Graphics Programs Reference
In-Depth Information
First iteration
Substituting
x
=
.
=
.
0
5,
y
1
5inEq. (c), we get
1
0
.
0 3
.
0
x
.
50
=
1
.
5 0
.
5
y
0
.
25
the solution of which is
x
=
y
=
0
.
125
.
Therefore, the improved coordinates of the
intersectionpoint are
x
=
0
.
5
+
0
.
125
=
0
.
625
y
=
1
.
5
+
0
.
125
=
1
.
625
Second iteration
Repeating the procedure using the latest values of
x
and
y
, we
obtain
1
.
250 3
.
250
x
−
0
.
031250
=
1
.
625 0
.
625
y
−
0
.
015625
which yields
x
=
y
=−
0
.
00694. Thus
x
=
0
.
625
−
0
.
00694
=
0
.
618 06
y
=
1
.
625
−
0
.
00694
=
1
.
618 06
Third iteration
Substitution of the latest
x
and
y
into Eq. (c) yields
1
.
.
−
.
236 12 3
23612
x
0
000 116
=
1
.
618 06 0
.
61806
y
−
0
.
000 058
The solutionis
x
=
y
=−
0
.
00003,sothat
x
=
0
.
618 06
−
0
.
000 03
=
0
.
618 03
=
.
−
.
=
.
y
1
618 06
0
000 03
1
618 03
Subsequent iterations wouldnot change the results within fivesignificant figures.
Therefore, the coordinates of the four intersectionpoints are
±
(0
.
618 03
,
1
.
618 03) and
±
(1
.
618 03
,
0
.
618 03)
Alternate solution
If there areonlyafew equations, it may be possible to eliminate
all butone of the unknowns. Thenwe wouldbe left with a single equationwhich
can be solvedbythe methods describedinArts. 4.2-4.5. In this problem, weobtain
fromEq. (b)
1
x
which upon substitutioninto Eq. (a) yields
x
2
y
=
x
2
+
1
/
−
3
=
0, or
x
4
3
x
2
−
+
=
1
0
=±
.
±
.
The solutionsofthis biquadraticequation:
x
0
618 03 and
1
618 03 agree with
the results obtainedbythe Newton-Raphsonmethod.
Search WWH ::
Custom Search