51 and
Qus. 1. Use Euclid's division algorithm to
find HCF of:
(i) 135 and 225
(iii) 867 and 255.
Sol • Sten I. Since 225 > 125 h
(ii) 196 and 38220
Si
TI
Answers
Answer:
equation involving finite differences), we expand y(x) into a Taylor series
around a point xn, up to fifth order:
yn−1 = yn − y
0
n∆x +
1
2
y
00
n
(∆x)
2 −
1
6
y
000
n
(∆x)
3 +
1
24 y
0000
n
(∆x)
4 −
1
120 y
00000
n
(∆x)
5
+O[(∆x)
6
]
yn+1 = yn + y
0
n∆x +
1
2
y
00
n
(∆x)
2 +
1
6
y
000
n
(∆x)
3 +
1
24 y
0000
n
(∆x)
4 +
1
120 y
00000
n
(∆x)
5
+O[(∆x)
6
].
(1.24)
If we sum the two equations, we obtain:
yn+1 + yn−1 = 2yn + y
00
n
(∆x)
2 +
1
12
y
0000
n
(∆x)
4 + O[(∆x)
6
]. (1.25)
Eq.(1.21) tells us that
y
00
n = −gnyn + sn ≡ zn. (1.26)
The quantity zn above is introduced to simplify the notations. The following
relation holds:
zn+1 + zn−1 = 2zn + z
00
n
(∆x)
2 + O[(∆x)
4
] (1.27)
(this is the simple formula for discretized second derivative, that can be obtained
in a straightforward way by Taylor expansion up to third order) and thus
y
0000
n ≡ z
00
n =
zn+1 + zn−1 − 2zn
(∆x)
2
+ O[(∆x)
2
]. (1.28)
By inserting back these results into Eq.(1.25) one finds
yn+1 = 2yn − yn−1 + (−gnyn + sn)(∆x)
2
+
1
12 (−gn+1yn+1 + sn+1 − gn−1yn−1 + sn−1 + 2gnyn − 2sn)(∆x)
2
+O[(∆x)
6
]
(1.29)
and finally the Numerov’s formula
yn+1 h
1 + gn+1
(∆x)
2
12 i
= 2yn
h
1 − 5gn
(∆x)
2
12 i
− yn−1
h
1 + gn−1
(∆x)
2
12 i
+(sn+1 + 10sn + sn−1)
(∆x)
2
12 + O[(∆x)
6
]
(1.30)
that allows to obtain yn+1 starting from yn and yn−1, and recursively the func-
tion in the entire box, as long as the value of the function is known in the first
two points (note the difference with “traditional” initial conditions, Eq.(1.22),
in which the value at one point and the derivative in the same point is speci-
fied). It is of course possible to integrate both in the direction of positive x and
in the direction of negative x. In the presence of inversion symmetry, it will be
sufficient to integrate in just one direction.
In our case—Schr¨odinger equation—the sn terms are absent. It is convenient
to introduce an auxiliary array fn, defined as
fn ≡ 1 + gn
(∆x)
2
12
, where gn =
2m
¯h
2
[E − V (xn)]. (1.31)
Within such assumption Numerov’s formula can be written as
yn+1 =
(12 − 10fn)yn − fn−1yn−1
fn+1
. (1.3
Answer:
make me as brainliest and thank me if the answer is useful.
Step-by-step explanation:
By Euclid's division lemma,
225=135×1+90
r=90
135=90×1+45
r=45
90=45×2+0
So, H.C.F of 135 and 225 is 45
(ii) By Euclid's division lemma,
38220=196×195+0r=0
So, H.C.F of 38220 and 196 is 196
(iii) By Euclid's division lemma,
867=255×3+102
r=10
255=102×2+51
r=51
102=51×2+0
So, H.C.F of 867 and 255 is 51
The highest HCF among the three is 196.