#### Newton Numeric Example 1

Let us consider simple 1-node circuit

```       --------+-------------
|        |             |
I=0.001   R=1000     i=V^3/1000
|        |             |
---      ---           ---     ```

Which is equivalent to equation:

F(V) = I + V/R + N(V) = V/1000 + 1/1000*V^3 - 0.001

J(V) = 1/1000 + 3/1000 * V^2

F'(V)= 3/1000*V^2*V + 0.001 - 1/1000*V^3

model code and circuit file for this example are here

Iteration sequence for this equation will be:

Iteration Newton classic (2,3) Gnucap's Newton Modification (5,6,7) Damped
0 V0=0 V0=0
J=0.001 J=0.001
F=-0.001 F' = 0.001
S=F/J=-1 R =F'/J=0.001/0.001=1;
V1= V0-S = 1 V1 = R = 1
1 V1=1 V1 =1
J=0.004 J = 0.004
F=0.001 F' =0.003
S=F/J=0.25 R=F'/J=0.75
V2=V1-S=0.75 V2=R=0.75
2 V2=0.75 V2=0.75
J=0.0026875 J=0.0026875
F=0.000171875 F'=0.00184375
S=F/J=0.063953 R=F'/J=0.686047
V3=V2-S=0.0686047 V3=R=0.686047
3 V3=0.0686047 V3=0.0686047
J=0.0024119815 J=0.0024119815
F=0.00000894221 F'=0.00164579043
S=F/J=0.003707 R=F'/J=0.68234
V4=V3-S=068234 V4=R=0.68234 