I want to verify the Taylor expansion of $\ln(1+x)$ does satisfy the properties of logarithm.


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:


$f(1+x)-f(1+y)= (x-y)-\frac{1}{2}(x^2-y^2)+\frac{1}{3}(x^3-y^3)-\frac{1}{4}(x^4-y^4)+\cdots$

But I couldn't make them equal. Please help. Thanks.

  • $\begingroup$ Just a tip on formatting with MathJax. When stuff inside parentheses are higher than the parentheses themselves, use \left( and \right) rather than just ( and ). Then it will look like $\left(\frac{x-y}{1+y}\right)$ rather than $(\frac{x-y}{1+y}).$ $\endgroup$ Sep 29 at 7:11
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    $\begingroup$ I would try expanding the powers in the first series using the Binomial Theorem, and do some re-arranging of terms. Also, you have all the $(1+y)$ denominators appearing in the first series, and not in the second, I'd try eliminating them first, using $1/(1+y)=\sum_j (-y)^j$. I think it will be unavoidable to manipulate some resulting double sum. $\endgroup$
    – DominikS
    Sep 29 at 7:28

1 Answer 1


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 $$f(1+\frac{x-y}{1+y})-f(1+x)=-f(1+y).$$

  • $\begingroup$ Great. This method may be the only way, because it seems too difficult to verify without calculus. The method also reveals the fundamental insight that logarithms must satisfy $f(kx)-f(x)$ being a constant, ie numbers that have the same ratio, their logarithms must have the same difference. Thanks. $\endgroup$ Sep 30 at 3:23

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