基于欧几里得度量的函数间距离构造及反求函数问题咨询

Hey there, let's tackle these two problems using Euclidean distance as requested. I'll break them down clearly with both mathematical reasoning and practical implementation tips.

Problem 1: Constructing $f(x)$ (Minimum Distance from Line $y=x$ to Curve $y=e^x$ at Point $x$) #

First, let's clarify what $f(x)$ represents: for a given $x$, we're looking for the shortest Euclidean distance between the point $(x, x)$ on the line $y=x$ and the curve $y=e^s$ (where $s$ is the independent variable for the exponential curve).

Mathematical Background #

The Euclidean distance between $(x, x)$ and a point $(s, e^s)$ on the curve is:
$$
D(s) = \sqrt{(x - s)^2 + (x - e^s)2}
$$
Minimizing $D(s)$ is equivalent to minimizing the squared distance (to avoid square roots and simplify differentiation):
$$
D^2(s) = (x - s)^2 + (x - e^s)2
$$
Taking the derivative with respect to $s$ and setting it to zero (for the minimum):
$$
\frac{d}{ds}D^2(s) = -2(x - s) - 2(x - e^s)es = 0
$$
Simplify to get the critical point condition:
$$
(x - s) + (x - e^s)es = 0 \implies x(1 + e^s) = s + e^{2s} \implies x = \frac{s + e^{2s}}{1 + e^s}
$$

Efficient Implementation Method #

Since this is an implicit equation (we can't solve for $s$ in terms of $x$ analytically), we use Newton-Raphson iteration—a fast, convergent numerical method—to find $s$ for any given $x$:

1. Initial Guess:

   • For large $x$: $s \approx \ln(x)$ (since $e^s$ dominates, $e^{2s}/es = e^s \approx x$)
   • For small $x$ (e.g., $x \leq 1$): start with $s = 0$ or $s = x$ (adjust based on initial test)

2. Newton Iteration:
   Define the function we want to root:
   $$
   g(s) = s + e^{2s} - x(1 + e^s)
   $$
   Its derivative is:
   $$
   g'(s) = 1 + 2e^{2s} - x e^s
   $$
   Iterate using:
   $$
   s_{n+1} = s_n - \frac{g(s_n)}{g'(s_n)}
   $$
   Stop when $|s_{n+1} - s_n| < 10^{-8}$ (or your desired precision).

3. Compute the Distance:
   Once you have the converged $s$, plug back into the distance formula:
   $$
   f(x) = \sqrt{(x - s)^2 + (x - e^s)2}
   $$

This method is efficient because Newton-Raphson typically converges in 3-5 iterations for most $x$ values, making it suitable for real-time or batch computations.

Problem 2: Finding Closed-Form $g(x)$ Given $e^x$ is the Shortest Distance Function #

First, let's clarify the interpretation: we assume $e^x$ is the shortest distance from the point $(x, g(x))$ on curve $g(x)$ to the line $y=x$. (If we interpreted it as the distance from $(x,x)$ to $g(x)$, no solution exists for $x > \ln(\sqrt{2})$ since the maximum possible shortest distance from a point on $y=x$ to any curve can't exceed $\sqrt{2}$, but $e^x$ grows beyond that.)

Mathematical Derivation #

The shortest distance from a point $(x, g(x))$ to the line $y=x$ (rewritten as $x - y = 0$) is given by the Euclidean distance formula for a point to a line:
$$
\text{Distance} = \frac{|x - g(x)|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - g(x)|}{\sqrt{2}}
$$
We know this distance equals $e^x$, so:
$$
\frac{|x - g(x)|}{\sqrt{2}} = e^x \implies |x - g(x)| = \sqrt{2}e^x
$$
Removing the absolute value gives two valid closed-form solutions:
$$
g(x) = x + \sqrt{2}e^x \quad \text{or} \quad g(x) = x - \sqrt{2}e^x
$$

To verify: take $g(x) = x + \sqrt{2}e^x$, the distance from $(x, x+\sqrt{2}e^x)$ to $y=x$ is $\frac{|x - (x+\sqrt{2}e^x)|}{\sqrt{2}} = \frac{\sqrt{2}e^x}{\sqrt{2}} = e^x$, which matches the requirement perfectly.

内容的提问来源于stack exchange，提问作者shai horowitz