Unsure how to simplify

Question

I have the following equation: $$\frac{0.8}{x+1}+\frac{1}{(x+1)^2}+\frac{1.2}{(x+1)^3} = 2.5$$

According to WolframAlpha, it can be simplified as follows, assuming $x$ is real: $$\frac{1.09047}{x + 1} = 1$$

Of course it's just a rounded value, but is there some method to get this form without solving the cubic equation? I'm just curious. The initial form is so neatly arranged, I thought there might be some trick here. I'm only interested in the real solution of the equation.

Answer 1

Using Newton's method

Let $y=x+1$. The given equation is equivalent to

$$p(y)=25y^3-8y^2 -10y-12=0.$$

The real root can be found using Newton's method, i.e. using the recurrence relation

$$y_{n+1} = y_n - \frac{p(y_n)}{p^\prime(y_n)}$$ starting with $y_0=1$. Using Python Sympy or whatever else, you'll find

$$\begin{aligned}y_1 &= \frac{54}{49} &= 1.10204081632653\\ y_2 &= \frac{4070958}{3732673} &= 1.09062808341368\\ y_3 &= \frac{1751267765388614226414}{1605973357978615747009} &= 1.09047124392703\\ \end{aligned}$$

Appendix - Python code

from sympy import poly, Rational, latex
from sympy.abc import x

p = poly(Rational(25, 1) * x**3 - Rational(8, 1) * x**2 - Rational(10, 1) * x - Rational(12, 1), )
pp = p.diff()

fx = x - p / pp
print(fx)

x1 = fx.subs(x, Rational(1))
print(f"y_1 &= {latex(x1)} &= {x1.evalf()}")

x2 = fx.subs(x, x1)
print(f"y_2 &= {latex(x2)} &= {x2.evalf()}")

x3 = fx.subs(x, x2)
print(f"y_3 &= {latex(x3)} &= {x3.evalf()}")

Answer 2

Working with whole numbers and letting $t=\frac 1{x+1}$ whe need to solve for $t$ the cubic $$12 t^3+10 t^2+8 t-25=0$$ By inspecion or graphing , we know that the root is close to $t=1$.

Consider that this is an expansion to $O(t-1)^4$ $$5+64 (t-1)+46 (t-1)^2+12 (t-1)^3+O\left((t-1)^4\right)$$ Use a series reversion to obtain $$t=1-\frac 5{64}-\frac{575}{131072}-\frac{54125}{134217728}=\frac{123089043}{134217728}$$ Back to $x$ $$x=\frac{11128685}{123089043}=0.0904117$$

Now, suppose that we consider that it is $$5+64 (t-1)+46 (t-1)^2+12 (t-1)^3+O\left((t-1)^6\right)$$ we shall have $$t= \frac{516245100589197}{562949953421312}$$ and then $$x=\frac{46704852832115}{516245100589197}=0.0904703$$

Edit

Zooming the graph, we know thet the solution is close to $\frac 9{10$}$. So, write again $$-\frac{119}{125}+\frac{1379}{25} \left(t-\frac{9}{10}\right)+\frac{212}{5} \left(t-\frac{9}{10}\right)^2+12 \left(t-\frac{9}{10}\right)^3+O\left(\left(t-\frac{9}{10}\right)^4\right)$$ Series reversion again gives $$t=\frac{9}{10}+\frac{17}{985}-\frac{61268}{267588055}+\frac{72063884}{14538754757093}=\frac{26665090332931}{29077509514186}$$ and then $$x=\frac{2412419181255}{26665090332931}=0.090471067$$

while the exact solution is $0.090471215$