Solutions or bounds for $x^{(1-\epsilon)/2}=1-x$

Question

Is there a known expression for the solution of the following simple equation, or at least good bounds on the solution? Assume $\epsilon \in [0,1)$ is a given parameter and $x \in (0,1)$.

$$x^{(1-\epsilon)/2}=1-x.$$

When $\epsilon=0$, we have a quadratic equation with explicit solution $x=(3-\sqrt{5})/2$.

So $x + x^\alpha - 1 = 0$ where $\alpha \in \left(0,\dfrac12\right]$? — Kenny Lau, Oct 11 '17 at 18:53
Experimentation shows that $f(a)$ is increasing is always in the interval $[0,f(0.5)]$ where $f(0.5)$ is the solution you calculated, namely $\dfrac{3-\sqrt5}2$. — Kenny Lau, Oct 11 '17 at 19:12
Yes, but is there a sharp estimate on $f(\epsilon)$ in terms of known functions? — MCH, Oct 11 '17 at 19:19
Leting $f(a) = ka^p$, from $f(0.2) = \dfrac 1 6 \left( 8-2^{2/3} \sqrt[3]{25-3\sqrt{69}} - 2^{2/3} \sqrt[3]{25+3\sqrt{69}} \right) \approx 0.24512$ and $f(0.5) = \dfrac{3-\sqrt5}2$ extrapolate $k \approx 0.53426$ and $p \approx 0.484098$, which turns out to be a good estimate, in my opinion. Someone smarter than me can probably derive a series with more terms. — Kenny Lau, Oct 11 '17 at 19:38
I note that $f'(x) + (1 - f(x)) \left(\ln f(x) + \dfrac x {f(x)} f'(x) \right) = 0$, whence $(1-f(x)) \ln f(x) + \left( \dfrac x{f(x)} - x + 1 \right) f'(x) = 0$. — Kenny Lau, Oct 11 '17 at 19:44
Thanks. But I'm interested in sandwiching the solution between reasonably sharp upper and lower bounds, rather than approximations. — MCH, Oct 11 '17 at 20:02
I believe one can prove that $x$ is a concave function of $\epsilon$ (by using the implicit function theorem and a careful analysis), which leads to some explicit bounds by linear functions (tangents and sections), sharp as long as $\epsilon$ is away from $1$. When $\epsilon$ converges to $1$, the derivative of $x$ w.r.t. $\epsilon$ becomes infinite, and a more detailed analysis is needed to get a sharp bound for the solution. — Mateusz Kwaśnicki, Oct 12 '17 at 09:55
I'd be even happy to know about the best techniques available to sandwich this kind of fixed point between upper and lower bounds in terms of elementary functions. — MCH, Oct 12 '17 at 10:31
Similar question has already been asked here https://mathoverflow.net/questions/249060/series-solution-of-the-trinomial-equation There is a series solution — yarchik, Oct 12 '17 at 12:38

score 2 · Answer 1 · answered Oct 11 '17 at 23:39

The function $g_e(x)=1-x^{(1-e)/2}$ is strictly decreasing on $[0,1]$ with $g_e(0)=1$ and $g_e(1)=0$, so there is a unique fixed point $h(e)$, and this is what we want to find. Put $x_0=h(0)=(3-\sqrt{5})/2$. Experiment makes it clear that $$ g_e(x_0) \leq g_e^3(x_0) \leq g_e^5(x_0) \leq \dotsb \leq h(e) \leq \dotsb \leq g_e^6(x_0) \leq g_e^4(x_0) \leq g_e^2(x_0) $$ for all $e$, and that $g_e^k(x_0)\to h(e)$ as $k\to\infty$. The above inequalities might or might not give bounds of the type that you need.

Robert Israel · Answer 2 · 2017-10-11T23:59:46.250

The minimax function in Maple's numapprox package gives this degree-$10$ polynomial approximation for the solution in $[0,1]$ of $x + x^\alpha = 1$ for $0 \le \alpha \le 1/2$, with maximum absolute error approximately $0.0009402242$:

$$0.0009400602+ 3.91194320\,\alpha- 77.4120170\,{\alpha}^{2}+ 1347.847663\,{\alpha}^{3}- 14566.47202\,{\alpha}^{4}+ 97487.90253\,{ \alpha}^{5}- 412006.0427\,{\alpha}^{6}+ 1100200.930\,{\alpha}^{7}- 1798435.690\,{\alpha}^{8}+ 1641636.824\,{\alpha}^{9}- 640721.0464\,{ \alpha}^{10} $$

EDIT: In view of Kenny Lau's comment, it might be better to allow half-integer powers. The following approximation has maximum absolute error approximately $0.0000307661$:

$$- 0.00003071092+ 0.028174589\, {\alpha^{1/2}}+ 5.33946514\,\alpha- 37.24787894\,{ \alpha}^{3/2}+ 211.7897251\,{\alpha}^{2}- 867.3432359\,{\alpha}^{5/2}+ 2385.793281\,{ \alpha}^{3}- 4267.365191\,{\alpha}^{7/2}+ 4745.954493\,{\alpha}^{4}- 2975.673043\,{\alpha} ^{9/2}+ 802.7705561\,{\alpha}^{5} $$

Thanks. But rather than a numerical approximation, I'm interested in explicit and clean upper and lower bounds that can be derived analytically. — MCH, Oct 12 '17 at 10:50

Solutions or bounds for $x^{(1-\epsilon)/2}=1-x$

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