Hello appleseed Originally Posted by

**appleseed** Hi Grandad!!!

Thank you for the reply.

I think i've heard of Newton-Raphson before, but i cant remember exactly how to find the answer using it. Could you briefly go over the steps please?

Thanks!

If you've not had any practice with this method, this is not the easiest example to start with.

If you want to get a solution to the equation$\displaystyle f(x) = 0$

and you have a first approximation, $\displaystyle x = a_1$, to the answer, then a second approximation, $\displaystyle a_2$, is given by$\displaystyle a_2=a_1-\frac{f(a_1)}{f'(a_1)}$

In this case you'll have to let$\displaystyle f(x) = 2\sin(\tfrac12x)-\frac{3}{3x+2}$

So$\displaystyle f'(x) = \cos(\tfrac12x) +\frac{9}{(3x+2)^2}$

From the graph, there's a solution close to $\displaystyle x = 1$. So with $\displaystyle a_1 = 1$ we get$\displaystyle a_2 = 1-\frac{f(1)}{f'(1)}$$\displaystyle \approx 0.7$

... and so on.

The next approximation, $\displaystyle a_3$, uses $\displaystyle 0.7$ as the starting value:$\displaystyle a_3 = 0.7 -\frac{f(0.7)}{f'(0.7)}$

$\displaystyle \approx 0.73$ (to 2 d.p.)

Repeating the process, to 4 d.p., I make the answer $\displaystyle 0.7314$.

The next positive solution is around $\displaystyle x = 6$. To 4 d.p., the solution converges very quickly to $\displaystyle 6.1361$.

Grandad