# verifying the Taylor expansion of ln(1+x) satisfies the properties of logarithm

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:

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

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

• 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 $(\frac{x-y}{1+y}).$ Sep 29 at 7:11
• 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. Sep 29 at 7:28

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

• 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. Sep 30 at 3:23