08
Approximate solutions to higher
order equations
f x ( )
Let
be a function continuous on the interval
f a ( )
[ , ]
a b
, with
and
that there must be at least one
f c ( )
=
0
.
If we divide the interval into two,
must belong to one of the intervals.
comparing the signs of
we can half the interval in which we search.
Repeating this procedure will yield an approximate
solution of arbitrary accuracy.
ON
MODE
MODE
Lbl 1:?→ A:?→ B: (
⇒ Goto 1: (
A
A:Lbl 2: (A + B)÷2→ X:X
3:X → B:Goto 2:Lbl 3:X → A:Goto 2 < 126 STEP >
Find solutions to the equation
Prog
1
1
EXE
1
EXE
14
f b ( )
having opposite signs, so
c
in
a
+
b
f a ( ) f
------------ -
,
, and
2
PRGM
MODE
3
A
-2 A
3
2
-2 A
-2 A +4
3
2
x
–
2x
–
5
EXE
(bisection method)
[ , ]
a b
with
c
Thus by
f b ( )
a
COMP
1
MODE
1
2
B
-2 A +4
) (
)>0⇒ Goto 2:B → C:A → B:C →
3
2
X
-2 X
-2 X +4
2x
+
4
=
0
. (Solutions are
S A
S A
S A
a+b
2
1
3
2
-2 B
-2 B +4
→ Y:Y >0⇒ Goto
±
2
2
and
)
P1 P1 P2 P3 P4
D R
G
P1 P1 P2 P3 P4
D R
G
P1 P1 P2 P3 P4
D R
G
b
)≧0
Disp
Disp