# 3. Greatest Common Divisor - Least Common Multiple

Save this PDF as:

Size: px
Start display at page:

Download "3. Greatest Common Divisor - Least Common Multiple"

## Transcription

1 3 Greatest Commo Divisor - Least Commo Multiple Defiitio 31: The greatest commo divisor of two atural umbers a ad b is the largest atural umber c which divides both a ad b We deote the greatest commo gcd ab, divisor of a ad b as ( ) Example 32: Fid gcd ( 24,90 ) Solutio: The atural umbers that divide 24 are 1, 2, 3, 4, 6, 12 ad 24 The atural umbers that divide 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30 ad 90 Therefore, gcd 24,90 = 6 ( ) Oe way to compute the greatest commo divisor of two atural umbers a ad b is to write their prime factorizatios For example, i the case of the umbers 24 ad 90 above, 3 2 we have 24 = ( 2) ( 3) ad 90 = ( 2)( 3) ( 5) Notice that we ca iclude all of the prime factors from both umbers by usig a expoet of = ( ) 2 90 = ( 2)( 3) ( 5) Now we take the smallest expoet that occurs for each prime factor ad obtai ( ) = ( ) = ( 2) ( 3) ( 5) = ( 2)( 3) mi(3,1) mi(1,2) mi(0,1) gcd 24, = 6 We geeralize this idea i the theorem below Theorem 33: Let ab, ad suppose p1, p2,, p are prime umbers so that α a= ( p1) ( p2 ) ( p ) ad β1 β2 β b= ( p1) ( p2 ) ( p ) where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p) ( p) ( p ) Proof: Let c= gcd( a, b) Sice p1, p2,, p are prime umbers, p divides both a ad b implies that p divides c

2 If p α divides a ad p β divides b, the the smaller of p α ad p β ad b ( 2 2 divides 12 ad 2 divides 30 implies 2 divides gcd(12,30) = 6 ) So, mi( α, β ) p the that divides both a ad b; ad hece c If p does ot divide c But whe we say p does ot divide c Hece, gcd (, ) ( ) 1 1 ( ) 2 2 ( ) p divides both a p divides a but does ot divide b, mi( α,0) 0 p mi( α, β ) mi( α, β ) mi( α, β) ab = p p p = = 1 divides c, we mea Example 34: Fid the greatest commo divisor of 1260 ad 600 Solutio: The prime factorizatio of these umbers is give by = ( 2) ( 3) ( 5)( 7) ad = ( 2) ( 3)( 5) So the combied prime factors are 2, 3, 5 ad 7 Notice that these are all factors of 1260 Also, 600 ca be writte usig these prime factors as The Theorem 33 implies = ( ) ( = ( ) ) ( ) = ( 2) ( 3) ( 5) ( 7) ( ) mi 2,3 mi 2,1 mi 2,1 mi 1,0 gcd 1260, = 60 Remark 35: The TI-83 fuctio gcd ca be used to fid the greatest commo divisor of two atural umbers This fuctio ca be foud by pressig MATH, selectig NUM, ad the scrollig dow to item 9 This will place gcd( o the calculator scree Typig 1260,600) ad pressig ENTER results i the scree shot below Gcd(1260,600) 60 How Do Calculators Compute The Greatest Commo Divisor? Calculators DO NOT use prime factorizatios to compute the greatest commo divisor of two atural umbers Istead, they use a very efficiet process called the Euclidea algorithm We give this process after the defiitio ad theorem below

3 Defiitio 36: Let ab, The iteger part of a/ b is the largest value c chose from { } 0,1, 2,3, so that c a/ b We deote the iteger part of a/ b by ipart(a/b) Remark 37: ipart is a built-i fuctio o the TI-83 You ca access this fuctio by pressig MATH ad selectig NUM The fuctio ipart is item 3 Theorem 38: Let ab, with a ad 0 r < b Proof: c ipart ( a/ b) > b, ad suppose c= ipart ( a/ b) The r = a cb = implies that a c k b = + where 0 k < 1 (otherwise c will ot be the largest iteger which is less tha or equal to a/ b) a c k a b( c k) bc bk a bc bk b = + = + = + = Sice 0 k < 1, we have; 0 bk < b Let r = bk The, r = a cb ad 0 r < b Remark 39: The value r i Theorem 38 is sometimes referred to as the remaider associated with the divisio a/ b, ad the iteger part c of a/ b is sometimes referred to as the quotiet associated with the divisio a/ b Example 310: The iteger part of 2/3 is 0, the iteger part of 9/4 is 2, ad the iteger part of 67/13 is 5 That is, ipart(2/3) = 0, ipart(9/4) = 2 ad ipart(67/13) = 5 Theorem 14 ca be realized with the equatios ad 1= 9 ( 2)( 4) 2= 67 ( 5)( 13) The Euclidea algorithm uses the Theorem 14 to fid the greatest commo divisor of two atural umbers This algorithm was origially proposed i Euclid s Elemets aroud 300 BC It is amazig that this process is still widely used i practice by high speed computig devices The process is give below Euclidea algorithm Let ab, with a > b r1 = a ipart( a/ b) b r = b ipart( b/ r) r r = r ipart( r / r ) r r = r ipart( r / r ) r Cotiue util the first time r = 0 i If i = 1 the gcd ( ab, ) = 1 Otherwise, gcd ( ab, ) ri 1 =

4 Theorem 311: Let ab, with a > b The Euclidea algorithm computes gcd ( ab, ) Proof: Let ab, with a b gcd ab, Suppose the remaider of the divisio of a by b is c The a= qb+ c, where q is the quotiet of the divisio Ay commo divisor of a ad b also divides c (sice c ca be writte as c= a qb); similarly ay commo divisor of b ad c will also divide a Thus, the greatest commo divisor of a ad b is the same as the greatest commo divisor of b ad c Therefore, it is eough if we cotiue the process with umbers b ad c Sice c is smaller i absolute value tha b, we will reach c = 0 after fiitely may steps > We are lookig for ( ) Example 312: Use the Euclidea algorithm to compute gcd ( 1260,600 ) Solutio: You ca verify that r1 = 1260 ipart ( 1260 / 600)( 600) = 1260 ( 2)( 600) = 60 r = 600 ipart 600 / = = 0 So, gcd ( 1260,600) = 60 2 Example 313: Use the Euclidea algorithm to compute gcd( 860,1375 ) Solutio: The algorithm results i r1 = 1375 ( 1)( 860) = 515 r2 = 860 ()( 1 515) = 345 r3 = 515 () = 170 r4 = 345 ( 2)( 170) = 5 r = = 0 Cosequetly, gcd( 860,1375) = 5 5 Example 314: Use the Euclidea algorithm to compute gcd( 1326,741 ) Solutio: You ca verify that ()( ) ()( ) r = = 585 r = = 156 r = = 117 r = = 39 r = = 0

5 Cosequetly, gcd( 1326,741) = 39 A quatity which is closely related to the greatest commo divisor is the least commo multiple Defiitio 315: Let ab, The least commo multiple of a ad b is the smallest atural umber which is a multiple of both a ad b The least commo multiple of a ad lcm ab, b is deoted by ( ) Remark 316: lcm ( ab, ) abwheever ab, Sice the lcm ( ab, ) is the smallest umber that is a multiple of both a ad b, it is also the smallest umber for which both a ad b are divisors As a result, all of the prime factors of a must divide lcm ( ab, ), ad all of the prime factors of b must divide lcm ( ab, ) This leads to the followig theorem Theorem 317: Let ab,, ad suppose the prime factorizatios of a ad b are give by α a= ( p1) ( p2 ) ( p ) ad β1 β2 β b= ( p1) ( p2 ) ( p ) where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) lcm ( ab, ) = ( p) ( p) ( p ) Proof: Let c= lcm( a, b) If a ad b are both multiples of p, the c is a multiple of too If a is a multiple of bigger of p α max( α, β ) p of p Usig ad p α p β ad b is a multiple of p β, the c is a multiple of the ( sice c is a multiple of both a ad b)hece, c is a multiple of Note that, if a is ot a multiple of p but b is; the c should be a multiple p max(0, β ) β max( α, β ) max( α, β ) max( α, β) = p solves this problem Thus, 1 2 lcm ab, = p p p Example 318: Use Theorem 17 to compute lcm ( 24,15 ) 3 Solutio: We ca see that 24 = ( 2) ( 3) ad 15 ( 3)( 5) commo form as = ( 2) ( 3)( 5) = These ca be writte i p

6 ad As a result, 0 15 = ( 2) ( 3)( 5) ( ( ) = ( ) ) ( ) = ( 2) ( 3) ( 5) ( ) max 3,0 max 1,1 max 0,1 lcm 24, = 120 Example 319: We show the value of lcm ( ab, ), gcd (, ) of ab, ab ad ab for several choices a, b lcm(a,b) gcd(a,b) ab 16, , , , Although it might ot be readily apparet, the table above gives some isight to the lcm, gcd ab, ad ab The table above has bee relatioship betwee ( ab ), ( ) reproduced below, alog with a additioal colum cotaiig the product of lcm ( ab, ) ad gcd ( ab, ) a, b lcm(a,b) gcd(a,b) ab lcm(a,b)gcd(a,b) 16, , , , For the choices of a ad b above, it appears that lcm ( ab, ) gcd ( ab, ) = ab This result is always true Theorem 320: Let, ab The lcm (, ) gcd (, ) ab ab = ab α Proof: Let = ( ) ad ( ) β β β a p1 p2 p b= p1 p2 p where the values α, α,, α, β, β,, β are either atural umbers or 0 The ad 1 2 ( ) = ( ) max( α, β ) max( α, β ) max( α, β) lcm ab, p p p

7 1 2 mi( α, β ) mi( α, β ) mi( α, β) gcd ab, = p p p max( 1, 1) max( 2, 2) mi( 1, 1) mi( 2, 2) ( ab) ( ab) ( p) α β ( p ) α β ( p ) ( p) α β ( p ) α β ( p ) max( α, β) mi( α, β) lcm, gcd, = max( α1, β1) + mi( α1, β1) max( α2, β2) + mi( α2, β 2) max( α, β) + mi( α, β) = ( p1) ( p2) ( p ) If max( α1, β1) = α1 the mi( α1, β1) = β1 Ad if mi( α1, β1) = α1, the max( α1, β1) = β1 Hece, i ay case max( α1, β1) + mi( α1, β1) = α1+ β1 So we have; Exercises α1+ β1 α2+ β2 α+ β ( ab) ( ab) = ( p1) ( p2) ( p ) ( p ) ( p ) ( p ) ( p ) ( p ) ( p ) lcm, gcd, α α α β β β = = ab 1 Use the prime factorizatios of 860 ad 1375 to compute gcd( 860,1375 ) 2 Use the prime factorizatios of 6300 ad 1584 to compute gcd ( 6300,1584 ) 3 Use the prime factorizatios of 1260 ad 2640 to compute gcd( 1260, 2640 ) 4 Use the prime factorizatios of 2373 ad 1374 to compute gcd( 2373,1374 ) 5 Use the divisio algorithm to compute gcd ( 6300,1584 ) 6 Use the divisio algorithm to compute gcd( 1260, 2640 ) 7 Use the divisio algorithm to compute gcd( 2373,1374 ) 8 Use the prime factorizatios of 75 ad 124 to fid lcm( 75,124 ) 9 Use the prime factorizatios of 236 ad 125 to fid lcm(236,125) 10 Use the prime factorizatios of 84 ad 118 to fid lcm ( 84,118 ) ab = ad lcm ( ab, ) = 300 Give gcd ( ab, ) 11 Suppose Let mab,, Show that gcd ( ma, mb) = m gcd ( a, b) ; ie the greatest commo divisor satisfies a distributive property 13 Let abc,, Show that gcd ( gcd( ab, ), c) = gcd ( a,gcd( bc, )) ; ie the greatest commo divisor satisfies a associative property 14 Let a Show that gcd( aa, ) = a; ie the greatest commo divisor is idempotet 15 Let, gcd( ab, ) = gcd ba, ; ie the greatest commo divisor ab Show that ( ) satisfies a commutative property 16 Let ab, Show that lcm( a,gcd( a, b)) = a ad gcd( a,lcm( a, b)) = a 17 Let a Show that lcm( aa, ) = a; ie the least commo multiple is idempotet 18 Let ab, Show that lcm( ab, ) = lcm ( ba, ) ; ie the least commo multiple satisfies a commutative property

8 19 Let,, abc Show that lcm( lcm( ab, ), c) lcm ( a,lcm( bc, )) = ; ie the least commo multiple satisfies a associative property 20 Let,, lcm ma, mb = mlcm a, b ; ie the least commo mab Show that multiple satisfies a distributive property Solutios: = ad 1375 = 5 11 Therefore; mi(0,2) mi(1,3) mi(0,1) mi(0,1) gcd(860,1375) = = ( 2) ( 5)( 11) ( 43) = = ad 1584 = Therefore; ( ) = ( ) ( ) ( ) ( ) ( ) mi(2,4) mi(0,2) mi(0,1) mi(0,1) gcd 6300, = = = ad 2640 = Therefore; ( ) = ( ) ( ) mi(2,4) mi(0,1) mi(0,1) mi(0,1) mi(0,1) gcd 1260, = 2 5 = = ad 1374 = Therefore; ( ) ( ) gcd 2373,1374 = = 3 5 gcd ( 6300,1584 ) 3 ()( ) ( ) r1 = = 1548 r2 = = 36 r = = 0

9 Thus; gcd( 6300,1584) = 36 6 r1 r2 r3 = 120 ( 2) 60 = 0 Cosequetly, gcd ( 1260, 2640) = 60 = = 120 = = 60 7 Hece, gcd( 2373,1374) = 3 7 ()( ) ( ) ()( ) ()( ) ()( ) r1 = = 999 r2 = = 375 r3 = = 249 r4 = = 126 r5 = r6 = = 123 = 3 r = = = 3 5 ad 124 = 2 31 Thus, ( ) = ( ) ( 2) ( 3) ( 5) ( 31) max 2,0 max 1,0 max 2,0 max(1,0) lcm 75, = = = 2 59 ad 125 = 5 Hece, ( ) = ( ) ( 2) ( 5) ( 59) max 2,0 max 3,0 max(1,0) lcm 236, = = = ad 118 = 2 59 Therefore,

10 ( ) = ( ) ( 2) ( 3) ( 7) ( 59) max 2,1 max 1,0 max 1,0 max(1,0) lcm 84, = = Suppose 900 So, ( ab) ab = ad lcm ( ab, ) = 300 We kow that lcm (, ) gcd (, ) ab 900 gcd, = = = 3 The greatest commo divisor is: lcm, 300 gcd ( ab, ) = 3 ( ab) ab ab = ab 12 Let mab,, α Suppose p1, p2,, p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad b= ( p1) ( p2 ) ( p ), where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p) ( p) ( p ) α We have: = ( ) ( ) ad ( ) ( ) ma m p1 p2 p So, the greatest commo divisor is: 1 2 ( ma mb) = ( m) ( p1) ( p2) ( p ) mi( α1, β1) mi( α2, β2) mi( α, β) = m( p ) ( p ) ( p ) mi(1,1) mi( α, β ) mi( α, β ) mi( α, β) gcd, = mgcd( a, b) β β β mb = m p1 p2 p Hece, the greatest commo divisor satisfies a distributive property 13 Let abc,, Suppose 1, 2,, α α α p p p are prime umbers so that a ( p1) ( p2 ) ( p ) β1 β2 β γ1 γ2 γ = ( ) ad ( ) =, b p1 p2 p c= p1 p2 p, where the values α, α,, α, β, β,, β, γ, γ,, γ are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) ( ab) ( p1) ( p2) ( p ) mi( β1, γ1) mi( β2, γ2) mi( β, γ) ( bc) ( p1) ( p2) ( p ) gcd, = ad gcd, = mi(mi( α, β ), γ ) mi(mi( α, β ), γ ) mi(mi( α, β), γ) gcd(gcd ab,, c) = p p p ad

11 ( a ( b c) ) ( p1) ( p2) ( p ) mi( α,mi( β, γ )) mi( α,mi( β, γ )) mi( α,mi( β, γ)) gcd,gcd, = Here, it is importat to observe that mi(mi( α, β), γ) = mi( α, mi( β, γ)) = mi( α, β, γ) The reaso behid this is; while comparig 3 atural umbers, you ca start comparig from whichever you wat If you wat to fid the smallest of 5, 9 ad 6, the arrage them i order; 5 < 6 < 9 The smallest is 5, this does ot chage So, it is ot importat if you compare 5 & 9 first ad the the result with 6; or if you compare 6 & 9 first ad the the result with 5 Therefore, gcd ( gcd( ab, ), c) gcd ( a,gcd( bc, )) = ; ie the greatest commo divisor satisfies a associative property 14 Let a Show that gcd( aa, ) = a; ie the greatest commo divisor is idempotet Suppose p1, p2,, p are prime umbers so that a= ( p1) ( p2 ) ( p ) where the values α1, α2,, α are either atural umbers or 0 The mi( α1, α1) mi( α2, α2) mi( α, α) gcd ( aa, ) = ( p1) ( p2) ( p ) where, of course, mi( α, α ) = α Hece, 1 2 ( aa) = ( p1) ( p2) ( p ) α = ( p ) ( p ) ( p ) = a mi( α, α ) mi( α, α ) mi( α, α) gcd, α α α It is easier to see this result if we restate the questio as what is the greatest umber that divides both a ad a? The aswer is, of course, a itself 15 Let, gcd( ab, ) = gcd ba, ; ie the greatest commo divisor satisfies a commutative property α Suppose p1, p2,, p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad b= ( p1) ( p2 ) ( p ), where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p1) ( p2) ( p ) ad mi( β1, α1) mi( β2, α2) mi( β, α) gcd ( ba, ) = ( p1) ( p2) ( p ) The coclusio follows from the observatio that mi( α, β) = mi( β, α) gcd( ab, ) = gcd ba, ab Show that ( ) Hece, ( )

12 16 Let ab, Suppose 1, 2,, α α α p p p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad ( ) b= p1 p2 p, where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) lcm ( ab, ) = ( p1) ( p2) ( p ) ad mi( α1, β1) mi( α2, β2) mi( α, β) gcd ( ab, ) = ( p) ( p) ( p ) ( ) max( α,mi( α, β )) max( α,mi( α, β )) max( α,mi( α, β)) lcm( a,gcd( a, b)) = p p p Let s fid max( α1, mi( α1, β 1)) If α1 < β1, the mi( α1, β1) = α1 ad max( α1, mi( α1, β1)) = max( α1, α1) = α1 If β1 < α1, the mi( α1, β1) = β1 ad max( α1, mi( α1, β1)) = max( α1, β1) = α1 Similarly, for ay, max( α, mi( α, β )) = α Therefore; ( 1) ( 2) ( ) α ( p ) ( p ) ( p ) a max( α,mi( α, β )) max( α,mi( α, β )) max( α,mi( α, β)) lcm( a,gcd( a, b)) = p p p = = ( ) mi( α,max( α, β )) mi( α,max( α, β )) mi( α,max( α, β)) gcd( a,lcm( a, b)) = p p p With the reasoig we used above we ca coclude that mi( α, max( α, β )) = α Thus, ( 1) ( 2) ( ) α ( p ) ( p ) ( p ) a mi( α,max( α, β )) mi( α,max( α, β )) mi( α,max( α, β)) gcd( a,lcm( a, b)) = p p p = = Hece, lcm( a,gcd( a, b)) = a ad gcd( a,lcm( a, b)) = a 17 Let a Suppose p1, p2,, p are prime umbers so that ( ) α α α a= p1 p2 p where the values α1, α2,, α are either atural umbers or 0 The 1 2 ( aa) = ( p1) ( p2) ( p ) α = ( p ) ( p ) ( p ) = a max( α, α ) max( α, α ) max( α, α) lcm,

13 Therefore, lcm( aa, ) = a; ie the least commo multiple is idempotet 18 Let ab, Suppose p1, p2,, p are prime umbers so that α = ( ) ad ( ) β β β a p1 p2 p b= p1 p2 p, where the values α, α,, α, β, β,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) ( ab) = ( p1) ( p2) ( p ) ad max( β1, α1) max( β2, α2) max( β, α) ( ba) = ( p1) ( p2) ( p ) Sice =, we ca coclude that lcm( ab, ) lcm ( ba, ) lcm, lcm, max( α, β ) max( β, α ) = 19 Let abc,, Suppose p1, p2,, p are prime umbers so that α = ( 1) ( 2 ) ( ), b ( p1) ( p2 ) ( p ) γ1 γ2 γ ( ) a p p p β β β = ad c= p1 p2 p, where the values α1, α2,, α, β1, β2,, β, γ1, γ2,, γ are either atural umbers or 0 The 1 2 ( ab) ( p1) ( p2) ( p ) 1 2 max( α, β ) max( α, β ) max( α, β) lcm, = ad max( β, γ ) max( β, γ ) max( β, γ) lcm bc, = p p p max(max( α, β ), γ ) max(max( α, β ), γ ) max(max( α, β), γ) lcm(lcm ab,, c) = p p p ad ( a ( b c) ) ( p1) ( p2) ( p ) max( α,max( β, γ )) max( α,max( β, γ )) max( α,max( β, γ)) lcm,lcm, = As we have poited out before, the order at which you start to compare atural umbers does ot matter; max(max( α, β ), γ ) = max( α, max( β, γ )) = max( α, β, γ ) Hece, lcm( lcm( ab, ), c) lcm ( a,lcm( bc, )) satisfies a associative property = ; ie the least commo multiple 20 Let mab,, Suppose 1, 2,, α α α p p p are prime umbers so that a= ( p1) ( p2 ) ( p ) β1 β2 β ad ( ) b= p1 p2 p, where the values α1, α2,, α, β1, β2,, β are either atural umbers or 0 The max( α1, β1) max( α2, β2) max( α, β) lcm ( ab, ) = ( p) ( p) ( p )

14 α We have: = ( ) ( ) ad ( ) ( ) ma m p1 p2 p So, the least commo multiple is: 1 2 ( ma mb) = ( m) ( p ) ( p ) ( p ) 1 2 ( ) β β β mb = m p1 p2 p max(1,1) max( α, β ) max( α, β ) max( α, β) lcm, max( α, β ) max( α, β ) max( α, β) = m p p p = mlcm( a, b) Thus, lcm ( ma, mb) mlcm ( a, b) distributive property = ; ie the least commo multiple satisfies a

15 Graphical Represetatio Of gcd(a,b) = 1 Numbers ab, are said to be relatively prime if ad oly if gcd( ab, ) = 1 It is possible to visualize all of the pairs ( ab, ) for which a ad b are relatively prime by plottig the poits ( ab, ) for a ad b betwee 1 ad 100 We give this plot below, ad the patter is very iterestig! Surprisigly, there are more pairs i this rage that are relatively prime tha those that are ot I fact, of the 10,000 pairs ( ab, ) for a ad b betwee 1 ad 100, there are 6,087 pairs which are relatively prime ( ab, ) for which gcd( ab, ) = 1 for ab, { 1,,100}

### 1. MATHEMATICAL INDUCTION

1. MATHEMATICAL INDUCTION EXAMPLE 1: Prove that for ay iteger 1. Proof: 1 + 2 + 3 +... + ( + 1 2 (1.1 STEP 1: For 1 (1.1 is true, sice 1 1(1 + 1. 2 STEP 2: Suppose (1.1 is true for some k 1, that is 1

More information

### Repeating Decimals are decimal numbers that have number(s) after the decimal point that repeat in a pattern.

5.5 Fractios ad Decimals Steps for Chagig a Fractio to a Decimal. Simplify the fractio, if possible. 2. Divide the umerator by the deomiator. d d Repeatig Decimals Repeatig Decimals are decimal umbers

More information

### In nite Sequences. Dr. Philippe B. Laval Kennesaw State University. October 9, 2008

I ite Sequeces Dr. Philippe B. Laval Keesaw State Uiversity October 9, 2008 Abstract This had out is a itroductio to i ite sequeces. mai de itios ad presets some elemetary results. It gives the I ite Sequeces

More information

### Example 2 Find the square root of 0. The only square root of 0 is 0 (since 0 is not positive or negative, so those choices don t exist here).

BEGINNING ALGEBRA Roots ad Radicals (revised summer, 00 Olso) Packet to Supplemet the Curret Textbook - Part Review of Square Roots & Irratioals (This portio ca be ay time before Part ad should mostly

More information

### Chapter 5: Inner Product Spaces

Chapter 5: Ier Product Spaces Chapter 5: Ier Product Spaces SECION A Itroductio to Ier Product Spaces By the ed of this sectio you will be able to uderstad what is meat by a ier product space give examples

More information

### SAMPLE QUESTIONS FOR FINAL EXAM. (1) (2) (3) (4) Find the following using the definition of the Riemann integral: (2x + 1)dx

SAMPLE QUESTIONS FOR FINAL EXAM REAL ANALYSIS I FALL 006 3 4 Fid the followig usig the defiitio of the Riema itegral: a 0 x + dx 3 Cosider the partitio P x 0 3, x 3 +, x 3 +,......, x 3 3 + 3 of the iterval

More information

### 1 Correlation and Regression Analysis

1 Correlatio ad Regressio Aalysis I this sectio we will be ivestigatig the relatioship betwee two cotiuous variable, such as height ad weight, the cocetratio of a ijected drug ad heart rate, or the cosumptio

More information

### CS103X: Discrete Structures Homework 4 Solutions

CS103X: Discrete Structures Homewor 4 Solutios Due February 22, 2008 Exercise 1 10 poits. Silico Valley questios: a How may possible six-figure salaries i whole dollar amouts are there that cotai at least

More information

### Infinite Sequences and Series

CHAPTER 4 Ifiite Sequeces ad Series 4.1. Sequeces A sequece is a ifiite ordered list of umbers, for example the sequece of odd positive itegers: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29...

More information

### Trigonometric Form of a Complex Number. The Complex Plane. axis. ( 2, 1) or 2 i FIGURE 6.44. The absolute value of the complex number z a bi is

0_0605.qxd /5/05 0:45 AM Page 470 470 Chapter 6 Additioal Topics i Trigoometry 6.5 Trigoometric Form of a Complex Number What you should lear Plot complex umbers i the complex plae ad fid absolute values

More information

### 2-3 The Remainder and Factor Theorems

- The Remaider ad Factor Theorems Factor each polyomial completely usig the give factor ad log divisio 1 x + x x 60; x + So, x + x x 60 = (x + )(x x 15) Factorig the quadratic expressio yields x + x x

More information

Chapter 5 O A Cojecture Of Erdíos Proceedigs NCUR VIII è1994è, Vol II, pp 794í798 Jeærey F Gold Departmet of Mathematics, Departmet of Physics Uiversity of Utah Do H Tucker Departmet of Mathematics Uiversity

More information

### Our aim is to show that under reasonable assumptions a given 2π-periodic function f can be represented as convergent series

8 Fourier Series Our aim is to show that uder reasoable assumptios a give -periodic fuctio f ca be represeted as coverget series f(x) = a + (a cos x + b si x). (8.) By defiitio, the covergece of the series

More information

### A probabilistic proof of a binomial identity

A probabilistic proof of a biomial idetity Joatho Peterso Abstract We give a elemetary probabilistic proof of a biomial idetity. The proof is obtaied by computig the probability of a certai evet i two

More information

### 0.7 0.6 0.2 0 0 96 96.5 97 97.5 98 98.5 99 99.5 100 100.5 96.5 97 97.5 98 98.5 99 99.5 100 100.5

Sectio 13 Kolmogorov-Smirov test. Suppose that we have a i.i.d. sample X 1,..., X with some ukow distributio P ad we would like to test the hypothesis that P is equal to a particular distributio P 0, i.e.

More information

### Chapter 7 Methods of Finding Estimators

Chapter 7 for BST 695: Special Topics i Statistical Theory. Kui Zhag, 011 Chapter 7 Methods of Fidig Estimators Sectio 7.1 Itroductio Defiitio 7.1.1 A poit estimator is ay fuctio W( X) W( X1, X,, X ) of

More information

### Factoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett <garrett@math.umn.edu>

(March 16, 004) Factorig x 1: cyclotomic ad Aurifeuillia polyomials Paul Garrett Polyomials of the form x 1, x 3 1, x 4 1 have at least oe systematic factorizatio x 1 = (x 1)(x 1

More information

### Sequences and Series

CHAPTER 9 Sequeces ad Series 9.. Covergece: Defiitio ad Examples Sequeces The purpose of this chapter is to itroduce a particular way of geeratig algorithms for fidig the values of fuctios defied by their

More information

### FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. 1. Powers of a matrix

FIBONACCI NUMBERS: AN APPLICATION OF LINEAR ALGEBRA. Powers of a matrix We begi with a propositio which illustrates the usefuless of the diagoalizatio. Recall that a square matrix A is diogaalizable if

More information

### WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER?

WHEN IS THE (CO)SINE OF A RATIONAL ANGLE EQUAL TO A RATIONAL NUMBER? JÖRG JAHNEL 1. My Motivatio Some Sort of a Itroductio Last term I tought Topological Groups at the Göttige Georg August Uiversity. This

More information

### Measures of Spread and Boxplots Discrete Math, Section 9.4

Measures of Spread ad Boxplots Discrete Math, Sectio 9.4 We start with a example: Example 1: Comparig Mea ad Media Compute the mea ad media of each data set: S 1 = {4, 6, 8, 10, 1, 14, 16} S = {4, 7, 9,

More information

### Convexity, Inequalities, and Norms

Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for

More information

### The following example will help us understand The Sampling Distribution of the Mean. C1 C2 C3 C4 C5 50 miles 84 miles 38 miles 120 miles 48 miles

The followig eample will help us uderstad The Samplig Distributio of the Mea Review: The populatio is the etire collectio of all idividuals or objects of iterest The sample is the portio of the populatio

More information

### Chapter 6: Variance, the law of large numbers and the Monte-Carlo method

Chapter 6: Variace, the law of large umbers ad the Mote-Carlo method Expected value, variace, ad Chebyshev iequality. If X is a radom variable recall that the expected value of X, E[X] is the average value

More information

### University of California, Los Angeles Department of Statistics. Distributions related to the normal distribution

Uiversity of Califoria, Los Ageles Departmet of Statistics Statistics 100B Istructor: Nicolas Christou Three importat distributios: Distributios related to the ormal distributio Chi-square (χ ) distributio.

More information

### SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES

SECTION 1.5 : SUMMATION NOTATION + WORK WITH SEQUENCES Read Sectio 1.5 (pages 5 9) Overview I Sectio 1.5 we lear to work with summatio otatio ad formulas. We will also itroduce a brief overview of sequeces,

More information

### Hypothesis testing. Null and alternative hypotheses

Hypothesis testig Aother importat use of samplig distributios is to test hypotheses about populatio parameters, e.g. mea, proportio, regressio coefficiets, etc. For example, it is possible to stipulate

More information

### 1. C. The formula for the confidence interval for a population mean is: x t, which was

s 1. C. The formula for the cofidece iterval for a populatio mea is: x t, which was based o the sample Mea. So, x is guarateed to be i the iterval you form.. D. Use the rule : p-value

More information

### Solving Logarithms and Exponential Equations

Solvig Logarithms ad Epoetial Equatios Logarithmic Equatios There are two major ideas required whe solvig Logarithmic Equatios. The first is the Defiitio of a Logarithm. You may recall from a earlier topic:

More information

### .04. This means \$1000 is multiplied by 1.02 five times, once for each of the remaining sixmonth

Questio 1: What is a ordiary auity? Let s look at a ordiary auity that is certai ad simple. By this, we mea a auity over a fixed term whose paymet period matches the iterest coversio period. Additioally,

More information

### Lecture 4: Cauchy sequences, Bolzano-Weierstrass, and the Squeeze theorem

Lecture 4: Cauchy sequeces, Bolzao-Weierstrass, ad the Squeeze theorem The purpose of this lecture is more modest tha the previous oes. It is to state certai coditios uder which we are guarateed that limits

More information

### 7.1 Finding Rational Solutions of Polynomial Equations

4 Locker LESSON 7. Fidig Ratioal Solutios of Polyomial Equatios Name Class Date 7. Fidig Ratioal Solutios of Polyomial Equatios Essetial Questio: How do you fid the ratioal roots of a polyomial equatio?

More information

### On Formula to Compute Primes. and the n th Prime

Applied Mathematical cieces, Vol., 0, o., 35-35 O Formula to Compute Primes ad the th Prime Issam Kaddoura Lebaese Iteratioal Uiversity Faculty of Arts ad cieces, Lebao issam.kaddoura@liu.edu.lb amih Abdul-Nabi

More information

### THE ARITHMETIC OF INTEGERS. - multiplication, exponentiation, division, addition, and subtraction

THE ARITHMETIC OF INTEGERS - multiplicatio, expoetiatio, divisio, additio, ad subtractio What to do ad what ot to do. THE INTEGERS Recall that a iteger is oe of the whole umbers, which may be either positive,

More information

### 1 Computing the Standard Deviation of Sample Means

Computig the Stadard Deviatio of Sample Meas Quality cotrol charts are based o sample meas ot o idividual values withi a sample. A sample is a group of items, which are cosidered all together for our aalysis.

More information

### Properties of MLE: consistency, asymptotic normality. Fisher information.

Lecture 3 Properties of MLE: cosistecy, asymptotic ormality. Fisher iformatio. I this sectio we will try to uderstad why MLEs are good. Let us recall two facts from probability that we be used ofte throughout

More information

### Elementary Theory of Russian Roulette

Elemetary Theory of Russia Roulette -iterestig patters of fractios- Satoshi Hashiba Daisuke Miematsu Ryohei Miyadera Itroductio. Today we are goig to study mathematical theory of Russia roulette. If some

More information

### I. Chi-squared Distributions

1 M 358K Supplemet to Chapter 23: CHI-SQUARED DISTRIBUTIONS, T-DISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad t-distributios, we first eed to look at aother family of distributios, the chi-squared distributios.

More information

### Asymptotic Growth of Functions

CMPS Itroductio to Aalysis of Algorithms Fall 3 Asymptotic Growth of Fuctios We itroduce several types of asymptotic otatio which are used to compare the performace ad efficiecy of algorithms As we ll

More information

### Basic Elements of Arithmetic Sequences and Series

MA40S PRE-CALCULUS UNIT G GEOMETRIC SEQUENCES CLASS NOTES (COMPLETED NO NEED TO COPY NOTES FROM OVERHEAD) Basic Elemets of Arithmetic Sequeces ad Series Objective: To establish basic elemets of arithmetic

More information

### Soving Recurrence Relations

Sovig Recurrece Relatios Part 1. Homogeeous liear 2d degree relatios with costat coefficiets. Cosider the recurrece relatio ( ) T () + at ( 1) + bt ( 2) = 0 This is called a homogeeous liear 2d degree

More information

### Discrete Mathematics and Probability Theory Spring 2014 Anant Sahai Note 13

EECS 70 Discrete Mathematics ad Probability Theory Sprig 2014 Aat Sahai Note 13 Itroductio At this poit, we have see eough examples that it is worth just takig stock of our model of probability ad may

More information

### Section 11.3: The Integral Test

Sectio.3: The Itegral Test Most of the series we have looked at have either diverged or have coverged ad we have bee able to fid what they coverge to. I geeral however, the problem is much more difficult

More information

### 4.3. The Integral and Comparison Tests

4.3. THE INTEGRAL AND COMPARISON TESTS 9 4.3. The Itegral ad Compariso Tests 4.3.. The Itegral Test. Suppose f is a cotiuous, positive, decreasig fuctio o [, ), ad let a = f(). The the covergece or divergece

More information

### Output Analysis (2, Chapters 10 &11 Law)

B. Maddah ENMG 6 Simulatio 05/0/07 Output Aalysis (, Chapters 10 &11 Law) Comparig alterative system cofiguratio Sice the output of a simulatio is radom, the comparig differet systems via simulatio should

More information

### Maximum Likelihood Estimators.

Lecture 2 Maximum Likelihood Estimators. Matlab example. As a motivatio, let us look at oe Matlab example. Let us geerate a radom sample of size 00 from beta distributio Beta(5, 2). We will lear the defiitio

More information

### Chapter 14 Nonparametric Statistics

Chapter 14 Noparametric Statistics A.K.A. distributio-free statistics! Does ot deped o the populatio fittig ay particular type of distributio (e.g, ormal). Sice these methods make fewer assumptios, they

More information

### Factors of sums of powers of binomial coefficients

ACTA ARITHMETICA LXXXVI.1 (1998) Factors of sums of powers of biomial coefficiets by Neil J. Cali (Clemso, S.C.) Dedicated to the memory of Paul Erdős 1. Itroductio. It is well ow that if ( ) a f,a = the

More information

### Descriptive Statistics

Descriptive Statistics We leared to describe data sets graphically. We ca also describe a data set umerically. Measures of Locatio Defiitio The sample mea is the arithmetic average of values. We deote

More information

### THE LEAST COMMON MULTIPLE OF A QUADRATIC SEQUENCE

THE LEAST COMMON MULTIPLE OF A QUADRATIC SEQUENCE JAVIER CILLERUELO Abstract. We obtai, for ay irreducible quadratic olyomial f(x = ax 2 + bx + c, the asymtotic estimate log l.c.m. {f(1,..., f(} log. Whe

More information

### Sequences and Series Using the TI-89 Calculator

RIT Calculator Site Sequeces ad Series Usig the TI-89 Calculator Norecursively Defied Sequeces A orecursively defied sequece is oe i which the formula for the terms of the sequece is give explicitly. For

More information

### A Note on Sums of Greatest (Least) Prime Factors

It. J. Cotemp. Math. Scieces, Vol. 8, 203, o. 9, 423-432 HIKARI Ltd, www.m-hikari.com A Note o Sums of Greatest (Least Prime Factors Rafael Jakimczuk Divisio Matemática, Uiversidad Nacioal de Luá Bueos

More information

### Section 8.3 : De Moivre s Theorem and Applications

The Sectio 8 : De Moivre s Theorem ad Applicatios Let z 1 ad z be complex umbers, where z 1 = r 1, z = r, arg(z 1 ) = θ 1, arg(z ) = θ z 1 = r 1 (cos θ 1 + i si θ 1 ) z = r (cos θ + i si θ ) ad z 1 z =

More information

### Lecture 13. Lecturer: Jonathan Kelner Scribe: Jonathan Pines (2009)

18.409 A Algorithmist s Toolkit October 27, 2009 Lecture 13 Lecturer: Joatha Keler Scribe: Joatha Pies (2009) 1 Outlie Last time, we proved the Bru-Mikowski iequality for boxes. Today we ll go over the

More information

### INFINITE SERIES KEITH CONRAD

INFINITE SERIES KEITH CONRAD. Itroductio The two basic cocepts of calculus, differetiatio ad itegratio, are defied i terms of limits (Newto quotiets ad Riema sums). I additio to these is a third fudametal

More information

### Here are a couple of warnings to my students who may be here to get a copy of what happened on a day that you missed.

This documet was writte ad copyrighted by Paul Dawkis. Use of this documet ad its olie versio is govered by the Terms ad Coditios of Use located at http://tutorial.math.lamar.edu/terms.asp. The olie versio

More information

### CME 302: NUMERICAL LINEAR ALGEBRA FALL 2005/06 LECTURE 8

CME 30: NUMERICAL LINEAR ALGEBRA FALL 005/06 LECTURE 8 GENE H GOLUB 1 Positive Defiite Matrices A matrix A is positive defiite if x Ax > 0 for all ozero x A positive defiite matrix has real ad positive

More information

### Lecture 5: Span, linear independence, bases, and dimension

Lecture 5: Spa, liear idepedece, bases, ad dimesio Travis Schedler Thurs, Sep 23, 2010 (versio: 9/21 9:55 PM) 1 Motivatio Motivatio To uderstad what it meas that R has dimesio oe, R 2 dimesio 2, etc.;

More information

### EGYPTIAN FRACTION EXPANSIONS FOR RATIONAL NUMBERS BETWEEN 0 AND 1 OBTAINED WITH ENGEL SERIES

EGYPTIAN FRACTION EXPANSIONS FOR RATIONAL NUMBERS BETWEEN 0 AND OBTAINED WITH ENGEL SERIES ELVIA NIDIA GONZÁLEZ AND JULIA BERGNER, PHD DEPARTMENT OF MATHEMATICS Abstract. The aciet Egyptias epressed ratioal

More information

### Chapter 7: Confidence Interval and Sample Size

Chapter 7: Cofidece Iterval ad Sample Size Learig Objectives Upo successful completio of Chapter 7, you will be able to: Fid the cofidece iterval for the mea, proportio, ad variace. Determie the miimum

More information

### Lesson 17 Pearson s Correlation Coefficient

Outlie Measures of Relatioships Pearso s Correlatio Coefficiet (r) -types of data -scatter plots -measure of directio -measure of stregth Computatio -covariatio of X ad Y -uique variatio i X ad Y -measurig

More information

### Case Study. Normal and t Distributions. Density Plot. Normal Distributions

Case Study Normal ad t Distributios Bret Halo ad Bret Larget Departmet of Statistics Uiversity of Wiscosi Madiso October 11 13, 2011 Case Study Body temperature varies withi idividuals over time (it ca

More information

### Approximating Area under a curve with rectangles. To find the area under a curve we approximate the area using rectangles and then use limits to find

1.8 Approximatig Area uder a curve with rectagles 1.6 To fid the area uder a curve we approximate the area usig rectagles ad the use limits to fid 1.4 the area. Example 1 Suppose we wat to estimate 1.

More information

### Math 114- Intermediate Algebra Integral Exponents & Fractional Exponents (10 )

Math 4 Math 4- Itermediate Algebra Itegral Epoets & Fractioal Epoets (0 ) Epoetial Fuctios Epoetial Fuctios ad Graphs I. Epoetial Fuctios The fuctio f ( ) a, where is a real umber, a 0, ad a, is called

More information

### CHAPTER 3 DIGITAL CODING OF SIGNALS

CHAPTER 3 DIGITAL CODING OF SIGNALS Computers are ofte used to automate the recordig of measuremets. The trasducers ad sigal coditioig circuits produce a voltage sigal that is proportioal to a quatity

More information

### Mann-Whitney U 2 Sample Test (a.k.a. Wilcoxon Rank Sum Test)

No-Parametric ivariate Statistics: Wilcoxo-Ma-Whitey 2 Sample Test 1 Ma-Whitey 2 Sample Test (a.k.a. Wilcoxo Rak Sum Test) The (Wilcoxo-) Ma-Whitey (WMW) test is the o-parametric equivalet of a pooled

More information

### Math C067 Sampling Distributions

Math C067 Samplig Distributios Sample Mea ad Sample Proportio Richard Beigel Some time betwee April 16, 2007 ad April 16, 2007 Examples of Samplig A pollster may try to estimate the proportio of voters

More information

### CHAPTER 3 THE TIME VALUE OF MONEY

CHAPTER 3 THE TIME VALUE OF MONEY OVERVIEW A dollar i the had today is worth more tha a dollar to be received i the future because, if you had it ow, you could ivest that dollar ad ear iterest. Of all

More information

### S. Tanny MAT 344 Spring 1999. be the minimum number of moves required.

S. Tay MAT 344 Sprig 999 Recurrece Relatios Tower of Haoi Let T be the miimum umber of moves required. T 0 = 0, T = 7 Iitial Coditios * T = T + \$ T is a sequece (f. o itegers). Solve for T? * is a recurrece,

More information

### THE REGRESSION MODEL IN MATRIX FORM. For simple linear regression, meaning one predictor, the model is. for i = 1, 2, 3,, n

We will cosider the liear regressio model i matrix form. For simple liear regressio, meaig oe predictor, the model is i = + x i + ε i for i =,,,, This model icludes the assumptio that the ε i s are a sample

More information

### SEQUENCES AND SERIES

Chapter 9 SEQUENCES AND SERIES Natural umbers are the product of huma spirit. DEDEKIND 9.1 Itroductio I mathematics, the word, sequece is used i much the same way as it is i ordiary Eglish. Whe we say

More information

### Department of Computer Science, University of Otago

Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS-2006-09 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly

More information

### Today s Topics. Primes & Greatest Common Divisors

Today s Topics Primes & Greatest Common Divisors Prime representations Important theorems about primality Greatest Common Divisors Least Common Multiples Euclid s algorithm Once and for all, what are prime

More information

### 5 Boolean Decision Trees (February 11)

5 Boolea Decisio Trees (February 11) 5.1 Graph Coectivity Suppose we are give a udirected graph G, represeted as a boolea adjacecy matrix = (a ij ), where a ij = 1 if ad oly if vertices i ad j are coected

More information

### 5.3. Generalized Permutations and Combinations

53 GENERALIZED PERMUTATIONS AND COMBINATIONS 73 53 Geeralized Permutatios ad Combiatios 53 Permutatios with Repeated Elemets Assume that we have a alphabet with letters ad we wat to write all possible

More information

### Theorems About Power Series

Physics 6A Witer 20 Theorems About Power Series Cosider a power series, f(x) = a x, () where the a are real coefficiets ad x is a real variable. There exists a real o-egative umber R, called the radius

More information

### . P. 4.3 Basic feasible solutions and vertices of polyhedra. x 1. x 2

4. Basic feasible solutios ad vertices of polyhedra Due to the fudametal theorem of Liear Programmig, to solve ay LP it suffices to cosider the vertices (fiitely may) of the polyhedro P of the feasible

More information

### Irreducible polynomials with consecutive zero coefficients

Irreducible polyomials with cosecutive zero coefficiets Theodoulos Garefalakis Departmet of Mathematics, Uiversity of Crete, 71409 Heraklio, Greece Abstract Let q be a prime power. We cosider the problem

More information

### GCSE STATISTICS. 4) How to calculate the range: The difference between the biggest number and the smallest number.

GCSE STATISTICS You should kow: 1) How to draw a frequecy diagram: e.g. NUMBER TALLY FREQUENCY 1 3 5 ) How to draw a bar chart, a pictogram, ad a pie chart. 3) How to use averages: a) Mea - add up all

More information

### Confidence Intervals. CI for a population mean (σ is known and n > 30 or the variable is normally distributed in the.

Cofidece Itervals A cofidece iterval is a iterval whose purpose is to estimate a parameter (a umber that could, i theory, be calculated from the populatio, if measuremets were available for the whole populatio).

More information

### Cooley-Tukey. Tukey FFT Algorithms. FFT Algorithms. Cooley

Cooley Cooley-Tuey Tuey FFT Algorithms FFT Algorithms Cosider a legth- sequece x[ with a -poit DFT X[ where Represet the idices ad as +, +, Cooley Cooley-Tuey Tuey FFT Algorithms FFT Algorithms Usig these

More information

### FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10

FOUNDATIONS OF MATHEMATICS AND PRE-CALCULUS GRADE 10 [C] Commuicatio Measuremet A1. Solve problems that ivolve liear measuremet, usig: SI ad imperial uits of measure estimatio strategies measuremet strategies.

More information

### Overview of some probability distributions.

Lecture Overview of some probability distributios. I this lecture we will review several commo distributios that will be used ofte throughtout the class. Each distributio is usually described by its probability

More information

### GCE Further Mathematics (6360) Further Pure Unit 2 (MFP2) Textbook. Version: 1.4

GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 4 MFP Tetbook A-level Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5

More information

### where: T = number of years of cash flow in investment's life n = the year in which the cash flow X n i = IRR = the internal rate of return

EVALUATING ALTERNATIVE CAPITAL INVESTMENT PROGRAMS By Ke D. Duft, Extesio Ecoomist I the March 98 issue of this publicatio we reviewed the procedure by which a capital ivestmet project was assessed. The

More information

### Running Time ( 3.1) Analysis of Algorithms. Experimental Studies ( 3.1.1) Limitations of Experiments. Pseudocode ( 3.1.2) Theoretical Analysis

Ruig Time ( 3.) Aalysis of Algorithms Iput Algorithm Output A algorithm is a step-by-step procedure for solvig a problem i a fiite amout of time. Most algorithms trasform iput objects ito output objects.

More information

### Mathematical goals. Starting points. Materials required. Time needed

Level A1 of challege: C A1 Mathematical goals Startig poits Materials required Time eeded Iterpretig algebraic expressios To help learers to: traslate betwee words, symbols, tables, ad area represetatios

More information

### Lecture 4: Cheeger s Inequality

Spectral Graph Theory ad Applicatios WS 0/0 Lecture 4: Cheeger s Iequality Lecturer: Thomas Sauerwald & He Su Statemet of Cheeger s Iequality I this lecture we assume for simplicity that G is a d-regular

More information

### How Euler Did It. In a more modern treatment, Hardy and Wright [H+W] state this same theorem as. n n+ is perfect.

Amicable umbers November 005 How Euler Did It by Ed Sadifer Six is a special umber. It is divisible by, ad 3, ad, i what at first looks like a strage coicidece, 6 = + + 3. The umber 8 shares this remarkable

More information

### Confidence Intervals

Cofidece Itervals Cofidece Itervals are a extesio of the cocept of Margi of Error which we met earlier i this course. Remember we saw: The sample proportio will differ from the populatio proportio by more

More information

### AP Calculus AB 2006 Scoring Guidelines Form B

AP Calculus AB 6 Scorig Guidelies Form B The College Board: Coectig Studets to College Success The College Board is a ot-for-profit membership associatio whose missio is to coect studets to college success

More information

### Week 3 Conditional probabilities, Bayes formula, WEEK 3 page 1 Expected value of a random variable

Week 3 Coditioal probabilities, Bayes formula, WEEK 3 page 1 Expected value of a radom variable We recall our discussio of 5 card poker hads. Example 13 : a) What is the probability of evet A that a 5

More information

### Incremental calculation of weighted mean and variance

Icremetal calculatio of weighted mea ad variace Toy Fich faf@cam.ac.uk dot@dotat.at Uiversity of Cambridge Computig Service February 009 Abstract I these otes I eplai how to derive formulae for umerically

More information

### Lesson 15 ANOVA (analysis of variance)

Outlie Variability -betwee group variability -withi group variability -total variability -F-ratio Computatio -sums of squares (betwee/withi/total -degrees of freedom (betwee/withi/total -mea square (betwee/withi

More information

### Confidence Intervals for One Mean

Chapter 420 Cofidece Itervals for Oe Mea Itroductio This routie calculates the sample size ecessary to achieve a specified distace from the mea to the cofidece limit(s) at a stated cofidece level for a

More information

### Bond Valuation I. What is a bond? Cash Flows of A Typical Bond. Bond Valuation. Coupon Rate and Current Yield. Cash Flows of A Typical Bond

What is a bod? Bod Valuatio I Bod is a I.O.U. Bod is a borrowig agreemet Bod issuers borrow moey from bod holders Bod is a fixed-icome security that typically pays periodic coupo paymets, ad a pricipal

More information

### Taking DCOP to the Real World: Efficient Complete Solutions for Distributed Multi-Event Scheduling

Taig DCOP to the Real World: Efficiet Complete Solutios for Distributed Multi-Evet Schedulig Rajiv T. Maheswara, Milid Tambe, Emma Bowrig, Joatha P. Pearce, ad Pradeep araatham Uiversity of Souther Califoria

More information

### *The most important feature of MRP as compared with ordinary inventory control analysis is its time phasing feature.

Itegrated Productio ad Ivetory Cotrol System MRP ad MRP II Framework of Maufacturig System Ivetory cotrol, productio schedulig, capacity plaig ad fiacial ad busiess decisios i a productio system are iterrelated.

More information

### Listing terms of a finite sequence List all of the terms of each finite sequence. a) a n n 2 for 1 n 5 1 b) a n for 1 n 4 n 2

74 (4 ) Chapter 4 Sequeces ad Series 4. SEQUENCES I this sectio Defiitio Fidig a Formula for the th Term The word sequece is a familiar word. We may speak of a sequece of evets or say that somethig is

More information