Inductive Step
Download
Report
Transcript Inductive Step
Problems to Solve Involving
Induction
n
ar
k
ar
n 1
a
r 1
k 0
,r 1 ?
Proof by Induction
Basis Step: Does it work for n=0?
0
ar
k
0
ar a
k 0
ar
0 1
a
r 1
ar a
r 1
a(r 1)
r 1
a
Inductive Step
Assume that for r1
n
ar
k
ar
k 0
n 1
a
r 1
We must show that
n 1
ar
k 0
k
ar
( n1 ) 1
r 1
a
n 1
ar
k
ar
n 1
n
k 0
ar
ar
k
ar
k 0
n 1
(r 1 ) ar
n 1
a
r 1
ar
n 2
ar
n1
ar
r 1
ar
( n1 ) 1
r 1
a
n1
a
n 1
ar
n1
a
r 1
n
Prove that (3 j 2 ) is
2
O(n )
j 1
First let’s find a closed form solution for the
sum.
n
n
n
j 1
j1
j1
(3 j 2 ) 3 j 2 1
= 3n(n+1)/2 - 2n = (3n2 + 3n -4n)/2 = (3n2 -n)/2
How can we be sure that our
arithmetic is correct?
n
Prove that
(3 j 2 )
2
3n n
2
j 1
Proof by Induction
Basis Step: 1
For n=1
(3 j 2 ) 3(1) 2 1
j 1
3(1) 2 1
2
2
2
1
Inductive Step
n
Assume that
(3 j 2 )
2
3n n
2
j 1
Then we must show that:
n1
(3 j 2 )
j 1
2
3( n 1) ( n 1)
2
n1
n
(3 j 2 ) (3 j 2) 3( n 1) 2
j 1
j 1
3n 2 n
2
3(n 1) 2
2
3n n 6( n 1) 4
2
3n 2 5n 2
2
3n 2 6n 3 n 1
3(n 2 2 n 1) ( n 1)
2
2
3( n 1) 2 ( n 1)
2
Big-Oh
(3n2 -n)/2 is clearly O(n2) since
(3n2 -n)/2 (3n2 -n) 3n2 since n is positive.
So chose C = 3 and k=1 and we have
| (3n2 -n)/2| (3) |n2 | when n > 1.
Does 1*1! + 2*2! + 3*3!+…+n*n!
= (n+1)! -1 n1?
Basis Step: n = 1
1*1! = 1, (1+1)!-1 = 2-1=1
Inductive Step:
Assume that
1*1! + 2*2! + 3*3!+…+n*n! = (n+1)! -1
We must show that
1*1!+2*2!+3*3!+…+n*n!+(n+1)*(n+1)! =(n+2)!-1
1*1!+2*2!+3*3!+…+n*n!+(n+1)*(n+1)!
= (n+1)! -1 + (n+1)*(n+1)!
= (n+1+1)(n+1)! -1
= (n+2)(n+1)! - 1
= (n+2)! - 1
n
2
n
2
Prove that every by
(n > 1)
chessboard can be tiled with T-ominos
Basis Case:
2
2
x
2
2 Board
Basis Case:
2
2
x
2
2 Board
Basis Case:
2
2
x
2
2 Board
Basis Case:
2
2
x
2
2 Board
Inductive Step
Assume that we can tile a board of size 2n by
2n. We must show that this implies that we
can tile a board of size 2n+1 by 2n+1.
Proof: Divide the 2n+1 by 2n+1 board into 4
parts, each of size 2n by 2n. Since we know
that each of these boards can be tiled, then
we can put them together to tile the 2n+1 by
2n+1 board.