What type of equation is this one?

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Hi,

I am struggling with a problem, where in the middle of my calculations I need to determine a function ##f(\alpha, \beta; x)##, namely the function of ##x## parametrized by ##\alpha## and ##\beta##, from the following equation

$$f\left(\alpha, \beta;\frac{1}{x}\right) = x^4f(\gamma, \delta; x).$$

So the ##f(1/x)## on the LHS and ##f(x)## on the RHS may differ in parametrization. Somehow I found that this equation admits the following solution

$$f(\alpha, \beta; x) = \left(\alpha + \beta x^a\right)^b $$

where ##a\cdot b = -4##. Although, this is probably not the most general solution.

So my question is what type of equation this one is? And how one should approach such equation?
 
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Some clarifying questions:

Is x a non-zero real number?

Are \alpha, \beta real numbers?

Are \gamma and \delta real number constants (independent of \alpha, \beta, x)?

In your example, you're assuming \gamma = \alpha = \delta = \beta?
 
disregardthat said:
Some clarifying questions:

Is x a non-zero real number?

Are \alpha, \beta real numbers?

Are \gamma and \delta real number constants (independent of \alpha, \beta, x)?

##f(x)## is well defined and restricted to either ##x>0## or ##x<0##.
##\alpha##, ##\beta##, ##\gamma##, ##\delta## in general may be complex and independent of ##x## (they parametrize function ##f(x)##).
However ##\gamma##, ##\delta## may be functions of ##\alpha##, ##\beta##.

disregardthat said:
In your example, you're assuming \gamma = \alpha = \delta = \beta?

In the example I provided previously we have

$$f\left(\alpha, \beta; x\right) = \left(\alpha + \beta x^{a} \right)^b \Rightarrow f\left(\alpha, \beta; \frac{1}{x}\right) = \left(\alpha + \beta x^{-a} \right)^b = x^{-ab}\left(\alpha x^a + \beta\right)^b = x^{4}f(\beta, \alpha; x)$$
Where the assumption ##ab = -4## was used. So in this case ##\alpha = \delta## and ##\beta = \gamma## and are not neccesarily equal.

Sorry for not making this clear enough earlier.
 
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Anyone?
 
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