I want to choose $f(x)$ such that the equation $$a + b e^{-x} - f(x) = 0$$ is analytically solvable. Ideally, I want $f(x)$ to be some function that is symmetric about 0 and everywhere positive, like $x^2$ or $|x|$. Is there a function with these properties that renders the equation solvable?
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... or, for that matter, any piecewise constant function. $;$ – Apr 20 '15 at 12:27
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Let $f$ be the function where $f(x)=be^{-x}-a$ everywhere :P – Akiva Weinberger Apr 22 '15 at 10:36
4 Answers
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$$f(x)=\cosh(x)$$satisfies your requirements.
2
There is a closed form solution to
$$ a + \exp(-x)= cx^n $$
with $n=2$ in terms of generalized Lambert functions.
This is no great surprise as Maple or Mathematica solves the problem with $n=1$:
Solve[a + Exp[-x] == c*x, x]
{{x -> (a + c*ProductLog[1/(c*E^(a/c))])/c}}
ProductLog being LambertW.
Now the $n=2$ case can be rewritten
$$ (x-s)(x+s)\exp(x)=\frac{1}{c} \qquad \text{with } s=\sqrt{\frac{a}{c}}, $$
and this is of the form treated in
I. Mező and A. Baricz,
On the generalization of the Lambert $W$ function with applications in theoretical physics.
arXiv:1408.3999v1.pdf (18 Aug 2014).
Antonio Vargas
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Grant Keady
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I'm not sure if this is in the spirit of the question but $f(x) = 1$
Also you can make a function symmetric by defining it piecewise: e.g. $f(x) = \text{if $x > 0$ then }e^{-x} \text{ else } e^x$.
wlad
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If $f(x)$ is a polynomial of degree at most four in $e^{-x}$, then the equation can be solved by the substitution $y = e^{-x}$ and then solving the equation.
This can also be done for $f(x)$ being various combinations of $e^x$ and $e^{-x}$ such as $\sinh(x)$, $\cosh(x)$, and $\tanh(h)$.
If the resulting equation is quadratic in $e^x$ or $e^{-x}$, the solution is easier.
If $f(x) = e^{2x}$, then the equation becomes a cubic in $e^x$, which is harder.
marty cohen
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