Xorpic, Zeros() – Texas Instruments PLUS TI-89 User Manual
Page 536
Appendix A: Functions and Instructions 519
8992APPA.DOC TI-89 / TI-92 Plus: Appendix A (US English) Susan Gullord Revised: 02/23/01 1:48 PM Printed: 02/23/01 2:21 PM Page 519 of 132
integer1
xor
integer2
⇒
integer
Compares two real integers bit-by-bit using
an
xor
operation. Internally, both integers are
converted to signed, 32-bit binary numbers.
When corresponding bits are compared, the
result is 1 if either bit (but not both) is 1; the
result is 0 if both bits are 0 or both bits are 1.
The returned value represents the bit results,
and is displayed according to the
Base
mode.
You can enter the integers in any number
base. For a binary or hexadecimal entry, you
must use the 0b or 0h prefix, respectively.
Without a prefix, integers are treated as
decimal (base 10).
If you enter a decimal integer that is too large
for a signed, 32-bit binary form, a symmetric
modulo operation is used to bring the value
into the appropriate range.
Note:
See
or
.
In Hex base mode:
0h7AC36 xor 0h3D5F ¸ 0h79169
In Bin base mode:
0b100101 xor 0b100 ¸0b100001
Note:
A binary entry can have up to 32
digits (not counting the 0b prefix). A
hexadecimal entry can have up to 8 digits.
XorPic
CATALOG
XorPic
picVar
[,
row
] [,
column
]
Displays the picture stored in
picVar
on the
current Graph screen.
Uses
xor
logic for each pixel. Only those pixel
positions that are exclusive to either the
screen or the picture are turned on. This
instruction turns off pixels that are turned on
in both images.
picVar
must contain a pic data type.
row
and
column
, if included, specify the pixel
coordinates for the upper left corner of the
picture. Defaults are (0, 0).
zeros()
MATH/Algebra menu
zeros(
expression
,
var
)
⇒
list
Returns a list of candidate real values of
var
that make
expression
=0.
zeros()
does this by
computing
exp
8
list(solve(
expression
=0,
var
)
,var
)
.
zeros(aù x^2+bù x+c,x) ¸
{
ë( bñ-4øaøc-+b)
2øa
bñ
-4øaøc-b
2øa
}
aù x^2+bù x+c|x=ans(1)[2] ¸
0
For some purposes, the result form for
zeros()
is more convenient than that of
solve()
. However, the result form of
zeros()
cannot express implicit solutions, solutions
that require inequalities, or solutions that do
not involve
var
.
Note:
See also
cSolve()
,
cZeros()
, and
solve()
.
exact(zeros(aù (e^(x)+x)
(sign (x)ì 1),x)) ¸
{}
exact(solve(aù (e^(x)+x)
(sign (x)ì 1)=0,x)) ¸
e
x
+ x = 0 or x>0 or a = 0
Important: Zero, not the letter O.