I have question on a derivation of the logarithmic series that I'm still unable to understand. I've already seen this post on it, but I'm still slightly confused. This derivation is from Higher Algebra by Henry Sinclair.
The derivation is like so:
$$(1 + x)^y = 1 + y\ln(1 + x) + \frac{y^2}{2!}\{\ln(1 + x)\}^2 + \frac{y^3}{3!}\{\ln(1 + x)\}^3 + ... (1)$$ By the Binomial theorem, when $x < 1$, we have $$(1 + x)^y = 1 + yx + \frac{y(y-1)}{2!}x^2 + \frac{y(y-1)(y-2)}{3!}x^3 + ... (2)$$ Now in (2), the coefficient of $y$ is $$x + \frac{-1}{1 \times 2}x^2 + \frac{(-1)(-2)}{1 \times 2 \times 3}x^3 + \frac{(-1)(-2)(-3)}{1 \times 2 \times 3 \times 4}x^4 + ...$$ That is, $$x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...(3)$$ Equate this to the coefficient of $y$ in (1); thus we have $$\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$$
Statement (1) is obtained by plugging in $1 + x$ into the Exponential Theorem. What I'm struggling to understand (from the previous post and this derivation directly from the book) is what it means by the "coefficient of $y$ in (2)", and how that is equivalent to (3).
Again, I've seen all 3 answers given to the original post, and I still don't precisely understand what's going on. If someone could explain it in clearer terms, that would be helpful.
Thanks.
By "the coefficient of $y$ in (2)", he means the coefficient on the $y$ term, i.e. the $y^1$ term, after you expand out all the polynomials in the numerators of the series.
What's going on is that there are two different series for $(1+x)^y$.
The first series is the one given in the first line. It doesn't say where it comes from, but in fact it comes from the series expansion for $\mathrm{e}^x$. That series is $$ \mathrm{e}^x = \sum_{k=0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$
$\mathrm{e}^x$ and $\ln{x}$ are inverses of each other, so you can use this fact to rewrite $(1+x)^y$ with a base of $\mathrm{e}$: that is, $(1+x)^y = \mathrm{e}^{\ln((1+x)^y)} = \mathrm{e}^{y\ln(1+x)}$. So you get (1) by plugging $y\ln(1+x)$ into the series for $\mathrm{e}^x$: $$ (1+x)^y = \mathrm{e}^{y\ln(1+x)} = 1 + y\ln(1+x) + \frac{y^2(\ln(1+x))^2}{2!} + \frac{y^3(\ln(1+x))^3}{3!} + \cdots \tag{1} $$
The second series comes from an entirely different source - it is the binomial expansion of $(1+x)^y$. \begin{align} (1+x)^y &= \sum_{k=0}^\infty\binom{y}{k}x^k \\ &= 1 + yx + \frac{y(y-1)}{2!}x^2 + \frac{y(y-1)(y-2)}{3!}x^3 + \cdots \tag{2} \end{align}
You have 2 different series and both of them independently equal $(1+x)^y$. So that means they must actually be the same series. The only way for this to happen, is if all the coefficients of each power of $y$ are equal, i.e. if $1, y, y^2, y^3, ...$ have the same coefficient in both series (1) and series (2).
In particular, the coefficient of $y$ alone (that is, disregarding whatever is going on with $y^2, y^3, ...$) has to be the same in series (1) and in series (2).
For series (1), it's simple: the coefficient can be read directly off the $y$ term, as $\ln(1+x)$.
For series (2), it's a little more complicated, but it's not too tricky to find a pattern, and see what would happen if were were to expand out all those polynomials in $y$ seen in the numerators, and collect all the coefficients of $y=y^1$ together. Let's look at the first few terms term-by-term, so that the pattern can emerge very clearly.
$yx$: the coefficient is $x$.
$\frac{y(y-1)x^2}{2!}$: if you expand it out, you get $y^2x^2/2! - yx^2/2!$, so the coefficient is $-x^2/2!$.
$\frac{y(y-1)(y-2)x^3}{3!}$: if you expand it out (this time, for reasons you're about to see, I'll use a very verbose sort of notation), you get
\begin{align} \frac{y(y-1)(y-2)}{3!}x^3 &= \frac{y(y+(-1))(y+(-2))}{3!}x^3 \\ &= \frac{(y^2+(-1)y)(y-2)}{3!}x^3 \\ &= \frac{(y^3+(-1)y^2+(-2)y^2+(-1)(-2)y)}{3!}x^3 \\ &= \frac{x^3}{3!}y^3 + \frac{(-1-2)x^3}{3!}y^2 + \frac{(-1)(-2)x^3}{3!}y \end{align}
The coefficient of interest on $y$ from this term is $(-1)(-2)x^3/3!$.
Hopefully this suffices to make the pattern clear - if you were to expand series (2) and collect terms in $y$, you would get exactly (3).
Finally, recalling that the coefficients of each of $y, y^2, y^3, ...$ have to be equal in both series for them to be both series for $(1+x)^y$, then we can focus just on the coefficient of $y$, setting them equal while disregarding the rest, to get the final result:
\begin{align} \ln(1+x) &= x + \frac{(-1)}{2!}x^2 + \frac{(-1)(-2)}{3!}x^3 + \frac{(-1)(-2)(-3)}{4!}x^4 + \cdots \\ \Rightarrow \ln(1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots = \sum_{k=1}^\infty \frac{(-1)^{k-1}x^k}{k} \end{align}