Proof by mathematical induction
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Transcript of Proof by mathematical induction
Proof by mathematicalinduction
Introductionβ’ Proof by mathematical induction is an extremely
powerful tool for proving mathematical statements
β’ As we know, proof is essential in Maths as although something may seem to work for a number of cases, we need to be sure it will work in every case
β’ You have seen some of the formulae used in the series chapter β the had to be proven to work for every case before mathematicians could confidently use them
Teachings for Exercise 6A
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
The way βproof by mathematical inductionβ works is often likened to
knocking dominoes over
If the dominoes are lined up, then you knock over the first one, every
domino afterwards will fall downiPhone Dominoes
Mathematically, if we want to prove that something is true for all possible
cases, we cannot do it numerically (as the numbers would just go on
forever)
However, if we show that if one case is true, and so is the next case, then we can therefore show it is true for
every caseβ¦
6A
How this works mathematically
BASIS Show that the statement to be proven works for the case n = 1
ASSUMPTION Assume that the statement is true for n = k (just replace the ns with ks!)
INDUCTIVE Show that if the statement is true for n = k, it is also true for n = k + 1 (ie β the next case)
This is harder to explain without an example. Essentially you find a way to express the next βcaseβ using k and show that it is equivalent to replacing k with βk + 1β
CONCLUSION You have shown that if the statement is true for one case, it must be true for the next
As it was true for n = 1, it must therefore be true for n = 2, 3, 4 and so on, PROVING the statement!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statementβ¦
6A
βπ=1
π
(2πβ1 )=π2
This means βn can be any
positive integerβ
This is the formula for the
sequence
This is the formula for the sum of the first n terms of the
sequence
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statementβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π
(2πβ1 )=π2
BASIS Show that the statement is true for n = 1
βπ=1
π
(2πβ1 )=π2
βπ=1
1
(2πβ1 )
ΒΏ1
(1)2
ΒΏ1
Replace n with 1 Replace n with 1
There will only be one term here, that we get by subbing n
= 1 into the expression
Calculate
The statement given is therefore true for n = 1
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statementβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π
(2πβ1 )=π2
ASSUMPTION Assume the statement is true for n = k
βπ=1
π
(2πβ1 )=π2
βπ=1
π
(2πβ1 )=ΒΏΒΏ1+3+5+7+9+.. ..+(2πβ1)ΒΏπ2
Write out the first few terms in the
sequence, and the last term, which will be in
terms of k
We are going to assume that this sequence is true for k, and hence the sum
will be equal to k2
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statementβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π
(2πβ1 )=π2
ASSUMPTION Assume the statement is true for n = k
βπ=1
π
(2πβ1 )=π2
βπ=1
π
(2πβ1 )=ΒΏΒΏ1+3+5+7+9+.. ..+(2πβ1)ΒΏπ2
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
βπ=1
π+1
(2πβ1 )=ΒΏΒΏ1+3+5+7+9+.. ..+(2πβ1 )ΒΏ1+3+5+7+9+ ....+(2πβ1 )+(2π+1)
ΒΏπ2+(2π+1)
You can replace the first part as we assumed it was equal to k2 earlier
ΒΏ (π+1 )2
+(2 (π+1 )β1)
The sequence will be the same, but with an extra term (sub in (k + 1)) for it!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
We will start by proving statements relating to the sum of a series.
Prove by mathematical induction that, for
So we need to use the steps from before to prove this statementβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π
(2πβ1 )=π2
βπ=1
π
(2πβ1 )=ΒΏπ2ΒΏ βπ=1
π+1
(2πβ1 )=ΒΏ(π+1)2 ΒΏ
We assumed that for n = k, the sum of the
series would be equal to k2
Using this assumption, we showed that the
summation for (k + 1) is equal to (k + 1)2
βπ=1
π
(2πβ1 )=π2
So if the statement is true for one value, it will therefore be true for the next value
As it is true for the next value, it will therefore be true for the value after that, and so onβ¦
However, this all relies on the assumption being correctβ¦
Remember for the BASIS step, we showed that the statement is true for n = 1?
Well because it is true for n = 1, it must therefore be true for n = 2, n = 3β¦β¦β¦ and so on!
The statement is therefore true for all values of n!
CONCLUSION
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2)=1
6π (π+1 ) (2π+1 )
BASIS
βπ=1
π(π 2)=1
6π (π+1 ) (2π+1 )
βπ=1
1
(π 2)
ΒΏ1
16 (1)(1+1 ) (2+1 )
ΒΏ1
Replace n with 1 Replace n with 1
There will only be one term
here, that we get by subbing n = 1
into the expression
Calculate
The statement given is therefore true for n = 1
Show that the statement is true for n = 1
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2)=1
6π (π+1 ) (2π+1 )
ASSUMPTION Assume the statement is true for n = k
βπ=1
π(π 2)=ΒΏ ΒΏ1+4+9+16β¦β¦β¦+π2ΒΏ 16 π (π+1 ) (2π+1 )
Write out the first few terms in the
sequence, and the last term, which will be in
terms of k
We are going to assume that this sequence is true for k, and hence the sum
will be equal to the expression above
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2)=1
6π (π+1 ) (2π+1 )
ASSUMPTION Assume the statement is true for n = k
βπ=1
π(π 2)=ΒΏ ΒΏ1+4+9+16β¦β¦β¦+π2ΒΏ 16 π (π+1 ) (2π+1 )
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
βπ=1
π+1
(π 2)=ΒΏ ΒΏ1+4+9+16β¦β¦β¦+π2+(π+1)2
The sequence will be the same, but with an extra term (sub in (k + 1)) for it!
ΒΏ16 π (π+1 ) (2π+1 )+(π+1)2
Replace the first part with the assumed formula from earlier!
This requires more simplification which will be shown on the next slide!!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2)=1
6π (π+1 ) (2π+1 )
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
βπ=1
π+1
(π 2)=ΒΏ ΒΏ1+4+9+16β¦β¦β¦+π2+(π+1)2
The sequence will be the same, but with an extra term (sub in (k + 1)) for it!
ΒΏ16 π (π+1 ) (2π+1 )+(π+1)2
Replace the first part with the assumed formula from earlier!
ΒΏπ (π+1 ) (2π+1 )
6+6 (π+1 )2
6
ΒΏπ (π+1 ) (2π+1 )+6 (π+1)2
6
ΒΏ(π+1)π(2π+1)+6 (π+1)ΒΏ ΒΏ6
ΒΏ(π+1)2π2+7π+6ΒΏ ΒΏ6
ΒΏ 6(π+1)(π+2)(2π+3)
Rewrite both as fractions over 6
Combine
βClever factorisationβ
method!Expand and simplify the inner brackets
Factorise the inner part
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
So we are now going to prove one of the formulae you have learnt in
chapter 5!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2)=1
6π (π+1 ) (2π+1 )
CONCLUSION Explain why it proves the original statement
ΒΏ16 π (π+1 ) (2π+1 ) ΒΏ
16 (π+1) (π+2 ) (2π+3 )
ΒΏ16 (π+1) (π+1+1 ) (2(π+1)+1 )
For n = k For n = (k + 1)
Rewrite some of the brackets
Written in this way, you can see that the kβs in the first statement have all been replaced with
βk + 1βs So the statement was true for n = 1
We also showed that if it is true for one statement, it is true for the next
Therefore the formula has been proven!
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2π )=2 [1+(πβ1)2π ]
BASIS
βπ=1
π(π 2π )=2 [1+(πβ1)2π ]
βπ=1
1
(π 2π )
ΒΏ2
2 [1+(1β1)21 ]
ΒΏ2
Replace n with 1 Replace n with 1
There will only be one term
here, that we get by subbing n = 1
into the expression
Calculate
The statement given is therefore true for n = 1
Show that the statement is true for n = 1
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2π )=2 [1+(πβ1)2π ]
ASSUMPTION Assume the statement is true for n = k
βπ=1
π(π 2π )=ΒΏΒΏ2+8+24+64β¦β¦+π(2π)ΒΏ2 [1+(πβ1 )2π ]
Write out the first few terms in the
sequence, and the last term, which will be in
terms of k
We are going to assume that this sequence is true for k, and hence the sum
will be equal to the expression above
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2π )=2 [1+(πβ1)2π ]
ASSUMPTION Assume the statement is true for n = k
βπ=1
π(π 2π )=ΒΏΒΏ2+8+24+64β¦β¦+π(2π)ΒΏ2 [1+(πβ1 )2π ]
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term The sequence will be the same, but with an extra
term (sub in (k + 1)) for it!
βπ=1
π(π 2π )=ΒΏΒΏ2+8+24+64β¦β¦+π(2π)+(π+1)2π+1
Replace the first part with the assumed formula from earlier!
ΒΏ2 [1+(πβ1 )2π ]+(π+1)2π+1
ΒΏ2+2(πβ1)2π+(π+1)2π+1
The simplification for this is difficult
You need to aim for the power of 2 to be βk + 1β (as it was βkβ originally)
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2π )=2 [1+(πβ1)2π ]
INDUCTIVE Show the statement is then true for (k + 1) ie) The next term
ΒΏ2+2(πβ1)2π+(π+1)2π+1
2 x 2k = 2k+1
(add the powers)ΒΏ2+(πβ1)2π+1+(π+1)2π+1
ΒΏ2+(πβ1+π+1)2π+1
In total, we have (k β 1) + (k + 1) 2k+1s
ΒΏ2+2π2π+1Simplify the bracket
ΒΏ2(1+π2π+1)Re-factorise the bracket
Proof by mathematical inductionYou can obtain a proof for the
summation of a series, by using the induction method
Prove, by mathematical induction, that for ,
This looks more complicated, but you just follow the same process as
you have seen already!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6A
βπ=1
π(π 2π )=2 [1+(πβ1)2π ]
CONCLUSION Explain why this shows the statement is true
2(1+π2π+1)2 [1+(πβ1 )2π ]
For n = k For n = (k + 1)
2(1+(π+1β1)2π+1)
Rewrite the first βkβ as βk + 1 β 1β
Written in this way, you can see that the kβs in the first statement have all been replaced with
βk + 1βs So the statement was true for n = 1
We also showed that if it is true for one statement, it is true for the next
Therefore the formula has been proven!
You will need to become familiar with manipulating powers in the way shown
here!
Teachings for Exercise 6B
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers
You follow the same steps as before!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
BASIS Show that the statement is true for n = 1π (π)=32π+11
π (1 )=32(1)+11
ΒΏ20
Sub in n = 1
Calculate
20 is divisible by 4, so the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers
You follow the same steps as before!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
ASSUMPTION Assume the statement is true for n = k
π (π )=32π+11ππ πππ£ππ ππππππ¦ 4 πππ πββ€+ΒΏ ΒΏ
INDUCTIVE Show that the statement is then true for n = (k + 1)
π (π+1 )=32(π+ 1)+11
π (π+1 )=32π+2+11
π (π+1 )=32πΓ32+11
π (π+1 )=9(3ΒΏΒΏ2π)+11ΒΏ
Multiply out the bracket
32k+2 = 32k x 32 (adding powers when multiplying)So we have 9
lots of 32k
At this point we will combine the expressions for f(k) and f(k + 1) in order to prove that the statement is always divisible
by 4
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that 32n + 11 is divisible by 4 for all positive integers
You follow the same steps as before!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
π (π )=32π+11
INDUCTIVE Show that the statement is then true for n = (k + 1)
π (π+1 )=9(3ΒΏΒΏ2π)+11ΒΏ
π (π+1 )β π (π)=ΒΏ[9 (32π )+11]β [32πβ11 ]
π (π+1 )β π (π)=ΒΏ8 (32π)
π (π+1 )β π (π)=ΒΏ4 [2(32π) ]
π (π+1 )=ΒΏπ (π )+4 [2(32π) ]
Subtract f(k) from f(k + 1), using the expressions above
Group terms on the right
sideTake out 4 as
a factor
Add f(k)
This shows that f(k + 1) is just f(k) with an expression added on
We assumed f(k) was divisible by 4 The expression to be added is divisible by 4
So the answer must be divisible by 4, if f(k) is! As the first case (n = 1) was divisible by 4, the
statement must be true!
CONCLUSION
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression βn3 β 7n + 9β is divisible
by 3 for all positive integers
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
BASIS Show that the statement is true for n = 1π (π)=π3β7π+9
Sub in n = 1π (1 )=(1)3β7(1)+9
ΒΏ3Calculate
3 is divisible by 3, so the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression βn3 β 7n + 9β is divisible
by 3 for all positive integers
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
ASSUMPTION Assume the statement is true for n = k
π (π )=π3β7π+9ππ πππ£ππ ππππππ¦ 3 πππ πββ€+ΒΏ ΒΏ
INDUCTIVE Show that the statement is then true for n = (k + 1)π (π+1 )=(π+1)3β7(π+1)+9
π (π+1 )=π3+3π2+3π+1β7πβ7+9
π (π+1 )=π3+3π2β4π+3
Multiply out the brackets
Group up terms
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression βn3 β 7n + 9β is divisible
by 3 for all positive integers
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
INDUCTIVE Show that the statement is then true for n = (k + 1)
π (π+1 )=π3+3π2β4π+3π (π )=π3β7π+9Subtract f(k) from f(k + 1), using the expressions above
π (π+1 )β π (π )=ΒΏ(π3+3π2β4π+3 )β (π3β7π+9 )
π (π+1 )β π (π )=ΒΏπ3+3π2β4π+3βπ3+7 πβ9
π (π+1 )β π (π )=ΒΏ3π2+3πβ6
π (π+1 )β π (π )=ΒΏ3 (π2+πβ2)
π (π+1 )=ΒΏπ (π )+3(π2+πβ2)
βRemoveβ the brackets
Group
termsTake out 3 as a
factor
Add f(k)
This shows that f(k + 1) is just f(k) with an expression added on
We assumed f(k) was divisible by 3 The expression to be added is divisible by 3
So the answer must be divisible by 3, if f(k) is! As the first case (n = 1) was divisible by 3, the
statement must be true!
CONCLUSION
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression β11n+1 + 122n-1β is divisible
by 133 for all positive integers
This example will require more manipulation as we work through it, but is essentially the same as the
previous twoβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
BASIS Show that the statement is true for n = 1π (π)=11π+1+122πβ1
Sub in n = 1π (1 )=112+121
ΒΏ133Calculate
133 is divisible by 133, so the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression β11n+1 + 122n-1β is divisible
by 133 for all positive integers
This example will require more manipulation as we work through it, but is essentially the same as the
previous twoβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
ASSUMPTION Assume the statement is true for n = k
π (π )=11π+1+122πβ1ππ πππ£ππ ππππππ¦ 133 πππ πββ€+ΒΏ ΒΏ
INDUCTIVE Show that the statement is then true for n = (k + 1)
π (π+1 )=11(π+1 )+1+122 (π+1)β 1
π (π+1 )=11π+2+122π+1
π (π+1 )=11Γ11π+1+122Γ122πβ1
π (π+1 )=11 (11π+1)+144 (122πβ1)
Simplify powers
11k+2 = 11 x 11k+1122k+1 = 122 x 122k-1
Simplify
Rewrite
*
*They have been re-written in this way so that, on
the next step, the 11s and 12s have the same powers as in the f(k) expression and therefore can
be grouped up!
Proof by mathematical inductionYou can use proof by induction to prove that an expression is
divisible by a given integer
Prove, by induction, that the expression β11n+1 + 122n-1β is divisible
by 133 for all positive integers
This example will require more manipulation as we work through it, but is essentially the same as the
previous twoβ¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6B
INDUCTIVE Show that the statement is then true for n = (k + 1)
π (π+1 )=11(11π+1)+144 (122πβ1)π (π )=11π+1+122πβ1
Subtract f(k) from f(k + 1), using the expressions above
π (π+1 )β π (π )=ΒΏ[11 (11π+1)+144 (122πβ1) ]β [11π+1+122πβ1 ]
π (π+1 )β π (π )=ΒΏ10(11π+1)+143(122πβ 1)
π (π+1 )β π (π )=ΒΏ10 (11π+1 )+10 (122πβ 1 )+133 (122πβ1 )
π (π+1 )β π (π )=10 π (π)+133 (122πβ1)
π (π+1 )=11 π (π )+133(122πβ 1)
Group terms
Split the 143 into 2
partsThe first 2 terms are
just 10 lots of f(k)
Add f(k)
If f(k) is divisible by 133, so is 11f(k) 133(122k-1) is divisible by 133 Therefore f(k+1) will also be divisible by 133
As f(1) was divisible by 133, the statement is therefore true!
Make sure you practise enough so you can spot how and when to manipulate in this way!
Teachings for Exercise 6C
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
You will have seen recurrence relations in C1. A recurrence relation
is a sequence where generating a term relies on a previous term.
It is very important that you understand the notation!
6C
ππ+1=ππ+5Example 1
π 1=3
The next term in the sequence
The current term
The first term
This is telling you that the first number in the sequence is 3
And to get the next number, you add on 5 The sequence will be: 3, 8, 13, 18, 23β¦β¦ and so onβ¦ As this is an arithmetic sequence, we can use the
formula from C1 for the βnthβ term.. (a + (n -1)d)
ππ+1=3ππβ1Example 2
π 1=1
This is telling you that the first number in the sequence is 1
To get the next number, you multiply the current number by 3 and subtract 1β¦
The sequence will be: 1, 2, 5, 14, 41, 122 β¦β¦and so onβ¦ This is NOT an arithmetic sequence, so we cannot use the
arithmetic sequence method for the nth term The βnthβ term for a sequence like this is far more
complicated! They can also be proven to be correct (once you think you
know what they are!) by use of induction!
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n β 2.
Before we start, letβs generate the first 5 terms in both the
ways shown aboveβ¦
You do not need to do this on an exam, this is just to show you
the two ways of generating the sequence give the same result!
So now, letβs prove this is the case!
6C
π’π+1=3π’π+4
π’1=1π’2=3 (1 )+4ΒΏ7
π’3=3 (7 )+4ΒΏ25
π’4=3 (25 )+4ΒΏ79
π’4=3 (79 )+4ΒΏ241
Using the recurrence relation
Using the βnthβ term formulaπ’π=3
πβ2
π’1=31β2ΒΏ1
π’2=32β2ΒΏ7
π’3=33β2ΒΏ25
π’4=34β2ΒΏ79
π’5=35β2ΒΏ241
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n β 2.
So we are being asked to show that, for the sequence with this
recurrence relation, that the nth term formula is 3n β 2β¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
BASIS Show that the statement is true for n = 1 and n = 2π’π+1=3π’π+4 π’1=1 There is a slight difference here. As we are given u1 already (n = 1), we need to also check the statement is true for n = 2β¦π’π+1=3π’π+4
π’1=1
π’2=3π’1+4π’2=7
π’π=3πβ2
π’1=31β2π’1=1
π’2=32β2π’2=7
The first 2 terms are both 1 and 7, so the statement is true for n = 1 and 2
We already know u1
Now use the recurrence relation to
find u2
Calculate
Sub in n = 1
Calculate
Now sub in n = 2
Calculate
π’π=3πβ2
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n β 2.
So we are being asked to show that, for the sequence with this
recurrence relation, that the nth term formula is 3n β 2β¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
ASSUMPTION Assume that the statement is true for n = kπΉππ π=π ,π’π=3
πβ2π ππ πββ€+ΒΏΒΏ
π’π+1=3π’π+4
INDUCTIVE Use the recurrence relation to create an expression for uk+1π’π+1=3π’π+4
π’π+1=3 (3πβ2)+4
π’π+1=3π+1β6+4
π’π+1=3π+1β2
Replace uk with the assumed expression
aboveMultiply out the brackets
Simplify
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+1 = 3un + 4, u1 = 1, prove by induction that un = 3n β 2.
So we are being asked to show that, for the sequence with this
recurrence relation, that the nth term formula is 3n β 2β¦
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
π’π=3πβ2
INDUCTIVE Use the recurrence relation to create an expression for uk+1
π’π+1=3π+1β2
The βkβ terms have all become βk + 1β
terms
CONCLUSION If the statement is true for βkβ, it is also true for βk + 1β
We showed in the basis that it is true for n = 1 and n = 2
Therefore the statement is true for all
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 β 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
π’π+2=5π’π+1β6π’π
The next term in the sequence
The current term
The previous
term
π’1=13π’2=35
The first term
The second term
So for this sequence, the next term is based on the current term AND the term before that!
This is why you have been given the first 2 termsβ¦
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 β 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
BASIS Show that the statement is true for n = 1, n = 2 and n = 3
π’π+2=5π’π+1β6π’π
π’1=13
The first 3 terms are 13, 35 and 97 for both sequences, so the statement has been shown to be true up to n = 3
We already know u1 and u2π’2=35
π’3=5(35)β6(13)
π’3=5π’2β6π’1 Sub in u2 and u1 to
find u3
π’3=97Calculate
π’π=2π+1+3π+1
π’1=22+32
π’1=13
π’2=23+33
π’2=35
π’3=24+34
π’3=97
Calculate
Calculate
Calculate
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 β 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
ASSUMPTION Assume that the statement is true for n = k AND n = k + 1πΉππ π=π ,π’π=2
π+1+3π+1π ππ πββ€+ ΒΏΒΏ
π’π+2=5π’π+1β6π’π
πΉππ π=π+1 ,π’π+1=2π+2+3π+2π ππ πββ€+ ΒΏΒΏ
INDUCTIVE Use the recurrence relation to create an expression for uk+2π’π+2=5π’π+1β6π’π
π’π+2=5 (2π+2+3π+2)β6 (2ΒΏΒΏπ+1+3π+1)ΒΏ
π’π+2=5 (2π+2)+5 (3π+2)β6 (2π+1 )β6 (3π+1 )
π’π+2=5 (2π+2)+5 (3π+2)β3 (2π+ 2 )β2 (3π+2 )
6 (2π+1 )ΒΏ3Γ2 (2π+1 )ΒΏ3 (2π+2 )
6 (3π+1 )ΒΏ2Γ3 (3π+1 )ΒΏ2 (3π+2 )
Sub in the assumed expressions for uk+1 and uk
from beforeSplit the bracketed
parts upRewrite all as
powers of βk + 2β(see below)
6 = 3 x 2
The 2 adds 1 to the power
6 = 2 x 3
The 3 adds 1 to the power
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 β 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
ASSUMPTION Assume that the statement is true for n = k AND n = k + 1πΉππ π=π ,π’π=2
π+1+3π+1π ππ πββ€+ ΒΏΒΏ
π’π+2=5π’π+1β6π’π
πΉππ π=π+1 ,π’π+1=2π+2+3π+2π ππ πββ€+ ΒΏΒΏ
INDUCTIVE Use the recurrence relation to create an expression for uk+2π’π+2=5π’π+1β6π’π
π’π+2=5 (2π+2+3π+2)β6 (2ΒΏΒΏπ+1+3π+1)ΒΏ
π’π+2=5 (2π+2)+5 (3π+2)β6 (2π+1 )β6 (3π+1 )
π’π+2=5 (2π+2)+5 (3π+2)β3 (2π+ 2 )β2 (3π+2 )
Sub in the assumed expressions for uk+1 and uk
from beforeSplit the bracketed
parts upRewrite all as
powers of βk + 2β
π’π+2=2 (2π+2)+3 (3π+2)
π’π+2=2π+3+3π+ 3
Group the βlikeβ terms
The 2 and 3 add 1 to the powers of 2 and 3 respectively
Proof by mathematical inductionYou can use mathematical
induction to produce a proof for a general term of a recurrence
relation
Given that un+2 = 5un+1 β 6un, and u1 = 13 and u2 = 35:
Prove by induction that un = 2n+1 + 3n+1
This sequence is slightly different to what you have seen!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6C
CONCLUSION
π’π=2π+1+3π+1
π’π+1=2π+2+3π+2
π’π+2=2π+3+3π+3
π’π+1=2(π+1 )+1+3 (π+1) +1
π’π+2=2(π+2) +1+3(π+2 )+1
These can both be written differently
As you can see, k is replaced with (k + 1), and then with (k + 2)
So we have shown that IF the statement is true for n = k and n = k + 1, then it must also be true for n = k + 2
As we showed in the basis that the statement is true for n = 1 and n = 2, then it must therefore be true for n = 3
And consequently it is then true for all values of n!
Teachings for Exercise 6D
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6D
[1 β10 2 ]
π
=[1 1β2π
0 2π ]πππ πββ€+ΒΏ ΒΏ
BASIS Show that the statement is true for n = 1
[1 β10 3 ]
π
=[1 1β2π
0 2π ]
[1 β10 2 ]
1
ΒΏ [1 β10 2 ]
[1 1β21
0 21 ]ΒΏ [1 β10 2 ]
Replace n with 1 Replace n with 1
Calculate
Calculate
So the statement is true for n = 1
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6D
[1 β10 2 ]
π
=[1 1β2π
0 2π ]πππ πββ€+ΒΏ ΒΏ
ASSUMPTION Assume the statement is true for n = k
[1 β10 2 ]
π
=[1 1β2π
0 2π ]ππ π‘ππ’π πππ πββ€+ ΒΏΒΏ
INDUCTIVE Show using the assumption, that the statement will also be true for n = k + 1
[1 β10 2 ]
π+1
ΒΏ [1 β10 2 ]
π
[1 β10 2 ]
1
ΒΏ [1 1β2π
0 2π ][1 β10 2 ]
Replace the power βkβ term with the assumed
matrix The second matrix doesnβt need the power!
Now we need to multiply these matrices using the skills from chapter 4!
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
6D
[1 β10 2 ]
π
=[1 1β2π
0 2π ]πππ πββ€+ΒΏ ΒΏ
INDUCTIVE Show using the assumption, that the statement will also be true for n = k + 1
[1 β10 2 ]
π+1
ΒΏ [1 β10 2 ]
π
[1 β10 2 ]
1
ΒΏ [1 1β2π
0 2π ][1 β10 2 ]
Replace the power βkβ term with the assumed
matrix The second matrix doesnβt need the power!
(1Γ1 )+((1β2π)Γ0) (1Γβ1 )+((1β2π)Γ2)(0Γ1 )+(2πΓ0) (0Γβ1 )+(2πΓ2)
1 β1+2β2(2π)0 2(2π)
ΒΏ [1 1β2π+1
0 2π+1 ]
1 1β2π+1
0 2π+1
Work out each term
Simplify (remember to
manipulate the powers)
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
As always, follow the same pattern as with the other induction
questions!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[1 β10 2 ]
π
=[1 1β2π
0 2π ]πππ πββ€+ΒΏ ΒΏ
CONCLUSION
[1 β10 2 ]
π
=[1 1β2π
0 2π ]ππ π‘ππ’π πππ πββ€+ΒΏΒΏ
We assumed that:
Using this, we showed that:
[1 β10 2 ]
π+1
=[1 1β2π+1
0 2π+1 ]As you can see, all the βkβ terms have been replaced with
βk + 1β terms
Therefore, IF the statement is true for one term, it will also be true for the next term, and so onβ¦
As we already showed that the statement is true for n = 1, it is therefore true for all values of n!
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]πππ πββ€+ΒΏ ΒΏ
BASIS Show that the statement is true for n = 1
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]
[β2 9β1 4 ]
1
ΒΏ [β2 9β1 4 ]
[β3 (1 )+1 9(1)β(1) 3 (1 )+1]
Replace n with 1 Replace n with 1
Calculate
Calculate
So the statement is true for n = 1
ΒΏ [β2 9β1 4 ]
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]πππ πββ€+ΒΏ ΒΏ
ASSUMPTION Assume the statement is true for n = k
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]ππ π‘ππ’π πππ πββ€+ΒΏΒΏ
INDUCTIVE Show using the assumption, that the statement will then be true for n = k + 1
[β2 9β1 4 ]
π+1
ΒΏ [β2 9β1 4 ]
π
[β2 9β1 4]
1
ΒΏ [β3π+1 9πβπ 3π+1 ][β2 9
β1 4 ]
Replace the power βkβ term with the assumed matrix
The second matrix doesnβt need the
power!
Now we need to multiply these matrices using the skills from chapter 4!
Simplify terms
(probably a good idea to
do in stagesβ¦)
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]πππ πββ€+ΒΏ ΒΏ
6D
INDUCTIVE Show using the assumption, that the statement will also be true for n = k + 1
((β3π+1 )Γβ2)+(9πΓβ1)
[β2 9β1 4 ]
π+1
ΒΏ [β2 9β1 4 ]
π
[β2 9β1 4]
1
ΒΏ [β3π+1 9πβπ 3π+1 ][β2 9
β1 4 ]
Replace the power βkβ term with the assumed matrix
The second matrix doesnβt need the
power!
((β3π+1 )Γ9)+(9πΓ4)(βπΓβ2)+((3π+1)Γβ1) (βπΓ9)+((3π+1)Γ4)
(6 πβ2)+(β9π)(2π)+(β3πβ1)
(β27π+9)+(36π)(β9π)+(12π+4)
β3 πβ2βπβ1
9π+93π+4
ΒΏ [β3πβ2 9π+9βπβ1 3 π+4 ]
Simplify fully
This is the answer to the multiplication
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]
Proof by mathematical inductionYou can use proof by induction to prove general statements
involving matrix multiplication
Use mathematical induction to prove that:
More complicated, but the same process!
BASISASSUMPTIONINDUCTIVE
CONCLUSION
[β2 9β1 4 ]
π=[β3π+1 9π
βπ 3π+1]πππ πββ€+ΒΏ ΒΏ
6D
CONCLUSION We assumed that:
Using this, we showed that:
ππ π‘ππ’π πππ πββ€+ΒΏΒΏ
[β2 9β1 4 ]
π+1
=[β3πβ2 9π+9βπβ1 3π+4 ]
ΒΏ [β3 (π+1 )+1 9(π+1)β(π+1) 3 (π+1 )+1]
Each part of the matrix can be written differentlyβ¦
(you will see why in a moment!)
If you compare this to the original matrix, you can see that all the βkβ terms have been replaced with
βk + 1β terms
So we have shown that if the statement is true for n = k, it will also be true for n = k + 1
As it was true for n = 1, it is also true for all positive values of k!
Summaryβ’ We have seen how to use proof by induction
β’ We have seen how to use it in situations regarding the summation of a series, tests of divisibility, recurrence relationships and matrices
β’ The four steps will always be the same, you will need to practice the βclever manipulationβ behind some of the inductive steps though!