# How to show that $x=e^y$, where $y=\sum \limits_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n$, not using logarithm?

Let $|x-1|<1$ and $y= \displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} (x-1)^n$.

How to show, without using the fact that $y=\ln x$, but using properties of absolutely convergent series, that $e^y=x$?

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Define $$f(x)=-\sum_{n=1}^\infty\frac{(1-x)^n}{n}\tag{1a}$$ so that $$f(1-x)=-\sum_{n=1}^\infty\frac{x^n}{n}\tag{1b}$$ After which, your question becomes: show that $x=e^{f(x)}$. \begin{align} f((1-u)(1-v)) &=f(1-(u+v-uv))\\ &=-\sum_{n=1}^{\infty}\frac{(u+v-uv)^n}{n}\\ &=-\sum_{n=1}^{\infty}\frac{(u(1-v)+v)^n}{n}\\ &=-\sum_{n=1}^{\infty}\sum_{k=0}^n\frac{1}{n}\binom{n}{k}(u(1-v))^kv^{n-k}\\ &=-\sum_{n=1}^\infty\frac{1}{n}v^n-\sum_{k=1}^\infty\sum_{n=k}^\infty\frac{1}{k}\binom{n-1}{k-1}(u(1-v))^kv^{n-k}\\ &=-\sum_{n=1}^\infty\frac{1}{n}v^n-\sum_{k=1}^\infty\frac{1}{k}\left(\frac{u(1-v)}{1-v}\right)^k\\ &=-\sum_{n=1}^\infty\frac{1}{n}v^n-\sum_{k=1}^\infty\frac{1}{k}u^k\\ &=f(1-u)+f(1-v)\tag{2} \end{align} Thus, $(2)$ says that $$f(xy)=f(x)+f(y)\tag{3}$$ Equation $(3)$ insures that $$f(1)=f(x\cdot1)-f(x)=0\tag{4}$$ Equations $(3)$ and $(4)$ yield \begin{align} f'(x) &=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\\ &=\lim_{h\to0}\frac{f(x(1+h/x))-f(x)}{h}\\ &=\frac{1}{x}\lim_{h\to0}\frac{f(1+h/x)-f(1)}{h/x}\\ &=\frac{1}{x}f'(1)\tag{5} \end{align} Let $g(x)=e^{f(x)}$, then $g(1)=1$ and \begin{align} g'(x) &=g(x)f'(x)\\ &=g(x)\frac{1}{x}f'(1)\tag{6} \end{align} Note that equation $(6)$ implies \begin{align} x^{f'(1)+1}\left(x^{-f'(1)}g(x)\right)' &=-f'(1)g(x)+xg'(x)\\ &=0\tag{7} \end{align} and equation $(7)$ implies $$x^{-f'(1)}g(x)=C\tag{8}$$ Since $g(1)=1$, $C=1$, and therefore, $$g(x)=x^{f'(1)}\tag{9}$$ Referring back to equation $(1)$, we see that $f'(1)=1$, so we have $$g(x)=x\tag{10}$$ as requested.

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Note that the only reference to $e^x$ or any of its properties is for equation $(6)$ :-) – robjohn Dec 11 '11 at 21:16
It is a great answer. Thanks a lot. – L.T Dec 11 '11 at 22:40

Given a power series, you can integrate (or) differentiate term by term within its radius of convergence, where it converges absolutely. The series \begin{align} y = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} (x-1)^n \end{align} converges absolutely for $|x-1| < 1$. Hence, term by term differentiation gives us \begin{align} \frac{dy}{dx} = \sum_{n=1}^{\infty} (-1)^{n-1} (x-1)^{n-1} = \sum_{n=1}^{\infty} (1-x)^{n-1} = \frac1{1-(1-x)} = \frac1x \end{align} for $|x-1| < 1$. Now integrate to get that $y = \log(x) + c$. Set $x=1$ in the initial given equation to get that $y=0$ and hence $c=0$. Hence, we get that, $$x = e^y$$

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Thanks a lot for answer. – L.T Dec 10 '11 at 14:03
Instead of integrating, we differentiate $e^y=x$ to get $y' e^y=1$, which gives $y'=1/x$. Adding the boundary condition $y(1)=0$, we are done since both expressions are solutions of the same differential equation with a provably unique solution. – Peteris Dec 10 '11 at 15:24
If you want to completely avoid using logs, you can cheat at the end. $\frac{dy}{dx}=\frac{1}{x}$ can be rewritten as $\frac{dx}{dy}=x$. Then use the substitution $z=\frac{x}{e^y}$ to get $\frac{dz}{dy}=0$. – N. S. Dec 10 '11 at 15:27
+1, cool and clear. – NoChance Dec 10 '11 at 15:50