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Explicit solution to this equation
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Simple problem: The equation below has two solutions for $x$ for a range of parameter values (see below). This is straightforward to see graphically. I need to find both of these solutions. Some additional information:
- Approximations are OK if necessary
- Ideally both solutions will be found, but ideas for finding either individually are also helpful
- In the equation below: $k approx 0.1$, $n approx 2.5$
The equation:
$frac{1}{n} e^{-n(1-e^{-x/n})}e^{frac{x}{n}} = 1-e^{ (x/n^2)(1-e^{-x/n})^{-1}e^{frac{x}{n}}k}$
polynomials systems-of-equations
asked Mar 21 at 21:21
JohnJohn
255
$endgroup$
add a comment |
$begingroup$
Simple problem: The equation below has two solutions for $x$ for a range of parameter values (see below). This is straightforward to see graphically. I need to find both of these solutions. Some additional information:
- Approximations are OK if necessary
- Ideally both solutions will be found, but ideas for finding either individually are also helpful
- In the equation below: $k approx 0.1$, $n approx 2.5$
The equation:
$frac{1}{n} e^{-n(1-e^{-x/n})}e^{frac{x}{n}} = 1-e^{ (x/n^2)(1-e^{-x/n})^{-1}e^{frac{x}{n}}k}$
polynomials systems-of-equations
asked Mar 21 at 21:21
JohnJohn
255
$endgroup$
$begingroup$
If approximations are OK, why don't you use en.wikipedia.org/wiki/Bisection_method?
$endgroup$
– Andrei Rykhalski
Mar 21 at 22:40
$begingroup$
Thanks for the response, @AndreiRykhalski. Solving numerically isn't the issue, though---I need the solution in terms of $k$ and $n$. I don't think I see hwo the bisection method could help with that.
$endgroup$
– John
Mar 22 at 2:58
add a comment |
$begingroup$
Simple problem: The equation below has two solutions for $x$ for a range of parameter values (see below). This is straightforward to see graphically. I need to find both of these solutions. Some additional information:
- Approximations are OK if necessary
- Ideally both solutions will be found, but ideas for finding either individually are also helpful
- In the equation below: $k approx 0.1$, $n approx 2.5$
The equation:
$frac{1}{n} e^{-n(1-e^{-x/n})}e^{frac{x}{n}} = 1-e^{ (x/n^2)(1-e^{-x/n})^{-1}e^{frac{x}{n}}k}$
polynomials systems-of-equations
asked Mar 21 at 21:21
JohnJohn
255
$endgroup$
Simple problem: The equation below has two solutions for $x$ for a range of parameter values (see below). This is straightforward to see graphically. I need to find both of these solutions. Some additional information:
- Approximations are OK if necessary
- Ideally both solutions will be found, but ideas for finding either individually are also helpful
- In the equation below: $k approx 0.1$, $n approx 2.5$
The equation:
$frac{1}{n} e^{-n(1-e^{-x/n})}e^{frac{x}{n}} = 1-e^{ (x/n^2)(1-e^{-x/n})^{-1}e^{frac{x}{n}}k}$
polynomials systems-of-equations
polynomials systems-of-equations
asked Mar 21 at 21:21
JohnJohn
255
asked Mar 21 at 21:21
JohnJohn
255
asked Mar 21 at 21:21
JohnJohn
255
asked Mar 21 at 21:21
JohnJohn
255
asked Mar 21 at 21:21
JohnJohn
255
255
$begingroup$
If approximations are OK, why don't you use en.wikipedia.org/wiki/Bisection_method?
$endgroup$
– Andrei Rykhalski
Mar 21 at 22:40
$begingroup$
Thanks for the response, @AndreiRykhalski. Solving numerically isn't the issue, though---I need the solution in terms of $k$ and $n$. I don't think I see hwo the bisection method could help with that.
$endgroup$
– John
Mar 22 at 2:58
add a comment |
$begingroup$
If approximations are OK, why don't you use en.wikipedia.org/wiki/Bisection_method?
$endgroup$
– Andrei Rykhalski
Mar 21 at 22:40
$begingroup$
Thanks for the response, @AndreiRykhalski. Solving numerically isn't the issue, though---I need the solution in terms of $k$ and $n$. I don't think I see hwo the bisection method could help with that.
$endgroup$
– John
Mar 22 at 2:58
$begingroup$
If approximations are OK, why don't you use en.wikipedia.org/wiki/Bisection_method?
$endgroup$
– Andrei Rykhalski
Mar 21 at 22:40
$begingroup$
If approximations are OK, why don't you use en.wikipedia.org/wiki/Bisection_method?
$endgroup$
– Andrei Rykhalski
Mar 21 at 22:40
$begingroup$
Thanks for the response, @AndreiRykhalski. Solving numerically isn't the issue, though---I need the solution in terms of $k$ and $n$. I don't think I see hwo the bisection method could help with that.
$endgroup$
– John
Mar 22 at 2:58
$begingroup$
Thanks for the response, @AndreiRykhalski. Solving numerically isn't the issue, though---I need the solution in terms of $k$ and $n$. I don't think I see hwo the bisection method could help with that.
$endgroup$
– John
Mar 22 at 2:58
add a comment |
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$begingroup$
If approximations are OK, why don't you use en.wikipedia.org/wiki/Bisection_method?
$endgroup$
– Andrei Rykhalski
Mar 21 at 22:40
$begingroup$
Thanks for the response, @AndreiRykhalski. Solving numerically isn't the issue, though---I need the solution in terms of $k$ and $n$. I don't think I see hwo the bisection method could help with that.
$endgroup$
– John
Mar 22 at 2:58