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Expert: - 9/5/2009


Question
Prove that if 3<x(sub 1)<5 and x(sub n+1)= 2+ sqrt x(sub n)- 2.  Then
3<x(sub n+1)< x(sub n) holds for all n(elements of)N.  

N meaning natural numbers, sorry about the notation but that is the best way to make it more understable.  I am thinking that I need to use proof by induction.  I am not sure though on how to get this one done.

Answer
If x0 = 5, x1 = 2 + √(5-2) = 2 + √3.
It is known that √3 is less than √4, and that √4 = 2.
Therefore 2 + √3 < 4, which is less than 5.

If x0 = 3, x1 = 2 + √(3-2) = 2 + √1 = 2 + 1 = 3, so at the endpoint, they are equal.

Consider the two function f(x) = x and g(x) = 2 + √(x-2).
It can be seen that f'(x) = 1 and g'(x) = 1/(2√(x-2)).
Between 3 and 5 the slope fof (x) is 1 and the slope of g(x) is always less than 1.

This means that the distance between f(x) and g(x) is always increasing from 3 to 5.

Since f(x) and g(x) are equal at 3 and the slope of f(x) is always greater than g(x),
and the two functions are continuous on (3,5), f(x) is always greater that g(x).  

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