# Newton Rhapson question

#### db5vry

This is of four parts:

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

#### TKHunny

First, why is this in "Algebra / PreAlgebra"?

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.
How?

I have f ' (x) = 0.25$$\displaystyle e^{sin(\frac{x}{2})}cos(\frac{x}{2})$$
I am quite puzzled by this. Why did you switch to 0.25, rather than staying with 1/4?

this gives me a value of less than 1 so this states that this is convergent.
Reality check. Read the definition again. It WILL converge or is has a chance to converge?

I've done this and I had values that were 0.2102871815 and then -0.2894231095
Something funny, here. Are you simply evaluating $$\displaystyle x_{i} = f(x_{i-1})$$ or are you trying to build $$\displaystyle x_{i} = x_{i-1} - \frac{g(x_{i-1})}{g'(x_{i-1})}$$ for some function g that has not been specified in your post?

REALLY BIG NOTE: When you created f(x), you should have noticed a Domain change. Did you? What is it? Why does it matter? Will it make any difference?

#### awkward

MHF Hall of Honor
This is of four parts:

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
Your problem title indicates that you have misunderstood this problem. It's not about Newton-Raphson iteration. It's about fixed-point iteration (or maybe you know it by some other name).

$$\displaystyle x_0 = 0.7$$
$$\displaystyle x_{n+1} = f(x_n)$$
to calculate $$\displaystyle x_1$$ and $$\displaystyle x_2$$.