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


 Neil Watson
 2 years ago
 Views:
Transcription
1 GCE Further Mathematics (660) Further Pure Uit (MFP) Tetbook Versio: 4
2 MFP Tetbook Alevel Further Mathematics 660 Further Pure : Cotets Chapter : Comple umbers 4 Itroductio 5 The geeral comple umber 5 The modulus ad argumet of a comple umber 6 4 The polar form of a comple umber 8 5 Additio, subtractio ad multiplicatio of comple umbers of the form + iy 9 6 The cojugate of a comple umber ad the divisio of comple umbers of the form +iy 0 7 Products ad quotiets of comple umbers i their polar form 8 Equatig real ad imagiary parts 9 Further cosideratio of z z ad arg(z z ) 4 0 Loci o Argad diagrams 5 Chapter : Roots of polyomial equatios Itroductio Quadratic equatios Cubic equatios 4 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets 7 5 Cubic equatios with related roots 8 6 A importat result 7 Polyomial equatios of degree 8 Comple roots of polyomial equatios with real coefficiets Chapter : Summatio of fiite series 8 Itroductio 9 Summatio of series by the method of differeces 40 Summatio of series by the method of iductio 45 4 Proof by iductio eteded to other areas of mathematics 48 Chapter 4: De Moivre s theorem ad its applicatios 5 4 De Moivre s theorem 54 4 Usig de Moivre s theorem to evaluate powers of comple umbers 56 4 Applicatio of de Moivre s theorem i establishig trigoometric idetities Epoetial form of a comple umber The cube roots of uity 7 46 The th roots of uity The roots of z, where is a oreal umber 77 cotiued overleaf
3 MFP Tetbook Alevel Further Mathematics 660 Further Pure : Cotets (cotiued) Chapter 5: Iverse trigoometrical fuctios 85 5 Itroductio ad revisio 86 5 The derivatives of stadard iverse trigoometrical fuctios 89 5 Applicatio to more comple differetiatio 9 54 Stadard itegrals itegratig to iverse trigoometrical fuctios Applicatios to more comple itegrals 96 Chapter 6: Hyperbolic fuctios 0 6 Defiitios of hyperbolic fuctios 0 6 Numerical values of hyperbolic fuctios 04 6 Graphs of hyperbolic fuctios Hyperbolic idetities Osbore s rule 0 66 Differetiatio of hyperbolic fuctios 67 Itegratio of hyperbolic fuctios 4 68 Iverse hyperbolic fuctios 5 69 Logarithmic form of iverse hyperbolic fuctios 6 60 Derivatives of iverse hyperbolic fuctios 9 6 Itegrals which itegrate to iverse hyperbolic fuctios 6 Solvig equatios 5 Chapter 7: Arc legth ad area of surface of revolutio 7 Itroductio 7 Arc legth 7 Area of surface of revolutio 7 Aswers to the eercises i Further Pure 4
4 MFP Tetbook Alevel Further Mathematics 660 Chapter : Comple Numbers Itroductio The geeral comple umber The modulus ad argumet of a comple umber 4 The polar form of a comple umber 5 Additio, subtractio ad multiplicatio of comple umbers of the form iy 6 The cojugate of a comple umber ad the divisio of comple umbers of the form iy 7 Products ad quotiets of comple umbers i their polar form 8 Equatig real ad imagiary parts 9 Further cosideratio of z z ad arg( z z) 0 Loci o Argad diagrams This chapter itroduces the idea of a comple umber Whe you have completed it, you will: kow what is meat by a comple umber; kow what is meat by the modulus ad argumet of a comple umber; kow how to add, subtract, multiply ad divide comple umbers; kow how to solve equatios usig real ad imagiary parts; uderstad what a Argad diagram is; kow how to sketch loci o Argad diagrams 4
5 MFP Tetbook Alevel Further Mathematics 660 Itroductio You will have discovered by ow that some problems caot be solved i terms of real umbers For eample, if you use a calculator to evaluate 64 you get a error message This is because squarig every real umber gives a positive value; both ( 8) ad ( 8) are equal to 64 As caot be evaluated, a symbol is used to deote it the symbol used is i i i It follows that i The geeral comple umber The most geeral umber that ca be writte dow has the form i y, where ad y are real umbers The term iy is a comple umber with beig the real part ad y the imagiary part So, both i ad 4i are comple umbers The set of real umbers, (with which you are familiar), is really a subset of the set of comple umbers, This is because real umbers are actually umbers of the form 0i 5
6 MFP Tetbook Alevel Further Mathematics 660 The modulus ad argumet of a comple umber Just as real umbers ca be represeted by poits o a umber lie, comple umbers ca be represeted by poits i a plae The poit P(, y) i the plae of coordiates with aes O ad Oy represets the comple umber iy ad the umber is uiquely represeted by that poit The diagram of poits i Cartesia coordiates represetig comple umbers is called a Argad diagram y r P(, y) O θ If the comple umber iy is deoted by z, ad hece z i, y z ( mod zed ) is defied as the distace from the origi O to the poit P represetig z Thus z OP r The modulus of a comple umber z is give by z y The argumet of z, arg z, is defied as the agle betwee the lie OP ad the positive ais usually i the rage (π, π) The argumet of a comple umber z is give y by arg z, where ta You must be careful whe or y, or both, are egative 6
7 MFP Tetbook Alevel Further Mathematics 660 Eample Fid the modulus ad argumet of the comple umber i Solutio The poit P represetig this umber, z, is show o the diagram y P, z ( ) ad ta θ Therefore, arg z π O Note that whe ta, θ could equal π or π However, the sketch clearly shows that θ lies i the secod quadrat This is why you eed to be careful whe evaluatig the argumet of a comple umber Eercise A Fid the modulus ad argumet of each of the followig comple umbers: (a) i, (b) i, (c) 4, (d) i Give your aswers for arg z i radias to two decimal places Fid the modulus ad argumet of each of the followig comple umbers: (a) i, (b) 4i, (c) 7i Give your aswers for arg z i radias to two decimal places 7
8 MFP Tetbook Alevel Further Mathematics The polar form of a comple umber y I the diagram alogside, y rsi rcos ad P(, y) If P is the poit represetig the comple umber z i, y it follows that z may be writte i the form rcos irsi This is called the polar, or modulus argumet, form of a comple umber O θ r A comple umber may be writte i the form z r(cos isi ), where z r ad arg z For brevity, r(cos isi ) ca be writte as (r, θ) Eercise B Write the comple umbers give i Eercise A i polar coordiate form Fid, i the form i y, the comple umbers give i polar coordiate form by: (a) z cosπ π isi, (b) 4 4 π π 4cos isi 8
9 MFP Tetbook Alevel Further Mathematics Additio, subtractio ad multiplicatio of comple umbers of the form + iy Comple umbers ca be subjected to arithmetic operatios Cosider the eample below Eample 5 Give that z 4i ad z i, fid (a) z z, (b) z z ad (c) zz Solutio (a) zz ( 4i) ( i) 4i (b) z z ( 4i) ( i) 6i (c) zz ( 4i)( i) 4i6i8i i 8 (sice i ) i I geeral, if z ai b ad z a i b, z z ( a a ) i( b b ) z z ( a a ) i( b b ) zz aa bb i( abab) Eercise C Fid z z ad z z whe: (a) z i ad z i, (b) z 6i ad z i 9
10 MFP Tetbook Alevel Further Mathematics The cojugate of a comple umber ad the divisio of comple umbers of the form + iy The cojugate of a comple umber z iy (usually deoted by z * or z ) is the comple umber z* i y Thus, the cojugate of i is i, ad that of i is i O a Argad diagram, the poit represetig the comple umber z * is the reflectio of the poit represetig z i the ais The most importat property of z * is that the product zz * is real sice zz* ( i y)( i y) iy iy i y y * zz z Divisio of two comple umbers demads a little more care tha their additio or multiplicatio ad usually requires the use of the comple cojugate Eample 6 z Simplify z, where z 4i ad z i Solutio 4i (4i)( i) i (i)(i) 4i6i8i i i 4i 50i 5 i z multiply the umerator ad deomiator of z by z *, ie ( i) so that the product of the deomiator becomes a real umber Eercise D z For the sets of comple umbers z ad z, fid z where (a) z 4 i ad z i, (b) z 6i ad z i 0
11 MFP Tetbook Alevel Further Mathematics Products ad quotiets of comple umbers i their polar form If two comple umbers are give i polar form they ca be multiplied ad divided without havig to rewrite them i the form i y Eample 7 Fid z Solutio z if z π π cos isi ad z π π 6 6 π π π π zz cos isi cos isi π π π π 6 6 π π 6 6 cos isi π π π π π π π π 6 cos cos isi cos isi cos i si si cos πcos π si πsi π i si πcos π cos πsi π cos isi 6cos isi Notig that arg z is π, it follows that the modulus of zz is the product of the modulus of 6 z ad the modulus of z, ad the argumet of zz is the sum of the argumets of z ad z Usig the idetities: cos( A B) cos Acos Bsi Asi B si( A B) si Acos Bcos Asi B Eercise E z (a) Fid cos isi z z (b) What ca you say about the modulus ad argumet of z? if z cosπ π isi ad z π π 6 6
12 MFP Tetbook Alevel Further Mathematics 660 Eample 7 If z r, ad z r,, show that zz rr Solutio zz r(cos isi ) r(cos isi ) cos( ) isi( ) rr coscos i(sicos cossi ) i sisi rr (cos cos si si ) i(si cos cos si ) rr cos( ) isi( ) If z ( r, ) ad z ( r, ) the zz ( rr, ) with the proviso that π may have to be added to, or subtracted from, if is outside the permitted rage for There is a correspodig result for divisio you could try to prove it for yourself z r (, ) ad (, ) the z, r same proviso regardig the size of the agle If z r z r with the
13 MFP Tetbook Alevel Further Mathematics Equatig real ad imagiary parts z Goig back to Eample 6, z ca be simplified by aother method 4i Suppose we let a ib The, i ( i)( ai b) 4i abi( b a) 4i Now, a ad b are real ad the comple umber o the left had side of the equatio is equal to the comple umber o the right had side, so the real parts ca be equated ad the imagiary parts ca also be equated: ab ad ba 4 Thus b ad a, givig i as the aswer to a ib as i Eample 6 While this method is ot as straightforward as the method used earlier, it is still a valid method It also illustrates the cocept of equatig real ad imagiary parts If aibc i d, where a, b, c ad d are real, the a c ad b d Eample 8 Fid the comple umber z satisfyig the equatio Solutio ( 4i) z( i) z* i Let z ( a i b), the z* ( a i b) Thus, ( 4i)( ai b) ( i)( ai b) i Multiplyig out, a4iaib4i baiaibi b i Simplifyig, abi( 5a4 b) i Equatig real ad imagiary parts, ab, 5a4b So, a ad b Hece, z i Eercise F If z π z π 6, ad,, fid, i polar form, the comple umbers (a) zz, (b) z z, (c) z, (d) z, (e) z z Fid the comple umber satisfyig each of these equatios: (a) ( i) z i, (b) ( z i)(i) 7i, (c) z iz*
14 MFP Tetbook Alevel Further Mathematics Further cosideratio of z z ad arg( z z) Sectio 5 cosidered simple cases of the sums ad differeces of comple umbers Cosider ow the comple umber z z z, where z iy ad z i y The poits A ad B represet z ad z, respectively, o a Argad diagram y A, ) B, ) ( y ( y O C The z z z ( ) i( y y) ad is represeted by the poit C (, y y) This makes OABC a parallelogram From this it follows that z z OC ( ) ( y y), that is to say z z is the legth AB i the Argad diagram Similarly arg( z z) is the agle betwee OC ad the positive directio of the ais This i tur is the agle betwee AB ad the positive directio If the comple umber z is represeted by the poit A, ad the comple umber z is represeted by the poit B i a Argad diagram, the z z AB, ad arg( z z) is the agle betwee AB directio of the ais ad the positive Eercise G Fid z z ad arg( z z) i (a) z i, z 7 5i, (b) z i, z 4 i, (c) z i, z 4 5i 4
15 MFP Tetbook Alevel Further Mathematics Loci o Argad diagrams A locus is a path traced out by a poit subjected to certai restrictios Paths ca be traced out by poits represetig variable comple umbers o a Argad diagram just as they ca i other coordiate systems Cosider the simplest case first, whe the poit P represets the comple umber z such that z k This meas that the distace of P from the origi O is costat ad so P will trace out a circle z k represets a circle with cetre O ad radius k If istead zz k, where z is a fied comple umber represeted by the poit A o a Argad diagram, the (from Sectio 9) z z represets the distace AP ad is costat It follows that P must lie o a circle with cetre A ad radius k zz k represets a circle with cetre z ad radius k Note that if zz k, the the poit P represetig z ca ot oly lie o the circumferece of the circle, but also aywhere iside the circle The locus of P is therefore the regio o ad withi the circle with cetre A ad radius k Now cosider the locus of a poit P represeted by the comple umber z subject to the coditios zz z z, where z ad z are fied comple umbers represeted by the poits A ad B o a Argad diagram Agai, usig the result of Sectio 9, it follows that AP BP because z z is the distace AP ad z z is the distace BP Hece, the locus of P is a straight lie zz z z represets a straight lie the perpedicular bisector of the lie joiig the poits z ad z Note also that if zz z z the locus of z is ot oly the perpedicular bisector of AB, but also the whole half plae, i which A lies, bouded by this bisector All the loci cosidered so far have bee related to distaces there are also simple loci i Argad diagrams ivolvig agles The simplest case is the locus of P subject to the coditio that arg z, where is a fied agle y O α P 5
16 MFP Tetbook Alevel Further Mathematics 660 This coditio implies that the agle betwee OP ad O is fied ( ) so that the locus of P is a straight lie arg z represets the half lie through O iclied at a agle to the positive directio of O Note that the locus of P is oly a half lie the other half lie, show dotted i the diagram above, would have the equatio arg z π, possibly π if π falls outside the specified rage for arg z I eactly the same way as before, the locus of a poit P satisfyig arg( zz), where z is a fied comple umber represeted by the poit A, is a lie through A arg( z z ) iclied at a agle to the positive directio of O represets the half lie through the poit z y P A α O Note agai that this locus is oly a half lie the other half lie would have the equatio arg( zz ) π, possibly π Fially, cosider the locus of ay poit P satisfyig arg( zz) This idicates that the agle betwee AP ad the positive ais lies betwee ad, so that P ca lie o or withi the two half lies as show shaded i the diagram below y A β α O 6
17 MFP Tetbook Alevel Further Mathematics 660 Eercise H Sketch o Argad diagrams the locus of poits satisfyig: (a) z, (b) arg( z ) π, (c) z i 5 4 Sketch o Argad diagrams the regios where: (a) z i, (b) π arg( z 4 i) 5π 6 Sketch o a Argad diagram the regio satisfyig both z i ad 0arg z π 4 4 Sketch o a Argad diagram the locus of poits satisfyig both z i z i ad z i 4 7
18 MFP Tetbook Alevel Further Mathematics 660 Miscellaeous eercises Fid the comple umber which satisfies the equatio ziz* 4 i, where z * deotes the comple cojugate of z [AQA Jue 00] The comple umber z satisfies the equatio ziz (a) Fid z i the form a i b, where a ad b are real (b) Mark ad label o a Argad diagram the poits represetig z ad its cojugate, z * (c) Fid the values of z ad z z* [NEAB March 998] The comple umber z satisfies the equatio zz* z z* i, where z * deotes the comple cojugate of z Fid the two possible values of z, givig your aswers i the form a i b [AQA March 000] 4 By puttig z i y, fid the comple umber z which satisfies the equatio z z* i, i where z * deotes the comple cojugate of z [AQA Specime] 5 (a) Sketch o a Argad diagram the circle C whose equatio is z i (b) Mark the poit P o C at which z is a miimum Fid this miimum value (c) Mark the poit Q o C at which arg z is a maimum Fid this maimum value [NEAB Jue 998] 8
19 MFP Tetbook Alevel Further Mathematics (a) Sketch o a commo Argad diagram (i) the locus of poits for which z i, (ii) the locus of poits for which arg z π 4 (b) Idicate, by shadig, the regio for which z i ad arg z π 4 [AQA Jue 00] 7 The comple umber z is defied by z i i (a) (i) Epress z i the form a i b (ii) Fid the modulus ad argumet of z, givig your aswer for the argumet i the form p π where p (b) The comple umber z has modulus ad argumet 7π The comple umber z is defied by z zz (i) Show that z π 4 ad arg z 6 (ii) Mark o a Argad diagram the poits P ad P which represet z ad z, respectively (iii) Fid, i surd form, the distace betwee P ad P [AQA Jue 000] 9
20 MFP Tetbook Alevel Further Mathematics (a) Idicate o a Argad diagram the regio of the comple plae i which 0arg π z (b) The comple umber z is such that 0arg π z ad π arg z π 6 (i) Sketch aother Argad diagram showig the regio R i which z must lie (ii) Mark o this diagram the poit A belogig to R at which z has its least possible value (c) At the poit A defied i part (b)(ii), z z A (i) Calculate the value of z A (ii) Epress z A i the form a i b [AQA March 999] 9 (a) The comple umbers z ad w are such that z 4i i ad w 4 i i Epress each of z ad w i the form a i b, where a ad b are real (b) (i) Write dow the modulus ad argumet of each of the comple umbers 4 i ad i Give each modulus i a eact surd form ad each argumet i radias betwee π ad π (ii) The poits O, P ad Q i the comple plae represet the comple umbers 0 0i, 4 i ad i, respectively Fid the eact legth of PQ ad hece, or otherwise, show that the triagle OPQ is rightagled [AEB Jue 997] 0
21 MFP Tetbook Alevel Further Mathematics 660 Chapter : Roots of Polyomial Equatios Itroductio Quadratic equatios Cubic equatios 4 Relatioship betwee the roots of a cubic equatio ad its coefficiets 5 Cubic equatios with related roots 6 A importat result 7 Polyomial equatios of degree 8 Comple roots of polyomial equatios with real coefficiets This chapter revises work already covered o roots of equatios ad eteds those ideas Whe you have completed it, you will: kow how to solve ay quadratic equatio; kow that there is a relatioship betwee the umber of real roots ad form of a polyomial equatio, ad be able to sketch graphs; kow the relatioship betwee the roots of a cubic equatio ad its coefficiets; be able to form cubic equatios with related roots; kow how to eted these results to polyomials of higher degree; kow that comple cojugates are roots of polyomials with real coefficiets
22 MFP Tetbook Alevel Further Mathematics 660 Itroductio You should have already met the idea of a polyomial equatio A polyomial equatio of degree, oe with as the highest power of, is called a quadratic equatio Similarly, a polyomial equatio of degree has as the highest power of ad is called a cubic equatio; oe with 4 as the highest power of is called a quartic equatio I this chapter you are goig to study the properties of the roots of these equatios ad ivestigate methods of solvig them Quadratic equatios You should be familiar with quadratic equatios ad their properties from your earlier studies of pure mathematics However, eve if this sectio is familiar to you it provides a suitable base from which to move o to equatios of higher degree You will kow, for eample, that quadratic equatios of the type you have met have two roots (which may be coicidet) There are ormally two ways of solvig a quadratic equatio by factorizig ad, i cases where this is impossible, by the quadratic formula Graphically, the roots of the equatio a b c 0 are the poits of itersectio of the curve y a b c ad the lie y 0 (ie the ais) For eample, a sketch of part of y 8 is show below y ( 4, 0) (, 0) (0, 8) The roots of this quadratic equatio are those of ( )( 4) 0, which are ad 4
23 MFP Tetbook Alevel Further Mathematics 660 A sketch of part of the curve y 4 4 is show below y (0, 4) O (, 0) I this case, the curve touches the ais The equatio ( ) 0 ad, a repeated root 44 0 may be writte as Not all quadratic equatios are as straightforward as the oes cosidered so far A sketch of part of the curve y 4 5 is show below y (0, 5) O (, ) This curve does ot touch the ais so the equatio 45 0 caot have real roots Certaily, 4 5 will ot factorize so the quadratic formula b b 4ac a has to be used to solve this equatio This leads to 4 60 ad, usig ideas from Chapter, this becomes 4 i or i It follows that the equatio 45 0 does have two roots, but they are both comple umbers I fact the two roots are comple cojugates You may also have observed that whether a quadratic equatio has real or comple roots depeds o the value of the discrimiat b 4 ac The quadratic equatio a b c 0, where a, b ad c are real, has comple roots if b 4ac 0
24 MFP Tetbook Alevel Further Mathematics 660 Eercise A Solve the equatios (a) 60 0, (b) 06 0 Cubic equatios As metioed i the itroductio to this chapter, equatios of the form a b c d 0 are called cubic equatios All cubic equatios have at least oe real root ad this real root is ot always easy to locate The reaso for this is that cubic curves are cotiuous they do ot have asymptotes or ay other form of discotiuity Also, as, the term a becomes the domiat part of the epressio ad a (if a 0), whilst a whe Hece the curve must cross the lie y 0 at least oce If a 0, the as, ad a as ad this does ot affect the result a 4
25 MFP Tetbook Alevel Further Mathematics 660 A typical cubic equatio, below y y a b c d with a 0, ca look like ay of the sketches The equatio of this curve has three real roots because the curve crosses the lie y 0 at three poits O y y O O I each of the two sketch graphs above, the curve crosses the lie y 0 just oce, idicatig just oe real root I both cases, the cubic equatio will have two comple roots as well as the sigle real root Eample (a) Fid the roots of the cubic equatio 0 (b) Sketch a graph of y Solutio y 4 (a) If f( ), the f() 0 Therefore is a factor of f() f( ) ( )( 4) ( )( )( ) Hece the roots of f() = 0 are, ad (b)
26 MFP Tetbook Alevel Further Mathematics 660 Eample Fid the roots of the cubic equatio Solutio Let f ( ) 4 6 The f () Therefore is a factor of f(), ad f ( ) ( )( 6 ) The quadratic i this epressio has o simple roots, so usig the quadratic formula o 6 0, 4 a i i b b ac Hece the roots of f ( ) 0 are ad i Eercise B Solve the equatios (a) 5 0, (b) (c) 4 0, 0 0 6
27 MFP Tetbook Alevel Further Mathematics Relatioship betwee the roots of a cubic equatio ad its coefficiets As a cubic equatio has three roots, which may be real or comple, it follows that if the geeral cubic equatio a b c d 0 has roots, ad, it may be writte as a ( )( )( ) 0 Note that the factor a is required to esure that the coefficiets of are the same, so makig the equatios idetical Thus, o epadig the right had side of the idetity, a b c d a( )( )( ) a a( ) a( ) a The two sides are idetical so the coefficiets of ad ca be compared, ad also the umber terms, ba( ) ca( ) d a If the cubic equatio a b c d 0 has roots, ad, the b, a c, a d a Note that meas the sum of all the roots, ad that meas the sum of all the possible products of roots take two at a time Eercise C Fid, ad for the followig cubic equatios: (a) 7 5 0, (b) The roots of a cubic equatio are, ad If, the cubic equatio 7 ad 5, state 7
28 MFP Tetbook Alevel Further Mathematics Cubic equatios with related roots The eample below shows how you ca fid equatios whose roots are related to the roots of a give equatio without havig to fid the actual roots Two methods are give Eample 5 The cubic equatio 4 0 has roots, ad Fid the cubic equatios with: (a) roots, ad, (b) roots, ad, (c) roots, ad Solutio: method From the give equatio, 0 4 (a) Hece From which the equatio of the cubic must be or 6 0 (b) 4 ( ) 66 ( )( ) (4) 00 ( )( )( ) Hece the equatio of the cubic must be 0 8
29 MFP Tetbook Alevel Further Mathematics 660 (c) So that the cubic equatio with roots, ad is or 4 0 9
30 MFP Tetbook Alevel Further Mathematics 660 The secod method of fidig the cubic equatios i Eample 5 is show below It is ot always possible to use this secod method, but whe you ca it is much quicker tha the first Solutio: method (a) As the roots are to be, ad, it follows that, if X, the a cubic equatio i X must have roots which are twice the roots of the cubic equatio i As the equatio i is 4 0, if you substitute X the equatio i X becomes as before or X X 4 0, X 6X 0 (b) I this case, if you put X i 4 0, the ay root of a equatio i X must be less tha the correspodig root of the cubic i Now, X gives X ad substitutig ito 4 0 gives ( X ) ( X ) 4 0 which reduces to X X 0 (c) I this case you use the substitutio X or X For 4 0 X X O multiplyig by X, this gives X 4X 0 or 4X X 0 as before 4 0 this gives Eercise D The cubic equatio 47 0 has roots, ad Usig the first method described above, fid the cubic equatios whose roots are (a), ad, (b), ad, (c), ad Repeat Questio above usig the secod method described above Repeat Questios ad above for the cubic equatio 6 0 0
31 MFP Tetbook Alevel Further Mathematics A importat result If you square you get ( ) ( )( ) So,, or for three umbers, ad This result is well worth rememberig it is frequetly eeded i questios ivolvig the symmetric properties of roots of a cubic equatio Eample 6 The cubic equatio has roots, ad Fid the cubic equatios with (a) roots, ad, (b) roots, ad [Note that the direct approach illustrated below is the most straightforward way of solvig this type of problem] Solutio (a) ( ) Hece the cubic equatio is (b) 5 6 usig the same result but replacig with with, ad with Thus 6 (5) 46 ( ) Hece the cubic equatio is 46 0
32 MFP Tetbook Alevel Further Mathematics Polyomial equatios of degree The ideas covered so far o quadratic ad cubic equatios ca be eteded to equatios of ay degree A equatio of degree has two roots, oe of degree has three roots so a equatio of degree has roots Suppose the equatio a b c d k 0 has roots,,, the b, a c, a d, a ( ) k util, fially, the product of the roots a Remember that is the sum of the products of all possible pairs of roots, is the sum of the products of all possible combiatios of roots take three at a time, ad so o I practice, you are ulikely to meet equatios of degree higher tha 4 so this sectio cocludes with a eample usig a quartic equatio Eample 7 4 The quartic equatio has roots,, ad Write dow (a), (b) (c) Hece fid Solutio (a) (b) (c) 4 6 ( Now ( This shows that the importat result i Sectio 6 ca be eteded to ay umber of letters Hece ( ) ( ) 0
33 MFP Tetbook Alevel Further Mathematics 660 Eercise E 4 The quartic equatio 58 0 has roots,, ad (a) Fid the equatio with roots,, ad (b) Fid 8 Comple roots of polyomial equatios with real coefficiets Cosider the polyomial equatio f( ) a b c k Usig the ideas from Chapter, if p ad q are real, f( pi q) a( pi q) b( pi q) k ui v, where u ad v are real Now, f( i ) ( i ) ( i ) p q a p q b p q k uiv sice i raised to a eve power is real ad is the same as i raised to a eve power, makig the real part of f ( p i q) the same as the real part of f ( p i q) But i raised to a odd power is the same as i raised to a odd power multiplied by, ad odd powers of i comprise the imagiary part of f ( p i q) Thus, the imagiary part of f ( p i q) is times the imagiary part of f ( p i q) Now if p iq is a root of f ( ) 0, it follows that u iv 0 ad so u 0 ad v 0 Hece, uiv 0 makig f ( p i q) 0 ad p iq a root of f ( ) 0 If a polyomial equatio has real coefficiets ad if p i q, where p ad q are real, is a root of the polyomial, the its comple cojugate, p i q, is also a root of the equatio It is very importat to ote that the coefficiets i f ( ) 0 must be real If f ( ) 0 has comple coefficiets, this result does ot apply
34 MFP Tetbook Alevel Further Mathematics 660 Eample 8 The cubic equatio k 0, where k is real, has oe root equal to i Fid the other two roots ad the value of k Solutio As the coefficiets of the cubic equatio are real, it follows that i is also a root Cosiderig the sum of the roots of the equatio, if is the third root, ( i) ( i), To fid k, k ( i)( i)( ) 5, k 5 Eample 8 The quartic equatio roots has oe root equal to i Fid the other three Solutio As the coefficiets of the quartic are real, it follows that i is also a root Hece (i) ( i) is a quadratic factor of the quartic Now, (i) (i) (i) (i) (i)(i) 5 4 Hece 5 is a factor of Therefore 45 ( 5)( a b) Comparig the coefficiets of, a a 4 Cosiderig the umber terms, 5b 5 b Hece the quartic equatio may be writte as ( 5)( 4) 0 ( 5)( )( ) 0, ad the four roots are i,i, ad 4
35 MFP Tetbook Alevel Further Mathematics 660 Eercise F A cubic equatio has real coefficiets Oe root is ad aother is i Fid the cubic equatio i the form a bc 0 The cubic equatio roots has oe root equal to i Fid the other two The quartic equatio three roots has oe root equal to i Fid the other 5
36 MFP Tetbook Alevel Further Mathematics 660 Miscellaeous eercises The equatio 4 0, p where p is a costat, has roots, ad, where 0 (a) Fid the values of ad (b) Fid the value of p [NEAB Jue 998] The umbers, ad satisfy the equatios ad (a) Show that 0 (b) The umbers, ad are also the roots of the equatio 0, p q r where p, q ad r are real (i) Give that 4i ad that is real, obtai ad (ii) Calculate the product of the three roots (iii) Write dow, or determie, the values of p, q ad r [AQA Jue 000] The roots of the cubic equatio are, ad 4 0 (a) Write dow the values of, ad (b) Fid the cubic equatio, with iteger coefficiets, havig roots, ad [AQA March 000] 4 The roots of the equatio are, ad (a) Write dow the value of (b) Give that i is a root of the equatio, fid the other two roots [AQA Specime] 6
37 MFP Tetbook Alevel Further Mathematics The roots of the cubic equatio 0, p q r where p, q ad r are real, are, ad (a) Give that, write dow the value of p (b) Give also that 5, (i) fid the value of q, (ii) eplai why the equatio must have two oreal roots ad oe real root (c) Oe of the two oreal roots of the cubic equatio is 4i (i) Fid the real root (ii) Fid the value of r [AQA March 999] 6 (a) Prove that whe a polyomial (b) The polyomial g is defied by 5 f is divided by a, g 6 p q, where p ad q are real costats Whe remaider is (i) Fid the values of p ad q [AQA Jue 999] (ii) Show that whe the remaider is a f g is divided by i, where i, the g is divided by i, the remaider is 6i 7
38 MFP Tetbook Alevel Further Mathematics 660 Chapter : Summatio of Fiite Series Itroductio Summatio of series by the method of differeces Summatio of series by the method of iductio 4 Proof by iductio eteded to other areas of mathematics This chapter eteds the idea of summatio of simple series, with which you are familiar from earlier studies, to other kids of series Whe you have completed it, you will: kow ew methods of summig series; kow which method is appropriate for the summatio of a particular series; uderstad a importat method kow as the method of iductio; be able to apply the method of iductio i circumstaces other tha i the summatio of series 8
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 informationREVISION SHEET FP2 (AQA) CALCULUS. x x π π. Standard Calculus of Inverse Trig and Hyperbolic Trig Functions = + arcsin x = + ar sinh x
the Further Mathematics etwork www.fmetwork.org.uk V 07 REVISION SHEET FP (AQA) CALCULUS The mai ideas are: Calculus usig iverse trig fuctios & hperbolic trig fuctios ad their iverses. Calculatig arc legths.
More informationModule 4: Mathematical Induction
Module 4: Mathematical Iductio Theme 1: Priciple of Mathematical Iductio Mathematical iductio is used to prove statemets about atural umbers. As studets may remember, we ca write such a statemet as a predicate
More informationSoving 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 informationSequences 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 informationIn 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 informationExample 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 information4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
More informationIncremental 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 informationChapter 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 informationCS103A Handout 23 Winter 2002 February 22, 2002 Solving Recurrence Relations
CS3A Hadout 3 Witer 00 February, 00 Solvig Recurrece Relatios Itroductio A wide variety of recurrece problems occur i models. Some of these recurrece relatios ca be solved usig iteratio or some other ad
More information1. 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 informationThe Euler Totient, the Möbius and the Divisor Functions
The Euler Totiet, the Möbius ad the Divisor Fuctios Rosica Dieva July 29, 2005 Mout Holyoke College South Hadley, MA 01075 1 Ackowledgemets This work was supported by the Mout Holyoke College fellowship
More informationHere 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 informationI. Chisquared Distributions
1 M 358K Supplemet to Chapter 23: CHISQUARED DISTRIBUTIONS, TDISTRIBUTIONS, AND DEGREES OF FREEDOM To uderstad tdistributios, we first eed to look at aother family of distributios, the chisquared distributios.
More information8.1 Arithmetic Sequences
MCR3U Uit 8: Sequeces & Series Page 1 of 1 8.1 Arithmetic Sequeces Defiitio: A sequece is a comma separated list of ordered terms that follow a patter. Examples: 1, 2, 3, 4, 5 : a sequece of the first
More informationTHE 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 informationEquation of a line. Line in coordinate geometry. Slopeintercept form ( 斜 截 式 ) Intercept form ( 截 距 式 ) Pointslope form ( 點 斜 式 )
Chapter : Liear Equatios Chapter Liear Equatios Lie i coordiate geometr I Cartesia coordiate sstems ( 卡 笛 兒 坐 標 系 統 ), a lie ca be represeted b a liear equatio, i.e., a polomial with degree. But before
More informationSEQUENCES 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 information7 b) 0. Guided Notes for lesson P.2 Properties of Exponents. If a, b, x, y and a, b, 0, and m, n Z then the following properties hold: 1 n b
Guided Notes for lesso P. Properties of Expoets If a, b, x, y ad a, b, 0, ad m, Z the the followig properties hold:. Negative Expoet Rule: b ad b b b Aswers must ever cotai egative expoets. Examples: 5
More informationINFINITE 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 informationGrade 7. Strand: Number Specific Learning Outcomes It is expected that students will:
Strad: Number Specific Learig Outcomes It is expected that studets will: 7.N.1. Determie ad explai why a umber is divisible by 2, 3, 4, 5, 6, 8, 9, or 10, ad why a umber caot be divided by 0. [C, R] [C]
More informationNATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 04 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages ad iformatio sheet. Please tur over Mathematics/P DBE/04 NSC Grade Eemplar INSTRUCTIONS
More information4 n. n 1. You shold think of the Ratio Test as a generalization of the Geometric Series Test. For example, if a n ar n is a geometric sequence then
SECTION 2.6 THE RATIO TEST 79 2.6. THE RATIO TEST We ow kow how to hadle series which we ca itegrate (the Itegral Test), ad series which are similar to geometric or pseries (the Compariso Test), but of
More informationSECTION 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 informationSection 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 informationWHEN 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 informationTAYLOR SERIES, POWER SERIES
TAYLOR SERIES, POWER SERIES The followig represets a (icomplete) collectio of thigs that we covered o the subject of Taylor series ad power series. Warig. Be prepared to prove ay of these thigs durig the
More informationDivide and Conquer. Maximum/minimum. Integer Multiplication. CS125 Lecture 4 Fall 2015
CS125 Lecture 4 Fall 2015 Divide ad Coquer We have see oe geeral paradigm for fidig algorithms: the greedy approach. We ow cosider aother geeral paradigm, kow as divide ad coquer. We have already see a
More informationInfinite 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 informationDepartment of Computer Science, University of Otago
Departmet of Computer Sciece, Uiversity of Otago Techical Report OUCS200609 Permutatios Cotaiig May Patters Authors: M.H. Albert Departmet of Computer Sciece, Uiversity of Otago Micah Colema, Rya Fly
More informationSequences II. Chapter 3. 3.1 Convergent Sequences
Chapter 3 Sequeces II 3. Coverget Sequeces Plot a graph of the sequece a ) = 2, 3 2, 4 3, 5 + 4,...,,... To what limit do you thik this sequece teds? What ca you say about the sequece a )? For ǫ = 0.,
More informationBasic Elements of Arithmetic Sequences and Series
MA40S PRECALCULUS 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 informationRepeating 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 informationThe second difference is the sequence of differences of the first difference sequence, 2
Differece Equatios I differetial equatios, you look for a fuctio that satisfies ad equatio ivolvig derivatives. I differece equatios, istead of a fuctio of a cotiuous variable (such as time), we look for
More informationLinear Algebra II. 4 Determinants. Notes 4 1st November Definition of determinant
MTH6140 Liear Algebra II Notes 4 1st November 2010 4 Determiats The determiat is a fuctio defied o square matrices; its value is a scalar. It has some very importat properties: perhaps most importat is
More informationAlgebra Vocabulary List (Definitions for Middle School Teachers)
Algebra Vocabulary List (Defiitios for Middle School Teachers) A Absolute Value Fuctio The absolute value of a real umber x, x is xifx 0 x = xifx < 0 http://www.math.tamu.edu/~stecher/171/f02/absolutevaluefuctio.pdf
More informationAsymptotic 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 information23 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 informationComplex Numbers. where x represents a root of Equation 1. Note that the ± sign tells us that quadratic equations will have
Comple Numbers I spite of Calvi s discomfiture, imagiar umbers (a subset of the set of comple umbers) eist ad are ivaluable i mathematics, egieerig, ad sciece. I fact, i certai fields, such as electrical
More informationx(x 1)(x 2)... (x k + 1) = [x] k n+m 1
1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,...,
More informationSUMS OF nth POWERS OF ROOTS OF A GIVEN QUADRATIC EQUATION. N.A. Draim, Ventura, Calif., and Marjorie Bicknell Wilcox High School, Santa Clara, Calif.
SUMS OF th OWERS OF ROOTS OF A GIVEN QUADRATIC EQUATION N.A. Draim, Vetura, Calif., ad Marjorie Bickell Wilcox High School, Sata Clara, Calif. The quadratic equatio whose roots a r e the sum or differece
More informationFOUNDATIONS OF MATHEMATICS AND PRECALCULUS GRADE 10
FOUNDATIONS OF MATHEMATICS AND PRECALCULUS GRADE 10 [C] Commuicatio Measuremet A1. Solve problems that ivolve liear measuremet, usig: SI ad imperial uits of measure estimatio strategies measuremet strategies.
More informationMATHEMATICS P1 COMMON TEST JUNE 2014 NATIONAL SENIOR CERTIFICATE GRADE 12
Mathematics/P1 1 Jue 014 Commo Test MATHEMATICS P1 COMMON TEST JUNE 014 NATIONAL SENIOR CERTIFICATE GRADE 1 Marks: 15 Time: ½ hours N.B: This questio paper cosists of 7 pages ad 1 iformatio sheet. Please
More informationDiscrete 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 informationM06/5/MATME/SP2/ENG/TZ2/XX MATHEMATICS STANDARD LEVEL PAPER 2. Thursday 4 May 2006 (morning) 1 hour 30 minutes INSTRUCTIONS TO CANDIDATES
IB MATHEMATICS STANDARD LEVEL PAPER 2 DIPLOMA PROGRAMME PROGRAMME DU DIPLÔME DU BI PROGRAMA DEL DIPLOMA DEL BI 22067304 Thursday 4 May 2006 (morig) 1 hour 30 miutes INSTRUCTIONS TO CANDIDATES Do ot ope
More informationNATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P NOVEMBER 0 MARKS: 50 TIME: 3 hours This questio paper cosists of 8 pages, diagram sheet ad iformatio sheet. Please tur over Mathematics/P DBE/November 0
More information3. Covariance and Correlation
Virtual Laboratories > 3. Expected Value > 1 2 3 4 5 6 3. Covariace ad Correlatio Recall that by takig the expected value of various trasformatios of a radom variable, we ca measure may iterestig characteristics
More information3. Greatest Common Divisor  Least Common Multiple
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
More informationLaws of Exponents Learning Strategies
Laws of Epoets Learig Strategies What should studets be able to do withi this iteractive? Studets should be able to uderstad ad use of the laws of epoets. Studets should be able to simplify epressios that
More informationOur aim is to show that under reasonable assumptions a given 2πperiodic 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 informationChapter 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 informationChapter 6: Variance, the law of large numbers and the MonteCarlo method
Chapter 6: Variace, the law of large umbers ad the MoteCarlo 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.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 informationLecture 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 BruMikowski iequality for boxes. Today we ll go over the
More information1.3 Binomial Coefficients
18 CHAPTER 1. COUNTING 1. Biomial Coefficiets I this sectio, we will explore various properties of biomial coefficiets. Pascal s Triagle Table 1 cotais the values of the biomial coefficiets ( ) for 0to
More informationSection 6.1 Radicals and Rational Exponents
Sectio 6.1 Radicals ad Ratioal Expoets Defiitio of Square Root The umber b is a square root of a if b The priciple square root of a positive umber is its positive square root ad we deote this root by usig
More informationCHAPTER 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 informationNPTEL STRUCTURAL RELIABILITY
NPTEL Course O STRUCTURAL RELIABILITY Module # 0 Lecture 1 Course Format: Web Istructor: Dr. Aruasis Chakraborty Departmet of Civil Egieerig Idia Istitute of Techology Guwahati 1. Lecture 01: Basic Statistics
More informationLecture 4: Cauchy sequences, BolzanoWeierstrass, and the Squeeze theorem
Lecture 4: Cauchy sequeces, BolzaoWeierstrass, 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 informationBuilding Blocks Problem Related to Harmonic Series
TMME, vol3, o, p.76 Buildig Blocks Problem Related to Harmoic Series Yutaka Nishiyama Osaka Uiversity of Ecoomics, Japa Abstract: I this discussio I give a eplaatio of the divergece ad covergece of ifiite
More informationEngineering 323 Beautiful Homework Set 3 1 of 7 Kuszmar Problem 2.51
Egieerig 33 eautiful Homewor et 3 of 7 Kuszmar roblem.5.5 large departmet store sells sport shirts i three sizes small, medium, ad large, three patters plaid, prit, ad stripe, ad two sleeve legths log
More informationSAMPLE 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 informationConvexity, 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 informationNUMBERS COMMON TO TWO POLYGONAL SEQUENCES
NUMBERS COMMON TO TWO POLYGONAL SEQUENCES DIANNE SMITH LUCAS Chia Lake, Califoria a iteger, The polygoal sequece (or sequeces of polygoal umbers) of order r (where r is r > 3) may be defied recursively
More informationTaylor Series and Polynomials
Taylor Series ad Polyomials Motivatios The purpose of Taylor series is to approimate a fuctio with a polyomial; ot oly we wat to be able to approimate, but we also wat to kow how good the approimatio is.
More informationCS103X: 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 sixfigure salaries i whole dollar amouts are there that cotai at least
More information7.1 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 informationProperties 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 informationMathematical 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 informationMath 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
Factoring x n 1: cyclotomic and Aurifeuillian polynomials Paul Garrett
(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 informationThe Field Q of Rational Numbers
Chapter 3 The Field Q of Ratioal Numbers I this chapter we are goig to costruct the ratioal umber from the itegers. Historically, the positive ratioal umbers came first: the Babyloias, Egyptias ad Grees
More informationFast Fourier Transform
18.310 lecture otes November 18, 2013 Fast Fourier Trasform Lecturer: Michel Goemas I these otes we defie the Discrete Fourier Trasform, ad give a method for computig it fast: the Fast Fourier Trasform.
More informationSection 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 informationWeek 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 informationRadicals and Fractional Exponents
Radicals ad Roots Radicals ad Fractioal Expoets I math, may problems will ivolve what is called the radical symbol, X is proouced the th root of X, where is or greater, ad X is a positive umber. What it
More informationSection 1.6: Proof by Mathematical Induction
Sectio.6 Proof by Iductio Sectio.6: Proof by Mathematical Iductio Purpose of Sectio: To itroduce the Priciple of Mathematical Iductio, both weak ad the strog versios, ad show how certai types of theorems
More informationTheorems 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 oegative umber R, called the radius
More informationAQA STATISTICS 1 REVISION NOTES
AQA STATISTICS 1 REVISION NOTES AVERAGES AND MEASURES OF SPREAD www.mathsbox.org.uk Mode : the most commo or most popular data value the oly average that ca be used for qualitative data ot suitable if
More information7. Sample Covariance and Correlation
1 of 8 7/16/2009 6:06 AM Virtual Laboratories > 6. Radom Samples > 1 2 3 4 5 6 7 7. Sample Covariace ad Correlatio The Bivariate Model Suppose agai that we have a basic radom experimet, ad that X ad Y
More information{{1}, {2, 4}, {3}} {{1, 3, 4}, {2}} {{1}, {2}, {3, 4}} 5.4 Stirling Numbers
. Stirlig Numbers Whe coutig various types of fuctios from., we quicly discovered that eumeratig the umber of oto fuctios was a difficult problem. For a domai of five elemets ad a rage of four elemets,
More information2.7 Sequences, Sequences of Sets
2.7. SEQUENCES, SEQUENCES OF SETS 67 2.7 Sequeces, Sequeces of Sets 2.7.1 Sequeces Defiitio 190 (sequece Let S be some set. 1. A sequece i S is a fuctio f : K S where K = { N : 0 for some 0 N}. 2. For
More informationCHAPTER 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 informationRecursion and Recurrences
Chapter 5 Recursio ad Recurreces 5.1 Growth Rates of Solutios to Recurreces Divide ad Coquer Algorithms Oe of the most basic ad powerful algorithmic techiques is divide ad coquer. Cosider, for example,
More informationDefinition. A variable X that takes on values X 1, X 2, X 3,...X k with respective frequencies f 1, f 2, f 3,...f k has mean
1 Social Studies 201 October 13, 2004 Note: The examples i these otes may be differet tha used i class. However, the examples are similar ad the methods used are idetical to what was preseted i class.
More informationCase 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 informationQuadratics  Revenue and Distance
9.10 Quadratics  Reveue ad Distace Objective: Solve reveue ad distace applicatios of quadratic equatios. A commo applicatio of quadratics comes from reveue ad distace problems. Both are set up almost
More informationClass Meeting # 16: The Fourier Transform on R n
MATH 18.152 COUSE NOTES  CLASS MEETING # 16 18.152 Itroductio to PDEs, Fall 2011 Professor: Jared Speck Class Meetig # 16: The Fourier Trasform o 1. Itroductio to the Fourier Trasform Earlier i the course,
More informationS. 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 informationBINOMIAL EXPANSIONS 12.5. In this section. Some Examples. Obtaining the Coefficients
652 (1226) Chapter 12 Sequeces ad Series 12.5 BINOMIAL EXPANSIONS I this sectio Some Examples Otaiig the Coefficiets The Biomial Theorem I Chapter 5 you leared how to square a iomial. I this sectio you
More informationGregory Carey, 1998 Linear Transformations & Composites  1. Linear Transformations and Linear Composites
Gregory Carey, 1998 Liear Trasformatios & Composites  1 Liear Trasformatios ad Liear Composites I Liear Trasformatios of Variables Meas ad Stadard Deviatios of Liear Trasformatios A liear trasformatio
More informationSum and Product Rules. Combinatorics. Some Subtler Examples
Combiatorics Sum ad Product Rules Problem: How to cout without coutig. How do you figure out how may thigs there are with a certai property without actually eumeratig all of them. Sometimes this requires
More informationNATIONAL SENIOR CERTIFICATE GRADE 11
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 007 MARKS: 50 TIME: 3 hours This questio paper cosists of pages, 4 diagram sheets ad a page formula sheet. Please tur over Mathematics/P DoE/Exemplar
More information3. If x and y are real numbers, what is the simplified radical form
lgebra II Practice Test Objective:.a. Which is equivalet to 98 94 4 49?. Which epressio is aother way to write 5 4? 5 5 4 4 4 5 4 5. If ad y are real umbers, what is the simplified radical form of 5 y
More informationLiteral Equations and Formulas
. Literal Equatios ad Formulas. OBJECTIVE 1. Solve a literal equatio for a specified variable May problems i algebra require the use of formulas for their solutio. Formulas are simply equatios that express
More informationModified Line Search Method for Global Optimization
Modified Lie Search Method for Global Optimizatio Cria Grosa ad Ajith Abraham Ceter of Excellece for Quatifiable Quality of Service Norwegia Uiversity of Sciece ad Techology Trodheim, Norway {cria, ajith}@q2s.tu.o
More informationArithmetic Sequences and Partial Sums. Arithmetic Sequences. Definition of Arithmetic Sequence. Example 1. 7, 11, 15, 19,..., 4n 3,...
3330_090.qxd 1/5/05 11:9 AM Page 653 Sectio 9. Arithmetic Sequeces ad Partial Sums 653 9. Arithmetic Sequeces ad Partial Sums What you should lear Recogize,write, ad fid the th terms of arithmetic sequeces.
More informationSolving equations. Pretest. Warmup
Solvig equatios 8 Pretest Warmup We ca thik of a algebraic equatio as beig like a set of scales. The two sides of the equatio are equal, so the scales are balaced. If we add somethig to oe side of the
More informationFactors 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