Newtons method and convergence

nobahar
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Hello!
In Newton's method, I am having difficulty understanding why the convergence is quadratic.
I can imagine the closer and closer approximations moving along the x-axis as the base of a right angle triangle, but I cannot see from here how I arrive at the convergence being quadratic.
Any one know of any sources that explain this in a straightforward manner? Or would anyone like to attempt to explain this for me?! It's really bugging me!
Thanks in advance.
 
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Should it be obvious that for, say, x2 - 2, the rate of convergence of the error should be quadratic? The examples I find on the internet are far from obvious, even those that proport to be geometrical interpretations!
Any help would be really appreciated!
Thanks.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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