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Solutions for Series of Exponential Equations in Terms of Lambert-W Function and Fundamental Constants

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*of were solved graphically, numerically and analytically. The analytical solution was derived in terms of Lambert-W function. A general numerical solution for any*

*y*is found in terms of

*n*or in base

*y.*A solution is close to the fine structure constant. The equation which provided the solution as the fine structure constant was derived in terms of the fundamental constants.

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1. Introduction

Exponential equations are widely used in natural and social sciences. In this paper, we considered series of exponential equations and solved them graphically, numerically, and analytically in terms of Lambert-W function. One equation connected to the fine structure constant, was derived in terms of the fundamental constants and led to a new equation. The Lambert-W function for real variables is defined by the equation $W\left(x\right)\mathrm{exp}\left[W\left(x\right)\right]=x$ [1] [2] [3] [4] and it has applications in Planks spectral distribution law [5] [6] , QCD renormalization [7] , solar cells [8] , bio-chemical kinetics [9] , optics [10] , population growth and water movement in soil [11] .

Considering the series of exponential equations defined by the following equation

$x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ (1.1)

where x, y, n are real variables.

Taking log_{y} on both sides of the Equation (1.1)

${\mathrm{log}}_{y}x=n+\frac{x}{{y}^{n+1}}$ (1.2)

Converting the Equation (1.2) to natural logarithm

$\frac{\mathrm{ln}x}{\mathrm{ln}y}=\frac{x}{{y}^{n+1}}+n$ (1.3)

The trivial solution of the Equations (1.1) to (1.3) is

$x={y}^{n+1}$ (1.4)

In this paper, we are focusing on the non-trivial solutions.

For n = 2, 1, 0, −1, −2, the Equations (1.1) and (1.3) become:

$x={y}^{2}{y}^{\frac{x}{{y}^{3}}}\text{or}\frac{\mathrm{ln}x}{\mathrm{ln}y}=\frac{x}{{y}^{3}}+2$ (1.5)

$x=y{y}^{\frac{x}{{y}^{2}}}\text{or}\frac{\mathrm{ln}x}{\mathrm{ln}y}=\frac{x}{{y}^{2}}+1$ (1.6)

$x={y}^{\frac{x}{y}}\text{or}\frac{\mathrm{ln}x}{\mathrm{ln}y}=\frac{x}{y}\text{or}{x}^{y}={y}^{x}$ (1.7)

$x={y}^{-1}{y}^{x}\text{or}\frac{\mathrm{ln}x}{\mathrm{ln}y}=x-1$ (1.8)

$x={y}^{-2}{y}^{xy}\text{or}\frac{\mathrm{ln}x}{\mathrm{ln}y}=xy-2$ (1.9)

2. Graphical Solutions

If y =10, the Equations (1.5) to (1.9) become

$x={10}^{2}\times {10}^{\frac{x}{{10}^{3}}}$ (2.1)

$x=10\times {10}^{\frac{x}{{10}^{2}}}$ (2.2)

$x={10}^{\frac{x}{10}}$ (2.3)

$x={10}^{-1}\times {10}^{x}$ (2.4)

$x={10}^{-2}\times {10}^{10x}$ (2.5)

The curves and the straight line to obtain the graphical solutions of the Equations (2.1) to (2.5) are shown in Figure 1. The intercepts of the curves and the straight line indicate the solutions.

Figure 1. Plots of the functions to obtain the graphical solutions for the Equations (2.1) to (2.5).

The intersecting points of 0.1, 1, 10, 100 and 1000 are the trivial solutions and the intersecting points at around 0.0137, 0.137, 1.37, 13.7 and 137 are the non-trivial solutions.

The non-trivial solutions imply the following equations:

${10}^{0.1371}=1.371$ (2.6)

$\frac{\mathrm{ln}1.371}{1.371}=\frac{\mathrm{ln}10}{10}=0.2302$ (2.7)

${10}^{1.371}={1.371}^{10}=23.5$ (2.8)

3. Numerical Solutions

Higher precision non-trivial numerical solutions were obtained for the series of equations $x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ using the iterative technique for n = 2, 1.5, 1, 0.5, 0, −0.5, −1, −2 and 1 ≤ y ≤ 15 (Table 1). The iterations do not converge on non-trivial solutions for y < e, and solutions in this range were obtained by trial and error.

The solutions in Table 1 for n = −2, −1, 0, 0.5, 1, 2 are plotted as x vs y with x axis in log scale (Figure 2). Sharp turning points in the plots are observed for y values in the range of 1 to 2.

4. Analytical Solution

Consider the Equation (1.3)

$\frac{\mathrm{ln}x}{\mathrm{ln}y}=\frac{x}{{y}^{n+1}}+n$

Table 1. Non-trivial numerical solutions for the series of equations $x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ .

Figure 2. Plots of x vs y for the series of equations, $x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ for n = 2, 1, 0, 0.5, −1 and −2.

Let

$t=-\mathrm{ln}x$

Then (1.3) becomes

$\frac{-t}{\mathrm{ln}y}=\frac{{\text{e}}^{-t}}{{y}^{n+1}}+n$

$\left(t+n\mathrm{ln}y\right){\text{e}}^{t}=\frac{-\mathrm{ln}y}{{y}^{n+1}}$

$\left(t+n\mathrm{ln}y\right){\text{e}}^{t+n\mathrm{ln}y}=\frac{-{\text{e}}^{n\mathrm{ln}y}\mathrm{ln}y}{{y}^{n+1}}$

$\left(t+n\mathrm{ln}y\right){\text{e}}^{t+n\mathrm{ln}y}=\frac{-\mathrm{ln}y}{y}$

$\left(t+n\mathrm{ln}y\right)=W\left(\frac{-\mathrm{ln}y}{y}\right)$

Substituting −lnx for t

$\left(-\mathrm{ln}x+n\mathrm{ln}y\right)=W\left(\frac{-\mathrm{ln}y}{y}\right)$

Using the Equation (1.3)

$-\frac{x\mathrm{ln}y}{{y}^{n+1}}=W\left(\frac{-\mathrm{ln}y}{y}\right)$

Hence the solution to Equation (1.3) is

$x=\frac{W\left(\frac{-\mathrm{ln}y}{y}\right)}{\left(-\frac{\mathrm{ln}y}{{y}^{n+1}}\right)}=\frac{{y}^{n}W\left(\frac{-\mathrm{ln}y}{y}\right)}{\left(-\frac{\mathrm{ln}y}{y}\right)}$ (3.1)

If n = 0, the Equation (1.1) $x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ becomes Equation (1.7) $x={y}^{\left(\frac{x}{y}\right)}$ .

Using the solution in the Equation (3.1), the analytical solution in terms of Lambert-W function is

$x=\frac{W\left(-\frac{\mathrm{ln}\left(y\right)}{y}\right)}{\left(-\frac{\mathrm{ln}\left(y\right)}{y}\right)}$ (3.2) [12] [13]

In Equation (3.2), if $y=\text{e}$ , $x=\frac{W\left(-\frac{1}{\text{e}}\right)}{\left(-\frac{1}{\text{e}}\right)}.$

But $W\left(\frac{1}{\text{e}}\right)=-1$ [6] .

Hence $x=\text{e}$ , the result in Table 1.

If n =0 and y =2 in Equation (3.2), the solutions in Table 1 and Equation (1.7) gives

${2}^{4}={4}^{2}$ and $\frac{\mathrm{ln}2}{2}=\frac{\mathrm{ln}4}{4}=-\mathrm{0.346.}$ (3.3)

Equation (3.1) gives

$4=\frac{W\left(-\frac{\mathrm{ln}\left(2\right)}{2}\right)}{\left(-\frac{\mathrm{ln}\left(2\right)}{2}\right)}=\frac{W\left(-0.346\right)}{-0.346}$

$2=\frac{W\left(-\frac{\mathrm{ln}\left(4\right)}{4}\right)}{\left(-\frac{\mathrm{ln}\left(4\right)}{4}\right)}=\frac{W\left(-0.346\right)}{-0.346}$

$W\left(-0.346\right)$ is double valued with−0.693 and −1.386.

If we substitute the solutions for n = 0 and y = 10 from Table 1 to Equation (3.2);

$W\left(-\frac{\mathrm{ln}\left(1.37129\right)}{1.37129}\right)=10\left(-\frac{\mathrm{ln}\left(1.37129\right)}{1.37129}\right)$ (3.4)

$W\left(-0.2302\right)=-2.302$ (3.5)

Since x and y are symmetric in Equation (1.7)

$W\left(-\frac{\mathrm{ln}\left(10\right)}{10}\right)=1.371289\left(-\frac{\mathrm{ln}\left(10\right)}{10}\right)$ (3.6)

$W\left(-0.2302\right)=-0.3157$ (3.7)

The W(x) has two real values for $-1/\text{e}\le x<0$ [1] .

If n = −1, the Equation (1.1) $x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ becomes Equation (1.8) $x={y}^{-1}{y}^{x}$ or $xy={y}^{x}$ .

Using the solution in the Equation (3.1), the analytical solution in terms of the Lambert-W function is

$x=\frac{W\left(-\frac{\mathrm{ln}y}{y}\right)}{-\mathrm{ln}y}$ (3.8) [12]

If

But $W\left(\frac{-1}{e}\right)=-1$ , Hence x = 1, the result in Table 1.

In Table 1, for any value of n $y=\text{e},x=\text{e}\times {\text{e}}^{n}$ , the trivial and nontrivial solutions coincide.

$y=10,x=1.371289\times {10}^{n}$

Using the solution in Equation (3.1), for any y the solution x can be written as

$x\left(n=0\right)\times {y}^{n}$ (3.9)

Plots of lnx vs n shown in Figure 3 are linear as expected from Equation (3.1).

The lnx vs n lines for different y values are crossing near the point (0.5, 1.4). This indicates the solutions for n = 0.5 have little dependency on y for y ≥ e. This is also evident in the numerical results for n = 0.5 in Table 1 and in the plot of $x={y}^{0.5}{y}^{\frac{x}{1.5}}$ in Figure 2.

5. Solutions x in Base y

The solutions x in Table 1 can be written in base y, (x_{y}) to indicate the general pattern.

For any valued of n, x_{y} can be written as

${x}_{y}\left(n=0\right)\times {10}^{n}$ (5.1)

For n = 2, the solutions
_{y} shown in Table 2.

For y > 11, the x_{y} are written using the hex notation.

There is a sharp change in the value of the x_{y} at y = 4.

For n = 2, plot x_{y} vs y, for 5 ≤ y ≤ 11 is shown in Figure 4.

6. Connection to the Fine Structure Constant

In Equation (1.1), when n = 2 and y = 10, the equation becomes $x={10}^{2}\times {10}^{\frac{x}{{10}^{3}}}$ and the solution is

Figure 3. Plots of lnx vs n.

Table 2. Solutions x in base y(x_{y}).

Figure 4. Plot of x_{y} vs y for n = 2.

$x=\frac{{10}^{2}W\left(\frac{-\mathrm{ln}10}{10}\right)}{\left(-\frac{\mathrm{ln}10}{10}\right)}=137.129$ (6.1)

The solution 137.129 is close to the inverse of the fine structure constant 137.036 [14] - [21] which is dimensionless.

The inverse of the fine structure constant ${\alpha}^{-1}$ is given by the expression

${\alpha}^{-1}=\frac{4\text{\pi}{\epsilon}_{o}\hslash c}{{e}^{2}}=137.036$ (6.2)

where;

$\hslash =1.0545718\times {10}^{-34}\text{\hspace{0.17em}}\text{J}\cdot \text{s}$ ,reduced Planck constant;

$c=2.99792458\times {10}^{8}\text{\hspace{0.17em}}\text{m}\cdot {\text{s}}^{-1}$ , speed of light in vacuum;

${\epsilon}_{0}=8.854187817\times {10}^{-12}\text{\hspace{0.17em}}\text{F}\cdot {\text{m}}^{-1}$ , electric constant;

$e=1.6021766208\times {10}^{-19}\text{\hspace{0.17em}}\text{C}$ , elementary charge;

${\alpha}^{-1}$ , dimensionless constant [22] .

In a recent publication Eaves [23] suggested an equation relating G and α;

$\frac{\alpha {q}^{2}}{8{\text{\pi}}^{2}G{m}_{e}^{2}}\approx \mathrm{exp}\left(\frac{2}{3\alpha}\right)$ (6.3)

where;

$G=6.67408\times {10}^{-11}\text{\hspace{0.17em}}{\text{m}}^{3}\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}$ , gravitational constant;

${m}_{e}=9.10938356\times {10}^{-31}\text{\hspace{0.17em}}\text{kg}$ , electron mass.

${q}^{2}=\frac{{e}^{2}}{4\pi {\epsilon}_{0}}$

By substituting the expression for α in Equation (6.3) we get

$\frac{{e}^{4}}{32{\text{\pi}}^{3}{\epsilon}_{0}^{2}G{m}_{e}^{2}\hslash c}\approx \mathrm{exp}\left(\frac{2}{3\alpha}\right)$ (6.4)

Using Equation (6.2), the Equation (6.4) becomes

${\alpha}^{-1}\approx \frac{64{\text{\pi}}^{4}{\epsilon}_{0}^{3}{\hslash}^{2}{c}^{2}G{m}_{e}^{2}}{{e}^{6}}\mathrm{exp}\left(\frac{{\alpha}^{-1}}{1.5}\right)$ (6.5)

Substituting numerical values for the pre-exponent,

${\alpha}^{-1}\approx 1.59947\times {10}^{-27}\mathrm{exp}\left(\frac{{\alpha}^{-1}}{1.5}\right)$ (6.6)

${\alpha}^{-1}\approx 1.59947\times {10}^{-27}\times {10}^{\left(\frac{{\alpha}^{-1}}{3.4538}\right)}$ (6.7)

By taking the power of (1/289.5) on both sides of the Equation (6.8) and writing the equation for ${\alpha}^{-1}$ yields

${\alpha}^{-1}\approx 106.6\times {10}^{\left(\frac{{\alpha}^{-1}}{1000}\right)}$ (6.8)

The Equation (6.8) is approximately the same as the equation
$x={10}^{2}\times {10}^{\frac{x}{{10}^{3}}}$ . The only difference is the 10^{2} is 106.6 in Equation (6.8). But the Equation (6.8) based on the Equation (6.3) is only an approximate equation.

The value $1.59947\times {10}^{-27}$ in Equation (6.6) is approximately equal to the ${\alpha}_{G}^{1/1.5},$ ${\alpha}_{G}$ defined by Jentschura [20] .

${\alpha}_{G}=\frac{G{m}_{e}{m}_{p}}{\hslash c}=3.21\times {10}^{-42}$ (6.9)

Hence the Equation (6.7) can be written as

${\alpha}^{-1}\approx {\alpha}_{G}^{1/1.5}\mathrm{exp}\left(\frac{{\alpha}^{-1}}{1.5}\right)$ (6.10)

7. Conclusions

An equation in the form of $x={y}^{n}{y}^{\left(\frac{x}{{y}^{n+1}}\right)}$ was solved graphically, numerically and analytically.

The plots of numerical solution x vs y indicate sharp turning points for y values in-between 1 to 2.

The analytical solution was found in terms of Lambert-W function as

$x=\frac{{y}^{n}W\left(\frac{-\mathrm{ln}y}{y}\right)}{\left(-\frac{\mathrm{ln}y}{y}\right)}$

The numerical solutions can be written as $x\left(n=0\right)\times {y}^{n}.$

The numerical solutions can also be written in base y as ${x}_{y}\left(n=0\right)\times {10}^{n}$ . For $y\ge 5{x}_{y}\left(n=0\right)$ is a universal number approximately equal to 1.37.

^{}If n = 2 and y = 10, the solution
$x=\frac{{10}^{2}W\left(\frac{-\mathrm{ln}10}{10}\right)}{\left(-\frac{\mathrm{ln}10}{10}\right)}=137.129$ (rounded) is close to the inverse of the fine structure constant value, 137.036.

The equation $x={10}^{2}\times {10}^{\frac{x}{{10}^{3}}}$ which gives the solution close to the fine structure constant can be derived from the equation $\frac{\alpha {q}^{2}}{8{\text{\pi}}^{2}G{m}_{e}^{2}}\approx \mathrm{exp}\left(\frac{2}{3\alpha}\right)$ suggested by Eaves.

The derivation resulted in an equation ${\alpha}^{-1}\approx {\alpha}_{G}^{1/1.5}\mathrm{exp}\left(\frac{{\alpha}^{-1}}{1.5}\right)$ .

Conflicts of Interest

The authors declare no conflicts of interest.

Cite this paper

*Journal of Applied Mathematics and Physics*,

**6**, 725-736. doi: 10.4236/jamp.2018.64065.

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