lesson 80 - mathematical induction hl2 - santowski
TRANSCRIPT
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Lesson 80 - Mathematical Induction
HL2 - Santowski
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Opening Exercise - Challenge
• Take a pile of n stones• Split the pile into two smaller piles of size r and s• Repeat until you have n piles of 1 stone each
• Take the product of all the splits• So all the r’s and s’s from each split
• Sum up each of these products
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Opening Exercise – An Example 310
3
14
7
1 12 2 2 1
23
1 1 1 1 1 1
21
12 2
4 2 1
1 1 1
10
2
9*104511112421221
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Opening Exercise - Challenge
• Take a pile of n stones• Split the pile into two smaller piles of size r and s• Repeat until you have n piles of 1 stone each
• Take the product of all the splits• So all the r’s and s’s from each split
• Sum up each of these products
• Prove that this product equals
4
2
)1( nn
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What is induction?• A method of proof
• It does not generate answers: it only can prove them
• Three parts:• Base case(s): show it is true
for one element• Inductive hypothesis: assume
it is true for any given element• Must be clearly labeled!!!
• Show that if it true for the next highest element
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Induction example #1
• Show that the sum of the first n odd integers is n2
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Induction example #1
• Show that the sum of the first n odd integers is n2
• Example: If n = 5, 1+3+5+7+9 = 25 = 52
• Formally, Show
• Base case: Show that P(1) is true
8
2i 1 n2
i1
n
11
11)(21
1
2
i
i
)P( where)P( nnn
)1P(
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Induction example #1, continued
• Inductive hypothesis: assume true for k• Thus, we assume that P(k) is true, or that
• Note: we don’t yet know if this is true or not!
• Inductive step: show true for k+1• We want to show that:
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2i 1 k2
i1
k
2i 1 (k 1)2
i1
k1
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Induction example #1, continued
• Recall the inductive hypothesis:
• Proof of inductive step:
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2(k 1) 1 2i 1i1
k
k2 2k 1
2(k 1) 1k2 k2 2k 1
2i 1 k2
i1
k
k2 2k 1 k2 2k 1
2i 1i1
k1
(k 1)2
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What did we show• Base case: P(1)• If P(k) was true, then P(k+1) is true
• i.e., P(k) → P(k+1)
• We know it’s true for P(1)• Because of P(k) → P(k+1), if it’s true for P(1), then it’s true for P(2)• Because of P(k) → P(k+1), if it’s true for P(2), then it’s true for P(3)• Because of P(k) → P(k+1), if it’s true for P(3), then it’s true for P(4)• Because of P(k) → P(k+1), if it’s true for P(4), then it’s true for P(5)• And onwards to infinity
• Thus, it is true for all possible values of n
• In other words, we showed that:
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)P()1P()P()1P( nnkkk
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The idea behind inductive proofs
• Show the base case
• Show the inductive hypothesis
• Manipulate the inductive step so that you can substitute in part of the inductive hypothesis
• Show the inductive step
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Why speling is not so important…
I cdnuolt blveieetaht I cluod aulaclty uesdnatnrd waht I was rdanieg. The phaonmneal pweor of thehmuan mind. Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in waht oredr the ltteers in a wrod are, the olny iprmoatnt tihng is taht thefrist and lsat ltteer be in the rghit pclae. The rset can be a taotl mses andyou can sitll raed it wouthit a porbelm. Tihs is bcuseae the huamn mnid deosnot raed ervey lteter by istlef, but the wrod as a wlohe. Amzanig huh? yaeh and I awlyas thought slpeling was ipmorantt.
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Second induction example
• Example #2:• Show the sum of the first n positive even integers is n2 + n
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Second induction example
• Example #2:• Show the sum of the first n positive even integers is n2 + n• Rephrased:
• The three parts:• Base case• Inductive hypothesis• Inductive step
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n P(n) where P(n) 2i n2 ni1
n
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Second induction example, continued
• Base case: Show P(1):
• Inductive hypothesis: Assume
• Inductive step: Show
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P(1) 2(i) 12 1i1
1
2 2
P(k) 2i k2 ki1
k
P(k 1) 2ii1
k1
(k 1)2 (k 1)
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Second induction example, continued
• Recall our inductive hypothesis:
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k2 3k 2 k2 3k 2
2(k 1)k2 k (k 1)2 k 1
2(k 1) 2ii1
k
(k 1)2 k 1
2ii1
k1
(k 1)2 k 1
P(k) 2i k2 ki1
k
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Notes on proofs by induction
• We manipulate the k+1 case to make part of it look like the k case
• We then replace that part with the other side of the k case
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P(k) 2i k2 ki1
k
k2 3k 2 k2 3k 2
2(k 1)k2 k (k 1)2 k 1
2(k 1) 2ii1
k
(k 1)2 k 1
2ii1
k1
(k 1)2 k 1
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Third induction example
• Show
19
6
)12)(1(
1
2
nnni
n
i
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Third induction example
• Show
• Base case: n = 1
• Inductive hypothesis: assume
20
6
)12)(1(
1
2
nnni
n
i
116
61
6
)12)(11(1
2
1
1
2
i
i
6
)12)(1(
1
2
kkki
k
i
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Third induction example
• Inductive step: show
21
6
)1)1(2)(1)1)((1(1
1
2
kkki
k
i
6
)32)(2)(1()1(
1
22
kkkik
k
i
6139261392 2323 kkkkkk
)32)(2)(1()12)(1()1(6 2 kkkkkkk
6
)32)(2)(1(
6
)12)(1()1( 2
kkkkkk
k
6
)1)1(2)(1)1)((1(1
1
2
kkki
k
i
6
)12)(1(
1
2
kkki
k
i
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Third induction again: what if your inductive hypothesis was wrong?
• Show:
22
6
)22)(1(
1
2
nnni
n
i
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Third induction again: what if your inductive hypothesis was wrong?
• Show:
• Base case: n = 1:
• But let’s continue anyway…
• Inductive hypothesis: assume
23
6
)22)(1(
1
2
nnni
n
i
6
71
6
71
6
)22)(11(1
2
1
1
2
i
i
6
)22)(1(
1
2
kkki
k
i
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Third induction again: what if your inductive hypothesis was wrong?
• Inductive step: show
24
6
)2)1(2)(1)1)((1(1
1
2
kkki
k
i
6
)42)(2)(1()1(
1
22
kkkik
k
i
816102614102 2323 kkkkkk
)42)(2)(1()22)(1()1(6 2 kkkkkkk
6
)42)(2)(1(
6
)22)(1()1( 2
kkkkkk
k
6
)2)1(2)(1)1)((1(1
1
2
kkki
k
i
6
)22)(1(
1
2
kkki
k
i
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Fourth induction example
• Show that n! < nn for all n > 1
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Fourth induction example
• Show that n! < nn for all n > 1
• Base case: n = 22! < 22
2 < 4
• Inductive hypothesis: assume k! < kk
• Inductive step: show that (k+1)! < (k+1)k+1
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Fourth induction example
• Show that n! < nn for all n > 1
• Base case: n = 22! < 22
2 < 4
• Inductive hypothesis: assume k! < kk
• Inductive step: show that (k+1)! < (k+1)k+1
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)!1( k !)1( kk kkk )1( kkk )1)(1( 1)1( kk
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Proving De Moivre’s Theorem by Induction
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Proving De Moivre’s Theorem by Induction
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Further Examples - #1 & #2
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Further Examples - #3
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IB Examples
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IB Examples
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IB Examples
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Mathematical Induction: Example
Show that any postage of ≥ 8¢ can be
obtained using 3¢ and 5¢ stamps.First check for a few particular values:
8¢ = 3¢ + 5¢
9¢ = 3¢ + 3¢ + 3¢
10¢ = 5¢ + 5¢
11¢ = 5¢ + 3¢ + 3¢
12¢ = 3¢ + 3¢ + 3¢ + 3¢How to generalize this?
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Mathematical Induction: Example
• Let P(n) be the sentence “n cents postage can beobtained using 3¢ and 5¢ stamps”.
• Want to show that “P(k) is true” implies “P(k+1) is true”
for any k ≥ 8¢.• 2 cases: 1) P(k) is true and
the k cents contain at least one 5¢. 2) P(k) is true and
the k cents do not contain any 5¢.
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Mathematical Induction: Example• Case 1: k cents contain at least one 5¢ coin.
• Case 2: k cents do not contain any 5¢ coin.
Then there are at least three 3¢ coins.
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5¢ 3¢ 3¢
Replace 5¢ coin by two 3¢ coinsk cents k+1 cents
3¢3¢
3¢ 5¢ 5¢
Replace three 3¢ coins by two 5¢ coins
k cents k+1 cents
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Opening Exercise - Solution
• We will show it is true for a pile of k stones, and show it is true for k+1 stones
• So P(k) means that it is true for k stones
• Base case: n = 1• No splits necessary, so the sum of the products = 0• 1*(1-1)/2 = 0• Base case proven
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Opening Exercise - Solution
• Inductive hypothesis: assume that P(1), P(2), …, P(k) are all true
• This is strong induction!
• Inductive step: Show that P(k+1) is true• We assume that we split the k+1 pile into a pile of i stones and a
pile of k+1-i stones• Thus, we want to show that
(i)*(k+1-i) + P(i) + P(k+1-i) = P(k+1)• Since 0 < i < k+1, both i and k+1-i are between 1 and k, inclusive
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kkkk 22
kkiikikkiiiiki 22222 2222
22
2
2
22222 kkiikikkiiiiki
)1P()1P()P()1(*)( kikiiki
22
)11)(1()1P(
2 kkkkk
2
2
2
)11)(1()1P(
22 iikikkikikik
2)P(
2 iii
Thus, we want to show that (i)*(k+1-i) + P(i) + P(k+1-i) =
P(k+1)