Calling
#f(x,y) = ((f_1),(f_2))=((x y -a),(y log_e x+x log_e y - b))#
with #a = 64, b = 2.5#
Expanding near the point #p_0 = {x_0,y_0}# in Taylor series for the linear term
#f(x,y) = f(x_0,y_0) + grad f_0 cdot(p - p_0) +O^2(p-p_0)# where
#grad f = (((df_1)/dx,(df_1)/dy),((df_2)/dx,(df_2)/dy))#
#p-p_0 = ((x-x_0),(y-y_0))#
#O^2(p-p_0)->{0,0}# as #abs(p-p_0)->0#
For #abs(p-p_0)# small then
#f(x,y) approx f(x_0,y_0) + grad f_0 cdot(p - p_0)#
If #abs(p-p_0)# is small then #f(x,y) approx {0,0}#
and
#p = p_0 - (grad f_0)^{-1}p_0#
or
#p_{k+1} = p_k - (grad f_k)^{-1}p_k#
Here
#grad f = ((y, x),(y/x + Log(y), x/y + Log(x)))#
#(grad f_0)^{-1} = (((x/y + Log(x))/(x - y + y Log(x) - x Log(y)),
x/(-x + y - y Log(x) + x Log(y))),((y + x Log(y))/(
x (-x + y - y Log(x) + x Log(y))), y/(x - y + y Log(x) - x Log(y))))#
Begining with
#p_0 = {1,20}# we obtain
#p_1 = {0.877443,66.4511}#
#p_2 = {0.970524,65.8899}#
#p_3 = {0.9761,65.5652}#
#p_4 = {0.976141,65.5643}#
#p_5 = {0.976141,65.5643}#
