Mathematical Induction Again

Transcription

1 Mathematical Induction Again James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University January 2, 207

2 Outline Mathematical Induction 2 Simple POMI Examples 3 More Abstract and Tricky POMI Arguments 4 Homework

3 Mathematical Induction Mathematical Induction Theorem The Principle of Mathematical Induction For each natural number n, let P(n) be a statement or proposition about the numbers n. P() is true: This is called the BASIS STEP P(k + ) is true when P(k) is true: This is called the INDUCTIVE STEP then we can conclude P(n) is true for all natural numbers n.

4 Mathematical Induction There are many alternative versions of this. Theorem The Principle of Mathematical Induction For each natural number n, let P(n) be a statement or proposition about the numbers n. If If there is a number n 0 so that P(n 0 ) is true: BASIS STEP P(k + ) is true when P(k) is true for all k n 0 : INDUCTIVE STEP then we can conclude P(n) is true for all natural numbers n n 0.

5 Mathematical Induction The POMI Template So in outline, we have Proposition Statement Body Proof: BASIS Verification INDUCTIVE Body of Argument Finishing Phrase Comment QED or Note we try to use pleasing indentation strategies and white space to improve readability and understanding.

6 Simple POMI Examples n = 2n(n + ), n Proof BASIS : P() is the statement = 2 ()(2) = which is true. So the basis step is verified. INDUCTIVE : We assume P(k) is true for an arbitrary k >. Hence, we know k = k(k + ) 2 Now look at P(k + ). We note (k + ) = { k} + (k + )

7 Simple POMI Examples Proof Now apply the induction hypothesis and let k = 2k(k + ) We find (k + ) = k(k + ) + (k + ) 2 { } = (k + ) 2 k + = (k + )(k + 2) 2 This is precisely the statement P(k + ). Thus P(k + ) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n.

8 Simple POMI Examples Recall the when you first encountered Riemann integration, you probably looked at taking the limit of Riemann sums using right side partitions. So for example, for f (x) = 2 + x on the interval [0, ] using a partition width of n, the Riemann sum is n f i= ( 0 + i ) n n n ( = 2 + i ) n n i= = 2 n + n n n 2 i. i= i=

9 Simple POMI Examples Recall the when you first encountered Riemann integration, you probably looked at taking the limit of Riemann sums using right side partitions. So for example, for f (x) = 2 + x on the interval [0, ] using a partition width of n, the Riemann sum is n f i= ( 0 + i ) n n n ( = 2 + i ) n n i= = 2 n + n n n 2 i. The first sum, n i= = n and so the first term is 2 n n = 2. To evaluate the second term, we use our formula from above: n i= i = n(n+) 2n(n + ) and so the second term becomes 2 n n which simplifies to 2 ( + n ). i= i=

10 Simple POMI Examples So the Riemann sum here is 2 + ( 2 + n) which as n gets large clearly approaches the value 2.5. The terms n form what is called a sequence and the limit of this sequence is 2.5. We will talk about this a lot more later.

11 Simple POMI Examples So the Riemann sum here is 2 + ( 2 + n) which as n gets large clearly approaches the value 2.5. The terms n form what is called a sequence and the limit of this sequence is 2.5. We will talk about this a lot more later. From your earlier calculus courses, you know 0 (2 + x)dx = (2x + 2 ) x 2 = which matches what we found with the Riemann sum limit. In MATH 4540, we discuss the theory of Riemann integration much more carefully, so consider this just a taste of that kind of theory!

12 Simple POMI Examples n 2 = 6n(n + )(2n + ) n Proof BASIS : P() is the statement = 6 ()(2)(3) = which is true. So the basis step is verified. INDUCTIVE : We assume P(k) is true for an arbitrary k >. Hence, we know k 2 = k(k + )(2k + ) 6 Now look at P(k + ). We note (k + ) 2 = { k 2 } +(k + ) 2

13 Simple POMI Examples Proof Now apply the induction hypothesis and let k 2 = 6k(k + )(2k + ) We find (k + ) 2 = k(k + )(2k + ) + (k + )2 6 { } = (k + ) k(2k + ) + (k + ) 6 = (k + ) {k(2k + ) + 6(k + )} 6 = 6 (k + ) { 2k 2 + 7k + 6) } = (k + )(k + 2)(2k + 3) 6 This is precisely the statement P(k + ). Thus P(k + ) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n.

14 Simple POMI Examples Now look at for f (x) = 2 + x 2 on the interval [0, ] using a partition width of n, the Riemann sum is ( n f i= ( 0 + i ) n n = = 2 n n i= 2 + ( ) ) i 2 n n n + n n 3 i 2. i= i=

15 Simple POMI Examples Now look at for f (x) = 2 + x 2 on the interval [0, ] using a partition width of n, the Riemann sum is ( n f i= ( 0 + i ) n n = = 2 n n i= 2 + ( ) ) i 2 n n n + n n 3 i 2. The first sum, n i= = n and so the first term is again 2 n n = 2. To evaluate the second term, we use our formula from above: n i= i 2 = 6n(n + )(2n + ) and so the second term ( becomes ) ( n(n+)(2n+) 6 n n n which simplifies to 6 + n 2 + n). i= i=

16 Simple POMI Examples So the Riemann sum here is 2 + ( ) ( 6 + n 2 + n) which as n gets large clearly approaches the value The terms 2 + ( ) ( 6 + n 2 + n) form a sequence and the limit of this sequence is 7/3.

17 Simple POMI Examples So the Riemann sum here is 2 + ( ) ( 6 + n 2 + n) which as n gets large clearly approaches the value The terms 2 + ( ) ( 6 + n 2 + n) form a sequence and the limit of this sequence is 7/3. From your earlier calculus courses, you know 0 (2 + x 2 )dx = (2x + 3 ) x 3 = which matches what we found with the Riemann sum limit.

18 More Abstract and Tricky POMI Arguments 2 n n 2, n 4 Proof BASIS : P(4) is the statement 2 4 = = 6 which is true. So the basis step is verified. INDUCTIVE : We assume P(k) is true for an arbitrary k > 4. Hence, we know 2 k k 2. Now look at P(k + ). We note 2 k+ = 2 2 k 2 k 2. We need to show 2 k+ (k + ) 2 so we must show 2 k 2 (k + ) 2. We can simplify this by multiplying both sides out to get 2k 2? k 2 + 2k + k 2? 2k +.

19 More Abstract and Tricky POMI Arguments Proof As discussed in the text, we can answer this question by doing another POMI inside this one or we can figure it out graphically. Draw the graph of x 2 and 2x + together and you can clearly see k 2 > 2k + when k > 3. You can look up the drawings in the notes. Thus P(k + ) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n 4.

20 More Abstract and Tricky POMI Arguments The number of objects in a set S is denoted by S. You can look up examples in the text. We call this the cardinality of the set S. For example, the cardinality of the set {Jim, Pauli, Qait, Quinn} is 4

21 More Abstract and Tricky POMI Arguments The number of objects in a set S is denoted by S. You can look up examples in the text. We call this the cardinality of the set S. For example, the cardinality of the set {Jim, Pauli, Qait, Quinn} is 4 Given a set S, S can have many subsets. For example, {Jim, Pauli} is a subset of the original S defined above. Since the original set has just 4 objects in it, there are 4 subsets with just one object, There are ( ( 4 2) ways to choose subsets of 2 objects. Recall 4 2) is 4! 2! 2! = 24 4 = 6. There are then ( 4 3) ways to choose 3 objects which 4! gives 3!! = 24 6 = 4. Finally there is just one way to choose 4 objects. So the total number of subsets is = 5. We always also add in the empty set = { } to get the total number of subsets is 6. Note this is the same as 2 4. Hence, we might conjecture that if S had only a finite number of objects in it, the number of subset of S is 2 S.

22 More Abstract and Tricky POMI Arguments The number of objects in a set S is denoted by S. You can look up examples in the text. We call this the cardinality of the set S. For example, the cardinality of the set {Jim, Pauli, Qait, Quinn} is 4 Given a set S, S can have many subsets. For example, {Jim, Pauli} is a subset of the original S defined above. Since the original set has just 4 objects in it, there are 4 subsets with just one object, There are ( ( 4 2) ways to choose subsets of 2 objects. Recall 4 2) is 4! 2! 2! = 24 4 = 6. There are then ( 4 3) ways to choose 3 objects which 4! gives 3!! = 24 6 = 4. Finally there is just one way to choose 4 objects. So the total number of subsets is = 5. We always also add in the empty set = { } to get the total number of subsets is 6. Note this is the same as 2 4. Hence, we might conjecture that if S had only a finite number of objects in it, the number of subset of S is 2 S. The collection of all subsets of a set S is denoted by 2 S for this reason and the cardinality of 2 S is thus 2 S.

23 More Abstract and Tricky POMI Arguments 2 S = 2 S, S Proof BASIS : When S is, there is only one element in S and so the number of subsets is just S itself and the empty set. So 2 S is 2 = 2. This shows the proposition is true for S =. So the basis step is verified. INDUCTIVE : We assume the proposition is true for an arbitrary S = k for any k >. We can thus label the objects in S as S = {a, a 2,..., a k }. Now consider a new set which has one more object in it. Call this set S. We see S = {S,, a k+ } and we can easily count or enumerate how many new subsets we can get. We can add or adjoin the object a k+ to each of the 2 S subsets of S to get 2 S additional subsets. Thus, the total number of subsets is 2 S + 2 S = 2 S +. But S + = S. Thus P( S = k + ) is true and we have verifed the inductive step. Hence, by the POMI, P( S = n) holds for all n.

24 More Abstract and Tricky POMI Arguments ( + x) n + nx, n N, x Proof BASIS : When n =, we are asking if + x + x when x which is actually true for all x. So the basis step is verified. INDUCTIVE : We assume the proposition is true for any x and for any k >. Thus, we assume ( + x) k + kx. Now look at the proposition at k +. We have ( + x) k+ = ( + x) ( + x) k ( + x) ( + kx). We must show ( + x) ( + kx) ( + (k + )x) for x. We have to show + (k + )x + kx 2? + (k + )x. We can cancel the + (k + )x on both sides which tells us we must check if kx 2 0 when x. This is true. Thus P(k + ) is true and we have verifed the inductive step. Hence, by the POMI, P(n) holds for all n.

25 Homework Homework 2 Use the POMI to prove the following propositions n (n+) = n n d dx x n = nx n, x, n N. You can assume you know the powers f (x) = x n are differentiable and that you know the product rule: if f and g are differentiable, then (fg) = f g + f g x + + x n = xn+ x, x, n N. 2.4 x n dx = n+ x n+, n N. You can assume you know integration by parts. The basis step is xdx = x 2 /2 which you can assume you know. After that it is integration by parts.