3
$\begingroup$

consider the equation

$ x^n - 4 x^{n-1} - 4 x^{n-2} - 4 x^{n-3} - \cdots-4x-4=0$

for $n = 1$ :: solution is : $x = 4$

for $n = 2$, ($x^2 - 4 x - 4 = 0$) :: solution is : $x = 4.8$

for $n = 3$, ($x^3 - 4x^2 - 4 x - 4 =0 $) :: solution is : x = $4.96$

how to prove that as $n \to\infty$ (what ever it means) the solution is $x = 5$.

  • 1
    For n=4 we have also a solution in x=0.907...2012-03-12
  • 0
    maybe you want to recheck that @Riccardo.Alestra2012-03-12
  • 0
    Those tags seem inappropriate to me. This is not a differential nor a diophantine equation. EDIT: retagged.2012-03-12

2 Answers 2