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