Question

Find the real root of f(x) = x+2-logx =0 by
Newton raphson method.​

Answer 1

Step-by-step explanation:

Let's look at the iterations you get from each.

With f(x)=x−e−xf(x)=x−e−x, you get

F(x)=x−f(x)f′(x)=x−x−e−x1+e−x=x+xe−x−x+e−x1+e−x=x+1ex+1.F(x)=x−f(x)f′(x)=x−x−e−x1+e−x=x+xe−x−x+e−x1+e−x=x+1ex+1.

That's a nice function, defined on all of RR, gets you close to the solution real fast when you start at a non-negative x0x0, but is not so good for negative xx, then it approaches the fixed point slowly at first.

With g(x)=xex−1g(x)=xex−1, you get

G(x)=x−g(x)