Module 4: Mathematical Induction


 Elmer Wade
 2 years ago
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1 Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate P () where the uiverse of discourse for is the set of atural umbers N f1; ;:::g. Example 1: Here are some examples of what we mea by P (): P () ( +1) ; 8 N; P () ( + 1)( +1) ; 8 N; 6 P () P () i0» ( +1) ; 8 N; a i a+1 1 ; 8 N; a 1 P ()! > for 4; P () < 1; 1 where meas logically equivalet. The first three expressios above provide closedform formulas for the sum of cosecutive positive itegers, the sum of squares of cosecutive positive itegers, ad the sum of cubes of cosecutive positive itegers, respectively. The fourth expressio is the sum of the first terms i the geometric series ad we studied it already i Module. The last two expressios are useful iequalities for factorial ad the sum of egative powers of. Every statemet P () above is about atural umbers or a subset of atural umbers (e.g., for 4). How ca we prove such statemets? Cosider the first example above regardig the sum of the first cosecutive positive itegers. We ca easily verify that P () is true for some selected. Ideed, P (1) is true sice 1 1 ; P () is true sice 1+ 3 ; P (3) is true sice : 1
2 But how ca we prove that P () is true for all N? The priciple of mathematical iductio (PMI) ca be used to prove statemets about atural umbers. The priciple of mathematical iductio: LetA be a set of atural umbers such that the followig two properties hold: (1) 1 A; () for every atural umber if A the +1 A: (1) The A N f1; ;:::g; that is, A cotais all atural umbers. How is it related to provig statemets like P () above? Let us defie A f : P () is true for g; that is, A is the set of atural umbers for which P is true. The goal is to show that A is the same as the set of all atural umbers, that is, A N. Imagie that oe verifies that P (1) is true. The we ca set A f1g. Let s ow assume that oe ca prove step () of PMI (that we shall call the iductive step). Thus sice we kow that 1 A, ad we kow the iductive step is valid, say for 1, we coclude that A. Therefore, A f1; g, thatis,p (1) ad P () are true. But usig agai the iductive step, we coclude that 3 A. Etc. Actually, PMI allows us to replace the imprecise etc by A N, thatis,p () is true for all atural umbers! But why is PMI true, i the first place? We demostrate its truth usig the proof by cotradictio. Suppose that (1) ad () of PMI hold but A is ot equal to N. Hece, it must be at least oe atural umber is omitted from N. Let 0 be the first umber (smallest) amog 1; ;:::omitted from N. We kow that 0 caot be 1 sice we assumed that 1 A by (1) of PMI. But by our costructio, 0 1 A. The by step () of PMI we must coclude that 0 A, which is the desired cotradictio. Therefore, A N. Let us itroduce some additioal otatio. The first step (1) of PMI is called the basis step, while the secod step is kow as the iductive step. It is usually trivial to verify the basis step, ad most work has to be doe to prove the iductive step. We shall illustrate it o the followig example. Example : Prove the first idetity above about the sum of cosecutive atural umbers, that is, ( +1) i ; 1: i1
3 I this case, the property P () is a predicate sayig that the above is true for,ad A f : idetity above is true for Ng: We prove P () is true for all (i.e., A N) usig PMI. We eed to go through the basis step ad the iductive step. Basis Step: We must prove P (1) is true, but this was already established before. Iductive Step: We ow assume that P () is true for a fixed but arbitrary. The above assumptio is called the iductive hypothesis ad i our case it takes the followig form ( +1) S : i i1 for arbitrary. (I the above symbol : meas equal by defiitio.) The reader must uderstad that the statemet immediately above ad the statemet that we wat to prove are ot the same, eve if they look alike. I the statemet above we assume that the idetity to be proved (about all sums of the first cosecutive atural umbers) is true for oe value of (but a arbitrary oe). We ow perform the iductive step. We must establish the iductive step, that is, to show that the formula for S above implies that +1 ( + 1)( +) S +1 i i1 is true, too. Observe that above we replaced by +1(o the lefthad side of the equatio as well as o the righthad side). Ideed, we have S ( +1) ( +1) +( +1) ( +1) +1 ( + 1)( +) ; i1 i +( +1) where i the secod lie above we ivoked the iductio hypothesis, i the third lie we factored out the term ( +1), ad the added what is left. This is exactly what we eed to prove the iductive step. But, there is actually aother, direct, proof P origially proposed by the 18th cetury mathematicia Carl Friedrich Gauss. Let, as before, S i1 i. We write the sum S twice oe startig the sum from 1 up to, ad the secod time startig from dow to 1. The, we add the idividual elemets vertically. Here is what comes out: 3
4 S ( 1) + S S ( +1) + ( +1) + + ( +1) + ( +1) Sice there are terms ( +1)i the bottom lie, we prove that Agai, we recover the same idetity. S ( +1): Exercise 4A: Usig mathematical iductio prove that i ( + 1)( +1) : 6 i1 Exercise 4B: Usig mathematical iductio prove that i1 i 3» ( +1) : Iductio o a Subset of Natural Numbers I the PMI discussed above i the first step we assumed that 1 A, however, if we start the iductio from aother atural umber, say k, the it holds for all k. This is show i the ext example. Example 3: Prove that! > for 4: We recall that! 1 3 ( 1)!. We are asked to prove the above iequality oly for 4. Thus let P () f! > is true for 4g: We first check that 4! 4 > 4 16, thus P (4) is true. (Observe that P (3) is ot true.) Now assume this statemet is true for arbitrary 4. We must prove that for 4. This is easy sice ( + 1)! > +1 ( + 1)!!( +1) > ( +1) > +1 : The first iequality follows from the iductio hypothesis! while the secod idetity is a cosequece of ( +1)> for 4. This proves the desired iequality for all 4. 4
5 The ext two examples require a little bit of work before the iductio ca be applied. Example 4: Beroulli s iequality. We shall prove the followig result. Theorem 1 If is a atural umber ad 1+x > 0,the (1 + x) 1+x: () Proof. The proof is by iductio. I the basis step, we assume 1ad verify that (1+x) 1+x is true for 1+x>0. Now, we assume (iductive hypothesis) that (1 + x) 1+x is true for a arbitrary, ad we must prove that (1 + x) +1 1+( +1)x for 1+x>0. To prove this we proceed as follows: (1 + x) +1 (1 + x) (1 + x) by iductio (1 + x)(1 + x) sice 1+x>0 1+( +1)x + x 1+( +1)x; where the first iequality is a cosequece of the iductio assumptio (i.e., we kow that (1 + x) 1+x so we ca replace (1 + x) by 1+x because 1+x>0; observe that if 1+x<0,thewe had to reverse the iequality sig 1 ). The ext step is simple algebra, while the last step follows from the fact that x is oegative; it does t matter what the value of x, because a + x a for ay a. This proves the theorem. Example 5: Let us prove that 1 < 1 (3) for 1. We prove it by iductio. The first step for 1is easy to check, so we cocetrate o the iductive step. We adopt the iductive hypothesis, which i this case is ad must prove that < 1; < 1: +1 A atural approach fails. If we ivoke the iductio hypothesis to the first terms of the above, we will get Thik of 3» 5 that after multiplyig by 3 becomes
6 which does ot imply that it is less tha or equal to 1 sice 1 +1 > 0. Here s how we proceed by iductio < » 1; + 1 where i the first lie o the righthad side we factor 1 ad observe that what is left i the parethesis must be smaller tha 1 by the iductive hypothesis. The rest is simple algebra. This proves the iequality. I some cases, we must use a geeralized mathematical iductio that we formulate i a little differet form tha before. If a statemet P () is true for 1, ad if for every >1, the truth of P () for all atural umbers < implies the truth of P () for, thep () is true for all atural umbers. The oly differece betwee the basis PMI ad the above is that i i the iductive step of the geeralized mathematical iductio we assume that the truth of P (1);P();:::;P( 1) implies the truth of P (). I other words, the secod step of the geeralized PMI ca be writte as where A is the set defied i the origial PMI. f1; ;:::; 1g A the A Recurreces We ow apply mathematical iductio to establish some facts about recurreces. We come back to recurreces i Theme 3. We start with a example that illustrates a applicatio of the geeralized mathematical iductio. Example 6: Let us defie T (0) 1 ad the 1 T () 1+ T (i); 1: (4) i0 This is a example of a recurrece that we shall study i some details later i this module. Observe that we ca compute cosecutive values T (1), T () ad so o from the recurrece itself. For example, T (1) 1+ T (0) 3; 1 T () + 1+ (T (0) + T (1)) 5; T (3) 7: This subsectio ca be omitted i the first readig. 6
7 But ca we guess how T () grows for arbitrary. I the table below we computed some umerical values of T () ad compared them to the growth of ad. T () From this table we should observe that T () grows faster tha admuchslowertha.letus the cojecture that 3 T ()» 4 log ; : (5) We ow use mathematical iductio to prove this guess. Observe that T () 5» 8 log 8 (sice log 1), but T (1) 3 > 4 log 10, therefore, we must start the iductio from. To carry out the iductive step we shall assume that for all j» 1 the above guess is is true. We ow prove that this guess is also true for. Ideed, T () T (j) j iductio j 4j log j log j»log log j j ( 1) log log 4 log +4 log» 1 4 log +4 log sice» 4 log ; ; 8» 8 log ; ; where (i) i the first step we use the recurrece ad extract the first two terms from the sum; (ii) i the secod lie we use the iductio assumptio i its geeral form ad boud every T (j) by 4j log j 3 We recall that y log b x is a logarithm to base b of x, that is, it is the expoet to which b must be raised to obtai x as show here b y x. 7
8 for» j < ; (iii) i the third lie we observe that log j» log (sice j» ) ad factor the costat term log i frot of the sum; (iv) i the fourth lie we apply the formula for the sum of the first 1 cosecutive itegers proved i Example ; (v) the fifth lie is simple algebra; (vi) i the sixth lie we observe that 8» 8 log for ad therefore cacel out the terms 8; fially the last iequality follows from the fact that 1 4 log» 0 for. (As we said at the begiig of this subsectio, if this derivatio is too ivolved i the first readig, the studet ca move forward to the ext sectio sice it will ot be used i the forthcomig discussio.) 8
9 Theme : Newto s Summatio Formula From high school we kow that (a + b) a +ab + b ; (a + b) 3 a 3 +3a b +3ab + b 3 ; (a + b) 4 a 4 +4a 3 b +6a b +4ab 3 + b 4 : But what about a formula for (a + b) for arbitrary. We shall derive it here, ad it is called Newto s summatio formula. Before we hadle the geeral case of (a+b), we must itroduce some ew otatio. I particular, biomial coefficiets also kow as Newto s coefficiets. We defie for atural k ad k! C(; k) : k!( k)! ( 1) ( k +1) ; k! where we remid that! 1 3 ( 1). By defiitio 0! 1. I literature the Newto coefficiets C(; k) are also deoted as From the defiitio we immediately fid C r : ψ k! : C(; 0) 1 But we shall also eed the followig lemma. Lemma 1 For atural k ad Proof. We give a direct proof. Observe that C(; 1) C( 1;k 1) + C( 1;k 1) C(; k) C(; k): C(; k) C( 1;k)+C( 1;k 1): (6) ( 1)! k!( k 1)! + ( 1)! (k 1)!( k)! ( 1)!( k) k!( k 1)!( k) + ( 1)! ( k + k) k!( k)!! k!( k)! C(; k) ( 1)!k (k 1)!k( k)! 9
10 where i the secod lie we multiply ad divide the first term by k ad the secod term by k. The we factorize ( 1)! ad after some simple algebra obtai the desired result. k!( k)! Now, we are ready to formulate ad prove the Newto summatio formula. Theorem For ay atural (a + b) k a k b k : (7) Proof. 4 The proof is by iductio with respect to. The basis step for 1is easy to check sice C(1; 0) C(1; 1) 1. We ow start the iductive step, ad postulate that if (7) is true for arbitrary,the must be true. We proceed as follows +1 (a + b) +1 C( +1;k)a k b +1 k (a + b) +1 (a + b) (a + b) C(; k)a k+1 b k + C( +1; 0)b +1 + ab [C(; 0) + C(; 1)] + a b 1 [C(; 1) + C(; )] + + a k b k+1 [C(; k 1) + C(; k)] + + a b [C(; ) +C(; 1)] + C( +1;+1)a +1 Lemma1 +1 C( +1;k)a k b +1 k : C(; k)a k b +1 k I the first lie above we use mathematical iductio ad the multiply out. I the ext few lies we group terms with the same power, that is a i b +1 i for all i. Fially, we applied Lemma 1 (i,e., C(; k) +C(; k 1) C( +1;k)) to fiish the derivatio. Exercise 4C: Apply Newto s formula to the followig (1 + x ) 4 : The above formula ca lead to surprisigly iterestig idetities. Here are two of them 0 4 The proof ca be omitted i the first readig. i i0 i i0 (8) ( 1) i (9) 10
11 The first idetity follows immediately from the Newto formula applied to (1 + 1) ; while the secod follows from (1 1) : We shall rederive these idetities usig combiatorial argumets i oe of the ext modules. 11
12 Theme 3: Recursio ad Recurreces Sometimes it is difficult to defie a object explicitly. I such cases, it is better to defie this object i terms of itself but of a smaller size. (Actually, we have see this priciple at work i mathematical iductio.) This process is called recursio ad ofte it is described mathematically by a recurrece. Example 7: Defiea 0 1ad for 0 a +1 a : Let s see what we get. We first compute some sample values: a 1 ; a a 1 4; a 3 a (a 1 )8 3 ; a 4 a 3 (a ) ((a 1 )) 4 : Based o this umerical evidece, we cojecture that a. We ca prove it usig mathematical iductio. But, i this it is easier to give a direct proof that is called telescopig. We proceed as follows: a +1 a a 1 a i+1 a ( i) a 0 : I the above we successively used the recurrece a i a i 1 util we reached the iitial value a 0 ad we kow that a 0 1. Observe that without kowig a 0 we ca either start the recurrece or fiish it. Exercise 4D: Derive a explicit formula for the followig recurrece for 1 a 3a 1 with a 0 1. We ca defie some other fuctios recursively. For example, F ()! ca be defied recursively as follows F (0) 1; F ( +1) ( +1)F (); 0: Furthermore, let S a k 1
13 where fa k g is a give sequece. For a computer to uderstad such a sum, we must defie it recursively. For example, we ca do it this way S 0 a 0 ; S +1 S + a +1 ; 0: But, let us cosider a more geeral recurreces. We uderlie that i order to start a recurrece we must defie some iitial values, ad to provide a method how to compute the ext value. Cosider the followig recurrece a 0 1; a a 1 + ; 1: This recurrece starts with 1; 3; 7; 15;::: but what is a geeral formula for a? Let us move the term a 1 to the other side of the recurrece ad write dow all the values as follows a 1 a 0 a a 1 a 3 a 3 ::: a i a i 1 i ::: a 1 a 1 a a 1 : Now, whe we add all these equatios together most of them will cacel out (we say that the sum telescopes) except a ad a 0 givig us which is the same as sayig a a 0 a i0 i1 i ; i : (10) Is this better tha the origial recurrece? Not yet sice we must compute the sum. Actually, i Module we defied the geometric progressio as follows b r ; 0; 1;::: 13
14 ad we derived S : i0 r i 1 r+1 ; r 6 1: (11) 1 r Actually, we shall reprove this formula usig mathematical iductio. It is easy to check its truth for 0(the basis step). Let us move to the iductive step. We first assume that the statemet above is true for arbitrary, ad we try to prove that this would imply that +1 S +1 r i 1 r+ 1 r : i0 We proceed as follows S +1 iductio +1 i0 r i r i + r +1 i0 1 r +1 + r +1 1 r 1 r +1 + r +1 r + ; 1 r where i the secod lie we extract the last term from the sum ad write it separately as r +1,the i the ext lie we apply to the first sum the iductive hypothesis, ad fially after some algebra we prove the desired formula. Now, we ca retur to (10) to coclude that a i0 +1 1: Let us solve some more recurreces. This is the oly way to lear how to hadle them. Let b 0 0 ad We do the followig telescopig b b 1 + ; 1: b b 1 + (b + 1) + b 3 +( ) + ( 1) + ::: b i +(i +1)+(i +)+ + ::: 14
15 b ( 1) + i i1 ( +1) where i the secod lie we substitute b 1 b +( 1), i the third lie we start observig a patter i which b i is followed by the sum of the first i +1atural umbers. The we apply the sum of cosecutive atural umbers derived i Example. I every step of the above derivatio we used the recurrece itself to reduce it util we reach the value that we kow, that is, b 0. We ca do it sice i step i we kow that b i b i 1 + i. Cosider ow a more complicated recurrece: 5 c 0 0; c c 1 + : Let us start the telescopig process ad try to fid a geeral patter. We have c c 1 + (c + 1) + c +( 1) + (c 3 + ) + ( 1) + 3 c 3 + ( ) + ( 1) + 3 (c 4 + 3) + ( ) + ( 1) + 4 c ( 3) + ( ) + ( 1) + ::: i c i + i ( i) + i 1 c i ( ) + ( 1) + ::: c [ ( 1)] + [ ( )] + + ( ) + ( 1) + 1 k ( k): I the secod lie above, we substitute c 1 by c + 1 ad observe that the additive term is ow +( 1) (the additive term is the oe that does ot ivolve c i ). After aother substitutio the additive term is elarged to ( )+( 1) +. Now you should be able to see the patter which becomes i ( i)+ i 1 c i ( ) + ( 1) +. After the last substitutio the additive term is fially 1 [ ( 1)]+ [ ( )] + + ( ) + ( 1) +. 5 The ext two examples ca be omitted i the first readig. 15
16 Now to fiish the recurrece we must fid a formula for the followig sum c 1 k ( k) 1 k 1 Observe that i the first sum we could factorize sice the summatio is over k, thus is fixed. After this observatio, the first sum is easy to estimate. We just foud above that it is equal to 1. But the secod oe is harder. To estimate it we first observe that k k : k+1 k k ( 1) k : The S (A) (B) k1 k1 k1 1 k k k1 k k k( k+1 k ) k k+1 k k+1 k k+1 k1 1 1 k k (k + 1) k+1 k k+1 k k (C) +1 0 ( +1 ) k+1 k k+1 k+1 ( 1) +1 +: I lie (A) we chage the idex of summatio from k to k +1, i lie (B) we expad the first sum ad observe that it cacels out the secod sum, fially i lie (C) we apply the geometric sum that we already discussed above. Comig back to the recurrece, puttig everythig together we have c +1 which is our fial aswer. Uffff :::it was ot that hard. Fially, we solve oe oliear recurrece. Cosider the followig 6 a 3a 1 ; 1 6 The forthcomig aalysis may be completely omitted, ad come back oly if a studet is iterested i a better uderstadig of oliear recurreces. 16
17 where a 0 1. It is a oliear recurrece sice a 1 is squared. Telescopig might be difficult for this recurrece. So we first simplify it. Defie b log a.thewehave b 0 1 b b 1 + log 3 sice log(xy) log x + log y. Now we are o familiar grouds. Usig telescopig we fid b ( 1) log 3; which implies a ( 1) log : Fially, we should say it is always a good idea to verify umerically our solutio by comparig its some iitial values to the values computed from the recurrece itself. 17
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