Sequence based on arithmetic and geometric means problem

Let 0 < y_{1} < x_{1} and set

x_{n+1} = (1/2)(x_{n} + y_{n}) and y_{n+1} = sqrt(x_{n}y_{n})

Prove that 0 < x_{n+1} - y_{n+1} < (x_{1} - y_{1})/2^{n} for n in **N**

Provided solution:

x_{n+1} - y_{n+1 }= (1/2)(x_{n} + y_{n}) - sqrt(x_{n}y_{n}) < (1/2)(x_{n} + y_{n}) - y = (1/2)(x_{n} - y_{n})

Hence by induction and by the fact that 0 < y_{n} < x_{n} for n in **N, **0 < x_{n+1} - y_{n+1} < (x_{1} - y_{1})/2^{n }

What I do not understand is the last part of this explanation. I understand that, by induction, x_{n+1} - y_{n+1} < (x_{1} - y_{1})/2, but why 2^{n}?

Thanks in advance.

Re: Sequence based on arithmetic and geometric means problem

Simple, \[x_{n+1} - y_{n+1} = (1/2)(x_{n} - y_{n})\], and since \[(x_{n} - y_{n}) = 1/2 * (x_{n-1} - y_{n-1})\], \[x_{n+1} - y_{n+1} = (1/2)(x_{n} - y_{n}) = 1/2* 1/2 * (x_{n-1} - y_{n-1}) = 1/4 * (x_{n-1} - y_{n-1} \], and so on. Got the idea?

Salahuddin

Maths online

Re: Sequence based on arithmetic and geometric means problem

Hey lm1988.

You have x_(n+1) - y_(n+1) < 1/2(x_n - y_n), but this means x_(n) - y(n) < 1/2(x_(n-1) - y_(n-1)) which means x_(n+1) - y_(n+1) < 1/2^2(x_(n-1) - y(n-1)) and if you repeat this until you get x1 - y1 you get that result.

So basically you have to chain these together.