What can I say about this simple functional equation?

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In summary, the function f(x) = e^{x} f(-x) with f(x) > 0 has a general solution of f(x) \propto \exp \left( \frac{x}{2} + \sum_{n=1}^{+ \infty} a_{2n} x^{2n} \right), where the coefficients for the even powers are arbitrary and the coefficient of highest power is even with negative coefficient. It can also be rewritten as f(x) = e^{x/2} \cdot h(x), where h(x) is an even function.
  • #1
nonequilibrium
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[itex]f(x) = e^{x} f(-x)[/itex] with f(x) > 0

Is there anything I can say about the general shape of this function (defined on the real axis)? For example the formula gives the derivative of f in zero in terms of f(0) (which is okay assuming I'm only interested in f up to a multiplicative constant).
 
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  • #2
EDIT: Note by the way that a non-trivial solution is [itex]f(x) = e^{x/2}[/itex] (although I'm more interested in normalizable solutions)

EDIT2: Another non-trivial solution is [itex]f(x) = e^{-(x-1)^2/4}[/itex]
 
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  • #3
Alright I think I've found the general solution.

Since f(x) > 0, we can write [itex]f(x) = e^{g(x)}[/itex]. The functional equation becomes [itex]e^{g(x) - g(-x)} = e^x[/itex] such that [itex]g(x) = g(-x) + x[/itex]. If we assume we can write g(x) as a power series, we have [itex]g(x) = \sum a_n x^n[/itex]. Substituting it in the functional equation: [itex]\left\{ \begin{array}{ll}
a_1 = \frac{1}{2} & \\
a_{n} = 0 & \textrm{for $n > 1$ odd}
\end{array} \right.
[/itex]
The coefficients for the even powers are arbitrary. If we want the solution to be normalizable, we only need to demand that the coefficient of highest power is even with negative coefficient.

More general: [itex]f(x) \propto \exp \left( \frac{x}{2} + \sum_{n=1}^{+ \infty} a_{2n} x^{2n} \right) [/itex]

EDIT: perhaps a more insightful formulation [itex]f(x) = e^{x/2} \cdot h(x)[/itex] with h > 0 an even function.
 
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  • #4
mr. vodka said:
[itex]g(x) = g(-x) + x[/itex]. If we assume we can write g(x) as a power series, …

easier is to notice the symmetry, and write g(x) - x/2 = g(-x) - (-x)/2 :wink:
 
  • #5


This functional equation represents a reflection symmetry about the y-axis, as evidenced by the fact that f(x) and f(-x) are equal. Additionally, the fact that f(x) is always positive suggests that the function is always increasing, with a minimum value at x=0. The formula also implies that the derivative of f at x=0 is equal to f(0), which can provide useful information about the behavior of the function near this point. However, without more context or information about the function, it is difficult to make any further conclusions about its general shape. Additional data or constraints would be needed to fully understand the behavior of this function.
 

1. What is a simple functional equation?

A simple functional equation is an equation that relates a function to itself or to another function in a straightforward and easily solvable way. It typically involves only one variable and simple operations such as addition, multiplication, and composition.

2. How is a simple functional equation different from a regular equation?

A simple functional equation is different from a regular equation in that it deals specifically with functions, which are mathematical objects that map inputs to outputs. Regular equations, on the other hand, can involve any type of mathematical expression or relationship.

3. What can I learn from studying a simple functional equation?

Studying a simple functional equation can help you understand the behavior and properties of different types of functions. It can also provide insights into more complex functional equations and real-world applications of functions.

4. Are there any real-life applications of simple functional equations?

Yes, simple functional equations have many real-life applications in fields such as economics, physics, and engineering. For example, they can be used to model natural processes, predict future trends, and optimize systems.

5. How can I solve a simple functional equation?

Solving a simple functional equation involves finding the function that satisfies the given equation. This can be done by manipulating the equation algebraically, graphically, or using other mathematical techniques such as substitution and iteration.

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