Module Summary Sheets. C3, Methods for Advanced Mathematics (Version B reference to new book) Topic 2: Natural Logarithms and Exponentials


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1 MEI Mthemtics in Ection nd Instry Topic : Proof MEI Structured Mthemtics Mole Summry Sheets C, Methods for Anced Mthemtics (Version B reference to new book) Topic : Nturl Logrithms nd Eponentils Topic : Functions Topic 4: Differentition Topic 5: Integrtion Topic 6: Methods of solution of equtions Purchsers hve the licence to mke multiple copies for use within single estblishment Februry, 5 MEI, Ok House, 9 Epsom Centre, White Horse Business Prk, Trowbridge, Wiltshire. BA4 XG. Compny No Englnd nd Wles Registered with the Chrity Commission, number 589 Tel: F:
2 Summry C Topic : Proof Chpter Pges  Eercise A Q. Proof by direct rgument Some proofs cn be constructed using known fcts (e.g. the squre of n even number is even.) Geometric proofs cn be constructed by direct rgument. E.g. Prove tht number is divisible by if the sum of digits is divisible by. Let the number be = + b+ c+... The sum of digits is s = + b+ c+... s = 9b+ 99c+ 999 d +... = s+ (b+ c+...) So if s is divisible by then so is the whole of the right hnd side nd so the left hnd side is lso divisible by. Chpter Pges 4 Eercise A Q. Chpter Pges 45 Eercise A Q. 5 Chpter Pges 67 Eercise A Q. 8 Proof by ehustion If there re finite number of possibilities then proving by ehustion involves testing the ssertion in every cse. E.g. proof tht if digit number is divisible by then the number obtined by reversing the digits is lso divisible by. This cn be done by ehustion s there re only smll number of such numbers ( in ll). Proof tht number is divisible by if the sum of the digits is divisible by cnnot be done by ehustion. Proof by contrdiction Either it is or it isn t. If you cn show tht it isn t is not correct then by defult it is must be right. Disproof by the use of counteremple For theorem to be true it must be true in every cse. To show tht it is not in just one cse is therefore sufficient to show tht the theorem is not true. Pure Mthemtics C Version B: pge Competence sttements p, p, p Prove tht there is only one prime number tht is less thn perfect squre. Consider ny number n. Its squre is n nd one less is n = (n )(n + ). Thus n is proct of two numbers nd cn therefore only be prime if one of those numbers is. The only possibility is when n = nd n =. So there is only one such number when n = nd one less thn its squre is. For ll other vlues of n, one less thn its squre is proct of two numbers nd is therefore not prime. E.g. prove tht is irrtionl. Assume tht is rtionl. I.e. tht it cn be epressed s frction where nd b re coprime b (tht is, they hve no common fctors). = = = b b b i.e. is even, which mens tht is even. So write = k = 4k = b b = k i.e. b is even, which mens tht b is even. So both nd b re even which contrdicts the ssertion tht nd b re coprime. So cnnot be written s frction nd so is not rtionl. E.g. Is n + n + 4 prime for ll positive n? Substituting n =,, nd gives 4, 4, 47, 5, ll of which re prime. It might therefore be ssumed tht the epression is prime for ll n. But when n = 4, n + n + 4 = = 4 4 which is not prime.
3 Summry C Topic : Logrithms nd Eponentils Chpter Pges 8  Chpter. Pges  Eercise A Q. Logrithms From C the following lws for logrithms were derived. log + log y = log y log log y = log y n log = nlog log =, log = log = log log = log Nturl Logrithms nd the eponentil Function The number e is clled the eponentil number nd the logrithm to the bse e is known s the nturl logrithm nd is denoted ln. It cn be seen tht = log e where e = e Then =. If y = ln then = e y So y = e is the inverse of y = ln (For the development of inverse functions see topic.) The grphs of y = ln nd y = e E.g. log + log = log6 Also ln + ln = ln6 E.g. log5 + log = log = Also ln5 + ln = ln E.g. log5 + log = log5 + log =log5 + log8 = log4 E.g. log + log9 = log + log = log + log = log E.g. log5 log = (log log) log = 4log = log6 = log.65 E.g. mke t the subject of the formul T = T + Ae T T A T T A kt kt kt = e = e e = A T T A kt = ln t = ln A ( T T ) k ( T T ) E.g. Sketch the grph of y = e nd its inverse. y = e = lny = ln ln lne = = e e So the inverse is y = ln y = e  kt y = e y = y = ln y = ln E.g. The popultion of rbbits on n islnd is modelled by the formul P = 5 4e .5t where t is yers fter the strt of counting. Eercise A Q. 7, Eponentil growth nd decy The grph of y = e is known s eponentil growth. The grph of y = e is known s eponentil decy. (i) Sketch grph of the function (ii) How mny rbbits re there initilly? (iii) Clculte the number of rbbits on the islnd fter 5 yers. (iv) Wht is the longterm popultion ccording to this model? (i) Pure Mthemtics, C Version B: pge Competence sttements,,, 4 (ii) When t =, P = 5 4 = (iii) When t = 5, P = 5 4e .5 = 5.8 = 467 (iv) When t, P 5 = 5
4 Summry C Topic : Functions Chpter Pges 9 Eercise A Q. (v), (i), 4(vi) Chpter Pges 58 Eercise B Q., 4 Chpter Pges  Terminology A mpping is rule which ssocites two sets of items. The object mps onto the imge. The set of possible objects forms the domin. The set of possible imges forms the codomin. The set of ctul imges is the rnge. The rnge is subset of the codomin. A mpping cn be one to one, mny to one, one to mny or mny to mny. If there is only one possible imge for ech object then the mpping is clled function. Nottion for functions : y = f = f : Using trnsformtions to sketch the curves of functions y= f ( ) is the curve y= f trnslted units in the +ve direction. y= f + is the curve y= f trnslted units in the +ve y direction. y= f for > is onewy stretch of the curve y= f of scle fctor prllel to the y is. f Rnge Domin Codomin f is mny to one mpping nd so is function. E.g. f: Rnge Domin : Rnge : y Domin E.g. f: g: ( + ) i.e. g() = f( + ) g f + Eercise C Q. y= f for > is onewy stretch of the curve y= f of scle fctor prllel to the is. E.g. f: g: + i.e. g() = f() + g y= f ( ) is reflection of the curve y= f in the y is. f Chpter Pges  Eercise C Q. 5 y= f is reflection of the curve y= f in the is. Qudrtic functions y= + b+ c ( t) + s So y= + b+ c is trnsltion of y = by t units in the +ve direction nd s units in the +ve y direction. This process of rewriting the qudrtic function is clled completing the squre. E.g. g: + y = ( ) + f: so g is trnsltion of f by unit in the +ve direction nd in the +ve y direction. g Pure Mthemtics, C Version B: pge 4 Competence sttements f, f, f, f4 f (, )
5 Summry C Topic : Functions Chpter Pges 68 Eercise D Q. (v),(vi) Chpter Pges 945 Eercise D Quest., 5 Chpter Pges Chpter Pges 495 Eercise E Q., 5, 7 Chpter Pges Composite functions A composite function is function of function. fg = f ( z) where z= g e.g. f = +, g = fg = f ( ) = + N.B. gf = g( + ) = ( + ) i.e. gf fg Inverse functions If the function f mps onto y then the inverse mpping is y onto. This mpping cn only be function if is uniquely defined for y. In other words the function f must be one to one. If y = f() then the inverse function y = f  () is the reflection in the line y =. The criterion is tht given mps onto y, y is the imge of only. Inverse trigonometricl functions Trigonometricl functions re mny to one nd so hve no inverse unless the domin is restricted. Even, Odd nd Periodic functions An even function hs the y is ( = ) s the is of symmetry. (i.e. f() = f().) An odd function hs rottionl symmetry of order bout the origin. ( f() = f().) A periodic function is one where f(+k) = f() where the minimum vlue of k is clled the period. The molus function If y = f() tkes negtive vlues s well s positive vlues then the function y = g() which tkes the positive numericl vlue is clled the molus function. We write y = f. E.g. f() =, g() = + Find (i) fg() nd (ii) gf(). (i) fg() = f( + ) = ( + ) (ii) gf() = g() = () + E.g. f = + f = + E.g. f = f = E.g.cos6 =.5 cos .5 hs infinitely mny solutions (e.g. 6,, 4 ) But in the rnge <<8 cos .5 = 6 (This unique vlue is clled the Principl Vlue.) E.g. y = cos is even. y = sin is odd. E.g. f = in hs no inverse s f() = f() = but f = in hs the inverse f = Both y = cos nd y = sin re periodic with k = 6. E.g. Solve 7 < 7 < < 7< 6< < 8 < < 4 This cn be seen grphiclly. E.g. < 5 5 < < 5 The grph of y = f is obtined from y = f by replcing vlues where f is negtive by equivlent positive vlues. E.g. y = + Does not cut either is It is n odd function = is n symptote, s is y = As, y Chpter Pges 66 Eercise F Q. Curve Sketching When sketching curve the following processes should be dopted. Find where the curve crosses the es Check for symmetry Find ny symptotes Emine the behviour when ± Look for ny sttionry points Pure Mthemtics, C Version B: pge 5 Competence sttements f5, f6, f7, f8, f9
6 Summry C Topic 4: Differentition Chpter 4 Pges 665 Eercise 4A Q. (i),(i), 4 The Chin Rule dz If y = f ( z) where z = g then =. dz E.g. If y = ( + ) then putting z = + gives y = z E.g. y= +. Putting z= + gives y= z dz dz =. nd = z nd = 4 dz dz = ( + ) 4= 8( + ) Chpter 4 Pges Eercise 4A Q. 6 Chpter 4 Pges 687 Eercise 4B Q. (i),(i), Chpter 4 Pges 77 Eercise 4B Q. (iv), (vii), 4 Chpter 4 Pges Eercise 4C Q. Rte of chnge y = f where nd y both vry with time, t. =. dt dt The Proct Rule If y= uv where u = f nd v= g then = u + v d d d e.g. If y= ( 4 ) then u= nd v= 4 The Quotient Rule u If y = where u = f nd v= g v v u then = d v e.g. If y = then u = nd v= 4 ( 4 ) Inverse functions When = f( y) = Sometimes it is possible to mke y the subject of the function, in which cse cn be found in the usul wy. This will usully not be possible, however, in emintion questions. Pure Mthemtics, C Version B: pge 6 Competence sttements c, c4, c5, c6, c E.g. A stone is dropped into pond of still wter. The ripples spred outwrds in circle t rte of cm/sec. Find the rte of increse of re of ripples when r = cm. da A = πr = πr dr da da dr =. = π r.. dt dr dt da When r =, = 6π dt E.g. y = + + Put u = + nd v = + = 4 nd = d y = ( + ) + ( + ) 4 = E.g.. y = ( + ) ( + ) Put u = + nd v = + Then = 4 nd = ( ) 4 + ( + ) = = E.g. ( + ) ( + ) E.g. = y + = y = = y Note here: = y + y= = = y + y = y+ = y+ Note here tht y cn be mde the subject of the function only with difficulty.
7 Summry C Topic 4: Differentition Chpter 4 Pges 884 Eercise 4D Q. (i),(iii), (i),(iii) Nturl logrithms nd eponentils y = e = e y = ln = n n y = ln = f ' y = ln ( f ) = f 4 4 E.g. y = e = 4e y= ln( 4) = 6 6 y= ln = N.B. by rule of logs, y= ln ( 4) = ln 4+ ln = Chpter 4 Pges 994 Eercise 4E Q. (i), (i), (i), 6 Differentition of trig functions f f ' f f ' sin cos sin cos cos sin cos sin tn tn cos cos is mesured in rdins. E.g. Differentite the following: (i) y = sin + cos 4 (ii) y = sin (using the Proct rule) (iii) y = (using the Quotient rule) sin (i) = cos 4sin 4 (ii) = sin + cos (iii) sin cos = sin Chpter 4 Pges Eercise 4F Q. 6 Implicit differentition This concerns functions where y is not the subject of the function The function is differentited using the chin rule. ( y ) d d g g( y ) =. Note tht when y f, will be function of but when y is not the subject of the function, will be function of nd y. E.g. y sin+ y = 4 y sin+ y cos + = y cos = ysin+ E.g. (See emple on previous pge) = = = = y= y y ( + y) ( y+ ) Pure Mthemtics, C Version B: pge 7 Competence sttements c, c4, c5, c6, c
8 Summry C Topic 5: Integrtion Chpter 5 Pges 7 Eercise 5A Q. (i)(vii), 7 Eercise 5A Q. (v) Integrtion by substitution (Chnge of vrible) The integrl is written in terms of new vrible u: f= g( u) Indefinite integrls should be chnged bck fter integrting to give n nswer in terms of. Definite integrls should hve the limits chnged to correspond to the new vrible. This method is most esily seen in two circumstnces: (i) When the function of function is liner function of. e.g. y = ( ) 4 In this cse you need to consider wht number to multiply or divide by. (ii) When the function to be integrted looks like proct, but one prt is the derivtive of the other. E.g. ( + ). Integrtion by inspection ( f ) n+ If I = f ' f d, then I = + c n + E.g. I = + d n where f = + nd f ' =. This cn be seen s the reverse of the Chin Rule. E.g. I = + d = d + where f = + nd f ' =. ( + ) I =. + c= c ( ) E.g. I = + d = + where f = + nd f ' = I = = = 6 Pure Mthemtics, C Version B: pge 8 Competence sttements c, c4, c5, c6, c 5 E.g. I = ( + 4) d ; Put u= + 4; = 6 ( + 4) 5 u I = u = = + c 8 8 E.g. I = ( + ) d ; Put u= + ; = When =, u= ; when =, u= u 6 49 I = u = = 5 = 6 6 E.g. I = + d [ ] where f = + nd f ' =. [ ] I = f ' f f = + c = ( + ) + c E.g. I = + 5 d where f = + 5 nd f ' = ( 5 ) d 6 ( + 5) ( + 5) I = = + c = + c 6 5 E.g. Differentite ln nd hence find ln d ( ln ) = ln +. = ln + ( ) ( + ) E.g. I = ln = ln + = ln + [ ] = ln = ln ln = ln where f( ) = + nd f ' =. I = = = = 9
9 Summry C Topic 5: Integrtion Chpter 5 Pges 4 Eercise 5B Q. (i),(v), (i),(v), 6 Integrtion involving eponentils nd logrithms e = e + c = ln + c Note tht the integrl represents the re under the curve. The re between the is nd the curve y = / for negtive cn be found if cre is tken over the signs. 5 E.g. = 5ln + c 5 5 E.g. e = e + c 5 E.g. = ln ( + ) + E.g. Find 9 = ( ln 9 ln 7) = ln = = = = + ln( + ) 9 ln6 6 ln5 = = ln 5 Chpter 5 Pges 4 Eercise 5C Q. (i), (i), 4(i) Integrtion of trig functions sin= cos+ c sin = cos+ c cos= sin+ c cosd = sin+ c ( + ) E.g. sin cos4 = cos + sin4+ c 4 π π 6 6 E.g. cos= sin = = Chpter 5 Pges 5 Eercise 5D Q. (i),(iii), 4 Chpter 5 Pges  Eercise 5E Q. (i),(iv), Integrtion by prts u = uv v This formul is used to integrte procts. e.g. sin d, e Definite integrtion by prts b b u = [ uv] v Pure Mthemtics, C Version B: pge 9 Competence sttements c, c4, c5, c6, c b E.g. I = cosd ; u = = = cos v = sin I = sin. sin = sin + cos + c E.g. e = e e u = = = v = 4 = ( e ) e = e ( e ) = e {, e e }
10 Summry C Topic 6: Numericl solution of equtions Note tht this topic is the subject of the component of coursework ttched to this mole. There will be no questions on this topic in the emintion. Chpter 6 Pges 554 Eercise 6A Q. (i),(iii), 4 Eercise 6B Q., 4 Eercise 6C Q., 5 Misconceptions nd common errors seen in C coursework Terminology Students commonly refer to n epression (e.g. + 7) or function (e.g. y = + 7) s n eqution. Wht you re doing is to solve the eqution f() = nd to illustrte it you re going to drw the grph of y = f(). Tke cre to use the correct words throughout your coursework! Numericl solutions There re some equtions tht cn be solved nlyticlly nd some which cnnot. Where n nlyticl solution is known to eist it should be employed to solve the eqution. When n eqution cnnot be solved nlyticlly numeric method my be employed. The one is not inferior to the other they re used in different circumstnces. Error bounds A numeric solution without error bounds is useless. Mny students will work numeric method to proce root such s =.456 nd then ssert tht it is correct to deciml plces or even not give ny error bounds. Error bounds re often stted nd justified by scrutiny of the digits within consecutive itertes. The deciml serch methods nd itertive methods which, when illustrted grphiclly, disply cobweb digrm, hve inbuilt error bounds (but even then need to be stted ppropritely!) but n itertive method which is illustrted by stircse digrm does not, nd in this cse error bounds need to be estblished by chnge of sign. This is true lso for root found by the NewtonRphson method. Filure of methods In ech cse you re sked to demonstrte filure. The NewtonRphson method requires s condition for use tht the initil vlue for be close to the root. If, therefore, the vlue of = yields root in the rnge [,] then it would not be pproprite to suggest tht the method hs filed if strting vlue of =, sy, does not yield the sme root. The importnce of grphicl illustrtions It is crucil tht you connect the grphicl understnding to wht is being done numericlly. The problems bout the misuse of terminology described bove my be overcome with cler understnding of the connection between the two. Indeed, if grphicl solutions re introced properly s method of solving equtions then much of the difficulty will be overcome. For instnce, to solve + 7 =, drw the grph of the function y = + 7. Where this grph crosses the is is the point where y = nd so gives n pproimte vlue for the root of the eqution. The employment of such process cn lso ct s check to the work being done. Often very ble students mke rithmetic (or lgebric) errors which they do not pick up becuse of their inbility to see visully wht is going on. Pure Mthemtics, C Version B: pge Competence sttements e, e, e, e4, e5, e6, e7
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