1. ## 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?

2. Check to make sure you're using radians. It sounds like you're in degrees.

3. you're right! how stupid of me. thanks!

4. 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 $\displaystyle f(x) = \arccos (x - 1) - \sqrt{x + 2}$ is continuous on [0,1]

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

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?