Finding a Formula for the General Term of a Sequence

[Jimbo57]

Homework Statement

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}.

Homework Equations

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 a_n= (a_(n-1))². Then it doesn't seem reasonable to show the limit of this sequence is = 0 as n increases, a_n decreases. Is there another way of finding a rule that is easier to work with?

 

[micromass]

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?

 

[Jimbo57]

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?

$a_1$=$(0.5)^2$

$a_2$=$(0.25)^2$

$a_3$=$(0.0625)^2$

$a_4$=$(0.00390625)^2$

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.

 

[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 $2$.

 

[Jimbo57]

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 $2$.

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

 

[LCKurtz]

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. $a_n=\frac 1 {2^{2^n}}$

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

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

 

[Jimbo57]

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

Thank you so much LCKurtz