k\right) \left(\sum_{k=1}^{\infty}s_k t^{k-1}\right). \tag{4} $$ For each $\,n>0\,$ this gives a system of equations linear in $\,s_1,s_2,\dots,s_n\,$ with coefficients using $\,a_1,a_2,\dots,a_n.\,$ Use Cramer's rule to solve for $\,s_n\,$ which leads to the MathWorld determinant $$(-1)^{n-1}s_n \!=\! \begin{vmatrix} 1\,a_1 & 1 & 0 & 0 &\dots & 0 \\ 2\,a_2 & a_1 & 1& 0 &\ddots & 0 \\ 3\,a_3 & a_2 & a_1 & 1 &\ddots & 0 \\ 4\,a_4 & a_3 & a_2 & a_1 &\ddots& 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots &1 \\ n\,a_n & a_{n-1} & a_{n-2} & a_{n-3} & \dots & a_1 \end{vmatrix}. \tag{5} $$
Here is an explicit example. Let $\,n=3.\,$ The linear equations are $$ 1s_1 \!=\! 1a_1,\;\; a_1s_1 \!+\! 1s_2 \!=\! 2a_2, \;\; a_2s_1 \!+\! a_1s_2 \!+\! 1s_3\!=\! 3a_3. \tag{6} $$ Setting this up as a matrix equation we get $$ \begin{bmatrix}1 & 0 & 0 \\a_1 & 1 & 0\\ a_2 & a_1 & 1 \end{bmatrix} \begin{bmatrix} s_1 \\ s_2 \\ s_3 \end{bmatrix} = \begin{bmatrix} 1\,a_1 \\ 2\,a_2 \\ 3\,a_3 \end{bmatrix}\!. \tag{7} $$ Cramer's rule solving for $\,s_3\,$ and using the denominator $3\times 3$ determinant $=1$ gives the solution in equation $(5)$ for $\,n=3\,$ and that is $$ s_3 = a_1^3 -3a_1a_2 + 3a_3 = (-1)^{3-1}s_3 = \begin{vmatrix} 1\,a_1 & 1 & 0 \\ 2\,a_2 & a_1 & 1 \\ 3\,a_3 & a_2 & a_1 \end{vmatrix}. \tag{8} $$