Another Take On Discrete

From the post "Quantized x, Discretely",

$\int { \cfrac { 1 }{ \psi  }  } d\, \psi +\int { \int { \cfrac { 1 }{ \psi ^{ 2 } }  }  } \left( d\, \psi  \right) ^{ 2 }=n^{ 2 }ln(x)+Ax+C$

where both $A$ and $C$ are constants for all $x$.  Simplifying further we have,

$n^{ 2 }ln(x)+Ax+C=0$ --- (*)

when $x=1$,

$C=-A$

So the constrain on $x$ is,

$x=e^{ -\cfrac { Ax+C }{ n^{ 2 } }  }$

$x=e^{ -\cfrac {A( x-1) }{ n^{ 2 } }  }$

This is wrong because $x=1$ will then always be a solution irrespective of $n$. And we have a unit circle onto which we can pack infinite number waves, as $n\rightarrow\infty$.  $x=1$ is not part of the final solution developed previously.

If we consider,

$x\rightarrow0$

$n\rightarrow0$

As,

$\lim\limits_{x\rightarrow0 \, n\rightarrow0}n^{ 2 }ln(x)=0^{ 2 }.\infty=0$

(*) becomes,

$0^{ 2 }.\infty+A.0+C=0$

$C=0$

And the constrain on $x$ is,

$x=e^{ -\cfrac { Ax}{ n^{ 2 } }  }$

And the constrain on $f$ is,

$f=\cfrac{nc}{2\pi}e^{ \cfrac { A }{ 2\pi\,n } \cfrac { c }{ f }  }$

by substituting $2\pi x=n\lambda$ and $\lambda=\cfrac{c}{f}$

A plot of y=n/(2*Pi)*(e^{1/(2*Pi*n*x)}) and y=n/(2*Pi)*(e^{8/(2*Pi*n*x)}) together with the line y=x, are shown below.

The value of $f$ for $n=1$ is folded up for larger values of $A=8$.  A plot for high value of $A=30$, y=n/(2*Pi)*(e^{30/(2*Pi*n*x)}) with y=x is shown below.

More intersections between $y=x$ and $y=\cfrac{nc}{2\pi}e^{ \cfrac { A }{ 2\pi\,n } \cfrac { c }{ x }  }$ are folded up producing pairs of closely spaced solutions of frequencies, $f$, when $A$ is large.

Note: In the plot for $f$, $c$ was set to $1$.  The effect is the same as scaling $f$ by $\cfrac{1}{c}$.