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
f x
( )
a b
[ , ]
f a
( )
f b
( )
c
a b
[ , ]
f c
( )
0
=
c
f a
( )
f
a
b
+
2
-------------
,
f b
( )
ON
MODE
MODE
MODE
1
PRGM
MODE
1
COMP
1
P1
Ａ
３
－２ Ａ
２
－２ Ａ ＋４
Ｂ
３
－２ Ｂ
２
－２ Ｂ ＋４
Ａ
３
－２ Ａ
２
－２ Ａ ＋４
Ｘ
３
－２ Ｘ
２
－２ Ｘ ＋４
x
3
2
x
2
–
2
x
–
4
+
0
=
2
±
2
Prog
1
S A
D R
P1
P1 P2 P3 P4
G
1
EXE
1
5
EXE
Disp
S A
D R
P1
P1 P2 P3 P4
G
EXE
Disp
S A
D R
P1
P1 P2 P3 P4
G
関数電卓事例集
.book
1
4
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