Order of convergence of Newton's method for $g(x) = xe ^{x/2} - 1 + x$

Question

The question is for a Matlab course and it is the following:

Study the order of convergence of Newton's method on the function $ g(x) = xe ^{x/2} -1 -x $ . Find the solution of accurate within $10^{-5}$, starting from $x_0 = 2.5$. Is it quadratic?

So I wrote a Matlab program for Newton's method. And it gave me an approximate root $x^*$ and the estimated error.

I'm having trouble understanding and applying the quadratic convergence definition.

My idea is to check $|x_{i-1}-x^*| / |x_{i}-x^*|^2$ in each step and see if its converging. Is that correct?

Answer 1

Yes that is correct and should result in

k        x[k]          f(x[k])         dx[k]      dx[k]/dx[k-1]^2

0    2.500000000000000 10.2259       -1.155037e+00  -1.155037e+00
1    1.344962880055704  2.97987      -6.967936e-01  -5.222906e-01
2    0.648169320860770  0.544435     -1.923188e-01  -3.961079e-01
3    0.455850509909319  0.0283948    -1.116912e-02  -3.019781e-01
4    0.444681389050070  8.70351e-05  -3.444617e-05  -2.761232e-01
5    0.444646942882529  8.23361e-10  -3.258703e-10  -2.746395e-01
6    0.444646942556658 -1.11022e-16   4.394048e-17   4.137855e+02

for the iteration formula in the title ($g(x)=...+x$) and

k        x[k]          f(x[k])         dx[k]      dx[k]/dx[k-1]^2

0  2.500000000000000  5.22586      -7.625347e-01  -7.625347e-01
1  1.737465307480700  1.40446      -4.065176e-01  -6.991336e-01
2  1.330947690963831  0.258294     -1.153082e-01  -6.977523e-01
3  1.215639529705251  0.0167891    -8.598166e-03  -6.466745e-01
4  1.207041363790715  8.83367e-05  -4.572034e-05  -6.184403e-01
5  1.206995643450354  2.48783e-09  -1.287697e-09  -6.160199e-01
6  1.206995642162657  4.44089e-16  -2.298597e-16  -1.386231e+02

for the iteration as formulated in the text ($g(x)=...-x$).

Both nicely show quadratic convergence until the number of available digits runs out.