How do I solve $x^x - 5x + 6 = 0$ using the Lambert W function?
EDIT:
I solved equations $2^x - 5x + 6 = 0$ and $3^x - 4x - 15 = 0$ using Lambert W function, but not able to solve $x^x - 5x + 6 = 0$ equation. How and which method can be used to solve this equation?
How to solve this equation without graphically plotting, which method can be used?
I am just preparing for my examination, I got this question after solving some Lambert W function equations and I don't know much about that.
My answer concerns solutions in closed form.
Because the equations mentioned here are polynomial equations of more than one algebraically independent monomials and without univariate factor, we don't know how we can solve the equations by rearranging for $x$ by applying only finite numbers of elementary functions we can read from the equation. But in some cases, we can try to guess solutions ($2$ is a solution of your first equation, 3 is a solution of your second equation, for example).
Therefore certain Special functions are needed. The Special functions mentioned here are not so widely known.
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If we want to see if an equation is solvable in terms of Lambert W, we can use the equation in its $\exp$-$\ln$ form.
Let $a,b,c\neq 0$ constants.
a) Equations of the form $$a^x+bx+c=0$$ can be solved in terms of LambertW: $$e^{\ln(a)x}+bx+c=0$$ We can rearrange the sum form to the product form: $$-(bx+c)e^{-\ln(a)x}=1.$$ We see, $x$ is in a linear function in both the coefficient and the exponent of the exponential term.
Multiplying by certain constants and constant exponentials yields adjusting the coefficient and the exponent of the exponential term to give the form $$f(x)e^{f(x)}=\text{constant}$$ which allows Lambert W to be applied.
$$x=-\frac{b\ W_k\left(\frac{\ln(a)\ e^{-\frac{\ln(a)c}{b}}}{b}\right)+\ln(a)\ c}{\ln(a)\ b}\ \ \ \ \ \ \ \ \ (k\in\mathbb{Z})$$
b) Equations of the form $$x^a+bx+c=0:$$ $$e^{\ln(x)a}+bx+c=0$$ $$(-bx-c)e^{-a\ln(x)}=1$$ cannot be solved in terms of Lambert W or Generalized Lambert W (see [Mező]) because $x$ isn't in a polynomial or rational function in the coefficient and the exponent.
But they can be solved as trinomial-like equations. (see [Belkić 2018])
c) Equations of the form $$x^x+bx+c=0:$$ $$e^{\ln(x)x}+bx+c=0$$ $$(-bx-c)e^{-\ln(x)x}=1$$ cannot be solved in terms of Lambert W or Generalized Lambert W (see [Mező]) because $x$ isn't in a polynomial or rational function in the coefficient and the exponent.
$c$ is disturbing the application of Hyper Lambert W. (see [Galidakis])
Using Lagrange reversion: $$x=a+bx^x\implies x=a+\sum_{n=1}^\infty\frac{b^n}{n!}\frac{d^{n-1}}{da^{n-1}}a^{na}$$ Now use a Stirling number of the first kind formula, possible to derive from this pattern $$\frac{d^n}{dx^n}f(\ln(x))=x^{-n}\sum_{k=1}^n S_n^{(k)}f^{(k)}(\ln(x)):\frac{d^{n-1}}{da^{n-1}}a^{na}=a^{-n}\sum_{k=1}^{n-1}S_{n-1}^{(k)}\left.\frac{d^k}{dt^k}e^{n t e^t}\right|_{\ln(a)},n>1$$ Also, the $n=1$ term is extracted out. Finally, use $e^x$’s Maclaurin series and general Leibniz rule $$\left.\frac{d^k}{dt^k}e^{n t e^t}\right|_{\ln(a)}=\sum_{m=0}^\infty\frac{n^m}{m!}\left.\frac{d^k}{dt^k}t^me^{mt}\right|_{\ln(a)}=\sum_{j=0}^k\binom kj\left.\frac{d^j}{dt^j}t^m\right|_{\ln(a)} \left.\frac{d^{k-j}}{dt^{k-j}}e^{tm}\right|_{\ln(a)}$$ summing over $j$ gives a Tricomi confluent hypergeometric function: $$x=a+bx^x\implies x=a+ba^a+\sum_{n=2}^\infty\sum_{m=1}^\infty\sum_{k=1}^{n-1}\frac{(-1)^kb^nn^m\ln^{m-k}(a)a^{m-n+1}S_{n-1}^{(k)}}{m!n!}\operatorname U(-k,m-k+1,-mt)$$
shown here:
However, this formula simplifies if $a=1$. Using the same derivation, one computes $\left.\frac{d^j}{dt^j}t^m\right|_0=\delta_{k,j}$ and uses this method to remove the $j$ sum to get:
$$x=1+bx^x\implies x=1+b+\sum_{n=1}^\infty\sum_{k=1}^{n-1}\sum_{m=0}^k\binom km\frac{b^nn^mm^kS_{n-1}^{(k)}}{m^m n!}$$
as shown here:
Both series work for $a,b\in\Bbb C$. For certain values, like small $|b|,|a|$, the series converges fine. However, for values outside these, the series converge more slowly, like for the asker’s example.
