I ask about a possible method to find the solution of algebraic equations of the form
$axⁿ+byⁿ+c=0$
where $a,b,c,x,y$ are real constants and $n$ is an integer. Maybe there is a simple method, but I cannot find it.
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If $\log_x(y) = j/k$ is rational, this reduces to a polynomial in $x^{1/k} = y^{1/j}$. Otherwise you're unlikely to get a closed form. You might use numerical methods, or a series expansion: if $y = x^r$, $$ n = \frac{\ln(-c/a)}{\ln(x)} + \sum_{k=0}^\infty \frac{(-c/a)^{kr}(b/c)^k}{k! \ln(x)} \prod_{j=1}^{k-1} (kr - j)$$
answered Nov 19, 2012 at 17:16
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$\begingroup$ What about the case where $a,b,c,x,y$ are real valued functions on certain w (the unknown variable) and $n$ is a fixed integer. $\endgroup$– SafwaneCommented Nov 19, 2012 at 17:35
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$\begingroup$ If you want a series solution, you'll want to expand around a known solution in powers of some parameter. Thus suppose you write your equation as $F(w) - \epsilon G(w) = 0$, where $F$ and $G$ are analytic, and $F(0) = 0$ so that $w = 0$ is a solution when $\epsilon = 0$, while $G(0) \ne 0$ and $F'(0) \ne 0$. Then the Lagrange inversion theorem gives a series expansion for $w$ in powers of $\epsilon$, convergent for sufficiently small $\epsilon$. $\endgroup$ Commented Nov 19, 2012 at 19:29
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$\begingroup$ Ok, Thank you very much. I will try to do this. $\endgroup$– SafwaneCommented Nov 19, 2012 at 20:14