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How to find the minimum distance between the two curves $e^x$ and $\ln (x)$ ?

No idea how to find the common normal in this case.Help!

juantheron
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    Consider that the two curves are reflections of one another across the line $y = x$. Then convince yourself that at the closest approach of the curves, the tangents to each curve should be parallel to this line (and the normals perpendicular). This should allow you to locate the pair of points on the curve that minimize the distance. – Dylan Murphy Oct 27 '15 at 16:50
  • Look at $e^x - \ln x$. The minimum of this function will be at $W(1)$, which is about $0.567$ – Shailesh Oct 27 '15 at 16:52
  • @DylanMurphy you mean $e^x=1$ and $1/x=1$ would provide the x coordinates of those points right? –  Oct 27 '15 at 16:52
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    @SanchayanDutta Huh? Not at all. – Did Oct 27 '15 at 16:55
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    Shailesh: That minimizes the difference between the functions, but not the distance in the plane between the two curves. The distance between $(0.567, \exp(0.567))$ and $(0.567, \log(0.567))$ is approximately 2.33, but the true minimum distance between the curves is $\sqrt{2}$. – Dylan Murphy Oct 27 '15 at 17:01
  • Sanchayan: Yes. See also Omnomnomnom's answer below, which gives the same conclusion by considering normals rather than tangents. – Dylan Murphy Oct 27 '15 at 17:01

2 Answers2

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Hint:

  • Each graph forms the boundary of a convex region. The line segment of minimal distance between the two curves must therefore be unique.

  • The picture is symmetric about the line $y = x$, so reflecting the line segment through this line must yield another line segment of minimal length.

We can deduce that the slope of the common normal must therefore be $-1$.

Ben Grossmann
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enter image description here

Minimum Distance between Two curve is Distance between two parallel tangents

drawn at point $P$ and $Q$ on the curves.

and Here $f(x)=e^x$ and $f(x)=\ln(x)$ are Inverse of each other .

So it is Symmetrical about $y=x$ Line.

Let We take any point $P(x_{1},y_{1})$ on $f(x) = \ln(x)\;,$ Then Slope of tangent at $P(x_{1},y_{1})$ to the curve

$f(x)=\ln(x)$ is $\displaystyle f'(x_{1})=\frac{1}{x_{1}}$ and Slope of line $f(x)=x$ is $f'(x)=1$

Now If line $f(x)=x$ and Tangent at point $P(x_{1},y_{1})$ to the curve are parallel.

Then $\displaystyle \frac{1}{x_{1}} = 1\Rightarrow x_{1}=1$ Now We get $y_{1}=f_{x_{1}}=\ln(1) =0$

So we Get Coordinate of $P(1,0)$

Now Distance between Line $x-y=0$ from $P(1,0)$ is $\displaystyle = \frac{1}{\sqrt{2}}$

So $PQ = 2\times $ of Distance of line from $\displaystyle P(1,0) = \frac{2}{\sqrt{2}}=\sqrt{2}$ Unit.

juantheron
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  • Beautiful.Just beautiful :-)! –  Oct 27 '15 at 17:15
  • Very nice, but it would be interesting to have an analytic (not geometric) proof that shows how we get $\sqrt{2}$ from $e^x$ and $\ln(x)$. Perhaps that should be another Question. – Keith McClary Oct 28 '15 at 04:49
  • @KeithMcClary Analytically, one wants to minimize $$f(x,u)=(e^x-\ln u)^2+(u-x)^2$$ over $x$ real and $u$ positive real. Computing the gradient of $f$ and reordering somewhat the equations, one sees that this can only happen when $$e^x=\frac1u\qquad \ln u=\frac{1-u^3}{u+u^2}.$$ The sign of the last fraction shows the second equation holds only when $u=1$, then the first equation yields $x=0$, then $f(0,1)=2$ is the square of the distance between the two curves, QED. – Did Oct 29 '15 at 10:53
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    Why did you equated the slope of $y=x$ to $\ln x$ ? You should have equated the slope of $\ln x$ to $e^x$ instead ? How did you already know their parallel tangents will also be parallel to $y=x$. Please give a rigorous proof of that. – An_Elephant Nov 30 '23 at 18:52