Finding a Formula for the General Term of a Sequence

1. Jul 31, 2013

Jimbo57

1. The problem statement, all variables and given/known data
A bored student enters the number 0.5 in her calculator, then repeatedly computes the square of the number in the display. Taking A0 = 0.5, find a formula for the general term of the sequence {An} of the numbers that appear in the display, and find the limit of the sequence {An}.

2. Relevant equations

3. The attempt at a solution

So, I'm having a difficult time finding a standard rule to this sequence. The only thing I came up with was an= (a(n-1))2. Then it doesn't seem reasonable to show the limit of this sequence is = 0 as n increases, an decreases. Is there another way of finding a rule that is easier to work with?

2. Jul 31, 2013

micromass

Staff Emeritus
So $a_0 = 0.5$. Can you give me $a_1$? $a_2$? $a_3$? $a_4$? (don't work it out completely, leave it in symbols, so don't say that $(0.5)^2 = 0.25$, but leave it as $(0.5)^2$). Do you notice a pattern?

3. Jul 31, 2013

Jimbo57

$a_1$=$(0.5)^2$

$a_2$=$(0.25)^2$

$a_3$=$(0.0625)^2$

$a_4$=$(0.00390625)^2$

Hmmm, I see that 0625 is recurring and I'm assuming that as n increases, the amount of decimal places increase by 2n places. Does that make sense?

EDIT: This is probably what you didn't want me to do eh Micromass? I improved my answer down below.

Last edited: Jul 31, 2013
4. Jul 31, 2013

LCKurtz

You are simplifying, which hides the pattern. And so do decimals. Start out with $a_0=\frac 1 2$ and try that. Look for un-multiplied out powers of $2$.

5. Jul 31, 2013

Jimbo57

Gotcha,

$a_0=\frac 1 {2}^{1}$
$a_1=\frac 1 {2}^{2}$
$a_2=\frac 1 {2}^{4}$
$a_3=\frac 1 {2}^{8}$

Bingo. $a_n={2}^{2}^{n}$

I can't seem to get latex to work but it's 22n

EDIT: I got excited, it's 1/22n

6. Jul 31, 2013

LCKurtz

I added parentheses which are necessary and fixed your latex. Right click on an expression to see how it was fixed.

7. Jul 31, 2013

Jimbo57

Thank you so much LCKurtz