Solve Exponential Fraction Equation for $α$ ($C$, $m$)

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SUMMARY

The discussion centers on solving the exponential fraction equation for α in the context of creating a binary matrix X with specific properties. The equation derived is $\frac{1-e^{(m+1)*\alpha}}{1-e^{\alpha}}= C + 1$, which can be transformed into a polynomial form: $x^{m+1} - (C+1)x + C = 0$, where $x = e^\alpha$. Exact solutions exist only for $m+1 \le 4$, with the trivial solution α = 0. For larger values of m, approximation algorithms are necessary to estimate α based on given C values.

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Stijn
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I want to create a binary matrix (${X}_{m*n}$) containing $C$ ones ($||X|| = C$). Additionally I want to have the number of elements of each row ($m$) to have an exponential form. This is: for each row the number elements needs to be equal to $e^{alpha * i}$ or in symbols: ||${X}_{i}$|| = $e^{alpha*i}$.

As the total number of ones is known ($C$), we can say that: $||X|| =\sum_{i=1}^{m}e^{alpha*i} = C$
Based on the function of the sum of geometric series ($\sum_{a=0}^{A-1} {b}^{a}=\d{1-{b}^{A}}{1-b}$) we can rewrite the equation as:

$\frac{1-e^{(m+1)*\alpha}}{1-e^{\alpha}}= C + 1$

Is there a way to solve alpha out of this function and write the equation as: $\alpha = f(C, m)$?

Thanks in advance!
 
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Hi Stijn! Welcome to MHB! ;)

Let me restrict myself to your immediate question:
Stijn said:
$\frac{1-e^{(m+1)*\alpha}}{1-e^{\alpha}}= C + 1$

Is there a way to solve alpha out of this function and write the equation as: $\alpha = f(C, m)$?

We can rewrite it as:
$$x^{m+1} - (C+1)x + C = 0$$
where $x = e^\alpha$.
Such a polynomial only has closed form solutions when $m+1 \le 4$.
We do have the solution $x=1$ or $\alpha=0$ though, meaning we can reduce the polynomial by one degree, so we can solve it up to $m=4$.
Beyond that we need approximation algorithms.
 
Hi,

Thank you for your quick response. As I was suspecting, no exact solution exists (other than $\alpha = 0$). I am able to estimate the values for different m and C values.
 

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