Tangent Mania

Let $|x| < 1$ .

$f(x) = \lim_{n \rightarrow \infty} \sum_{j = 1}^{n} (\dfrac{j}{n})^n (\dfrac{1}{e^x})^{n - j} =$ $\lim_{n \rightarrow \infty} \sum_{j = 0}^{n - 1} (1 - \dfrac{j}{n})^n (\dfrac{1}{e^x})^{j} =$ $\sum_{n = 0}^{\infty} (\dfrac{1}{e^{x + 1}})^n = \dfrac{e^{x + 1}}{e^{x + 1} - 1}$ .

$g(x) = \sum_{n = 0}^{\infty} (-1)^{n} (\dfrac{1}{e^{x + 1}})^n = -\sum_{n = 0}^{\infty} (\dfrac{1}{e^{1 - x}})^n = -\dfrac{e^{1 - x}}{e^{1 - x} - 1}$ .

$\dfrac{d}{dx}(f(x)) = \dfrac{-e^{x + 1}}{(e^{x + 1} - 1)^2}$ and $\dfrac{d}{dx}(g(x)) = \dfrac{-e^{1 - x}}{(e^{1 - x} - 1)^2}$ .

Let $(0 \leq a < 1) \implies \dfrac{d}{dx}(f(x))|_{x = a} = -\dfrac{e^{a + 1}}{(e^{a + 1} - 1)^2} = \dfrac{d}{dx}(g(x))|_{x = -a}$ .

For $a = 0$ :

$A: (0, \dfrac{e}{e - 1})$ and $B: (0, -\dfrac{e}{e - 1})$

$(e - 1)^2 y + ex = e(e - 1)$ is tangent to $f(x)$ at point $A$ and $(e - 1)^2 y + ex = -e(e - 1)$ is tangent to $g(x)$ at point $B$ .

The line passing thru $B: (0, -\dfrac{e}{e - 1})$ and $\perp$ to $AC$ has slope $m_{\perp} = \dfrac{(e - 1)^2}{e}$ and $\overleftrightarrow{BC}$ is $ey - (e - 1)^2 x = -\dfrac{e^2}{e - 1}$ .

Let $D: (x_{0},y_{0})$ be intersection point of the lines:

$ey - (e - 1)^2 x = -\dfrac{e^2}{e - 1}$

$(e - 1)^2 y + ex = e(e - 1)$

Solving the system we obtain:

$x_{0} = \dfrac{2e^2(e - 1)}{(e - 1)^4 + e^2}$ and $y_{0} =\dfrac{e((e - 1)^4 - e^2)}{(e - 1)(e^2 + (e - 1)^4)}$

Using $B: (0, -\dfrac{e}{e - 1})$ and $D: (x_{0},y_{0})$

$\implies d^2 = (BD)^2 = \dfrac{4e^4(e - 1)^2}{((e - 1)^4 + e^2)^2} + \dfrac{e^2}{(e - 1)^2}(\dfrac{4(e - 1)^8}{((e - 1)^4 + e^2)^2}) = \dfrac{4e^4 (e - 1)^4 + 4(e - 1)^8 e^2}{(e - 1)^2 ((e - 1)^4 + e^2)^2} =$ $\dfrac{4e^2(e - 1)^2}{((e - 1)^4 + e^2)^2}((e - 1)^4 + e^2) \implies d = \dfrac{2e(e - 1)}{\sqrt{e^2 + (e - 1)^4}} \approx \boxed{2.327670}$ .