# Solution for $(4+2r) x^{(1+r)}−x−1=0$

I am trying to find an analytic solution for the following equation:

$$(4+2r) x^{(1+r)}−x−1=0$$

for

$$r \in \mathbb{N}$$

and

$$\frac{1}{2}

I am trying to solve for $$x$$. I have been unable to find a straightforward solution and I am wondering if an analytic solution exists for this equation. I understand that solving such equations can be quite complex and may not always have a simple, closed-form solution.

I found a similar post which discusses an analytic solution for a somewhat similar equation, but I am unsure how to apply the methods discussed there to my equation.

Here's the plot of numerical solution: Any help or guidance would be greatly appreciated.

• Like mentioned in the other discussion you linked, you can find analytical solutions for r = 0,1,2,3,4, or numerical approximation. Generally, such equations are difficult to solve Commented Feb 4 at 15:34
• You accept that there isn't a simple closed form solution, and you have the numerical solution in your post, so what are you're actually asking for?
– Sal
Commented Feb 4 at 15:35
• Is it hard for you to solve the equation because of the extra $(4+2r)$ factor? Commented Feb 4 at 15:39
• yes, that's exactly my problem @ТymaGaidash Commented Feb 4 at 15:41
• Interesting problem since it is the inverse of $$r+2=\frac 1 {\log(x)}\,W_{-1}\left(\frac{1}{2} x (x+1) \log (x)\right)$$ Commented Feb 6 at 10:35

$$(4+2r)x^{1+r}-x-1=0\ \ \ \ \ \ (r\in\mathbb{N})$$

Your equation is a polynomial equation and an algebraic equation, and you can use the known solution formulas and methods for algebraic equations.

For $$r\in\{0,1,2,3\}$$, you can use the solution formulas for polynomial equations and the solutions are radical expressions.

Also, the equation is a trinomial equation. See [Guldberg 1902], [Szabo 2010].

Moreover:

$$-x+(4+2r)x^{r+1}-1=0$$ $$x\to -t$$: $$t+(4+2r)(-t)^{r+1}-1=0$$ $$t+(4+2r)(-1)^{r+1}t^{r+1}-1=0$$ $$t-(4+2r)(-1)^rt^{r+1}-1=0$$ $$r\to\alpha-1$$: $$-(2\alpha+2)(-1)^{\alpha-1}t^\alpha+t-1=0$$ $$(2\alpha+2)(-1)^{\alpha-1}\to y$$: $$t-yt^\alpha-1=0$$

Now the equation is in the form of equation 8.1 of [Belkic 2019]. Solutions in terms of Bell polynomials, Pochhammer symbols or confluent Fox-Wright Function $$\ _1\Psi_1$$ can be obtained therefore.
$$\$$

Guldberg, A. S.: Sur la résolution des équations trinomes. Vidensk.-Selskab. Skrift. Math.-Naturv. 10 (1902) 1-39

Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104

Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106

If you look for approximate but accurate approximations, consider instead the function $$f(x)=\log \left(2( r+2)\, x^{r+1}\right)-\log (x+1)$$which is much closer to linearity than the original one.

Starting at $$x=1$$, the first iterate of Newton method gives

$$\color{blue}{x_0=1-\frac{2 \log (r+2)}{2 r+1}}$$

What we have is $$f(x_0)<0$$ and $$f''(x_0)<0$$. So by Darboux theorem, Newton method will converge without any overshoot to the solution.

Now, compute the first iterate of a Newton-like method of order $$n$$; the solution is explicit.

Here are the results for $$n=2,3,4$$ corresponding to "named" methods $$\left( \begin{array}{ccccc} r &\text{Newton} &\text{Halley} &\text{Householder} & \text{solution} \\ 10 & 0.7893256697 & 0.7897861894 & 0.7897835658 & 0.7897836101 \\ 20 & 0.8600829302 & 0.8601537395 & 0.8601535887 & 0.8601535896 \\ 30 & 0.8926124870 & 0.8926347324 & 0.8926347063 & 0.8926347078 \\ 40 & 0.9118567862 & 0.9118663299 & 0.9118663227 & 0.9118663230 \\ 50 & 0.9247493155 & 0.9247542027 & 0.9247542001 & 0.9247542002 \\ 60 & 0.9340638411 & 0.9340666479 & 0.9340666468 & 0.9340666468 \\ 70 & 0.9411457486 & 0.9411474958 & 0.9411474953 & 0.9411474953 \\ 80 & 0.9467327605 & 0.9467339152 & 0.9467339149 & 0.9467339149 \\ 90 & 0.9512656601 & 0.9512664593 & 0.9512664591 & 0.9512664591 \\ 100 & 0.9550251563 & 0.9550257301 & 0.9550257300 & 0.9550257300 \\ \end{array} \right)$$

Using the next order, we have more than ten exact decimals.

Edit

It is possible to generate better and better $$x_0$$'s using higher order methods. For example, Halley method would give $$x_0=1-\frac{4 (2 r+1) \log (r+2)}{2 (2 r+1)^2+(4 r+3) \log (r+2)}$$.

We still have $$f(x_0^{(n)})<0$$ and $$f''(x_0^{(n)})<0$$

Playing with the order of the method, for $$r=10$$, this would give

$$\left( \begin{array}{cc} n & x_0^{(n)} & \text{method} \\ 2 & 0.7633422238 &\text{Newton} \\ 3 & 0.7889144456 &\text{Halley} \\ 4 & 0.7896940032 &\text{Householder} \\ 5 & 0.7897718056 &\text{no name}\\ 6 & 0.7897818574 &\text{no name}\\ 7 & 0.7897833306 &\text{no name}\\ 8 & 0.7897835634 &\text{no name}\\ \cdots & \cdots \\ \infty & 0.7897836101 &\text{solution}\\ \end{array} \right)$$

The problem is that the formulae become quickly messy. What can be done is to let $$t=(r+2)$$ and expand as a series for large values of $$t$$. This gives, without any loss of accuracy, $$\color{blue}{x_0=1-\frac{T}{t}\sum_{n=0}^\infty (-1)^n\, \frac{P_n(T)}{b_n\,t^n}}\qquad \text{with}\qquad T=\log(t)$$ where the first $$b_n$$ are $$\{1,2,24,48,960,2880\}$$ and the polynomials $$\left( \begin{array}{cc} n & P_n(T) \\ 0 & 1 \\ 1 & T-3 \\ 2 & 4 T^2-39 T+54 \\ 3 & 2 T^3-42 T^2+189 T-162 \\ 4 & 8 T^4-295 T^3+2910 T^2-8100 T+4860 \\ 5 & 4 T^5-225 T^4+4050 T^3-24975 T^2+48600 T-21870 \\ \end{array} \right)$$