Bertrand s Postulate
|
|
- Horace Elliott
- 5 years ago
- Views:
Transcription
1 Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
2 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a prime betwee ad. -Joseph Bertrad, cojectured i 1845 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
3 Bertrad s Postulate Bertrad did ot prove his postulate. He verified the statemet (by had for all positive itegers up to 6,000,000. I 1850, Chebyshev proved Bertrad s Postulate. For this reaso, it is also referred to as Chebyshev s Theorem. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
4 Bertrad s Postulate May other proofs have bee foud i the time sice Chebyshev first proved this theorem. We will follow a proof due to Ramauja ad Erdös. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
5 Outlie 1 Fu With Biomial Coefficiets 2 A Few New Arithmetic Fuctios 3 A Upper Boud for ψ(x 4 Provig Bertrad s Postulate The Setup: ABC = ( A Lower Boud for ( A Upper Boud for C A Upper Boud for B Puttig Everythig Together 5 Geeralizatios 6 Some Neat Applicatios Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
6 Fu With Biomial Coefficiets Defiitio! = # of ways of orderig elemets Defiitio ( k = # of ways of choosig k elemets from a set cotaiig objects, without worryig about order. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
7 Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
8 Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. If we write dow Pascal s Triagle, the first few rows are: Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
9 Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. If we write dow Pascal s Triagle, the first few rows are: We ca obtai the umbers i the ext row by addig adjacet pairs of umbers from the previous row: Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
10 Fu With Biomial Coefficiets There is a curious relatioship betwee ( k ad Pascal s Triagle. If we write dow Pascal s Triagle, the first few rows are: We ca obtai the umbers i the ext row by addig adjacet pairs of umbers from the previous row: The elemets i the 4 th row are precisely ( 4 0 through ( 4 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
11 Fu With Biomial Coefficiets Aother iterestig observatio that you might make is that, at least i the examples above, the sum of all of the elemets i a row is a power of 2. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
12 Fu With Biomial Coefficiets Aother iterestig observatio that you might make is that, at least i the examples above, the sum of all of the elemets i a row is a power of 2. For example, = 2 2 ad = 2 3. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
13 Fu With Biomial Coefficiets Aother iterestig observatio that you might make is that, at least i the examples above, the sum of all of the elemets i a row is a power of 2. For example, = 2 2 ad = 2 3. ( Exercise: k = 2. k=0 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
14 Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
15 Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Which primes divide (? Let s look at some examples. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
16 Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Which primes divide (? Let s look at some examples. ( 4 ( 2 = 6. Which primes divide 6? What about 6 ( 3? 8 ( 4? 10 5? Ay cojectures about whe a prime has to divide (? Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
17 Fu With Biomial Coefficiets Oe particular type of biomial coefficiet will be of iterest to us as we prove Bertrad s Postulate. That coefficiet is (, the coefficiet i the ceter of the th row of Pascal s Triagle. Which primes divide (? Let s look at some examples. ( 4 ( 2 = 6. Which primes divide 6? What about 6 ( 3? 8 ( 4? 10 5? Ay cojectures about whe a prime has to divide (? Exercise: p ( if there exists a positive iteger j with < p j Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
18 Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
19 Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Proof We kow that each prime p with < p j divides.sice the primes are distict, they are pairwise relatively ( prime.thus, their product must divide (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
20 Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Proof We kow that each prime p with < p j divides.sice the primes are distict, they are pairwise relatively ( prime.thus, their product must divide (. Lemma (2 ( 4 for all 0. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
21 Fu With Biomial Coefficiets Assumig that those exercises are true, we ca prove two results: Lemma (1 p p:<p j (. Proof We kow that each prime p with < p j divides.sice the primes are distict, they are pairwise relatively ( prime.thus, their product must divide (. Lemma (2 ( 4 for all 0. Proof Look at the th row of Pascal s Triagle: ( is i the ceter, 2 = sum of all terms i the row. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
22 A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
23 A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Examples: Λ(2 = log 2, Λ(4 = log 2, Λ(6 = 0. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
24 A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Examples: Λ(2 = log 2, Λ(4 = log 2, Λ(6 = 0. Defiitio ψ(x = x Λ( Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
25 A Few New Arithmetic Fuctios We re already familiar with σ, τ ad ϕ. Let s defie some ew arithmetic fuctios: Defiitio { log p, = p k, k > 0 Λ( = 0, otherwise Examples: Λ(2 = log 2, Λ(4 = log 2, Λ(6 = 0. Defiitio ψ(x = x Λ( Example: ψ(6 = 0 + log 2 + log 3 + log 2 + log = log( = log(60. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
26 A Upper Boud for ψ(x Lemma ψ(x x log 4 For ow, let s preted that x Z. This will allow us to use the Well-Orderig Priciple. Base Case: If = 1, the ψ(1 = 1 Λ( = Λ(1 = 0. (This is certaily 1 log 4 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
27 Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
28 Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k = 2k Λ( k Λ( (from defiitio of ψ Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
29 Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k = Λ( Λ( (from defiitio of ψ 2k k = Λ( (cacelig terms k< 2k Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
30 Proof of Upper Boud for ψ(x Let S = {x Z + ψ(x > x log 4}. Assume that S is o-empty. The, by Well-Orderig Priciple, S has a least elemet. Call it l. Case 1: l = 2k ψ(2k ψ(k = Λ( Λ( (from defiitio of ψ 2k k = Λ( (cacelig terms k< 2k log ( 2k k (sice Λ( = p if = p j ad 0 otherwise, so this lie follows from Lemma (1 after takig log of both sides Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
31 Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
32 Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
33 Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
34 Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. So ψ(2k ψ(k + k log 4 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
35 Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. So ψ(2k ψ(k + k log 4 k log 4+k log 4 (sice l = 2k was the smallest elemet i S, so k / S ψ(k k log 4 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
36 Proof of Upper Boud for ψ(x From the previous slide: ψ(2k ψ(k log ( 2k k. From Lemma (2, ( 2k k 4 k, i.e. ψ(2k ψ(k log(4 k. From high school math: log(4 k = k log 4. So ψ(2k ψ(k + k log 4 k log 4+k log 4 (sice l = 2k was the smallest elemet i S, so k / S ψ(k k log 4 ψ(2k 2k log 4, ie. l caot be eve. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
37 Proof of Upper Boud for ψ(x Case 2: l = 2k + 1. The argumet for the case where l is odd is similar to the case where l is eve. The result is the same, i.e. it shows that l caot be odd. Sice the least elemet of S is either eve or odd the S must empty. Remark: The iequality ψ(x x log 4 holds for ALL x 1 (ot just itegers. Why? Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
38 Proof of Upper Boud for ψ(x Case 2: l = 2k + 1. The argumet for the case where l is odd is similar to the case where l is eve. The result is the same, i.e. it shows that l caot be odd. Sice the least elemet of S is either eve or odd the S must empty. Remark: The iequality ψ(x x log 4 holds for ALL x 1 (ot just itegers. Why? ψ(x = ψ( x x log4 x log 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
39 Provig Bertrad s Postulate We will use what we have leared about ( ad ψ(x i order to prove our mai result: Theorem (Bertrad s Postulate For every Z +, there exists a prime p such that < p. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
40 The Setup : ABC = ( Let A = <p p. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
41 The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
42 The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. By the Uique Factorizatio Theorem, we ca factor m ito a product of primes (uiquely. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
43 The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. By the Uique Factorizatio Theorem, we ca factor m ito a product of primes (uiquely. Let B = cotributio to ( from primes p (, ]. Let C = cotributio to ( from primes p. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
44 The Setup : ABC = ( Let A = <p p. From Lemma (1, we kow that A (. So, there exists m Z st. A m = (. By the Uique Factorizatio Theorem, we ca factor m ito a product of primes (uiquely. Let B = cotributio to ( from primes p (, ]. Let C = cotributio to ( from primes p. The ABC = (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
45 The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
46 The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. But A = p p, so there must be a prime dividig A. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
47 The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. But A = p p, so there must be a prime dividig A. I other words, there must be a prime betwee ad dividig (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
48 The Setup : ABC = ( Goal: We wat to show that BC < (. Why does this imply that Bertrad s Postulate holds? If BC < ( the A 1. But A = p p, so there must be a prime dividig A. I other words, there must be a prime betwee ad dividig (. This proves Bertrad s Postulate because it proves the existece of a prime betwee ad. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
49 The Setup : ABC = ( I order to show that BC < (, we will fid upper bouds for B ad C ad we will fid a lower boud for (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
50 A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
51 A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Sice the two ed terms i a row of Pascal s triagle are both 1, the 1 = 2 +. j=0 ( j j=1 ( j Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
52 A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Sice the two ed terms i a row of Pascal s triagle are both 1, the 1 = 2 +. j=0 ( j j=1 ( j We kow that the middle term, (, is the largest term i the sum. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
53 A Lower Boud for ( Lemma ( 4 Proof I a earlier exercise, we showed that j=0 ( j = 4. Sice the two ed terms i a row of Pascal s triagle are both 1, the 1 = 2 +. j=0 ( j j=1 ( j We kow that the middle term, (, is the largest term i the sum. 1 ( Thus, 2 + ( j ( sice we are summig terms. j=1 Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
54 Aother Pair of Useful Exercises The followig exercises will also be useful i boudig BC: Exercise Prove or disprove ad salvage if possible: If x R, the 2x 2 x = 0 or 1. Exercise (a What power of the prime p appears i the prime factorizatio of!? (b What power of p appears i the factorizatio of ( k? Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
55 A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
56 A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j The sum o the right couts the umber of j that satisfy p j. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
57 A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j The sum o the right couts the umber of j that satisfy p j. I order to determie that umber, we eed to solve the equatio p x. But this is the same as solvig x log p log. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
58 A Upper Boud for C Lemma C ( 1 Proof Let k be the highest power of p dividig (. Have you solved both of the exercises o the previous slide? Oce you have, you will see that k 1. j:p j The sum o the right couts the umber of j that satisfy p j. I order to determie that umber, we eed to solve the equatio p x. But this is the same as solvig x log p log. Thus, x log log p. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
59 A Upper Boud for C Sice the j that satisfy p j must be itegers, the log 1 = log p. j:p j Recall that C = p p ( p. Suppose that a prime p is ot icluded i C, i.e. p > ad p. log The log p = 1. So all of the primes p such that p k ( for k > 1 must occur i C. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
60 A Upper Boud for C Recallig that C = C p p log p log p log p log p p p ( p, we have Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
61 A Upper Boud for C Recallig that C = C p p = p log p log p log p log p p p ( p, we have e log (log rule: p α = e α log p Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
62 A Upper Boud for C Recallig that C = C p p = p = p log p log p log p log p p p ( p, we have e log (log rule: p α = e α log p Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
63 A Upper Boud for C But = ( π(, where π( = # of primes. p Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
64 A Upper Boud for C But = ( π(, where π( = # of primes. p Sice 1 is ot prime, the π( 1. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
65 A Upper Boud for C But = ( π(, where π( = # of primes. p Sice 1 is ot prime, the π( 1. Thus, we see that C ( 1. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
66 A Upper Boud for B Lemma B Proof Recall that B = p. <p p ( Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
67 A Upper Boud for B Lemma B Proof Recall that B = <p p ( For all > 4.5, < 2 3 (square both sides, the divide both sides by p. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
68 A Upper Boud for B Lemma B Proof Recall that B = <p p ( For all > 4.5, < 2 3 (square both sides, the divide both sides by We will separate B ito two products: p. <p 2 3 p ad p. 2 3 <p Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
69 A Upper Boud for B Let p ( 2 3, ]. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
70 A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
71 A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Sice p ( 2 3, ], we have 1 p < 3 2. Thus p = 1 + r, with 0 r < 1 2. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
72 A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Sice p ( 2 3, ], we have 1 p < 3 2. Thus p = 1 + r, with 0 r < 1 2. Usig this fact with the formula for k that you foud i the problem set yields k = 0. So, the product p does t cotribute ay 2 3 <p primes to B. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
73 A Upper Boud for B Let p ( 2 3, ]. We will show that k = 0 whe p is i this rage, where k is the highest power of p dividig (. Sice p ( 2 3, ], we have 1 p < 3 2. Thus p = 1 + r, with 0 r < 1 2. Usig this fact with the formula for k that you foud i the problem set yields k = 0. So, the product p does t cotribute ay 2 3 <p primes to B. B = p 2 p 4 3 (sice ψ(x x log 4. <p p 2 3 p ( Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
74 Puttig Everythig Together What we have show: ( = ABC ( 4 C ( 1 B So, A = ( /(BC Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
75 Puttig Everythig Together What we have show: ( = ABC ( 4 C ( 1 B So, A = ( /(BC 4 ( Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
76 Puttig Everythig Together What we have show: ( = ABC ( 4 C ( 1 B So, A = ( /(BC 4 ( = ( Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
77 Puttig Everythig Together Remember that i order for Bertrad s Postulate to hold, we eed to show that A > 1. Thus, we eed to determie whe > (. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
78 Puttig Everythig Together Remember that i order for Bertrad s Postulate to hold, we eed to show that A > 1. Thus, we eed to determie whe > (. Usig high school math, we ca show that > ( holds whe > 450. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
79 Puttig Everythig Together Remember that i order for Bertrad s Postulate to hold, we eed to show that A > 1. Thus, we eed to determie whe > (. Usig high school math, we ca show that > ( holds whe > 450. Bertrad s Postulate holds for all > 450. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
80 Fiishig Up I order to coclude that Bertrad s Postulate is true for all Z +, we just eed to check values of 450. Remember that Bertrad checked all up to 6,000,000, so if you believe him the we re doe! If ot, cosider the list of primes 2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, where each is less tha twice the precedig. This proves Bertrad s Postulate for all < 631, sice ay such ca be squeezed betwee two umbers o the list. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
81 Geeralizatios Oe way that a mathematicia fids ew problems to solve is by lookig at a result that has already bee prove ad askig Does this hold i a more geeral settig? Of course, more geeral ca mea may differet thigs. For example, we showed that ( 4 for all 0. Perhaps we could have show a similar result for ay iteger. Aother geeralizatio would be to try to boud ( k, k Z +. Ca you thik of a more geeral statemet of Bertrad s Postulate? Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
82 Geeralizatios Here are a few well-kow geeralizatios of Bertrad s Postulate: Theorem (Sylvester The product of k cosecutive itegers greater tha k is divisible by a prime greater tha k. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
83 Geeralizatios Here are a few well-kow geeralizatios of Bertrad s Postulate: Theorem (Sylvester The product of k cosecutive itegers greater tha k is divisible by a prime greater tha k. Theorem (Erdös For ay positive iteger k, there is a atural umber N such that for all > N, there are at least k primes betwee ad. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
84 Geeralizatios Here are a few well-kow geeralizatios of Bertrad s Postulate: Theorem (Sylvester The product of k cosecutive itegers greater tha k is divisible by a prime greater tha k. Theorem (Erdös For ay positive iteger k, there is a atural umber N such that for all > N, there are at least k primes betwee ad. Cojecture (Legedre For every > 1, there is a prime p such that 2 < p < ( Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
85 Some Neat Applicatios Usig Bertrad s Postulate, we ca also prove may other iterestig results, icludig: Every iteger > 6 ca be writte as a sum of distict primes. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
86 Some Neat Applicatios Usig Bertrad s Postulate, we ca also prove may other iterestig results, icludig: Every iteger > 6 ca be writte as a sum of distict primes. N N, there exists a eve iteger k > 0 for which there are at least N prime pairs p, p + k. Lola Thompso (Ross Program Bertrad s Postulate July 3, / 33
and each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.
MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More informationZeros of Polynomials
Math 160 www.timetodare.com 4.5 4.6 Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered with fidig the solutios of polyomial equatios of ay degree
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationMath 155 (Lecture 3)
Math 55 (Lecture 3) September 8, I this lecture, we ll cosider the aswer to oe of the most basic coutig problems i combiatorics Questio How may ways are there to choose a -elemet subset of the set {,,,
More informationLecture 1. January 8, 2018
Lecture 1 Jauary 8, 018 1 Primes A prime umber p is a positive iteger which caot be writte as ab for some positive itegers a, b > 1. A prime p also have the property that if p ab, the p a or p b. This
More informationSquare-Congruence Modulo n
Square-Cogruece Modulo Abstract This paper is a ivestigatio of a equivalece relatio o the itegers that was itroduced as a exercise i our Discrete Math class. Part I - Itro Defiitio Two itegers are Square-Cogruet
More informationInfinite Sequences and Series
Chapter 6 Ifiite Sequeces ad Series 6.1 Ifiite Sequeces 6.1.1 Elemetary Cocepts Simply speakig, a sequece is a ordered list of umbers writte: {a 1, a 2, a 3,...a, a +1,...} where the elemets a i represet
More information3.2 Properties of Division 3.3 Zeros of Polynomials 3.4 Complex and Rational Zeros of Polynomials
Math 60 www.timetodare.com 3. Properties of Divisio 3.3 Zeros of Polyomials 3.4 Complex ad Ratioal Zeros of Polyomials I these sectios we will study polyomials algebraically. Most of our work will be cocered
More informationThe Structure of Z p when p is Prime
LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z
More informationSummer High School 2009 Aaron Bertram
Summer High School 009 Aaro Bertram 3 Iductio ad Related Stuff Let s thik for a bit about the followig two familiar equatios: Triagle Number Equatio Square Number Equatio: + + 3 + + = ( + + 3 + 5 + + (
More informationInduction: Solutions
Writig Proofs Misha Lavrov Iductio: Solutios Wester PA ARML Practice March 6, 206. Prove that a 2 2 chessboard with ay oe square removed ca always be covered by shaped tiles. Solutio : We iduct o. For
More informationRADICAL EXPRESSION. If a and x are real numbers and n is a positive integer, then x is an. n th root theorems: Example 1 Simplify
Example 1 Simplify 1.2A Radical Operatios a) 4 2 b) 16 1 2 c) 16 d) 2 e) 8 1 f) 8 What is the relatioship betwee a, b, c? What is the relatioship betwee d, e, f? If x = a, the x = = th root theorems: RADICAL
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationTHE ASYMPTOTIC COMPLEXITY OF MATRIX REDUCTION OVER FINITE FIELDS
THE ASYMPTOTIC COMPLEXITY OF MATRIX REDUCTION OVER FINITE FIELDS DEMETRES CHRISTOFIDES Abstract. Cosider a ivertible matrix over some field. The Gauss-Jorda elimiatio reduces this matrix to the idetity
More informationNICK DUFRESNE. 1 1 p(x). To determine some formulas for the generating function of the Schröder numbers, r(x) = a(x) =
AN INTRODUCTION TO SCHRÖDER AND UNKNOWN NUMBERS NICK DUFRESNE Abstract. I this article we will itroduce two types of lattice paths, Schröder paths ad Ukow paths. We will examie differet properties of each,
More informationMath 4400/6400 Homework #7 solutions
MATH 4400 problems. Math 4400/6400 Homewor #7 solutios 1. Let p be a prime umber. Show that the order of 1 + p modulo p 2 is exactly p. Hit: Expad (1 + p) p by the biomial theorem, ad recall from MATH
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More information1 Generating functions for balls in boxes
Math 566 Fall 05 Some otes o geeratig fuctios Give a sequece a 0, a, a,..., a,..., a geeratig fuctio some way of represetig the sequece as a fuctio. There are may ways to do this, with the most commo ways
More informationMath F215: Induction April 7, 2013
Math F25: Iductio April 7, 203 Iductio is used to prove that a collectio of statemets P(k) depedig o k N are all true. A statemet is simply a mathematical phrase that must be either true or false. Here
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationIn number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.
Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers
More information2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.
CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.
More informationCALCULATION OF FIBONACCI VECTORS
CALCULATION OF FIBONACCI VECTORS Stuart D. Aderso Departmet of Physics, Ithaca College 953 Daby Road, Ithaca NY 14850, USA email: saderso@ithaca.edu ad Dai Novak Departmet of Mathematics, Ithaca College
More informationSequences, Mathematical Induction, and Recursion. CSE 2353 Discrete Computational Structures Spring 2018
CSE 353 Discrete Computatioal Structures Sprig 08 Sequeces, Mathematical Iductio, ad Recursio (Chapter 5, Epp) Note: some course slides adopted from publisher-provided material Overview May mathematical
More informationACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER / Statistics
ACCESS TO SCIENCE, ENGINEERING AND AGRICULTURE: MATHEMATICS 1 MATH00030 SEMESTER 1 018/019 DR. ANTHONY BROWN 8. Statistics 8.1. Measures of Cetre: Mea, Media ad Mode. If we have a series of umbers the
More information1. n! = n. tion. For example, (n+1)! working with factorials. = (n+1) n (n 1) 2 1
Biomial Coefficiets ad Permutatios Mii-lecture The followig pages discuss a few special iteger coutig fuctios You may have see some of these before i a basic probability class or elsewhere, but perhaps
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More informationProperties and Tests of Zeros of Polynomial Functions
Properties ad Tests of Zeros of Polyomial Fuctios The Remaider ad Factor Theorems: Sythetic divisio ca be used to fid the values of polyomials i a sometimes easier way tha substitutio. This is show by
More informationLecture Overview. 2 Permutations and Combinations. n(n 1) (n (k 1)) = n(n 1) (n k + 1) =
COMPSCI 230: Discrete Mathematics for Computer Sciece April 8, 2019 Lecturer: Debmalya Paigrahi Lecture 22 Scribe: Kevi Su 1 Overview I this lecture, we begi studyig the fudametals of coutig discrete objects.
More informationROSE WONG. f(1) f(n) where L the average value of f(n). In this paper, we will examine averages of several different arithmetic functions.
AVERAGE VALUES OF ARITHMETIC FUNCTIONS ROSE WONG Abstract. I this paper, we will preset problems ivolvig average values of arithmetic fuctios. The arithmetic fuctios we discuss are: (1)the umber of represetatios
More informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More information62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +
62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of
More informationBertrand s Postulate. Theorem (Bertrand s Postulate): For every positive integer n, there is a prime p satisfying n < p 2n.
Bertrad s Postulate Our goal is to prove the followig Theorem Bertrad s Postulate: For every positive iteger, there is a prime p satisfyig < p We remark that Bertrad s Postulate is true by ispectio for,,
More informationClassroom. We investigate and further explore the problem of dividing x = n + m (m, n are coprime) sheep in
Classroom I this sectio of Resoace, we ivite readers to pose questios likely to be raised i a classroom situatio. We may suggest strategies for dealig with them, or ivite resposes, or both. Classroom is
More informationLecture 23 Rearrangement Inequality
Lecture 23 Rearragemet Iequality Holde Lee 6/4/ The Iequalities We start with a example Suppose there are four boxes cotaiig $0, $20, $50 ad $00 bills, respectively You may take 2 bills from oe box, 3
More informationSOLUTIONS TO PRISM PROBLEMS Junior Level 2014
SOLUTIONS TO PRISM PROBLEMS Juior Level 04. (B) Sice 50% of 50 is 50 5 ad 50% of 40 is the secod by 5 0 5. 40 0, the first exceeds. (A) Oe way of comparig the magitudes of the umbers,,, 5 ad 0.7 is 4 5
More informationFermat s Little Theorem. mod 13 = 0, = }{{} mod 13 = 0. = a a a }{{} mod 13 = a 12 mod 13 = 1, mod 13 = a 13 mod 13 = a.
Departmet of Mathematical Scieces Istructor: Daiva Puciskaite Discrete Mathematics Fermat s Little Theorem 43.. For all a Z 3, calculate a 2 ad a 3. Case a = 0. 0 0 2-times Case a 0. 0 0 3-times a a 2-times
More informationRecurrence Relations
Recurrece Relatios Aalysis of recursive algorithms, such as: it factorial (it ) { if (==0) retur ; else retur ( * factorial(-)); } Let t be the umber of multiplicatios eeded to calculate factorial(). The
More informationChapter 7 COMBINATIONS AND PERMUTATIONS. where we have the specific formula for the binomial coefficients:
Chapter 7 COMBINATIONS AND PERMUTATIONS We have see i the previous chapter that (a + b) ca be writte as 0 a % a & b%þ% a & b %þ% b where we have the specific formula for the biomial coefficiets: '!!(&)!
More informationThe Growth of Functions. Theoretical Supplement
The Growth of Fuctios Theoretical Supplemet The Triagle Iequality The triagle iequality is a algebraic tool that is ofte useful i maipulatig absolute values of fuctios. The triagle iequality says that
More informationP1 Chapter 8 :: Binomial Expansion
P Chapter 8 :: Biomial Expasio jfrost@tiffi.kigsto.sch.uk www.drfrostmaths.com @DrFrostMaths Last modified: 6 th August 7 Use of DrFrostMaths for practice Register for free at: www.drfrostmaths.com/homework
More information18th Bay Area Mathematical Olympiad. Problems and Solutions. February 23, 2016
18th Bay Area Mathematical Olympiad February 3, 016 Problems ad Solutios BAMO-8 ad BAMO-1 are each 5-questio essay-proof exams, for middle- ad high-school studets, respectively. The problems i each exam
More information7 Sequences of real numbers
40 7 Sequeces of real umbers 7. Defiitios ad examples Defiitio 7... A sequece of real umbers is a real fuctio whose domai is the set N of atural umbers. Let s : N R be a sequece. The the values of s are
More informationA sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationIntermediate Math Circles November 4, 2009 Counting II
Uiversity of Waterloo Faculty of Mathematics Cetre for Educatio i Mathematics ad Computig Itermediate Math Circles November 4, 009 Coutig II Last time, after lookig at the product rule ad sum rule, we
More informationDr. Clemens Kroll. Abstract
Riema s Hypothesis ad Stieltjes Cojecture Riema s Hypothesis ad Stieltjes Cojecture Dr. Clemes Kroll Abstract It is show that Riema s hypothesis is true by showig that a equivalet statemet is true. Eve
More information2.4 Sequences, Sequences of Sets
72 CHAPTER 2. IMPORTANT PROPERTIES OF R 2.4 Sequeces, Sequeces of Sets 2.4.1 Sequeces Defiitio 2.4.1 (sequece Let S R. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For each
More informationMathematical Induction
Mathematical Iductio Itroductio Mathematical iductio, or just iductio, is a proof techique. Suppose that for every atural umber, P() is a statemet. We wish to show that all statemets P() are true. I a
More informationSection 5.1 The Basics of Counting
1 Sectio 5.1 The Basics of Coutig Combiatorics, the study of arragemets of objects, is a importat part of discrete mathematics. I this chapter, we will lear basic techiques of coutig which has a lot of
More informationInjections, Surjections, and the Pigeonhole Principle
Ijectios, Surjectios, ad the Pigeohole Priciple 1 (10 poits Here we will come up with a sloppy boud o the umber of parethesisestigs (a (5 poits Describe a ijectio from the set of possible ways to est pairs
More informationLINEAR ALGEBRAIC GROUPS: LECTURE 6
LINEAR ALGEBRAIC GROUPS: LECTURE 6 JOHN SIMANYI Grassmaias over Fiite Fields As see i the Fao plae, fiite fields create geometries that are uite differet from our more commo R or C based geometries These
More informationSequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
More informationModern Algebra 1 Section 1 Assignment 1. Solution: We have to show that if you knock down any one domino, then it knocks down the one behind it.
Moder Algebra 1 Sectio 1 Assigmet 1 JOHN PERRY Eercise 1 (pg 11 Warm-up c) Suppose we have a ifiite row of domioes, set up o ed What sort of iductio argumet would covice us that ocig dow the first domio
More informationBeurling Integers: Part 2
Beurlig Itegers: Part 2 Isomorphisms Devi Platt July 11, 2015 1 Prime Factorizatio Sequeces I the last article we itroduced the Beurlig geeralized itegers, which ca be represeted as a sequece of real umbers
More informationEnumerative & Asymptotic Combinatorics
C50 Eumerative & Asymptotic Combiatorics Stirlig ad Lagrage Sprig 2003 This sectio of the otes cotais proofs of Stirlig s formula ad the Lagrage Iversio Formula. Stirlig s formula Theorem 1 (Stirlig s
More informationx c the remainder is Pc ().
Algebra, Polyomial ad Ratioal Fuctios Page 1 K.Paulk Notes Chapter 3, Sectio 3.1 to 3.4 Summary Sectio Theorem Notes 3.1 Zeros of a Fuctio Set the fuctio to zero ad solve for x. The fuctio is zero at these
More informationPower Series Expansions of Binomials
Power Series Expasios of Biomials S F Ellermeyer April 0, 008 We are familiar with expadig biomials such as the followig: ( + x) = + x + x ( + x) = + x + x + x ( + x) 4 = + 4x + 6x + 4x + x 4 ( + x) 5
More informationMath 61CM - Solutions to homework 3
Math 6CM - Solutios to homework 3 Cédric De Groote October 2 th, 208 Problem : Let F be a field, m 0 a fixed oegative iteger ad let V = {a 0 + a x + + a m x m a 0,, a m F} be the vector space cosistig
More informationMA131 - Analysis 1. Workbook 2 Sequences I
MA3 - Aalysis Workbook 2 Sequeces I Autum 203 Cotets 2 Sequeces I 2. Itroductio.............................. 2.2 Icreasig ad Decreasig Sequeces................ 2 2.3 Bouded Sequeces..........................
More informationDiscrete Mathematics for CS Spring 2007 Luca Trevisan Lecture 22
CS 70 Discrete Mathematics for CS Sprig 2007 Luca Trevisa Lecture 22 Aother Importat Distributio The Geometric Distributio Questio: A biased coi with Heads probability p is tossed repeatedly util the first
More informationCombinatorics and Newton s theorem
INTRODUCTION TO MATHEMATICAL REASONING Key Ideas Worksheet 5 Combiatorics ad Newto s theorem This week we are goig to explore Newto s biomial expasio theorem. This is a very useful tool i aalysis, but
More informationSequences I. Chapter Introduction
Chapter 2 Sequeces I 2. Itroductio A sequece is a list of umbers i a defiite order so that we kow which umber is i the first place, which umber is i the secod place ad, for ay atural umber, we kow which
More informationOn Divisibility concerning Binomial Coefficients
A talk give at the Natioal Chiao Tug Uiversity (Hsichu, Taiwa; August 5, 2010 O Divisibility cocerig Biomial Coefficiets Zhi-Wei Su Najig Uiversity Najig 210093, P. R. Chia zwsu@ju.edu.c http://math.ju.edu.c/
More information6.3 Testing Series With Positive Terms
6.3. TESTING SERIES WITH POSITIVE TERMS 307 6.3 Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial
More informationProblems from 9th edition of Probability and Statistical Inference by Hogg, Tanis and Zimmerman:
Math 224 Fall 2017 Homework 4 Drew Armstrog Problems from 9th editio of Probability ad Statistical Iferece by Hogg, Tais ad Zimmerma: Sectio 2.3, Exercises 16(a,d),18. Sectio 2.4, Exercises 13, 14. Sectio
More informationUSA Mathematical Talent Search Round 3 Solutions Year 27 Academic Year
/3/27. Fill i each space of the grid with either a or a so that all sixtee strigs of four cosecutive umbers across ad dow are distict. You do ot eed to prove that your aswer is the oly oe possible; you
More informationNotes #3 Sequences Limit Theorems Monotone and Subsequences Bolzano-WeierstraßTheorem Limsup & Liminf of Sequences Cauchy Sequences and Completeness
Notes #3 Sequeces Limit Theorems Mootoe ad Subsequeces Bolzao-WeierstraßTheorem Limsup & Limif of Sequeces Cauchy Sequeces ad Completeess This sectio of otes focuses o some of the basics of sequeces of
More informationLecture 2: April 3, 2013
TTIC/CMSC 350 Mathematical Toolkit Sprig 203 Madhur Tulsiai Lecture 2: April 3, 203 Scribe: Shubhedu Trivedi Coi tosses cotiued We retur to the coi tossig example from the last lecture agai: Example. Give,
More informationMATH 112: HOMEWORK 6 SOLUTIONS. Problem 1: Rudin, Chapter 3, Problem s k < s k < 2 + s k+1
MATH 2: HOMEWORK 6 SOLUTIONS CA PRO JIRADILOK Problem. If s = 2, ad Problem : Rudi, Chapter 3, Problem 3. s + = 2 + s ( =, 2, 3,... ), prove that {s } coverges, ad that s < 2 for =, 2, 3,.... Proof. The
More informationThe Binomial Theorem
The Biomial Theorem Robert Marti Itroductio The Biomial Theorem is used to expad biomials, that is, brackets cosistig of two distict terms The formula for the Biomial Theorem is as follows: (a + b ( k
More informationWeek 5-6: The Binomial Coefficients
Wee 5-6: The Biomial Coefficiets March 6, 2018 1 Pascal Formula Theorem 11 (Pascal s Formula For itegers ad such that 1, ( ( ( 1 1 + 1 The umbers ( 2 ( 1 2 ( 2 are triagle umbers, that is, The petago umbers
More information4.3 Growth Rates of Solutions to Recurrences
4.3. GROWTH RATES OF SOLUTIONS TO RECURRENCES 81 4.3 Growth Rates of Solutios to Recurreces 4.3.1 Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer.
More informationProblem 4: Evaluate ( k ) by negating (actually un-negating) its upper index. Binomial coefficient
Problem 4: Evaluate by egatig actually u-egatig its upper idex We ow that Biomial coefficiet r { where r is a real umber, is a iteger The above defiitio ca be recast i terms of factorials i the commo case
More information6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
More informationBINOMIAL COEFFICIENT AND THE GAUSSIAN
BINOMIAL COEFFICIENT AND THE GAUSSIAN The biomial coefficiet is defied as-! k!(! ad ca be writte out i the form of a Pascal Triagle startig at the zeroth row with elemet 0,0) ad followed by the two umbers,
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More informationTennessee Department of Education
Teessee Departmet of Educatio Task: Comparig Shapes Geometry O a piece of graph paper with a coordiate plae, draw three o-colliear poits ad label them A, B, C. (Do ot use the origi as oe of your poits.)
More informationMath 2112 Solutions Assignment 5
Math 2112 Solutios Assigmet 5 5.1.1 Idicate which of the followig relatioships are true ad which are false: a. Z Q b. R Q c. Q Z d. Z Z Z e. Q R Q f. Q Z Q g. Z R Z h. Z Q Z a. True. Every positive iteger
More informationExercises 1 Sets and functions
Exercises 1 Sets ad fuctios HU Wei September 6, 018 1 Basics Set theory ca be made much more rigorous ad built upo a set of Axioms. But we will cover oly some heuristic ideas. For those iterested studets,
More informationAssignment 5: Solutions
McGill Uiversity Departmet of Mathematics ad Statistics MATH 54 Aalysis, Fall 05 Assigmet 5: Solutios. Let y be a ubouded sequece of positive umbers satisfyig y + > y for all N. Let x be aother sequece
More informationHomework 9. (n + 1)! = 1 1
. Chapter : Questio 8 If N, the Homewor 9 Proof. We will prove this by usig iductio o. 2! + 2 3! + 3 4! + + +! +!. Base step: Whe the left had side is. Whe the right had side is 2! 2 +! 2 which proves
More informationRiemann Sums y = f (x)
Riema Sums Recall that we have previously discussed the area problem I its simplest form we ca state it this way: The Area Problem Let f be a cotiuous, o-egative fuctio o the closed iterval [a, b] Fid
More informationMA131 - Analysis 1. Workbook 7 Series I
MA3 - Aalysis Workbook 7 Series I Autum 008 Cotets 4 Series 4. Defiitios............................... 4. Geometric Series........................... 4 4.3 The Harmoic Series.........................
More informationProblem Set 2 Solutions
CS271 Radomess & Computatio, Sprig 2018 Problem Set 2 Solutios Poit totals are i the margi; the maximum total umber of poits was 52. 1. Probabilistic method for domiatig sets 6pts Pick a radom subset S
More informationMetric Space Properties
Metric Space Properties Math 40 Fial Project Preseted by: Michael Brow, Alex Cordova, ad Alyssa Sachez We have already poited out ad will recogize throughout this book the importace of compact sets. All
More informationMath 609/597: Cryptography 1
Math 609/597: Cryptography 1 The Solovay-Strasse Primality Test 12 October, 1993 Burt Roseberg Revised: 6 October, 2000 1 Itroductio We describe the Solovay-Strasse primality test. There is quite a bit
More informationSEQUENCES AND SERIES
Sequeces ad 6 Sequeces Ad SEQUENCES AND SERIES Successio of umbers of which oe umber is desigated as the first, other as the secod, aother as the third ad so o gives rise to what is called a sequece. Sequeces
More informationSequences. Notation. Convergence of a Sequence
Sequeces A sequece is essetially just a list. Defiitio (Sequece of Real Numbers). A sequece of real umbers is a fuctio Z (, ) R for some real umber. Do t let the descriptio of the domai cofuse you; it
More informationMath 299 Supplement: Real Analysis Nov 2013
Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality
More information(ii) Two-permutations of {a, b, c}. Answer. (B) P (3, 3) = 3! (C) 3! = 6, and there are 6 items in (A). ... Answer.
SOLUTIONS Homewor 5 Due /6/19 Exercise. (a Cosider the set {a, b, c}. For each of the followig, (A list the objects described, (B give a formula that tells you how may you should have listed, ad (C verify
More informationOn a Smarandache problem concerning the prime gaps
O a Smaradache problem cocerig the prime gaps Felice Russo Via A. Ifate 7 6705 Avezzao (Aq) Italy felice.russo@katamail.com Abstract I this paper, a problem posed i [] by Smaradache cocerig the prime gaps
More informationAnalysis of Algorithms. Introduction. Contents
Itroductio The focus of this module is mathematical aspects of algorithms. Our mai focus is aalysis of algorithms, which meas evaluatig efficiecy of algorithms by aalytical ad mathematical methods. We
More informationInduction proofs - practice! SOLUTIONS
Iductio proofs - practice! SOLUTIONS 1. Prove that f ) = 6 + + 15 is odd for all Z +. Base case: For = 1, f 1) = 41) + 1) + 13 = 19. Sice 19 is odd, f 1) is odd - base case prove. Iductive hypothesis:
More informationSOLVED EXAMPLES
Prelimiaries Chapter PELIMINAIES Cocept of Divisibility: A o-zero iteger t is said to be a divisor of a iteger s if there is a iteger u such that s tu I this case we write t s (i) 6 as ca be writte as
More informationThe Random Walk For Dummies
The Radom Walk For Dummies Richard A Mote Abstract We look at the priciples goverig the oe-dimesioal discrete radom walk First we review five basic cocepts of probability theory The we cosider the Beroulli
More informationRegent College Maths Department. Further Pure 1. Proof by Induction
Reget College Maths Departmet Further Pure Proof by Iductio Further Pure Proof by Mathematical Iductio Page Further Pure Proof by iductio The Edexcel syllabus says that cadidates should be able to: (a)
More informationMATH 147 Homework 4. ( = lim. n n)( n + 1 n) n n n. 1 = lim
MATH 147 Homework 4 1. Defie the sequece {a } by a =. a) Prove that a +1 a = 0. b) Prove that {a } is ot a Cauchy sequece. Solutio: a) We have: ad so we re doe. a +1 a = + 1 = + 1 + ) + 1 ) + 1 + 1 = +
More information