# Is there a specific named function that is the inverse of $x+x^a$ for $x$ real?

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!

• 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. Nov 4, 2021 at 19:29
• 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 $(0,1)$. Nov 4, 2021 at 21:18
• Here it is: mathoverflow.net/questions/249060/… Nov 4, 2021 at 21:43
• Does this answer your question? Series solution of the trinomial equation Nov 4, 2021 at 21:47
• It's so known as proximal map of the function $x^{a+1}/(a+1)$.
– Dirk
Nov 5, 2021 at 6:13

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)

• 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. Nov 10, 2021 at 14:56
• @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). Nov 12, 2021 at 2:57
• 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? Dec 31, 2021 at 14:42
• @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) Jan 2 at 1:16

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 $$\begin{equation} u_*\vee u_*^{1/a} where $$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 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}

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 $$\{(y,\log_{10}(v_n(y)-u_n(y)))\colon0 for $$a=7/10$$ with $$n=1$$ (blue), $$n=2$$ (orange), and $$n=3$$ (green): 