# Recurrence Relation

• Mar 16th 2011, 09:17 AM
turtlejacks
Recurrence Relation
Here is a question that I have been stuck on.

A sequence is defined by the following recursive equation.
Xn+1=√(2-xn) X0=0

NOTE: the n+1 is supposed to be subscript and so is the last n as well as the 0 after the x.

(a) Calculate the values of xn for n = 1, 2, 3 and 4. Report an exact answer and a decimal
approximation.
(b) Use mathematical induction to show that the even-numbered terms are increasing, that is,
that: x0 < x2 < x4 < x6 … etc.
(c) Show that the even-numbered terms are all less than
• Mar 16th 2011, 11:14 AM
topsquark
Quote:

Originally Posted by turtlejacks
Here is a question that I have been stuck on.

A sequence is defined by the following recursive equation.
Xn+1=√(2-xn) X0=0

NOTE: the n+1 is supposed to be subscript and so is the last n as well as the 0 after the x.

(a) Calculate the values of xn for n = 1, 2, 3 and 4. Report an exact answer and a decimal
approximation.
(b) Use mathematical induction to show that the even-numbered terms are increasing, that is,
that: x0 < x2 < x4 < x6 … etc.
(c) Show that the even-numbered terms are all less than

How far have you gotten? And your last question got cut off.

-Dan
• Mar 16th 2011, 11:51 AM
turtlejacks
oh its supposed to say less then 1. And I solved part a.
Im confused how to do part b
• Mar 16th 2011, 12:42 PM
Soroban
Hello, turtlejacks!

Quote:

$\text{A sequence is de{f}ined by the following recursive equation:}$

. . . . $x_{n+1}\:=\:\sqrt{2-x_n},\;\;x_0\,=\,0$

$\text{(a) Calculate the values of }x_n\text{ for }n = 1, 2, 3,4.$
. . . . $\text{Report an exact answer and a decimal
approximation.}$

$\begin{array}{ccc}x_n & \text{Exact} & \text{Approx.} \\ \hline \\[-4mm]
x_0 & 0 & 0 \\ \\[-3mm]
x_1 & \sqrt{2} & 1.414213 \\ \\[-3mm]
x_2 & \sqrt{2\!-\!\sqrt{2}} & 0.761367 \\ \\[-3mm]
x_3 & \sqrt{2\!-\!\sqrt{2\!-\!\sqrt{2}} & 1.111140 \\ \\[-3mm]
x_4 & \sqrt{2\!-\!\sqrt{2\!-\!\sqrt{2\!-\!\sqrt{2}}} & 0.942793\end{array}$

Quote:

$\text{(b) Use mathematical induction to show that the even-numbered}$
. . . . $\text{terms are increasing; that is, that: }\:x_0 < x_2 < x_4 < x_6\;\hdots$

$\text{(c) Show that the even-numbered terms are all less than 1.}$

I have not solved these parts yet, but I have some observations.

We have: . $x \:=\:\sqrt{2 + \sqrt{2 + \sqrt{2 + \hdots }}}$

That is: . $x \:=\:\sqrt{2 + x} \quad\Rightarrow\quad x^2 \:=\:2+x \quad\Rightarrow\quad x^2 + x - 2 \:=\:0$

. . . $(x - 1)(x + 2) \:=\:0 \quad\Rightarrow\quad x \:=\:1,-2$

Since $\,x$ is obviously positive: . $\sqrt{2 + \sqrt{2+ \sqrt{2 + \hdots}}} \:=\: 1$

Extend and examine the table in (a).

We find that the odd-numbered terms are approaching 1 from above,
. . and that the even-numbered terms are approaching 1 from below.

Does that give anyone a jump-start?

• Mar 16th 2011, 02:01 PM
Plato
For the induction part.
Suppose that we know that $a_{2K}.

Which implies that $-a_{2K}>-a_{2K+2}$

By adding 2 and taking the square root we get:
$\sqrt{2-a_{2K}}>\sqrt{2-a_{2K+2}}$

Or $a_{2K+1}>a_{2K+3}$

Now consider $a_{2K+2}=\sqrt{2-a_{2K+1}} <\sqrt{2-a_{2K+3}}=a_{2K+4}$
• Mar 16th 2011, 06:22 PM
topsquark
Thank you, Plato. That one was bugging me. :)

-Dan