I was recently studying exponential and logarithmic series on Higher Algebra by Hall and Knight. While reading it I encountered a solved example:—
To expand $\log _{e}(1+x)$ in ascending powers of $x$.
Let us suppose $e^c=a$, so that $c=\log _{e} a$; by substituting for c we obtain
$a^{y}=1+y \log _{e} a+\frac{y^{2}\left(\log _{e} a\right)^{2}}{2 !}+\frac{y^{3}\left(\log _{e} a\right)^{3}}{3 !}+\ldots$
In this series write $1+x$ for $a$; thus
$(1+x)^{y}=1+y \log _{e}(1+x)+\frac{y^{2}}{2 !}\left\{\log _{e}(1+x)\right\}^{2}+\frac{y^{3}}{3 !}\left\{\log _{e}(1+x)\right\}^{3}+\text{ }\ldots .(1)$
Also by Binomial Theorem when $x<1$ we have
$(1+x)^{y}=1+y x+\frac{y(y-1)}{2 !} x^{2}+\frac{y(y-1)(y-2)}{3 !} x^{3}+\text{ }\ldots .(2)$
Now in $(2)$ the coefficient of y is
$x+\frac{(-1)}{1 \cdot 2} x^{2}+\frac{(-1)(-2)}{1 \cdot 2 \cdot 3} x^{3}+\frac{(-1)(-2)(-3)}{1 \cdot 2 \cdot 3 \cdot 4} x^{4}+\ldots$
That is, $x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots$
Equate this to the coefficient of y in $(1)$; thus we have
$\log _{e}(1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots$
Can anyone please explain me how in this "Now in $(2)$ the coefficient of y is" step the coefficient of $y$ is found to be the series given below?
$x+\frac{(-1)}{1 \cdot 2} x^{2}+\frac{(-1)(-2)}{1 \cdot 2 \cdot 3} x^{3}+\frac{(-1)(-2)(-3)}{1 \cdot 2 \cdot 3 \cdot 4} x^{4}+\ldots$
Can anyone provide a proof of this?
Any help would be appreciated.
Edit:— Thanks to AkivaWeinberger for explaining me.
In the later articles the book is repeatedly stating about the coefficient of $y^2$ without showing what it is or how it is to be found out.
Can anyone please show me how to find the coefficient of$y^2$ from (2)?