Is there a way to calculate directly the root of a polynomial of the form:
$$a x^{n+1} + b x^{n} - 1 = 0, \quad a,b,x \in \mathbb{R}^{+}, n \in \mathbb{N}$$
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Note that this is equivalent to $z^n=bz+a$, the intersection of a power law with a straight line. – May 28 '20 at 10:33
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1If "directly" means "in radicals", see this example. – metamorphy May 28 '20 at 10:44
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With "directly", that I can write it down in some kind of expression without starting my numerical math engine. – thinkingeye May 28 '20 at 10:49
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See this question out of many other similar ones for solving $x^y-ax+b=0$ – Тyma Gaidash Jun 02 '23 at 17:51
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Here: let $x=1/z$ to obtain $$a+bz-z^{n+1}=0.$$ Then you can go several ways. One is $$z=\sqrt[n+1]{a+b\sqrt[n+1]{a+b\sqrt[n+1]{a+b(\cdots)}}}$$ provided this continued radical converges. The other such method is Series Reversion. Thus $$z=-b\sum_{n\geqslant 1}\frac{(2n-2)!}{n!(n-1)!}\frac{a^n}{b^{2n}}=-b(a/b^2+a^2/b^4+2a^3/b^6+\cdots)$$
Alexander Conrad
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