# iterative formula

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• January 27th 2008, 06:54 AM
Oranges&Lemons
iterative formula
a) given that A is the root of the equation

arccos(x-1)= square root of x+2

show that 0<A<1

b) use the iterative formula

x(n+1) = 1+cos of the square root of xn +2

sorry, the n+1 and n are supposed to be subscript.

when i tried to do part b, i got 1.998....which is not between 0 and 1 so i must be doing it wrong!

can anyone help?
• January 27th 2008, 08:10 AM
Henderson
Check to make sure you're using radians. It sounds like you're in degrees.
• January 27th 2008, 09:19 AM
Oranges&Lemons
you're right! how stupid of me. thanks!
• January 27th 2008, 09:25 AM
Jhevon
Quote:

Originally Posted by Oranges&Lemons
a) given that A is the root of the equation

arccos(x-1)= square root of x+2

show that 0<A<1

note that the function $f(x) = \arccos (x - 1) - \sqrt{x + 2}$ is continuous on [0,1]

now $f(0) > 0$ and $f(1) < 0$. thus by the intermediate value theorem, there is some $x \in (0,1)$ such that $f(x) = 0$. this $x$ is a root. we choose $A$ to be this $x$

Quote:

b) use the iterative formula

x(n+1) = 1+cos of the square root of xn +2

sorry, the n+1 and n are supposed to be subscript.
use the iterative formula to do what? show that 0 < A < 1?