www.mymathscloud.com©MyMathsCloudA Level FurtherFormula SheetSummationResultsProperties(3)(These work similar to integrals)•Can take constants out: ∑"#!="∑#!"!#$"!#$•Can split up: ∑(#!±(!)"!#$=∑#!"!#$±∑((!)"!#$•inclusive exclusive principle ∑(#!)"!#%=∑(#!)"!#$−∑(#!)%&$!#$Results(5)∑1"'#$=,∑""'#$=",∑-"'#$="(")$)+∑-+"'#$="(")$)(+")$),∑--"'#$="!(")$)!.Matrix TransformationsReflection in the line!=(#$%')).cos23sin23sin23−cos236Horizontal stretch by scale factor *.70016Vertical stretch byscale factor *.10076Enlargement by scale factor *centre (0,0).70076Anti-clockwise rotation of angle 'about origin.cos3−sin3sin3cos36, 3>0Clockwise rotation of angle 'about origin.−cos3+sin3sin3cos36,3>0MatricesDeterminant<×<:?.#("@6?=#@−("A×A:BC#("@DEFℎ-HB=#?DEℎ-?−(I@EF-I+"I@DFℎI=#(D-−Eℎ)−((@-−EF)+"(@ℎ−DF)Inverse1,-.-/01232.×3,5673.-<×<:.#("@6&$=$/0&12.@−"−(#6A×A:C#("@DEFℎ-H&3To find adjugate:Step 1: Find matrix of minors(cross of corresponding row and column for each element andthedeterminant of each remaining part forms the new element)Step 2: Find matrix of cofactors(get the correct signs)Step 3: Transpose all the elements. In other words swap their positions over the diagonal (the diagonal stays the same)3 types of solutionsfor systemsof linear equations•Consistent/unique -one solution (a point)To find unknowns: Use basic elimination or solve det ≠0(ifunknowns onLHS) •Consistent/non-unique –infinite solutions (det =0)To find unknowns: Use basic elimination with 2 pairs of the same variable eliminated and look to get 0=0•Inconsistent/non unique –no sol (det =0)To find unknowns: Use basic elimination with 2 pairs of the same variable eliminated and look to get an inconsistencyRootsQuadratics, Cubics and QuarticsQuadratic: •form: (K−L)(K−M)Land Mas the roots•form: K++(K+"sum roots = −(=L+Mproduct roots ="=LM•form: #K++(K+"sum: −1/=L+,product:2/="=LMCubic •form: (K−L)(K−M)(K−N)L, Mand Nare the roots•form: aK-+(K++"K+@=0sum:L+M+N= −1/, product: LMN= −0/sum of all possible products of pairs of roots:LM+MN+LN=2/Quartic:•form : (K−L)(K−M)(K−N)(K−O)L, M, N#,@Oare the roots•form: aK.+(K-+"K++@K+D=0sum :L+M+N+O=−1/, product:LMNO=4/sum of all possible products of pairs of roots:LM+LN+LO+MN+MO+NO=2/sum of all possible products of triples of roots:LMN+MNO+NLO+OLM=−0/form ∑$!)!"!#$= 0Sum=&/"#$/", Product=(&$)"/%/"Polar CoordinatesArea of sector12PQ+@3Complex NumbersDefinition√−1=-, -+=−1Cartesian FormS=#+(-ModulusArgumentFormS=Q("TU3+-U-,3)=Q"-U3Q=√#++(+#QF=3=V#,&$WI(#IXThen draw the angle 3in the quadrant where the complex number #+(-lies. Read off 3by starting on the positive Kaxis (like when you solve for trig using the CAST diagram). Remember that −Y≤3<Y)Or you can use the followingto get 3:3=⎩⎪⎨⎪⎧tan&$W(#X(-Ebc#@Q#,V1TQ4)tan&$W(#X+Y(-Ebc#@Q#,V2)tan&$W(#X−Y(-Ebc#@Q#,V3)Eulers FormS=QD'5De Moivres’ TheoremS"=Q"("TU,3+-U-,,3)=Q""-U,3Roots of 9"=;S=D+6!'",ETQ7=0,1,2,...,,−1Linear AlgebraEigenvaluesSet characteristic polynomial which is det(A-gh)or det(g−ih)equal to zero 0, where A =matrix and I=identity matrix. solve thisforthe eigenvalueEigenvectorsAv=gjwhere g=eigenvector &v=eigenvectorAv−gj=ki.e. A.786=g.786. Solve with value of gaboveQuickway: Put. ginto (i−gh)and you’ll obviously have a matrix. Multiply this matrix by .786and set equal to 0. Solve for Kin termsof lConicsEllipseParabolaHyperbolaRectangularHyperbolaStandard FormK+#++l+(+=1l+=4#KK+#+−l+(+=1Kl="+Parametric Form(acos3,(sin3)(#V+,2#V)(asec3,(V#,3(±acosh3,(sinh3)."V,"V6EccentricityD<1(+=#+(1−D+)D=1D>1(+=#+(D+−1)D=√2Foci(±#D,0)(#,0)(±#D,0)o±√2",±√2"pDirectricesK=±#DK=−#K=±#DK+l=±√2"AsymptotesnonenoneK#=±l(K=0,l=0GroupsOrderof a group: Number of elements in the groupOrder of an element:Least positive integer n such that F"=D(how many times before you get e in modular arithmetic). If g∈rthen g......g=e i.e. F"=D. If no n such that K"=ethen we say element has infinite orderDefinition:A set G with a binary operation * on G such that i.G is associativeii.G is closed under *iii.G has an identity element (usually denoted e)iv.Each element of G has an inverseVectorsNotationssD"VTQ=t, #, uvwwwwww⃗distance=OAVector Form#y+(z+"{≡}#("~Properties(addition/subtraction, multiplication and scalar product)}#("~±C@DEH=C#±@(±D"±EHg}#("~=Cg#g(g"H}#("~.C@DEH=#@+(D+"EMagnitude of a vectorNotation is ||Å}#("~Å=√#++(++"+Unit VectorUnit vector of ./126=$√/!)1!)2!./1"6Parallel and Perpendicular toParallel means vectors are a multiple of eachotherPerpendicular means scalar product equalszeroAngle Between 2 vectorsAlways use the direction vectors3=cos&$⎝⎜⎜⎛}#("~.C@DEHÅ}#("~ÅBC@DEHB⎠⎟⎟⎞Vector Equation of a lineTo find this we need:Point and direction(if given 2 points find the directions and use either point)Q=C#("H+gà@DEâ./126=äTU-V-T,,.04:6@-QD"V-T,(ä#Q#ããDãVT)Cartesian Equationof a lineK−#@=l−(D=S−"EParametric Form of a lineK=#+g@,l=(+gD,S="+gEEquation of a planeå.ç=}#("~.çwhere n is the normal vectorVector Equation of a planeTo find this we need: a point in plane and perp direction. If not given perp direction take the cross product of 2 directionvectors. Remember to find a direction we subtract 2 position vectors.78;6=./126+gW04:X+é.<=>6./126=äTU-V-T,.04:6#,@.<=>6=@-QD"V-T,U(ä#Q#ããDãVT)Cartesian Equationof aplane#K+(l+"S=@@=@-UV#,"DETQèTQ-F-,VTäã#,DC#("H=@-QD"V-T,sD"VTQ(äDQäD,@-"cã#QVT)Scalar ProductNote: 3is the angle between }#("~and C@DEH}#("~.C@DEH=Å}#("~ÅBC@DEHB"TU3Vector ProductNote: 3is the angle between }#("~and C@DEH}#("~×C@DEH=C(E−D"−(#E−"@)#D−(@HorB}#("~×C@DEHB=Å}#("~ÅBC@DEHBsin3Area of a Parallelogrami=B}#("~×C@DEHB}#("~and C@DEHform 2 adjacent sides of a parallelogramPerp Distance between point and planefrom (L,M,N)to #K+(l+"S=@|#(L)+((M)+"(N)+@|√#++(++"+ScalarProduct Properties 0.t=tt.ê=ê.t(−t).ê=−(t.ê)(7t).ê=7(t.ê)t.(ê+ë)=t.ê+t.ëIf a and b are parallel: t.ê=|t||ê|, moreover t.t=|t|+CrossProduct Propertiesí×t=0t×0=0×t=0g(t×ê)=(gt)×ê=t×(gê)t×(ê+ë)=(t×ê)+(t×ë)t×ê=−(ê×t)ê.(ë×t)=ë.(t×ê)TrigonometryIf V=tan$+K⟹U-,K=+>$)>!and "TUK=$&>!$)>!HyperbolicsDefinitionsU-,ℎK=4&&4#&+coshK4&)4#&+V#,ℎK=?@AB7DE?B7=4&&4#&4&)4#&cschK=$?@AB7=+4&&4#&sechK=$DE?B7=+4&)4#&"TVℎK=$FGAB7=4&)4#&4&&4#&Identitiescosh+K−sinh+K=1tanh+K+sech+K=1coth+K−csch+K=1tanhK=sinhKcoshKsinh2K=2sinhK"TUℎKcosh2K=cosh+K+sinh+KInverse#Q"TUℎK=cosh&$K=ln(K+ïK+−1), K≥1#QU-,ℎK=sinh&$K==ln(K+√K++1)artanhK=tanh&$K==$+ã,.$)7$&76, |K|<1Number TheoryFermat’s Theorem#H≡#(èT@ä)if äis prime and #is any integerMechanicsCentres Of Mass For Uniform Bodies•Triangular Lamina: +-along median from vertex•Circular arc, radius Q,angle at centre 2L≔<?@AIIfrom centre•Sector of circle, radius Q,angle at centre 2L:+<?@AI-Ifrom centre•Solid hemisphere, radius Q:-JQfrom centre•Hemispherical Shell, radius Q:$+Qfrom centre•Solid cone or pyramid of height ℎ: $.ℎabove the base on the line from centre to base of vertex•Solid cone or pyramid of height ℎ: $.ℎabove the base on the line from centre to base of vertex•Solid cone or pyramid of height ℎ: -.ℎfrom vertex•Conical shell of height ℎ:$.ℎabove the base on the line from centre to base of vertexMotion In A CircleTransverse velocity: s=Q3̇Transverse acceleration: ṡ=Q3̈Radial acceleration: −Q3+̇=−K!<. Note: Mag =Q3+̇TQK!<Motion of a ProjectileEquation of a trajectory: l=Ktan3−L7!+M!DE?!5Elastic Strings and SpringsF=ForceneededtoextendorcompressT=tensionK=lengthofextension/compressionk=stiffnessconstant(springconstantmeasuredinN/mg=modulus of elasticity (spring modulus)measured in Newtonsã=natural length of the springHooke’s Lawú=−7KTension in elastic spring/string :ù=NOK=N7OEnergy Note: if answeris negative then means a lossKinetic Energy:$+ès+Gravitational Potential Energy:èFℎElastic Potential Energy:N7!+OChange in kinetic energy: $+è$s$+−$+è+s++Change in potential energy: è$Fℎ$−è+Fℎ+Work Done•W=work done•F=magnitude of the force•@=distance moved IN THE DIRECTION of the force•3=angle between the force and the displacement•Total energy = Kinetic + Potential + ElasticIf work done against an opposing force:Final total energy=Initial total energy –work done against forcewhere work is done against opposing force is Wû=üú@@cos3VTV#ãD,DQFlãTUV(#7#†TQ7D,DQFläQ-,"-äãD)äTVD,V-#ãD,DQFl(-EV#ã7-,F#(TcV#F-#,UVFQ#s-Vl)Note: Total energy lost = change in total energy = change in kinetic+ change in potentialIf no work done against an opposing forceFinal total energy = Initial total energyInductionTemplateLet °"be the proposition ...•Let ,=1: Plug in ,=1 to both the LHS and RHS. Show that LHS=RHS⇒°$true•Assume ,=7true i.e. °!true: Replace ,with 7. There is nothing to prove here, we just assumethis to be true. •Let ,=7+1:Replace ,with 7+1. Usually only need work on the LHS by simplifying and using assumed £Pstep to show that what we get for LHS is equal to RHS (sometimes we may need to work on the RHS also) ⇒°!)$trueSo °!true ⇒°!)$true ∵°$true then °+,°-,°.,...true∴true ETQ#ãã,∈⋯Typesof questionsto know: sigmaresults, divisibilityand matricesStatistics & ProbabilityBinomial Distribution(discrete)K~®(,,ä)E(X)=Mean=np,Var(X)=,ä(1−ä), °©ú=°(™=K)=o"7pä7(1−ä)"&7Calculator: =use pd, ≤cUD"@PoissonDistribution(discrete)(happeningataverage rate)K~°T(g)Mean=g,variance=g, °©ú=°(™=K)=D&NN&7!Calculator: =use pd, ≤cUD"@Uniform Distribution(discrete)K~ ́[#,(]Mean=$+(#+(),variance=$$+((−#)+, PMF=°(™=K)=$1&/To find unknowns: use fact that area of rectangle is the probabilityTo find probabilities: Find area of rectangleGeometric(discrete)(how long until 1st success)™~rDT(ä)Mean=$H,variance=$&HH!PMF=P(X=K)=ä(1−ä)7&$P(X>Q)=(1−ä)7By hand: Need to turn all into =or >and use formulae aboveCalculator: =use pd, ≤cUD"@Negative Binomial(discrete)(how long until r successes)K~Æ®(Q,ä)°©ú=°(K=7)=.7−1Q−16ä<(1−ä)7&<Mean=<Hvariance=<($&H)H!Exponential(waiting time between poisson events)K~DKä(g)°Øú=°(K=7)=gD&N7Mean=$Nvariance=$N!Normal Distribution(continuous)K~Æ(é,∞+)Mean=é,variance=∞+, °Øú=°(™=K)=$R√+6D&$!S&#'(T!Standardised variable S=7&URTo find probabilities: use invnorm on calculatorTo find x, é,TQ∞: use invnorm on calculatorExpected ValueDiscreteE(X)=∑K°(™=K)For a function: ±(F(™))=∑F(K)°(™=K)Expected Value Continuous±(™)=PKE(K)@KFor a function: ±(F(™))=∫F(K)E(K)@KV&VVarianceDiscreteVar(™)=∑K+°(™=K)−±(™)+=∑K+°(™=K)−é+VarianceContinuous≥#Q(™)=PK+E(K)@K−±(™)+=PK+E(K)@K−é+PDF and CDF°Øú:E(KW)=P(™=KW)¥Øú:ú(KW)=°(™≤KW)=∫E(V)@V7%&V•PDF to CDF i.e. f to F ⟹integrate (or find areas undergraph)Careful with integration if more than 2 functions, always start from the beginning•CDF to PDF i.e.F to f ⟹differentiate (find gradients of graph)•To find probabilities: find areas orintegrate E(K)or plug value straight into F(x)•Median=Find m such that ∫E(K)@K%OXY4<O'%'>=0.5•Mode: Solve EZ(K)=0or if piecewise graph and see which Kgives you highest point•Finding unknowns or showing valid PDF:Use fact that ∫E(K)@K=1Probability Generating Function(PGF)PGF is forDRV’s only and represents the PMF as a power series i.e.you can find certain (=) probabilities from itBinomial: (1−ä+äV)"Poisson: DN(>&$)Geometric:H>$&($&H)>Negative Binomial:.H>$&($&H)>6<Definition of PGF:∏(S)TQr(S)TQr7(S)=±(S7)=∑∂!°(™=7V!#WX<$)Note: Sometimes uselettert instead of zProperties: r(1)=1, Mean=E(X)=G’(1)Variance=Var(X)=G’’(1)+G’(1)−(r′(1))+=G’’(1)−é(é−1)If Z=™+∏, where X and Y independent: r[(V)=r\(V)×r](V)Finding probabilities: P(X=,)=$"!r(")(0)PMF to PGF: Want answer as a power seriesFill °(™=7)into ∑∂!°(™=7)V!#WX<$using PMF andgetrid of sigma notationusing geometric sum orbinomial expanbackwardsPMF to PGF: PGF should be of the form P(™=0)+SP(X=1)+S+°(X=2)+...Moment Generating Function©7(V)=±(D>7)using either discrete or continuous Expected value defUniform=4)*&4+*(1&/)>, Exponential=NN&>, Normal=DU>)$!R!>!Binomial=[(1−ä)+äD>]", Poisson =DN(4*&$), Geometric=H4*$&($&H)4*Goodness of FitK+2/O2=∑(^&_)!_. Reject if K+2/O2>K+2<'>'2/OSpearman’s Rank1−6∑@+,(,+−1)Expectation Algebra±(#™±()=#±(™)±(, ≥i∫(#™±()=#+≥#Q(™)hEX and Y independent:±(™∏)=±(™)±(∏), ≥#Q(#™±(∏)=#+≥#Q(™)+(+≥#Q(∏)Unbiased EstimatorsK=∑K,,U=ª∑K+,−1−(∑K)+,(,−1)=ª∑(K−K̅)+,−1=Ω,,−1∞Central LimitTheoremK~Æ}é,∞+,~Sample Proportionä̂~ÆWä,ä(1−ä),XTest StatisticsZ=H`&Ha,($#,)",Z=(H$b&H!b)&(H$&H!)H`($&H`)a$"$)$"!, Z/T=7̅&U/√", Z =\$dddd&\!dddd&(U$&U!)a/$!"$)/!!"!, T=\$dddd&\!dddd&(U$&U!)=,a$"$)$"!Where ä$ø=7$"$,ä+¿=7!"!,ä̂=7$)7!"$)"!UH=Ω("$&$)=$!)("!&$)=!!"$)"!&+Differentiation and IntegrationDerivativessin&$E(K)⇒:1(7)a$&e:(7)f!cos&$E(K)⇒−:1(7)a$&e:(7)f!tan&$E(K)⇒:1(7)$)e:(7)f!sec&$E(K)⇒:1(7):(7)ae:(7)f!&$"TUD"&$E(K)⇒−:1(7):(7)ae:(7)f!&$cot&$E(K)⇒−:1(7)$)e:(7)f!sinhE(K)=fZ(K)coshf(K)coshE(K)⇒f′(K)sinhf(K)tanhE(K)⇒f′(K)sech+f(K)sinh&$E(K)⇒E′(K)ï1+E(K)+cosh&$E(K)⇒E′(K)ïE(K)+−1tanh&$E(K)⇒E′(K)1−E(K)+Integrals•∫$g/!&(17)!@K=$1sin&$.17/6+"•∫$&g/!&(17)!@K=$1cos&$.17/6+"•∫$/!)(17)!@K=$/1tan&$.17/6+"•∫$g/!)(17)!@K=$1sinh&$.17/6+"=$1lno(K+ï((K)++#+p+"•∫$g(17)!&/!@K=$1cosh&$.17/6+"= $1lno(K+ï((K)+−#+p+",K>a•∫$/!&(17)!@K=$/1tanh&$.17/6+"=$+/1ã,?/)17/&17?+", |K|<1•∫$(17)!&/!@K=$+/1ã,?17&/17)/?+"Volume of RevolutionAbout K#K-U∶≥=∫Yl+@K1/, About laxis: ≥=∫YK+@l1/Surface area of revolutionCartesian: 2Y∫lΩ1+.08076+@KParametric: 2Y∫lΩ.070>6++.080>6+@VPolar: 2Y∫QU-,3ΩQ++.0<056+@3Arc LengthCartesian:∫Ω1+.08076+@KParametric:∫Ω.070>6++.080>6+@VPolar: ∫ΩQ++.0<056+@3Differential EquationsIntegrating FactorlZ+°(K)l=√(K)⟹h,VDFú#"VTQ=D∫i(7)07Homogeneous and Non-Homogeneous(Second Order)Solution Form:l=l2:+lHcomplementary function l2:(set LHS 0)•real roots:è$,è+⇒l2:=iD%$7+®D%!7=DH7[i"TUℎ(èK)+iU-,ℎ(èK)]if roots are ä±è•repeated roots:è⇒l2:=iD%7+®KD%7=D%7(i+®K)•complex roots:ä+-b⇒l2:=DH7(i"TUbK+®U-,bK)Non-homogeneous lH(plug intoRHS ofequation&solve for unknowns)•l=7(constant) ⟹use l=7•l=K⟹l=ä+bK•l=#K"⟹l=iK"+®K"&$+⋯+©K+Æ•l=#D17⟹l="D17l="KD17(if b matches rootof CF)l="K+D17(if b matches repeated rootof CF)•l=#"TU(KTQ#U-,(Kor l=any linear combo of U-,or "TU⟹l=Ø"TU(K+±U-,(Kl=K(Ø"TU(K+±U-,(K)if b matches repeated rootof CF)Maclaurin SeriesE(K)=E(0)+KEZ(0)+7!+!EZZ(0)+⋯Taylor’s SeriesE(K)=E(#)+(K−#)EZ(#)+(7&/)!+!EZZ(#)+...Maclaurin and Taylors Series Common Results(5)•D7=1+K+7!+!+⋯for all K•ln(1+K)=K−7!++--+⋯for −1<K≤1•U-,K=K−72-!+73j!+⋯+(−1)<7!45$(+<)$)!for all K•"TUK=1−7!+!+76.!+⋯+(−1)<7!4(+<)!for all K•arctanK=K−72-+73j+⋯+(−1)<7!45$(+<)$)!for −1≤K≤1•sinhK=K+72-!+73j!+⋯+7!45$(+<)$)!for all K•coshK=1+7!+!+76.!+⋯+7!4(+<)!for all K•tanhK=K+72-+73j+⋯+7!45$(+<)$)!for −1≤K≤1
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