Answer to Question #92186 in Quantitative Methods for lee

Question

The equation e^(-Lx)=Lx has a root/zero between x=0 and x=1. This root is approximately at the midpoint of these two x-values. Find this root using a suitable algebra method.
L is 1.55

Answer 1

Let Lx = y, equation e^-y = y or ye^y = 1 have exact solution y = W(1), where W is Lambert W function, W(1) = Ω = Omega constant. Ω is transcendental number.

Recurrent relation for Ω is

$$\varOmega_{n+1}=\frac{1+\varOmega_n}{1+e^{\varOmega_n}},\ \varOmega_1=1$$

Using this recurrent sequence first 5 values for Ω_i are: Ω₁ = 1, Ω₂ = 0.537883, Ω₃ = 0.566987,

Ω₄ = 0.567143, Ω₅ = 0.567143.

Root of the equation f(y) = ye^y - 1 = 0 can be found using Newton's method.

$$f'(y) = (ye^y-1)'=e^y+ye^y$$

recurrent relation for Newton's method is

$$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}=y_n-\frac{y_ne^{y_n}-1}{e^{y_n}+y_ne^{y_n}}, y_1=1$$

Using this recurrent sequence first 5 values for y are y₁ = 1, y₂ = 0.68394, y₃ = 0.577454,

y₄ = 0.56723, y₅ = 0.567143.

If L = 1.55, x = Ω/L = 0.567143/1.55 = 0.36589

Answer: If L = 1.55, x = 0.36589.