mathematical induction notes
TRANSCRIPT
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FURTHER MATHEMATICAL INDUCTION
TOPIC OUTLINE
ā¢ Proof by mathematical induction
ā¢ Summation and Product
ā¢ Divisibility
ā¢ Inequalities
ā¢ Geometry, Trigonometry, Combinatorics, Calculus, ā¦
ā¢ Recursive
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MATHEMATICAL INDUCTION
The process of mathematical induction is as follows:
ā¢ Show the statement is true for š 1 (an initial value)
ā¢ Assume it is true some š ā¤
ā¢ Prove that it is true for š š 1
ā¢ Since it is true for š 1 and proven true for š 1 1 2 and so on, therefore it is true for all š
SUMMATION TYPE
Prove by induction that, for integers š 1,
1 212
312
āÆ š12
4š 22
Step 1: Show true for š 1.
šæš»š 1
š š»š 43
21
Therefore, true for š 1.
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SUMMATION TYPE
Prove by induction that, for integers š 1,
1 212
312
āÆ š12
4š 22
Step 2: Assume true for some š ā ā¤ .
1 212
312
āÆ š12
4š 22
SUMMATION TYPE
Prove by induction that, for integers š 1,
1 212
312
āÆ š12
4š 22
Step 3: Prove true for š š 1.
š šš: 1 212
312
āÆ š 112
4š 3
2
5
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SUMMATION TYPE
š šš: 1 212
312
āÆ š 112
4š 3
2
š š š
4š 22
š 112
šš¦ šš š š¢ššš”ššš
42š 4
2š
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SUMMATIONTYPE
š šš: 1 212
312
āÆ š 112
4š 3
2
š 42š 4
2š
21
2
42š 4 š 1
2
4š 3
2Therefore, true for š š 1. Therefore, true for all š.
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PRODUCT TYPE
By the process of induction, prove that for all integers š 2:
11š
š 12š
Step 1: Show true for š 2.
šæš»š 114
34
š š»š2 1
434
Therefore, true for š 2.
PRODUCT TYPE
By the process of induction, prove that for all integers š 2:
11š
š 12š
Step 2: Assume true for some š ā ā¤ .
Let š be the product of š terms.
šš 1
2š
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PRODUCT TYPE
By the process of induction, prove that for all integers š 2:
11š
š 12š
Step 3: Prove true for š š 1.
š šš: šš 2
2 š 1
PRODUCT TYPE
š šš: šš 2
2 š 1
š š š
š 12š
11
š 1 šš¦ šš š š¢ššš”ššš
š 12š
š 1 1š 1
š 2š 1 12š š 1
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PRODUCT TYPE
š šš: šš 2
2 š 1
šš 2š 1 1
2š š 1
š š 22š š 1
š 22 š 1
Therefore, true for š š 1. Therefore, true for all š.
DIVISIBILITY PROOFS
Prove that 1 š„ is divisible by 1 š„ for all š ā ā¤ .
Step 1: Show true for š 1.
1 š„ 1 š„ which is divisible by 1 š„.
Step 2: Assume true for some š ā ā¤ .
1 š„ 1 š„ š where š is an integer.
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DIVISIBILITY PROOFS
Prove that 1 š„ is divisible by 1 š„ for all š.
Step 3: Prove true for š š 1.
1 š„ 1 š„ . š„
1 š„ 1 1 š„ š šš¦ šš š š¢ššš”ššš
1 š„ š„ 1 š„ š
1 š„ 1 šš„ which is divisible by 1 š„
Therefore, true for š š 1. Therefore, true for all š.
INEQUALITY PROOFS
With induction proofs involving inequalities, we often make use of the
following:
If š š then š š š where š ā ā
If š š š then š š where š ā ā
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INEQUALITY PROOFS
Use the principle of mathematical induction to show that
4 7 7š 0 for all integers š 2.
Step 1: Show true for š 3.
4 7 21 36 0
Step 2: Assume true for some š ā ā¤ .
4 7 7š 0
INEQUALITY PROOFS
Use the principle of mathematical induction to show that
4 7 7š 0 for all integers š 2.Step 3: Prove true for š š 1.
4 7 7 š 1 4 . 4 7 7š 7
4 4 7 7š 14 21š
0 since 4 7 7š 0 (by assumption)
and 14 2š 0Therefore, true for š š 1. Therefore, true for all š.
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INEQUALITY PROOFS
Use mathematical induction to prove that 2š ! 2 š! for all
positive integers š.
Step 1: Show true for š 1.
šæš»š 2! 2
š š»š 2 1! 2
Step 2: Assume true for some š ā ā¤ .
2š ! 2 š!
2š ! 2 š! 0
INEQUALITY PROOFS
Use mathematical induction to prove that 2š ! 2 š! for all
positive integers š.
Step 3: Prove true for š š 1.
š šš: 2 š 1 ! 2 š 1 !
2 š 1 ! 2 š 1 ! 0
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INEQUALITY PROOFS
2š 2 ! 2 š 1 !
2š 2 2š 1 2š ! 2. 2 š 1 š!
4š 6š 2 2š ! 2 š 2š 1 2 š!
4š 6š 2 2š ! 2š 4š 2 2 š!
4š 6š 2 2š ! 2 š! 2š 2š 2 š!
0 by assumption and since 4š 6š 2 0 and
2š 2š 2 š! 0
GEOMETRICAL PROOFS
Prove that the angle sum of an šāsided polygon is š 2 180 for all integers š 3.
Step 1: Show true for š 3.
For š 3 we have a triangle.
The angle sum is 3 2 180 180 which is true.
Step 2: Assume true for some š ā ā¤ .
A šāsided polygon has an angle sum of š 2 180 .
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GEOMETRICAL PROOFS
Prove that the angle sum of an šāsided polygon is š 2 180 for all integers š 3.
Step 3: Prove true for š š 1.
Taking the š 1 āsided polygon, split it into two: a šāsided polygon and a triangle.
The šāsided polygon has an angle sum of š 2 180 and the triangle has an angle sum of
180 . Total is š 1 180 .
TRIGONOMETRIC PROOFS
It is given that 2cosš“sinšµ sin š“ šµ sin š“ šµ . Prove by induction that, for integers š 1,
cosš cos3š āÆ cos 2š 1 šsin 2šš
2sinšStep 1: Show true for š 1.
šæš»š cosš š š»šsin2š2sinš
2sinšcosš2sinš
cosš
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TRIGONOMETRIC PROOFS
It is given that 2cosš“sinšµ sin š“ šµ sin š“ šµ . Prove by induction that, for integers š 1,
cosš cos3š āÆ cos 2š 1 šsin 2šš
2sinšStep 2: Assume true for some š ā ā¤ .
cosš cos3š āÆ cos 2š 1 šsin 2šš
2sinš
TRIGONOMETRIC PROOFS
It is given that 2cosš“sinšµ sin š“ šµ sin š“ šµ . Prove by induction that, for integers š 1,
cosš cos3š āÆ cos 2š 1 šsin 2šš
2sinšStep 3: Prove true for š š 1.
RTP: cosš cos3š āÆ cos 2š 1 šsin2 š 1 š
2sinš
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TRIGONOMETRIC PROOFS
RTP: cosš cos3š āÆ cos 2š 1 šsin2 š 1 š
2sinšš š š
sin 2šš2sinš
cos 2š 1 š šš¦ šš š š¢ššš”ššš
sin 2šš 2sinĪøcos 2š 1 š2sinš
sin 2šš sin 2š 1 š š sin 2š 1 š š2sinš
TRIGONOMETRIC PROOFS
RTP: cosš cos3š āÆ cos 2š 1 šsin2 š 1 š
2sinš
šsin 2šš sin 2š 1 š š sin 2š 1 š š
2sinš
sin 2šš sin 2 š 1 š sin 2šš2sinš
sin2 š 1 š2sinš
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COMBINATORICS
A 2 by š grid is made up of two rows of š square tiles, as shown.
The tiles of the 2 by š grid are to be painted so that tiles sharing an
edge are painted using different colours. There are š„ different colours
available, where š„ 2.
It is NOT necessary to use all the colours.
COMBINATORICS
Consider the case of the 2 by 2 grid with tiles labelled A, B, C and D, as
shown:
There are š„ š„ 1 ways to choose colours for the first column
containing tiles A and B. Do NOT prove this.
i) Assume the colours for tiles A and B have been chosen. There are two
cases to consider when choosing colours for the second column. Either
C is the same colour as tile B or tile C is a different colour from B.
By considering these two cases, show that the number of ways of
choosing colours for the second column is š„ 3š„ 3.
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COMBINATORICS
There are š„ š„ 1 ways to choose colours for the first column
containing tiles A and B.
Either C is the same colour as tile B:
There is one way to choose C and š„ 1 ways to choose the colour for D
Or C is a different colour to B:
There are š„ 2 ways to choose C and then š„ 2 ways to choose D
Altogether there are š„ 1 š„ 2 š„ 3š„ 3
COMBINATORICS
ii) Prove by mathematical induction that the number of ways in which the
2 by š grid can be painted is š„ š„ 1 š„ 3š„ 3 , for š 1
Step 1: Show true for š 1
š„ š„ 1 š„ 3 3 š„ š„ 1 true for š 1
Step 2: Assume true for some š ā ā¤
Let š be the number of ways to paint a 2 by š grid
š š„ š„ 1 š„ 3š„ 3
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COMBINATORICS
ii) Prove by mathematical induction that the number of ways in which the
2 by š grid can be painted is š„ š„ 1 š„ 3š„ 3 , for š 1
Step 3: Prove true for š š 1
š šš:š š„ š„ 1 š„ 3š„ 3
š š š„ 3š„ 3
š„ š„ 1 š„ 3š„ 3 š„ 3š„ 3
š„ š„ 1 š„ 3š„ 3
Therefore, true for š š 1. Therefore, true for all š.
COMBINATORICS
iii) In how many ways can a 2 by 5 grid be painted if 3 colours are
available and each colour must now be used at least once?
š š„ š„ 1 š„ 3š„ 3
š 3 2 9 9 3
486
but this includes the cases where only two colours are used.
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COMBINATORICS
iii) In how many ways can a 2 by 5 grid be painted if 3 colours are
available and each colour must now be used at least once?
We need to subtract these cases.
There are six arrangements where only two colours are used:
A B C A B C
B A A C C B
Therefore, total arrangements = 486 ā 6 = 480
A B A B A
B A B A B
CALCULUS PROOF
Use mathematical induction to prove that the derivative of š¦ š„ is
šš„ for all š 1.
Step 1: Show true for š 1.
That is, for š¦ š„, š¦ā² 1 š„ 1.
We have to prove this by first principles.
šā² š„ limā
š š„ ā š š„ā
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CALCULUS PROOF
Use mathematical induction to prove that the derivative of š¦ š„ is
šš„ for all š 1.
Step 1: Show true for š 1
šā² š„ limā
š š„ ā š š„ā
limā
š„ ā š„ā
limā
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CALCULUS PROOF
Use mathematical induction to prove that the derivative of š¦ š„ is
šš„ for all š 1.
Step 2: Assume true for some š ā ā¤
For š¦ š„ ,šš¦šš„
šš„
Step 3: Prove true for š š 1
For š¦ š„ ,šš¦šš„
š 1 š„
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CALCULUS PROOF
š šš: For š¦ š„ , š¦ā² š 1 š„
How is š¦ š„ related to š¦ š„ ?
š„ š„ š„
ššš„
š„ššš„
š„ š„
šš„ š„ š„ 1 (using assumption, initial case)
šš„ š„
š 1 š„
REFLECTION QUESTIONS
Why did we have to use first principles
to show true for š 1?
Why were we allowed to use the
product rule?
Do we have to use induction to prove
this or is there an easier proof?
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RECURSION
A recursive formula is one where a
term is defined by referring to one
or more previous terms.
eg. A sequence š is defined by
š 2š š for š 2 with
š š 2.
RECURSIVE PROOFS
A sequence š is defined by š 2š š for š 2 with
š š 2. Use mathematical induction to prove that:
š 1 2 1 2 for all š 0.
Step 1: Show true for š 0 and 1.
š 1 2 1 2 1 1 2
š 1 2 1 2 1 1 2
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RECURSIVE PROOFS
A sequence š is defined by š 2š š for š 2 with
š š 2. Use mathematical induction to prove that:
š 1 2 1 2 for all š 0.
Step 2: Assume true for some š and š 1 ā ā¤
š 1 2 1 2
š 1 2 1 2
RECURSIVE PROOFS
A sequence š is defined by š 2š š for š 2 with
š š 2. Use mathematical induction to prove that:
š 1 2 1 2 for all š 0.
Step 3: Prove true for š š 2.
š šš: š 1 2 1 2
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RECURSIVE PROOFS
š šš: š 1 2 1 2
We were given that š 2š š .
š 2š š
2 1 2 1 2 1 2 1 2
1 2 2 1 2 1 1 2 2 1 2 1
1 2 3 2 2 1 2 3 2 2
RECURSIVE PROOFS
1 2 3 2 2 1 2 3 2 2
Now: 1 2 1 2 2 2 3 2 2
Similarly 1 2 1 2 2 2 3 2 2
Therefore 1 2 3 2 2 1 2 3 2 2
1 2 1 2 1 2 1 2
1 2 1 2
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RECURSIVE PROOFS
Let š§ 1 š and for š 2 let š§ š§ 1š
š§.
Use mathematical induction to prove that š§ š for all integers š 2.
Step 1: Show true for š 2.
š§ 1 1 2
RECURSIVE PROOFS
Let š§ 1 š and for š 2 let š§ š§ 1š
š§.
Use mathematical induction to prove that š§ š for all integers š 2.
Step 2: Assume true for some š ā ā¤ .
š§ š
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RECURSIVE PROOFS
Let š§ 1 š and for š 2 let š§ š§ 1š
š§.
Use mathematical induction to prove that š§ š for all integers š 2.
Step 3: Prove true for š š 1.
š šš: š§ š 1
RECURSIVE PROOFS
We are given š§ š§ 1š
š§.
š§ š§ 1šš§
š§ 1š
š by assumption
š§ š§ 1š
š
š§ 1š
š
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RECURSIVE PROOFS
š§ š§ 1š
š
š 11š
š1 šš
š 1
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