Question

If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0$, show that $\log (a-bx+c{x}^2)=\log a+(\alpha +\beta )x-\frac{{\alpha }^2+{\beta }^2}{2}{x}^2+\frac{{\alpha }^3+{\beta }^3}{3}{x}^3-...$

Answer 1

Since $\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$, we have

$a-bx+c{x}^2=a\{1+(\alpha +\beta )x+\alpha \beta x^2\}$
$=a(1+\alpha x)(1+\beta x)$

Thus $a-bx+c{x}^2$ $=a(1+\alpha x)(1+\beta x)$

Taking $\log$ on both sides, we get

$\log (a-bx+c{x}^2)$ $=\log \left[a\right(1+\alpha x\left)\right(1+\beta x\left)\right]$

$\log (a-bx+c{x}^2)=\log a+\log (1+ax)+\log (1+\beta x)$

$=\log a+ax-\beta \frac{{\alpha }^2{x}^2}{2}+\frac{{\alpha }^3{x}^3}{3}-...+\beta x-\frac{{\beta }^2{x}^2}{2}+\frac{{\beta }^3{x}^3}{3}-...$

$=\log a+(\alpha +\beta )x-\frac{{\alpha }^2+{\beta }^2}{2}{x}^2+\frac{{\alpha }^3+{\beta }^3}{3}{x}^3-...$