Consider a real number $a>0$. Show that the Newton method to solve the equation $x^2-a=0$ is given by
$$x^{k+1} = \frac{1}{2}\left(x^k + \frac{a}{x^k}\right)$$
finding a zero of $f'$ is the same as finding a minimu of $f$, so that's what I did:
$$f'(x) = x^2 -a \implies f(x) = \frac{x^3}{3}-ax $$
$$\nabla^2 f(x) d = -\nabla f(x)\implies 2xd = -(x^2-a)\implies d = \frac{-x^2+a}{2x}$$
so we get
$$x^{k+1}=x^k + \lambda_k\left(\frac{-x^2+a}{2x}\right)$$
Now I must find $\lambda_k$ minium of $f(x^{k+1})=\frac{\left(x^k + \lambda_k\left(\frac{-x^2+a}{2x}\right)\right)^3}{3}-a\left(x^k + \lambda_k\left(\frac{-x^2+a}{2x}\right)\right)$
how should I find $\lambda_k$ that minimizes $f$ above?
This looks far too complicated: Just plugin your function $f(x) = x^2-a$ in the formula for Newton's method $$x_{k+1} = x_k -\frac{f(x_k)}{f'(x_k)} = x_k -\frac{x_k^2-a}{2x_k}$$ $$= x_k -\frac{x_k^2}{2 x_k} + \frac{a}{2x_k} = x_k -\frac{1}{2}x_k + \frac{a}{2x_k}$$ and finally $$ x_{k+1} = \frac{1}{2}\left(x_k + \frac{a}{x_k}\right)$$