Texas Instruments TITANIUM TI-89 User Manual page 852

expression1
part(
Simplifies
or operand, where
top-level arguments or operands returned by
expression1
part(
By combining the variations of
extract all of the sub-expressions in the simplified
result of
the right, you can store an argument or operand
and then use
expressions.
Note: When using
particular order in sums and products.
Expressions such as (x+y+z) and (xì yì z) are
represented internally as (x+y)+z and (xì y)ì z.
This affects the values returned for the first and
second argument. There are technical reasons
why
part(
Similarly, xù yù z is represented internally as
(xù y)ù z. Again, there are technical reasons why
the first argument is returned as yøx instead of
xøy.
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.
The example Program Editor function to the right
uses
getType()
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.
852
) ⇒ ⇒ ⇒ ⇒
n expression
,
and returns the
expression1
is > 0 and  the number of
n
. Otherwise, an error is returned.
)
part()
. As shown in the example to
expression1
to extract further sub-
part()
, do not rely on any
part()
x+y+z,1 returns y+x instead of x+y.
)
and to partially
part()
part(cos( pù x+3),1) ¸
argument
n
th Note: Simplification changed the order of the
argument.
part(cos( pù x+3)) ¸
, you can
part(cos( pù x+3),0) ¸
part(cos( pù x+3),1) ! temp ¸
temp ¸
part(temp,0) ¸
part(temp) ¸
part(temp,2) ¸
part(temp,1) ! temp ¸
part(temp,0) ¸
part(temp) ¸
part(temp,1) ¸
part(temp,2) ¸
part(x+y+z) ¸
part(x+y+z,2) ¸
part(x+y+z,1) ¸
part(x ù y ù z) ¸
part(x ù y ù z,2) ¸
part(x ù y ù z,1) ¸
part([a,b,c;x,y,z],0) ¸
part([a,b,c;x,y,z]) ¸
part([a,b,c;x,y,z],2) ! temp ¸
part(temp,0) ¸
part(temp) ¸
part(temp,3) ¸
delVar temp ¸
: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 ,
: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
Appendix A: Functions and Instructions
3+ pø x
1
"cos"
3+ pø x
pø x+3
"+"
2
3
pø x
" ù "
2
p
x
2
z
y+x
2
z
y ø x
"{"
2
{x y z}
"{"
3
z
Done
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