If $\alpha$ be a root of
$x^{x-\sqrt{x}} = \sqrt{x}+1$.
We need to find the value(s) of $(\alpha + \frac{1}{\alpha})$. WolframAlpha shows two possible roots of the equation.
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Aroonalok
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1what is the question here? – Dr. Sonnhard Graubner Apr 06 '17 at 13:35
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We need to find a+1/a if a is root of the equation. – Aroonalok Apr 06 '17 at 13:39
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2Hitting this with Mathematica's numerical techniques suggests that $\alpha + 1/\alpha \approx 3.0000...$ for the larger of the two roots, out to at least 100 decimal places. But I have no idea how to prove it. – Michael Seifert Apr 06 '17 at 13:56
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I tried converting it to a form suitable to use Lambert W Function by using $t = \sqrt{x} + 1$. The equation transformed into $t = e^{2(t-1)(t-2)ln(t-1)}$. But, I cannot proceed further with this. – Aroonalok Apr 06 '17 at 14:05
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1In what context did this problem arise? Given what @Michael Seifert found, I would find it very amazing that you would have stumbled on this by randomly looking at various transcendental equations. – Dave L. Renfro Apr 06 '17 at 14:53
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A friend of mine gave it to me. Don't know where he got it from. – Aroonalok Apr 06 '17 at 14:57
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@MichaelSeifert. Even with 5000 decimal places – Claude Leibovici Sep 19 '22 at 13:13
1 Answers
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One solution of the equation is:
$x=\frac{3+\sqrt{5}}{2}$.
Therefore
$\alpha+\frac{1}{\alpha}=3$.
Setting $\sqrt{x}+1=t$, the equation can be written as
$t=x^{x-\sqrt{x}}$.
We have an equivalence between two powers.
We know that two powers are equal if they have the same base and the same exponent, so we can assume for a moment that the base $x=t$ and write:
$t=t^{x-\sqrt{x}}$.
This equality of powers implies that since they have the same base ($t$), they must have the same exponent:
$1=x-\sqrt{x}$.
Solving this equation we get
$x=\frac{3+\sqrt{5}}{2}$.
giuseppe mancò
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“ We know that two powers are equal if they have the same base and the same exponent, so we can assume for a moment that the base =”
This logic is not working here.
– Aroonalok Sep 07 '21 at 03:47