Transcript FP1 Chapter 5 - Series

FP1 Chapter 6 – Proof By
Induction
Dr J Frost ([email protected])
Last modified: 31st March 2015
What is Proof by Induction?
We tend to use proof by induction whenever we want to show some property holds
for all integers (usually positive).
Example
Show that
𝑛
𝑖=1 𝑖
1
= 2 𝑛 𝑛 + 1 for all 𝑛 ∈ ℕ.
Showing it’s true for a few values of 𝑛 is not a proof.
We need to show it’s true for ALL values of 𝑛.
If it’s true for 𝑛 = 1, it must be true for 𝑛 = 2.
𝒏=𝟏
𝒏=𝟐
Suppose we could prove the
above is true for the first
value of 𝑛, i.e. 𝑛 = 1.
𝒏=𝟑
If it’s true for 𝑛 = 3, it must be true for 𝑛 = 4.
𝒏=𝟒
𝒏=𝟓
…
If it’s true for 𝑛 = 2, it must be true for 𝑛 = 3.
And so on. Hence the statement must be true for all 𝑛.
Suppose that if we assumed it’s true
for 𝑛 = 𝑘, then we could prove it’s
true for 𝑛 = 𝑘 + 1. Then:
Chapter Overview
We will use Proof of Induction for 4 different types of proof:
𝑛
𝑖=1 𝑖
1
1
Summation Proofs
Show that
2
Divisibility Proofs
Prove that 𝑛3 − 7𝑛 + 9 is divisible by 3 for all 𝑛 ∈ ℤ+ .
3
Recurrence Relation
Proofs
Given that 𝑢𝑛+1 = 3𝑢𝑛 + 4,
prove that 𝑢𝑛 = 3𝑛 − 2
Matrix Proofs
1
Prove that
0
4
= 2 𝑛 𝑛 + 1 for all 𝑛 ∈ ℕ.
−1
3
𝑛
1
=
0
𝑢1 = 1,
1 − 2𝑛
for all 𝑛 ∈ ℤ+ .
𝑛
2
Bro Tip: Recall that ℤ is the set of all integers, and ℤ+ is the set of all
positive integers. Thus ℕ = ℤ+ .
The Four Steps of Induction
For all these types the process of proof is the same:
! Proof by induction:
Step 1: Basis:
Prove the general statement is true for 𝑛 = 1.
Step 2: Assumption: Assume the general statement is true for 𝑛 = 𝑘.
Step 3: Inductive: Show that the general statement is then true for 𝑛 = 𝑘 + 1.
Step 4: Conclusion: The general statement is then true for all positive integers 𝑛.
(Step 1 is commonly known as the ‘base case’ and Step 3 as the ‘inductive case’)
Type 1: Summation Proofs
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Show that
𝑛
𝑖=1(2𝑖
− 1) = 𝑛2 for all 𝑛 ∈ ℕ.
1 Basis Step
When 𝑛 = 1, 𝐿𝐻𝑆 = 1𝑖=1 2𝑖 − 1 = 2 1 − 1 = 1
𝑅𝐻𝑆 = 12 = 1. 𝐿𝐻𝑆 = 𝑅𝐻𝑆 so?summation true for 𝑛 = 1.
2 Assumption
Assume summation true for 𝑛 = 𝑘.
?
So 𝑘𝑖=1(2𝑖 − 1) = 𝑘 2
3 Inductive
When 𝑛 = 𝑘 + 1:
𝑘+1
𝑘
2𝑖 − 1 =
𝑖=1
2𝑖 − 1
𝑖=1
= 𝑘 2 + 2𝑘 + 1
= 𝑘+1 2
Hence true when 𝑛 = 𝑘 + 1.
4 Conclusion
?
+ 2 𝑘+1 −1
Since true for 𝑛 = 1 and if true for 𝑛 = 𝑘, true for 𝑛 = 𝑘 + 1,
?
∴ true for all 𝑛.
More on the ‘Conclusion Step’
“As result is true for 𝑛 = 1, this implies true for all positive integers and hence
true by induction.”
I lifted this straight from a mark scheme, hence use this exact wording!
The mark scheme specifically says:
(For method mark)
Any 3 of these seen anywhere in the proof:
• “true for 𝒏 = 𝟏”
• “assume true for 𝒏 = 𝒌”
• “true for 𝒏 = 𝒌 + 𝟏”
• “true for all 𝒏/positive integers”
Test Your Understanding
Edexcel FP1 Jan 2010
?
Exercise 6A
All questions except Q2.
Type 2: Divisibility Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
Prove by induction that 32𝑛 + 11 is divisible by 4 for all positive integers 𝑛.
1 Basis Step
Let 𝑓 𝑛 = 32𝑛 + 11, where 𝑛 ∈ ℤ+ .
𝑓 1 = 32 + 11 = 20 which is divisible
? by 4.
2 Assumption
Assume that for 𝑛 = 𝑘, 𝑓 𝑘 = 32𝑘 + 11 is divisible by 4 for 𝑘 ∈ ℤ+ .
?
…
Type 2: Divisibility Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
Prove by induction that 32𝑛 + 11 is divisible by 4 for all positive integers 𝑛.
3 Inductive
Method 1 (the textbook’s method, which I don’t like)
Find the difference between successive terms and show this is
divisible by the number of interest.
𝑓 𝑘 + 1 = 32 𝑘+1 + 11
Bro Tip: Putting in terms of 32𝑘 allows us to
better compare with the assumption case
= 32𝑘 ⋅ 32 + 11
since the power is now the same.
= 9 32𝑘 + 11
?
∴ 𝑓 𝑘 + 1 − 𝑓 𝑘 = 9 32𝑘 + 11 − 32𝑘 + 11
= 8 32𝑘 = 4 2 32𝑘
Method: You need to explicitly factor out 4 to show divisibility by 4.
32𝑘
Method: Then write 𝑓 𝑘 + 1 back in term of 𝑓(𝑘) and the
difference. We know both the terms on the RHS are divisible by 4.
∴𝑓 𝑘+1 =𝑓 𝑘 +4 2
Therefore 𝑓(𝑛) is divisible by 4 when 𝑛 = 𝑘 + 1.
4 Conclusion
Since true for 𝑛 = 1 and if true for 𝑛 = 𝑘, true for 𝑛 = 𝑘 + 1,
?
∴ true for all 𝑛.
Type 2: Divisibility Proofs
Prove by induction that 32𝑛 + 11 is divisible by 4 for all positive integers 𝑛.
3 Inductive
Method 2 (presented as ‘Alternative Method’ in mark schemes)
Directly write 𝑓(𝑘 + 1) in form 𝑓 𝑘 + 𝑑(… ) by ‘splitting off’ 𝑓(𝑘)
or some multiple of it, where 𝑑 is the divisibility number.
𝑓 𝑘 + 1 = 32 𝑘+1 + 11
= 32𝑘 ⋅ 32 + 11
= 9 32𝑘 + 11
= 32𝑘 + 11 + 8 32𝑘
?
Method: We have separated 𝟗 𝟑𝟐𝒌 into
32𝑘 + 8 32𝑘 because it allowed us to then
replace 32𝑘 + 11 with 𝑓(𝑘)
= 𝑓 𝑘 + 4 2 32𝑘
Therefore 𝑓(𝑛) is divisible by 4 when 𝑛 = 𝑘 + 1.
This method is far superior for two reasons:
a) The idea of ‘separating out 𝑓(𝑘)’ is much more conceptually similar to both summation
proofs and (when we encounter it) matrix proofs, whereas using 𝑓 𝑘 + 1 − 𝑓(𝑘) is
specific to divisibility proofs. Thus we needn’t see divisibility proofs as a different
method from the other types.
b) We’ll encounter a question next where we need to get in the form 𝑓 𝑘 + 1 =
𝑎 𝑓 𝑘 + 𝑑(… ), i.e. we in fact separate out a multiple of 𝑓(𝑘). Thus causes absolute
havoc for Method 1 as our difference 𝑓 𝑘 + 1 − 𝑓(𝑘) has to refer to 𝑓(𝑘), which is
mathematically inelegant.
Type 2: Divisibility Proofs
Prove by induction that 8𝑛 − 3𝑛 is divisible by 5.
2 Assumption
Assume that for 𝑛 = 𝑘, 𝑓 𝑘 = 8𝑘 −
? 3𝑘 is divisible by 5 for 𝑘 ∈ ℤ+.
3 Inductive
When 𝑛 = 𝑘 + 1:
𝑓 𝑘 + 1 = 8𝑘+1 − 3𝑘+1
= 8 8𝑘 − 3 3𝑘
?
= 3 8𝑘 − 3𝑘 + 5 8𝑘
Using Method 2
= 3𝑓 𝑘 + 5(8𝑘 )
therefore true for 𝑛 = 𝑘 + 1.
Method: We have 3 lots of 8𝑘 and
3𝑘 we can split off.
Test Your Understanding
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
Prove by induction that 𝑛3 − 7𝑛 + 9 is divisible by 3 for all positive integers 𝑛.
1 Basis Step
Let 𝑓 𝑛 = 𝑛3 − 7𝑛 + 9, where 𝑛 ∈ ℤ+ .
𝑓 1 = 3 which is divisible by 3. ?
2 Assumption
Assume that for 𝑛 = 𝑘, 𝑓 𝑘 = 𝑘 3 − 7𝑘 + 9 is divisible by 3 for 𝑘 ∈
?
ℤ+ .
3 Inductive
Method 1:
𝑓 𝑘+1 = 𝑘+1 3−7 𝑘+1 +9
=⋯
= 𝑘 3 + 3𝑘 2 − 4𝑘 + 3
∴ 𝑓 𝑘 + 1 − 𝑓 𝑘 = ⋯ = 3 𝑘2 + 𝑘 − 2
∴ 𝑓 𝑘 + 1 = 𝑓 𝑘 + 3 𝑘2 + 𝑘 − 2
Therefore 𝑓(𝑛) is divisible by 3 when 𝑛 = 𝑘 + 1.
?
4 Conclusion
Method 2:
𝑓 𝑘+1 = 𝑘+1 3−7 𝑘+1 +9
=⋯
= 𝑘 3 + 3𝑘 2 − 4𝑘 + 3
= 𝑘 3 − 7𝑘 + 9 + 3𝑘 2 + 3𝑘 − 6
= 𝑓 𝑘 + 3 𝑘2 + 𝑘 − 2
Therefore 𝑓(𝑛) is divisible by 3 when 𝑛 = 𝑘 + 1.
Since true for 𝑛 = 1 and if true for 𝑛 = 𝑘, true for 𝑛 = 𝑘 + 1,
?
∴ true for all 𝑛.
Exercise 6B
Page 130.
All questions.
Type 3: Recurrence Relation Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
Notice that we’re trying
to prove a position-toterm formula from a
term-to-term one.
Given that 𝑢𝑛+1 = 3𝑢𝑛 + 4 and 𝑢1 = 1, prove by induction that 𝑢𝑛 = 3𝑛 − 2
1 Basis Step
When 𝑛 = 1, 𝑢1 = 31 − 2 = 1 as given.
?
IMPORTANT NOTE: The textbook gets this wrong (bottom of Page 131),
saying that 𝒖𝟐 is required. I checked a mark scheme – it isn’t.
Bro Tip: You need to carefully reflect on what statement you are trying to prove, i.e. 𝑢𝑛 = 3𝑛 − 2, and what
statement is GIVEN (and thus already know to be true), i.e.𝑢𝑛+1 = 3𝑢𝑛 + 4
2 Assumption
Assume that for 𝑛 = 𝑘, 𝑢𝑘 = 3𝑘 − 2 is true for 𝑘 ∈ ℤ+ .
3 Inductive
Then 𝑢𝑘+1 = 3𝑢𝑘 + 4
?
3𝑘
=3
−2 +4
?
= 3𝑘+1 − 6 + 4
= 3𝑘+1 − 2
Method: Sub in your
assumption expression
into the recurrence.
Bro Tip: For harder questions, write out what you’re trying to prove in advance, i.e.
that we want to get 3𝑘+1 − 2, , to help guide your manipulation.
4 Conclusion
Since true for 𝑛 = 1 and if true for 𝑛 = 𝑘, true for 𝑛 = 𝑘 + 1, ∴ true for all 𝑛.
?
Test Your Understanding
Edexcel FP1 Jan 2011
?
Recurrence Relations based on two previous terms
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
Given that 𝑢𝑛+2 = 5𝑢𝑛+1 − 6𝑢𝑛 and 𝑢1 = 13, 𝑢2 = 35, prove by induction that
𝑢𝑛 = 2𝑛+1 + 3𝑛+1
1 Basis Step
When 𝑛 = 1, 𝑢1 = 22 + 32 = 13 as given.
When 𝑛 = 2, 𝑢2 = 23 + 33 = 35 as
? given.
We have two base cases so had to check 2 terms!
2 Assumption
3 Inductive
Assume that for 𝑛 = 𝑘 and 𝑛 = 𝑘 + 1, 𝑢𝑘 = 2𝑘+1 + 3𝑘+1 and
𝑢𝑘+1 = 2𝑘+2 + 3𝑘+2 is true for 𝑘 ∈?ℤ+ . We need two previous instances of
Let 𝑛 = 𝑘 + 1
𝑛 in our assumption case now.
(Brominder: it’s worth noting in advance we’re aiming for 𝑢𝑘+1 = 2
𝑘+1 +1
Then 𝑢𝑘+2 = 5𝑢𝑘+1 − 6𝑢𝑘
= 5 2𝑘+1 + 3𝑘+1 ?
− 6 2𝑘+1 + 3𝑘+1
=⋯
= 2𝑘+3 + 3𝑘+3
= 2𝑘+2+1 + 3𝑘+2+1
4 Conclusion
+3
𝑘+1 +1
)
Since true for 𝑛 = 1 and 𝑛 = 2 and if true for 𝑛 = 𝑘 and 𝑛 = 𝑘 + 1,
?
true for 𝑛 = 𝑘 + 2,
∴ true for all 𝑛. (Warning: slightly different to before)
Exercise 6C
Page 132.
Q1, 3, 5, 7
Type 4: Matrix Proofs
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
1
Prove by induction that
0
1 Basis Step
2 Assumption
3 Inductive
4 Conclusion
−1
2
𝑛
1
=
0
1 − 2𝑛
for all 𝑛 ∈ ℤ+ .
𝑛
2
1 −1 1
1 −1
When 𝑛 = 1, 𝐿𝐻𝑆 =
=
0 2
0 2
1
1 −1 ?
𝑅𝐻𝑆 = 1 1 −12 =
. As 𝐿𝐻𝑆 = 𝑅𝐻𝑆, true for 𝑛 = 1.
0
2
0
2
𝑘
1 −1 𝑘
1
1
−
2
Assume true 𝑛 = 𝑘, i.e.
=
0 3 ?
0
2𝑘
When 𝑛 = 𝑘 + 1,
𝑘
1 −1 𝑘+1
1 −1 𝑘 1 −1
1 −1
1
1
−
2
=
=
0 2
0 2
0 2
0 2
0
2𝑘
𝑘+1
2
? Method: As with summation and divisibility
= ⋯ = 1 1 −𝑘+1
0
2
proofs, it’s just a case of ‘separating off’ the
Therefore true when 𝑛 = 𝑘 + 1.
assumption expression.
Since true for 𝑛 = 1 and if true for 𝑛 = 𝑘, true for 𝑛 = 𝑘 + 1,
?
∴ true for all 𝑛.
Test Your Understanding
Step 1: Basis:
Step 2: Assumption:
Step 3: Inductive:
Step 4: Conclusion:
Prove the general statement is true for 𝑛 = 1.
Assume the general statement is true for 𝑛 = 𝑘.
Show that the general statement is then true for 𝑛 = 𝑘 + 1.
The general statement is then true for all positive integers 𝑛.
−2 9
Prove by induction that
−1 4
1 Basis Step
2 Assumption
3 Inductive
4 Conclusion
𝑛
=
−3𝑛 + 1
9𝑛
for all 𝑛 ∈ ℤ+ .
−𝑛
3𝑛 + 1
−2 9 1
−2 9
When 𝑛 = 1, 𝐿𝐻𝑆 =
=
−1 4
−1 4
−3 + 1
9
−2 ? 9
𝑅𝐻𝑆 =
=
. As 𝐿𝐻𝑆 = 𝑅𝐻𝑆, true for 𝑛 = 1.
−1
3+1
−1 4
−2 9 𝑘
−3𝑘 + 1
9𝑘
Assume true 𝑛 = 𝑘, i.e.
=
−𝑘
3𝑘 + 1
−1 4 ?
When 𝑛 = 𝑘 + 1,
−2 9 𝑘+1
−2 9 𝑘 −2 9
−3𝑘 + 1
9𝑘
−2 9
=
=
−𝑘
3𝑘 + 1 −1 4
−1 4
−1 4
−1 4
−3
? 𝑘 + 1 + 1 9(𝑘 + 1)
−3𝑘 − 2 9𝑘 + 9
=⋯=
=
−(𝑘 + 1)
3 𝑘+1 +1
−𝑘 − 1 3𝑘 + 4
Therefore true when 𝑛 = 𝑘 + 1.
Since true for 𝑛 = 1 and if true for 𝑛 = 𝑘, true for 𝑛 = 𝑘 + 1,
?
∴ true for all 𝑛.
Exercise 6D
Page 134.
All questions.