14
08
Approximate solutions to higher
order equations
(bisection method)
Let be a function continuous on the interval
, with and having opposite signs, so
that there must be at least one in with
. If we divide the interval into two,
must belong to one of the intervals. Thus by
comparing the signs of , and
we can half the interval in which we search.
Repeating this procedure will yield an approximate
solution of arbitrary accuracy.
Program
Lbl 1:?→ A:?→ B:( )( )≧0
⇒ Goto 1:(
)>0⇒ Goto 2:B → C:A → B:C →
A:Lbl 2:(A + B)÷2→ X:X
→ Y:Y >0⇒ Goto
3:X → B:Goto 2:Lbl 3:X → A:Goto 2 < 126 STEP >
Execution Example:
Find solutions to the equation . (Solutions are and )
a
b
a+b
2
x()
ab[,]
a()
b()
cab[,]
fc() 0= c
a()f
ab+
2
-------------
,
b()
ON
MODE MODE MODE
1
PRGM
MODE
1
COMP
1
P1
A
3
-2 A
2
-2A+4
B
3
-2 B
2
-2B+4
A
3
-2 A
2
-2A+4
X
3
-2 X
2
-2X+4
x
3
2x
2
–2x–4+0=2± 2
Prog
1
S A
D R
P1
P2 P3 P4
G
1
EXE
1 5
EXE
Disp
S A
D R
P1
P2 P3 P4
G
EXE
Disp
S A
D R
P1
P2 P3 P4
G
関数電卓事例集 .book 14 ページ 2002年9月2日 月曜日 午後6時51分