# Convergence of Sequence

• Mar 9th 2010, 09:53 PM
harish21
Convergence of Sequence
For c>0, consider the following quadratic equation:

$\displaystyle x^2$-$\displaystyle x$-$\displaystyle c$=0 , x>0

Define the sequence {$\displaystyle x_n$} recursively fixing $\displaystyle x_1$ > 0 and then, if n is an index for which {$\displaystyle x_n$} has been defined, defining

$\displaystyle x_{n+1}$ = sqrt{c+x_n}

Prove that the sequence {$\displaystyle x_n$} converges monotonically to the solution of the above equation!

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I tried to attempt this problem by setting $\displaystyle x_1$ = 1, and creating a sequence. It turns out to be a monotonic incereasing sequence. But how do we show that the sequence converges to the solution of the quadratic equation?

Also I am unclear about the starting value of $\displaystyle x_1$. Is it ok to start with $\displaystyle x_1$ = 1, or does the starting value have to be another number?
• Mar 9th 2010, 10:04 PM
Prove It
Quote:

Originally Posted by harish21
For c>0, consider the following quadratic equation:

$\displaystyle x^2$-$\displaystyle x$-$\displaystyle c$=0 , x>0

Define the sequence {$\displaystyle x_n$} recursively fixing $\displaystyle x_1$ > 0 and then, if n is an index for which {$\displaystyle x_n$} has been defined, defining

$\displaystyle x_{n+1}$ = sqrt{c+x_n}

Prove that the sequence {$\displaystyle x_n$} converges monotonically to the solution of the above equation!

--------------------

I tried to attempt this problem by setting $\displaystyle x_1$ = 1, and creating a sequence. It turns out to be a monotonic incereasing sequence. But how do we show that the sequence converges to the solution of the quadratic equation?

Also I am unclear about the starting value of $\displaystyle x_1$. Is it ok to start with $\displaystyle x_1$ = 1, or does the starting value have to be another number?

Isn't it because for a monotonically convergent sequence, then as $\displaystyle n \to \infty, x_{n + 1} \approx x_n$?

So if $\displaystyle x_{n + 1} = \sqrt{c + x_n}$

$\displaystyle x_{n} = \sqrt{c + x_n}$ for large $\displaystyle n$

$\displaystyle x_n^2 = c + x_n$

$\displaystyle x_n^2 - x_n - c = 0$, the same Quadratic Equation given above...
• Mar 9th 2010, 10:15 PM
harish21
Quote:

Originally Posted by Prove It
Isn't it because for a monotonically convergent sequence, then as $\displaystyle n \to \infty, x_{n + 1} \approx x_n$?

So if $\displaystyle x_{n + 1} = \sqrt{c + x_n}$

$\displaystyle x_{n} = \sqrt{c + x_n}$ for large $\displaystyle n$

$\displaystyle x_n^2 = c + x_n$

$\displaystyle x_n^2 - x_n - c = 0$, the same Quadratic Equation given above...

got it. thanks
• Mar 10th 2010, 12:59 AM
chisigma
The difference equation that generates the sequence is...

$\displaystyle \Delta_{n} = x_{n+1} - x_{n} = \sqrt{x_{n} + c} - x_{n} = f(x_{n})$ (1)

... and its solution depends from $\displaystyle c$ and $\displaystyle x_{0}$. The 'fixed point' [if any] are the solution of the equation...

$\displaystyle f(x) = \sqrt{x + c} - x = 0$ (2)

... that is, in the case we consider the positive value of the 'square root' ...

$\displaystyle x_{+} = \frac{1 + \sqrt{1+4c}}{2}$ (3)

From (3) we deduce that $\displaystyle x_{+}$ exists only if is $\displaystyle c > -\frac{1}{4}$ and from (2) that in that case any $\displaystyle x_{0} > -c$ will produce a sequence converging at $\displaystyle x_{+}$. The function $\displaystyle f(x)$ is represented in the figure for $\displaystyle c=2\rightarrow x_{+}=2$...

http://digilander.libero.it/luposabatini/MHF49.bmp

The point $\displaystyle x_{+}$ is an 'actratting fixed point' and, because the 'slope' of $\displaystyle f(x)$ is negative and in absolute value less than 1 [see red line...] for any $\displaystyle x_{0}> -2$ the generated sequence will converge to $\displaystyle x_{+}$ without oscillations...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$