Isiqondiso kumsebenzisi

x
y
x > 0: -1 × 10
100
< ylogx < 100
x = 0: y > 0
x < 0: y = n,
m
2n + 1
(m, n bayizinombolo
eziphelele)
Nokho: -1 × 10
100
< ylog |x| < 100
x
√y
y > 0: x ≠ 0, -1 × 10
100
< 1/x logy < 100
y = 0: x > 0
y < 0: x = 2n+1,
2n + 1
m
(m ≠ 0; m, n
bayizinombolo eziphelele)
Nokho: -1 × 10
100
< 1/x log |y| < 100
a
b
/
c
Isamba senombolo ephelele, inombolo
engenhla eqhezwini, nenombolo
engezansi eqhezwini kumelwe sibe imivo
engu-10 noma ngaphansi (sekuhlangene
nophawu lokuhlukanisa).
RanInt#(a; b) a < b; |a|, |b| < 1 × 10
10
; b - a < 1 × 10
10
GCD(a; b)
|a|, |b| < 1 × 10
10
(a, b bayizinombolo
eziphelele)
LCM(a; b)
0 ≦ a, b < 1 × 10
10
(a, b bayizinombolo
eziphelele)
• Ukunemba kuyafana nalokho okuchazwe ngaphansi kwesithi
"Umkhawulo Nokunemba Kokubala", ngenhla.
• x
y
,
x
√y,
3
√ , x!, nPr, nCr idinga ukubala kwangaphakathi okulandelanayo,
okungenza ukuthi amaphutha enzeka esibalweni ngasinye
anqwabelane.
• Iphutha liyanqwabelana bese liba into enkulu endaweni lapho kwenzeka
khona i-inflekshini.
•
Umkhawulo wemiphumela yokubala ongavezwa ngesimo sika π uma
usebenzisa Isimo Sobuso Semvelo ungu-|x| < 10
6
. Nokho, phawula
ukuthi iphutha lokubala elenzeka ngaphakathi lingenza ingaveli eminye
imiphumela ngesimo sika-π. Lingabangela nokuba imiphumela yokubala
obekufanele ivele ngesimo sedesimali, ivele ngesimo se-π.
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