This question was asked in GATE exam and It's answer is 1. I could not understand this question.
Suppose that the Newton-Raphson method is applied to the equation $2x^2+1-e^{x^2}=0$ with an initial approximation $x_0$ sufficiently close to zero. Then, for the root $x= 0$, the order of convergence of the method is equal to _________(Fill in the blanks)
In the case of a simple zero Newton's method converges quadratically. But not so if we have a zero of higher order. Assume that we want to find the zero of the function $f(x):=x^2$ using this method. We then obtain the recursion $$x_{n+1}=x_n-{f(x_n)\over f'(x_n)}=x_n-{x_n^2\over 2x_n}={x_n\over2}\qquad(n\geq0)\ .$$ This shows that at each step the distance to the zero of $f$ is halved, in other words: We only have linear convergence.
In the example at hand the situation is similar: We are told to find the zero of $$g(x):=1+2x^2-e^{x^2}=1+2x^2-\left(1+x^2+{x^4\over2}+?x^6\right)=x^2-{x^4\over2}+?x^6\ .$$ Newton tells us to look at the auxiliary function $$\psi(x):=x-{g(x)\over g'(x)}=x-{x^2-x^4/2+?x^6\over 2x-2x^3+?x^5}={x^2-3x^4/2+?x^6\over 2x(1-x^2+?x^4)}\ .$$ It follows that $$\psi(x)={x\over2}\left(1-{x^2\over2}+?x^4\right)\ ,$$ so that at each step the distance to the zero is essentially halved when $x_n\to0$.
(In the above each question mark represents a certain power series $a_0+a_1x+a_2 x^2+\ldots$ convergent in a neighborhood of $0$.)