Newton's method: 2 variables

The following functions are predefined:

	f1 = x**2+y**2-1, f2 = x*y-0.25 This has four solutions on the unit circle. Proposed first box is [-4,4]x[-4,4]
	f1 = x**2-y-2, f2 = x*y+1 Here three solutions exist, the point (1,-1) and the two points +/-((1-sqrt(5))/2,2/(1-sqrt(5)) Proposed first box is [-4,4]x[-4,4]
	f1 = 2*x+y+1+(1/12)*tanh(x-y), f2 = 2*x+3*y+1+(1/8)*atan(x+y) The solution of this problem is globally unique. Proposed first box [-10,10]x[-10,10]
	f1 = (1/2)*x*sin((1/2)*pi*x)-y, f2 = y**2-x+1 Here we have an infinite but countable set of solutions. Proposed first box is [-6,10]x[1,10]
	Input of functions of your own Please type the evaluation program of your function here using FORTRAN rules. This function has the form d22usr(x,y,f1,f2) with input x, y and output f1, f2. Your final statement must be f1= f2= You may use the constants pi, e(=exp(1)), sqrt2(=1.414...), the integer variables i,j,k, the logicals bool1,bool2,bool3 and the double precision variables sum,h1,h2,h3,h4,u(100),v(100),a(100,100) which are all intialized with zero resp. .false. . never change x or y!. first is a local integer and set 0 before calling the function the first time. You may use this in order to initialize some local data and set it 1 afterwards to avoid multiple such initialization. Your settings of the local variables are preserved during program execution. c**** it computes the truncated McLaurin series of f1=sin(pi*x)-cos(pi*y) c**** and f2 such that a solution exists at the line x=y c**** with a bounding box [0,5]x[0,5] and the guess (1,2) it works, but be c**** careful because of the exp! if ( first .ne. 0 ) goto 100 u(1)=1.d0 do 50 i=2,80 c****** coefficients of exp(pi*z) series u(i)=u(i-1)*pi/dble(i-1) 50 continue 100 continue first=1 f1=0.d0 h1=1.d0 do 200 i=40,1,-1 k=i+i f1=f1+h1*(u(k)*x**(k-1)-u(k-1)*y**(k-2)) h1=-h1 200 continue f2=y-x+dexp(x-y)-1.d0

Please specify the bounding rectangle here
If the iteration leaves this, we terminate the computation.
Of course :
xmin <= xmax und ymin <= ymax.

xmin =		xmax =
ymin =		ymax =

Please specify the termination criterion : correction 1-norm less than ε × box-size:
ε =
ε must be in the interval [1.0e-12,1.0e-2]. Other values will be projected on this interval.