I'm currently interested in this function: $y$ = $(1+1/y)^x$ which can also be written as $y^{x+1}$ = ${(y+1)}^x$. I've checked Wolfram Alpha and I couldn't find any way to write in an explicit form of $y$ in terms of $x$. I want to ask whether: 1) there is a way to write the function in an explicit form or if not, then 2) why it cannot be written in an explicit form.
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2$x$ can be written as an elementary function in dependence of $y$: $x=\frac{\ln(y)}{\ln(y+1)-\ln(y)}$. – IV_ Dec 01 '19 at 18:39
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For future: you can probably get a closed form using $t=\frac 1y\iff 1=(1+t)t^{\frac1x}=t^\frac1x+t^{\frac1x+1}$ and series solution to general trinomial – Тyma Gaidash Jul 23 '23 at 14:03
1 Answers
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I tried MAPLE, MATHEMATICA and SAGEmath. They too don't know a solution.
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My answer is only for the elementary functions and Lambert W function.
Your equation cannot be solved for $y$ by applying only such functions. That means, expressing $y$ in dependence of $x$ only by those functions is not possible.
The elementary functions are generated from complex constants and/or one variable by applying only a finite number of algebraic operations, $\exp$ and/or $\ln$.
The branches of Lambert W are the inverse relations of the functions $x\mapsto xe^x$.
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The relationship between the two variables of your equation can be a function, or, in the general case, a binary relation. Because we are interested in functions, we need, in the general case, branches of functions. Inverse means the compositional inverse function. Because only bijective functions have an inverse, we need partial inverses. A partial inverse of a function is the inverse of a restriction of that function on a domain where the restriction is bijective. Solving an equation of two variables for one of the two variables means looking for an inverse or partial inverses.
Regarding the wanted variable $y$, you have a transcendental equation. Because not further specified, we have to assume that $x$ is from a non-discrete domain.
Consider that functions with the same function term but different domains are different functions.
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You have an equation in $y$. Try to transform it into the form just described:
$$y^{(x+1)}=(y+1)^x$$
$$e^{ln(y)(x+1)}=e^{ln(y+1)x}$$
$$e^{ln(y)(x+1)}-e^{ln(y+1)x}=0$$
The left-hand side of the equation is the function term of an elementary function in dependence of $y$.
This expression is an algebraic expression of the two transcendental terms $e^{ln(y)(x+1)}$ and $e^{ln(y+1)x}$. It is not possible by applying only elementary functions to transform this expression into an algebraic expression of only one transcendental term.
According to the theorem in [Ritt 1925] that is proved also in [Risch 1979], this elementary function therefore cannot have elementary partial inverses. That means, you cannot solve your equation by applying only elementary partial inverses of the elementary functions you can read from the equation.
Also, the expression is not in a form required for applying Lambert W as presented in my answer to Algebraic solution to natural logarithm equations like $1-x+x\ln(-x)=0$ .
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Another but related explanation:
You have an equation of the two variables $x$ and $y$. Solving the equation for $y$ would give you functions $f$ with $y=f(x)$. We already found, this functions are not in the set of the elementary functions and Lambert W.
Solve your equation for $x$:
$$x=\frac{\ln(y)}{\ln(y+1)-\ln(y)}$$
Now you have functions $\phi$ with $x=\phi(y)$. That are the partial inverses of $f$. $\phi$ is an elementary function. The function term of $\phi$ depends on the two transcendental terms $\ln(y)$ and $\ln(y+1)$. By transforming the equation only by elementary functions and/or Lambert W, they cannot be combined into a single term. Ritt's theorem says, the elementary function $\phi$ doesn't have an elementary inverse therefore.
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IV_
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