Texas Instruments TI-89 Voyage 200, User Manual

Page: 852 / 1008

852 Appendix A: Functions and Instructions
part(
expression1
,
n
)
⇒ ⇒ ⇒ ⇒
expression
Simplifies
expression1
and returns the
n
th
argument
or operand, where
n
is > 0 and

the number of
top-level arguments or operands returned by
part(
expression1
)
. Otherwise, an error is returned.
part(cos(
p ù
x+3),1)
¸
3+
p ø
x
Note: Simplification changed the order of the
argument.
By combining the variations of
part()
, you can
extract all of the sub-expressions in the simplified
result of
expression1
. As shown in the example to
the right, you can store an argument or operand
and then use
part()
to extract further sub-
expressions.
Note: When using
part()
, do not rely on any
particular order in sums and products.
part(cos(
p ù
x+3))
¸
1
part(cos(
p ù
x+3),0)
¸
"cos"
part(cos(
p ù
x+3),1)
!
temp
¸
3+
p ø
x
temp
¸ p ø
x+3
part(temp,0)
¸
"+"
part(temp)
¸
2
part(temp,2)
¸
3
part(temp,1)
!
temp
¸ p ø
x
part(temp,0)
¸
"
ù
"
part(temp)
¸
2
part(temp,1)
¸ p
part(temp,2)
¸
x
Expressions such as (x+y+z) and (x
ì
y
ì
z) are
represented internally as (x+y)+z and (x ì y) ì z.
This affects the values returned for the first and
second argument. There are technical reasons
why
part(
x+y+z,1
)
returns y+x instead of x+y.
part(x+y+z)
¸
2
part(x+y+z,2)
¸
z
part(x+y+z,1)
¸
y+x
Similarly, x ù y ù z is represented internally as
(x
ù
y)
ù
z. Again, there are technical reasons why
the first argument is returned as y
ø
x instead of
x
ø
y.
part(x
ù
y
ù
z)
¸
2
part(x
ù
y
ù
z,2)
¸
z
part(x
ù
y
ù
z,1)
¸
y
ø
x
When you extract sub-expressions from a matrix,
remember that matrices are stored as lists of lists,
as illustrated in the example to the right.
part([a,b,c;x,y,z],0)
¸
"{"
part([a,b,c;x,y,z])
¸
2
part([a,b,c;x,y,z],2)
!
temp
¸
{x y z}
part(temp,0)
¸
"{"
part(temp)
¸
3
part(temp,3)
¸
z
delVar temp
¸
Done
The example Program Editor function to the right
uses
getType()
and
part()
to partially
implement symbolic differentiation. Studying and
completing this function can help teach you how
to differentiate manually. You could even include
functions that the cannot differentiate, such as
Bessel functions.
:d(y,x)
:Func
:Local f
:If getType(y)="VAR"
: Return when(y=x,1,0,0)
:If part(y)=0
: Return 0
¦
y=
p
,
ˆ
,
i
,numbers
:part(y,0)
!
f
:If f="
L
"
¦
if negate
: Return
ë
d(part(y,1),x)
:If f=" − "
¦
if minus
: Return d(part(y,1),x)
ì
d(part(y,2),x)
:If f="+"
: Return d(part(y,1),x)
+d(part(y,2),x)
:If f="
ù
"
: Return part(y,1)
ù
d(part(y,2),x)
+part(y,2)
ù
d(part(y,1),x)
:If f="{"
: Return seq(d(part(y,k),x),
k,1,part(y))
:Return undef
:EndFunc