# how to prove that a solution of the equation is

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 $x = 5$.

-
For n=4 we have also a solution in x=0.907... – Riccardo.Alestra Mar 12 '12 at 11:48
maybe you want to recheck that @Riccardo.Alestra – Tomarinator Mar 12 '12 at 12:29
Those tags seem inappropriate to me. This is not a differential nor a diophantine equation. EDIT: retagged. – Giuseppe Negro Mar 12 '12 at 13:04

Step 1 (Recast the problem in a simpler form). Let $p_n(x) := x^n-4x^{n-1}-4x^{n-2}-\cdots -4x-4$ and observe that neither $x=0$ nor $x=1$ solve the equation $p_n(x)=0$ for any value of $n$, hence you can assume $x\neq 0,1$. Now, rewrite: $$p_n(x) = x^n-4\sum_{k=0}^{n-1} x^k\; ;$$ from the rule for the sum of a geometric progression you get: $$\begin{split} p_n(x) &= x^n -4\frac{1-x^n}{1-x} &= \frac{x^{n+1}-5x^n+4}{x-1} \; , \end{split}$$ thus $p_n(\bar{x})=0$ iff $f_n(\bar{x})=0$, where: $$f_n(x) := x^{n+1}-5x^n+4\; .$$

Step 2 (Existence and properties of the solutions of $f_n(x)=0$). By the IVT, function $f_n$ has some zeros in the oper interval $I:=]4,5[$ because $f_n(4)<0<f_n(5)$; more precisely, there is only one zero in $I$ and it lies in the subinterval $]\frac{5n}{n+1} , 5[$.

Let $x_n$ be the zero of $f_n$ (and a fortiori of $p_n$) which lies in $I$. Since the sequence $f_n(x)$ is strictly decreasing for any $x\in I$, the sequence $x_n$ is strictly increasing and therefore it has a limit $\bar{x}$; from the upper bound $x_n<5$, you infer the bound $\bar{x} \leq 5$ thus $\bar{x}$ is finite.

Step 3 (Evaluation of $\bar{x}$). Now, you have to evaluate $\bar{x}$. From $f_n(x_n)=0$ and $x\neq 0,1$ you get: $$= \frac{1}{x_n^n\ (x_n-1)}\ \left( x_n-5+\frac{4}{x_n^n}\right) =0 \qquad \Leftrightarrow \qquad x_n=5-\frac{4}{x_n^n} \; ;$$ since $4<x_n<5$, the sequence $x_n^n$ tends to zero, hence finally: $$\bar{x} = \lim_{n\to \infty} x_n =\lim_{n\to \infty} 5-\frac{4}{x_n^n} =5$$ as you claimed.

-

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

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

Using the property

$1+x+x^2......x^n-1$= $\frac{1-x^n}{1-x}$

$x^n$ - 4{$\frac{1-x^n}{1-x}$} =0

$x^{n+1} - 5x^n + 4$= 0

$x - 5 + \frac{4}{x^n}$= 0

now if n tends to infinity x will tend to five.

-
Yo don't really want the $\}$ after the "=0" do you? – Thomas Andrews Mar 12 '12 at 12:07
In the second one, it should be "+" before "$\dots$". And you should use "\frac" instead of "/" e.g. $\frac{1-x^n}{1-x}$ – Gigili Mar 12 '12 at 12:09
Thanks @ThomasAndrews ,@gigili for pointing those things out. – Tomarinator Mar 12 '12 at 12:18
If n tends to infinity x will tend to 5 provided you are looking for a solution with x greater than some fixed number greater than 1 (say a solution greater than 4). If n is even there will be at least one more real solution between $x=0$ where the polynomial expression comes out at $-4$ and $x=-1$ where the value is $+1$. (This is because complex roots come in pairs.) – Mark Bennet Mar 12 '12 at 15:22