Answer to Question #158009 in Real Analysis for Cypress

Question

Go from 0 to 1 on the x-axis, then back half-way to 1/2, then forward half as far to 3/4, then back as half as far to 5/8, then forward half as far, and so on. Give an explicit formula for your position after the nth step. Where do you end up?

Answer 1

1) A_n = 0 + 2⁰ - 2^-1 + 2^-2 - 2^-3 + ... + 2^-(n-1) , where n = $2i-1$, $i\in\N$ (for odd).

Because at the odd step we move forward, so "plus".

A_n = (2⁰ - 2^-1) + (2^-2 - 2^-3) + ... + (2^-(n-3) - 2^-(n-2)) + 2^-(n-1)

A_n = (2*2^-1 - 2^-1) + (2*2^-3 - 2^-3) + ... + (2*2^-(n-2) - 2^-(n-2)) + 2^-(n-1)

A_n = (2^-1) + (2^-3) + ... + (2^-(n-2)) + 2^-(n-1)

2) A_n = 0 + 2⁰ - 2^-1 + 2^-2 - 2^-3 + ... - 2^-(n-1) , where n = $2i$, $i\in\N$ (for even).

Because at an even step we move backward, so "minus".

A_n = (2⁰ - 2^-1) + (2^-2 - 2^-3) + ... + (2^-(n-4)) - 2^-(n-3)) + (2^-(n-2)) - 2^-(n-1))

A_n = (2*2^-1 - 2^-1) + (2*2^-3 - 2^-3) + ...+ (2*2^-(n-3)) - 2^-(n-3)) + (2*2^-(n-1) - 2^-(n-1))

A_n = (2^-1) + (2^-3) + ...+ (2^-(n-3)) + (2^-(n-1))

If it is necessary to bring to a common denominator, then :

1)

A_n $=\frac{2^{-1}*2^{n-1}+2^{-3}*2^{n-1}+...+2^{-(n-2)}*2^{n-1}+1}{2^{n-1}}$

A_n $=\frac{2^{n-1-1}+2^{n-1-3}+...+2^{n-1-(n-2)}+1}{2^{n-1}}$

A_n $=\frac{2^{n-2}+2^{n-4}+...+2^1+1}{2^{n-1}}$ , for odd.

2)

A_n $=\frac{2^{-1}*2^{n-1}+2^{-3}*2^{n-1}+...+2^{-(n-3)}*2^{n-1}+1}{2^{n-1}}$

A_n $=\frac{2^{n-2}+2^{n-4}+...+2^2+1}{2^{n-1}}$ , for even.