Download X10c

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts

Basis (linear algebra) wikipedia, lookup

Eisenstein's criterion wikipedia, lookup

Fundamental theorem of algebra wikipedia, lookup

Algebraic number field wikipedia, lookup

Factorial wikipedia, lookup

Transcript
Recursive Definitions
Rosen, 3.4
Recursive (or inductive) Definitions
• Sometimes easier to define an object in
terms of itself.
• This process is called recursion.
– Sequences
• {s0,s1,s2, …} defined by s0 = a and sn = 2sn-1 + b for
constants a and b and n Z+
– Sets
• 3  S and x+y  S if xS and yS
– Functions
• Example: f(n) = 2n, f(n) = 2f(n-1) and f(0) = 1
Recursively Defined Functions
To define a function with the set of
nonnegative integers as its domain
1. Specify the value of the function at zero
(or sometimes, it first k terms).
2. Give a rule for finding its value at an
integer from its values at smaller integers.
Examples of Recursively Defined
Functions
• Factorial Function n!
 n! = n(n-1)(n-2)….(1)
 f(0) = 1, f(n) = n(f(n-1))
• an
 f(0) = 1, f(n+1) = f(n)*a
• Fibonacci Numbers
 f0=0, f1=1, f n+1 = fn + f n-1
 {0,1,1,2,3,5,8,13,...}
Prove that the nth term in the Fibonacci
n 2
sequence is 1   f i when n2
i 0
Induction Proof:
2 2
Basic Step: Let n = 2, then f2 = 1 =
1   f i =1+0
i 0
Inductive Step: Consider k2 and assume that the expression is
true for 2  n  k. We must show
that the expression is true for
k1
 (1   f i )
n = k+1, i.e., that fk+1
i 0
fk+1
= fk +fk-1 by definition
k 2
 (1   f i )  f k 1
i 0
by the inductive hypothesis
Fibonacci Proof (cont.)
k 2
 (1   f i )  f k 1
i 0
k1
 (1   f i )
i 0
Since f2 is true and [fn is true for 2  n  k  fk+1] is
true, then fn is true for all positive integers n2.
Recursively Defined Set Example
Let the set S be defined recursively as 3  S
and x+y  S if xS and yS.
Prove that S is equal to the set of positive
integers divisible by 3 (set A).
Proof:
To show that S = A, we must show that AS
and that SA.
Recursive Set Example (cont.)
First we will show that A S. To do this we must show that
every positive integer divisible by 3 is in S. This is the
same as saying that 3n is in S for nZ+. We will do an
inductive proof.
Let p(n) be the statement that 3nS for n a positive integer.
Basis Step: p(1) = 3(1) = 3
Inductive Step: We will show that p(n)  p(n+1)
Let p(n)  S. Then p(n+1) = 3(n+1) = 3n + 3. Since 3nS
and 3S, then 3n+3S.
Therefore every element in A must be in S.
Recursive Set Example (cont.)
Now we will show that SA. To do this we must show that
every element in S is divisible by 3, i.e., that it can be
represented as 3(j) where j is a positive integer.
Basis Step: 3A since 3 = 3(1)
Inductive Step:
Assume x,y are in A (where x,y  S). x and y are divisible by
3 since they are both in A. Therefore x = 3j and y = 3k for
j,k  Z+. Then z = x+y = 3j+3k = 3(j+k), which is
divisible by 3, so z A (and by definition, z  S).
Therefore every element in S must be in A
Prove fn > n-2 where  = (1+5)/2
whenever n3
Inductive Proof
Basis Step: First we will show fn > n-2 for f3 and
f 4.
3-2 = 1 = (1+5)/2 < (1+9)/2 = 4/2 = 2 = f3
4-2 = 2 = [(1+5)/2]2 = (1+ 25 + 5)/4 = (6+ 25)/4
= 3/2 + 5/2 < 3/2 + 9/2 = 3 = f4
fn > n-2 (cont.)
Inductive Step:
Assume that fk > k-2 is true for all
nonnegative integers 3  k  n when n>4 .
We must show that fn+1> n-1.
Lemma: 2 = [(1+5)/2]2 = (1+25 +5)/4 =
3/2 + (5)/2 = 1 + (1+5)/2 = 1+
fn > n-2 (cont.)
n-1 = 2n-3 = (+1)(n-3) = n-2+n-3
By our inductive hypothesis
fn-1 > n-3 and fn > n-2
fn+1= fn + fn-1 > n-2+n-3 = n-1
Since f3 and f3 are true and fk is true for 3  k  n  fn+1 is
true, then fn is true for all positive integers n3
Find a closed form solution to the recursive
equation: T(n) = T(n-1) + c1. T(0) = c0
One approach is to write down several terms, guess
what the equation is, and then prove that your
guess is correct (or not) using an induction proof.
T(0) = c0
T(1) = T(0) + c1 = c0 + c1
T(2) = T(1) + c1 = c0 + 2c1
T(3) = T(2) + c1= c0 + 3c1
T(4) = T(3) + c1= c0 + 4c1
Guess that T(n) = c0 + nc1
Induction proof
Basis Step: for n = 0, T(0) = c0 + (0)c1 = c0
Inductive Step: Assume that T(n) = c0 + nc1,
we must show that T(n+1) = c0 + (n+1)c1.
T(n+1) = T(n) + c1 = c0 + nc1 + c1 = c0 +
(n+1)c1.
Find a closed form solution to
T(1) = c0, T(n) = 2T(n-1)+c1
T(1) = c0
T(2) = 2T(1) + c1 = 2c0 + c1
T(3) = 2T(2)+c1 = 2(2c0+c1)+c1 = 4c0+3c1
T(4) = 2T(3)+c1 = 2(4c0+3c1)+c1 = 8c0+7c1
T(5) = 2T(4)+c1 = 2(8c0+7c1)+c1 = 16c0+15c1
Guess that T(n) = 2n-1c0 + (2n-1-1)c1
Prove that T(1) = c0, T(n) = 2T(n-1)+c1has
closed form solution T(n) = 2n-1c0 + (2n-1-1)c1
Basis Step: T(1) = 21-1c0 + (21-1-1)c1 = c0
Induction Step: Assume that T(n) = 2n-1c0 +
(2n-1-1)c1. We must show that T(n+1) =
2(n+1)-1c0 + (2(n+1)-1-1)c1 = 2nc0 + (2n-1)c1.
T(n+1) = 2T(n) + c1 = 2[2n-1c0 + (2n-1-1)c1]+
c1 = 2nc0 + (2n-2)c1 + c1 = 2nc0 + 2nc1 - c1 =
2nc0 + (2n-1)c1.