Background:
I want to verify the Taylor expansion of $\ln(1+x)$ does satisfy the properties of logarithm.
Question:
Let $f(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$. Without using the fact that $f(1+x)$ is $\ln(1+x)$, because that is what I want to verify, I want to manipulate the polynomials to show that:
$f(\frac{1+x}{1+y}) = f(1+x)-f(1+y)$.
My attempt:
$f(\frac{1+x}{1+y})=f(1+\frac{x-y}{1+y})=(\frac{x-y}{1+y})-\frac{1}{2}(\frac{x-y}{1+y})^2+\frac{1}{3}(\frac{x-y}{1+y})^3-\frac{1}{4}(\frac{x-y}{1+y})^4+\cdots$
$f(1+x)-f(1+y)= (x-y)-\frac{1}{2}(x^2-y^2)+\frac{1}{3}(x^3-y^3)-\frac{1}{4}(x^4-y^4)+\cdots$
But I couldn't make them equal. Please help. Thanks.
The derivative of $f(1+x)$ within the circle of convergence is clearly $\frac{1}{1+x}$.
So now fix $y$ and note that the derivative of $f(1+\frac{x-y}{1+y})-f(1+x)$ is $$\frac{1}{1+\frac{x-y}{1+y}}\frac{1}{1+y}-\frac{1}{1+x}=0.$$ Hence $f(1+\frac{x-y}{1+y})-f(1+x)$ is constant; evaluating at $x=y$ we see that $$f(1+\frac{x-y}{1+y})-f(1+x)=-f(1+y).$$