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(i) Truefor
n
n
.
6
6, falsefor 7
(ii) Truefor
n
39, falsefor
n
40, or a multipleof 41, or 40
k
.
P
(
n
) does not imply
P
(
n
1).
7
6
5
n
1 divisibleby 5
6
5(
n
1)
1 divisibleby 5.
But both statements are false because the first statement is false when
n
1.
P
(1) is false.
8
(i) (1
x
)
dy
/
dx
1, or (1
x
)
y
1, so (1
x
)
y
y
0, and
theproposition holds for
n
1.
Suppose(1
x
)
y
ny
0, differentiating we have
x
)
y
(
n
1)
y
(1
0, and theproposition for
n
implies the proposition for
n
1.
When
x
0,
y
1 and
y
ny
.So
y
(
1)
n
!
y
(
1)
(
n
1)! and thersult holds
by induction.
(ii) Differentiate (1
x
)
y
1 twice to get result for
n
1.
Differentiate result for
n
to obtain result for
n
1. Result follows
by induction.
When
x
0,
y
1,
y
0 and
y
n
(
n
1)
y
0. So
result holds for
n
1.
Also
y
0
y
0, and
y
(
1)
(2
n
)!
y
(2
n
1)(2
n
2)(
1)
(2
n
)!
(
1)
(2
n
2)!.
So result holds by induction.
