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This seems such a simple question that I fear I must have missed some elementary maths.

I am looking for a way to solve $x+x^a = y$ by reference to an already defined function, $a,x,y > 0$ are real.

Failing that a reasonable approximation for $a$ in $(0,1)$.

Many thanks!

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    $\begingroup$ I'm assuming you are trying to solve for $x$ in terms of $a,y$. For $a=5$ you are looking at Bring radical, which has no closed form in terms of radicals. For general $a$, especially noninteger ones, I doubt there is a sensible solution, though one might exist in terms of hypergeometric functions. $\endgroup$
    – Wojowu
    Nov 4, 2021 at 19:29
  • $\begingroup$ If by approximation you mean numerical approximation, you might try iterating $f(x)=y-x^a$. I believe generically the sequence $f(0),f(f(0)),f(f(f(0))),...$ converges to a solution for $a$ in $(0,1)$. $\endgroup$ Nov 4, 2021 at 21:18
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    $\begingroup$ Here it is: mathoverflow.net/questions/249060/… $\endgroup$ Nov 4, 2021 at 21:43
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    $\begingroup$ Does this answer your question? Series solution of the trinomial equation $\endgroup$ Nov 4, 2021 at 21:47
  • $\begingroup$ It's so known as proximal map of the function $x^{a+1}/(a+1)$. $\endgroup$
    – Dirk
    Nov 5, 2021 at 6:13

2 Answers 2

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The answer is yes indeed. It is a special case of Fox-H function, a variation of the confluent Fox-Wright $_{1}\Psi_{1}$ function (a generalization of the confluent hypergeometric function $_{1}F_{1}$) providing the inverse function. See a previous answer here for details and references. For this particular case solution is

(Setting $\alpha = a$), for $\alpha>1$

$$x = y\cdot\,_{1}\Psi_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})$$ which can be set as $$x=y+\sum_{n=1}^\infty\binom{n\alpha}{n-1}\frac{(-1)^ny^{n(\alpha-1)+1}}{n}$$ whose convergence region is $|y^{\alpha-1}|<|(\alpha-1)^{\alpha-1}\alpha^{-\alpha}|$. For non integer $\alpha>1$ binomials must be set in terms of $\Gamma$ function.

Since Fox-Wright's generalized function can be expressed in terms of Fox-H function we have $$\,_{1}\Psi_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})=H_{1,2}^{1,1}([(0,\alpha)];[(0,1),(-1,\alpha-1)];y^{\alpha-1})$$ $$\,_{1}\Psi_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})=H_{2,1}^{1,1}([(1,1),(2,\alpha-1)];[(1,\alpha)];y^{1-\alpha})$$ for this particular case, Wolfram's Mathematica 12.3 provides an explicit inverse as

$x=y\cdot$FoxH[{{{0,$\alpha$}},{{}}},{{{0,1}},{{-1,$\alpha$-1}}},$y^{\alpha-1}$]

For $0<\alpha<1$ the solution is $$x = y\cdot(1-\,_{1}\Psi_{1}([1,\alpha^{-1}];[2,\alpha^{-1}-1];-y^{\alpha^{-1}-1}))$$

and the above relationships are turned in replacing $\alpha$ by $\alpha^{-1}$ and Fox-Wright function $\,_{1}\Psi_{1}$ by $1-\,_{1}\Psi_{1}$. In this case Mathematica's expression is

$x=y\cdot(1-$FoxH[{{{0,$\alpha^{-1}$}},{{}}},{{{0,1}},{{-1,$\alpha^{-1}$-1}}},$y^{\alpha^{-1}-1}$])

Finally, just to complement this answer, general trinomial equation solutions are developed in section 4 of the following

Reference

Miller A.R., Moskowitz I.S. Reduction of a Class of Fox-Wright Psi Functions for Certain Rational Parameters. Computers Math. Applic. Vol. 30, No. 11, pp. 73-82, (1995). Pergamon

A preprint can be found here. (Document has mis-embedded fonts, isolated commas must be replaced by $\Gamma$ symbol)

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  • $\begingroup$ This is very helpful. Now can one use the Fox-Wright to solve y=x^𝛼-x? I am finding it difficult to hit on a reference book here as Fox-Wright doesn't appear in the DLMF. $\endgroup$
    – J.Ham
    Nov 10, 2021 at 14:56
  • $\begingroup$ @j-ham, An introduction to (Extended) Generalized Hypergeometric Functions (MeijerG, Fox-Wright and Fox-H) can be found in the link Fox-H function above. Several reference books are found at the bottom of this document. Fox-Wright function, as a special case of Fox-H function, can be computed through Wolfram's Mathematica (version 12.3). $\endgroup$ Nov 12, 2021 at 2:57
  • $\begingroup$ I tried to find solution to equation: $x^2+x=1$ in Mathematica code:N[y*FoxH[{{{0, \[Alpha]}}, {{}}}, {{{0, 1}}, {{-1, \[Alpha] - 1}}}, y^(\[Alpha] \[Minus] 1)] /. \[Alpha] -> 2 /. y -> 1] ,but I can't get numeric value? $\endgroup$ Dec 31, 2021 at 14:42
  • $\begingroup$ @Mariusz_Iwaniuk. Since FoxH is a new function, I think this is a question for Mathematica & Wolfram Language StackExchange Site. I will ask to check it. These expressions come from Fox-Wright function. There are slightly different formulae using FoxH function for all roots of trinomial equations in Wolfram's site reference.wolfram.com/language/ref/FoxH.html (Applications Section) $\endgroup$ Jan 2 at 1:16
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If $a$ is rational, then the root, say $x_*$, of your equation is algebraic, and (say) Mathematica will find for you with any degree of accuracy. Otherwise, one can approximate $a$ by rational numbers.


Another way to get bounds on $x_*$ is to use a combination of the Newton and secant methods to bracket the root, as follows. For $a\in(0,1)$, using the substitution $u=x^a$, rewrite your equation as \begin{equation} f(u):=u^{1/a}+u=y, \tag{1} \end{equation} so that the function $f$ is convex and increasing. Let $u_*$ be the positive root of equation (1), so that $x_*=u_*^{1/a}$ and $f(u_*)=y$.

Note that \begin{equation} u_*\vee u_*^{1/a}<y=f(u_*)<2(u_*\vee u_*^{1/a}), \end{equation} where $u\vee v:=\max(u,v)$ and $u\wedge v:=\min(u,v)$. So, letting \begin{equation} u_0:=u_0(y):=\frac y2\wedge\Big(\frac y2\Big)^a,\quad v_0:=v_0(y):=y\wedge y^a, \end{equation} we get the initial bracketing of $u_*$: \begin{equation} u_0<u_*<v_0. \end{equation} Use now the following combination of the secant and Newton methods for every natural $n$: \begin{equation} u_n:=u_n(y):=U(u_{n-1},v_{n-1}),\quad v_n:=v_n(y):=V(v_{n-1}), \end{equation} where \begin{equation} U(u,v):=u+\frac{y-f(u)}{f(v)-f(u)}\,(v-u), \end{equation} \begin{equation} V(v):=v-\frac{f(v)-y}{f'(v)}. \end{equation} Then $u_n$ and $v_n$ bracket the root $u_*$ and monotonically converge to it (very fast): \begin{equation} u_{n-1}<u_n<u_*<v_n<v_{n-1},\quad u_n\uparrow u_*,\quad v_n\downarrow u_*. \end{equation}

The bracketing $u_*\in[u_1,v_1]$ can be already pretty good, while providing almost digestible explicit lower and upper bounds on $u_*$.

As an illustration, here are the graphs $\{(y,\log_{10}(v_n(y)-u_n(y)))\colon0<y<3\}$ for $a=7/10$ with $n=1$ (blue), $n=2$ (orange), and $n=3$ (green):

enter image description here

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