1] show that the equation ln2x = sin(x/2) has one root \(\displaystyle \alpha\) that lies between 0.6 and 0.8. I've done this and it was simple enough.

2] An iterative sequence, based on rearranging this equation, is given by

\(\displaystyle x_0 = 0.7\),

\(\displaystyle x_{n+1} = f(x_n)\)

where f(x) = \(\displaystyle \frac{1}{2}e^{sin(\frac{x}{2})}\).

evaluate f ' (0.7) and state why the result suggests that the sequence is convergent.

I have f ' (x) = 0.25\(\displaystyle e^{sin(\frac{x}{2})}cos(\frac{x}{2})\)

and when I put x=0.7 I get 0.331 to 3dp, and dividing this by 0.7 gives me 0.473 to 3dp. In the modulus, this gives me a value of less than 1 so this states that this is convergent.

My work above is uncertain and I'm not sure if I'm right.

3] Calculate \(\displaystyle x_1\) and \(\displaystyle x_2\) recording the answers as accurately as possible.

I've done this and I had values that were 0.2102871815 and then -0.2894231095, and seeing as the root is between 0.6 and 0.8 I'm not quite sure why this is. Either I've used the Newton-Rhapson method incorrectly or maybe I should do something else?

4] Round the value of \(\displaystyle x_2\) to three decimal places and determine whether this is the root correct to 3 decimal places.

I will be able to do this once I get a value that resembles something like what I'm looking for.

Can anyone help me with this iteration? Thanks if you can guide me in the right direction with this