MEI Mthemtics in Eduction nd Industry MEI Structured Mthemtics Module Summry Sheets FP, Further Applictions of Advnced Mthemtics (Version B: reference to new ook) Option : Vectors Option : Multivrile Clculus Option : Differentil Geometry Option 4: Groups Option 5: Mrkov Chins Purchsers hve the licence to mke multiple copies for use within single estlishment, April, 6 MEI, Ok House, 9 Epsom Centre, White Horse Business Prk, Trowridge, Wiltshire. BA4 XG. Compny No. 6549 Englnd nd Wles Registered with the Chrity Commission, numer 589 Tel: 5 776776. Fx: 5 775755.
Summry FP Option : Vectors - Vector Product Chpter Pges - Chpter Pges 4-7 Exmple.4 Pge 7 The vector product sin θ nˆ where θ is the ngle etween nd nd nˆ is the unit vector perpendiculr to nd such tht, nd nˆ form right-hnded set of vectors. θ Note tht Properties The vector product is nti-commuttive. (N.B. Commuttive mens tht either order of inry opertion gives the sme result. So ddition is commuttive, since + +, ut sutrction is nti-commuttive since ( ). ) If nd re prllel then. If either or oth nd re then. Note tht does not men tht either or re they my e prllel. (m) (n) mn( ) ( + c) + c This is the distriutive lw the vector product is distriutive over ddition nd sutrction. Bse Vectors The unit vectors prllel to the coordinte xes re i, j nd k. E.g. Find the mgnitude of when,, sinθn ˆ 6 5 sinθnˆ From the sclr product,. cosθ + + 9 7 cosθ sinθ 7 sinθ E.g. If c prove tht + c is prllel to. c c + c ( + c) i.e. + c is prllel to. E.g. Prove tht ( ) ( + ). ( ) ( + ) ( ) + ( ) + + since + E.g. Clculte. 4 4 4 4 5 Chpter Pges -4 Exmple. Pge 6 Exercise A Q. (i), 5(i), 8 i j k, j k i, k i j i i j j k k Component Form If,, FP; Further Applictions of Advnced Mthemtics Version B: pge Competence sttements v, v, v E.g. Find the eqution of the plne ABC, where the coordintes of A, B nd C re (,,), (,,) nd (,, ) respectively. n AB AC where AB nd AC n So eqution of plne is x + y z + d nd is stisfied y A, giving + d d x + y z + ( You cn check tht B nd C lso stisfy this eqution.)
Summry FP Option : Vectors - Plnes Chpter Pges 9- Exmple.5 Pge 9 Exercise B Q. (i), (iii) Chpter Pge Exercise C Q. (i), (ii),, 6 Intersection of two plnes The line of intersection of two plnes lies in oth plnes nd is therefore perpendiculr to the perpendiculr of oth plnes If the norml direction of L is n nd the norml direction of L is n then the direction of the intersecting line is n n. It remins to find one point common to oth plnes, which cn e done y putting, sy, z nd solving the equtions simultneously. If z nowhere on the plnes then the resulting equtions will e inconsistent nd so you should try, sy, x. Angle etween two plnes The ngle etween two vectors is found from the Sclr Product (covered in C4.) The ngle etween two plnes is the ngle etween their norml directions. Exmple.6 Pge If the norml direction of L is n nd the norml direction of L is n then the ngle etween the Exercise B Q. (i), (iii) Chpter Pge Exercise B Q. 4 Chpter Pges 4-6 Exmple.8 Pge 5 Intersection of two lines If the lines re l nd l then the line l will e defined y one point, p, nd direction, n. Any point on this line is then given y p + λn. Similrly, ny point on l is given y q + μn. Equting these two will give three equtions in two unknowns. Find the vlues of λ nd μ from the first two nd check for consistency in the third. If they re consistent then the vlues of λ nd μ will give the point of intersection; if they re not consistent then the lines do not meet. FP; Further Applictions of Advnced Mthemtics Version B: pge Competence sttements v4, v5 plnes is the ngle etween these directions which is given y n. n n n cosθ Fmily (or shef) of plnes If π is the eqution of one plne nd π nother such tht the line of intersection is l, Then the eqution λπ + μπ is the generl eqution of fmily of plnes with the common line l. This is ecuse ny point on the line stisfies π nd π nd therefore λπ + μπ for ll vlues of λ nd μ. E.g. Find the line of intersection of the plnes: L :x+ y z 4 L :x y+ z 6 n, n n n 5 Sustitute z x+ y 4, x y 6 5x x, y x y+ z 5 E.g. Find the ngle etween the plnes defined in the exmple ove. L: x+ y z 4 L :x y+ z 6 n, n n. n 6 n 9+ +, n 4+ + 9 4 cosθ.6 8.7 4 E.g. Find the eqution of the plne tht psses through the point (,, ) nd the line found in the first exmple ove. π :x+ y z 4 π :x y+ z 6 Any plne through the line of intersection of these plnes is given y λπ+ μπ 4λ + 6μ ( λ+ μ) x+ ( λ μ) y+ ( λ+ μ) z 4λ+ 6μ This is stisfied y the point (,, ) λ+ μ+ λ μ λ+ 9μ 4λ+ 6μ λ+ μ A solution to this eqution is μ, λ x+ y+ z 4 x y+ z E.g. You re given l : x+ 4 y z x y z+ l: l: Show tht l nd l intersect ut l nd l do not. Any point on l is ( + λ, + λ,λ) Any point on l is ( 4+ μ, μ,+ μ) These two points re the sme if + λ 4+μ nd +λ μ λ μ 5 nd λ+ μ 4 λ, μ The xnd y vlues re the sme nd sustituting into the z coordintes lso gives the sme vlue. So the lines meet t (,,). Any point on l is ( + ν, + ν, + ν) 8 Solving for equl xnd y vlues gives λ, ν 5 5 ut the z vlues re not the sme so the lines do not intersect.
Summry FP Option : Vectors - Distnces Chpter Pges 9- Exmple. Pge Exercise D Q. (i), (ii) Distnce of point from line. If P is point not on line l nd A is ny point on the line nd M is the closest point on the line from P, then the distnce is the length of the line PM. If the direction of l is defined y the unit vector, dˆ, then PM APsinPAM. Since sin θ, tke AP nd dˆ. Then d ˆ AP dˆ AP sin PAM AP sin PAM PM So PM d ˆ AP A θ ˆd M P E.g. Find the distnce of the point (,, ) from x y+ z+ the line ˆ d d 4 d 4 Tke A (,, ) nd P (,, ) AP 5 4 9 ˆ d AP 4 5 4 4 9 56 4 4 ˆ d AP 9 + + 9 Chpter Pge 6-7 Exercise D Q. (i), 4(ii) Distnce of point from plne. If M is the foot of the perpendiculr from P(x, y, z ) to plne then the distnce of P from the plne is PM. The direction of PM is the norml direction of the plne, n. Let the eqution of the plne e x + y + cz + d Tke ny point, R, on the plne. In generl this cn e (x, y, z ), ut if c is not zero then this cn e (,, d / c ). Let the ngle etween PR nd PM e θ. Then the sclr product gives PR. nˆ PR cos θ. nd PM PRcosθ PM PR.ˆ n x x RP, y y n z z c n x x y y c z z RP. ( ) + ( ) + ( ) x + y + cz + d since x + y + cz + d x + y + cz + d n RP.ˆ distnce PR.ˆ n Note tht: If two distnces re opposite signs then the points re on opposite sides of the plne. If the distnce is then the point lies on the plne. FP; Further Applictions of Advnced Mthemtics Version B: pge 4 Competence sttements v6, v7 n x + y + cz + d + + c R P M E.g. Find the distnce of the point (,, ) from the plne x + y z 4 x+ y+ cz+ d + 4 Distnce + + c + + 4 E.g. Show tht the points (6,, ) nd (, -4, 4) re equidistnt from the plne x + y z x+ y+ cz+ d Distnce + + c 6+ 4 For (6,, ), d + + 4 4 4 4 For (, 4, 4), d + + 4 So sme distnce ut opposite sides. E.g. Find the foot of the perpendiculr from the point P(, 5, 4) to the line r + λ. Any point, A, on the line is ( + λ, λ, + λ). + λ λ PA λ 5 λ. + λ 4 λ This direction is perpendiculr to the line. λ λ. λ 4λ 4+ λ+ + λ λ A is,,
Summry FP Option : Vectors - Sclr Triple Product Chpter Pges -6 Exercise E Q. (i), (i) Exmple. Pge Chpter Pges 4-5 Exercise F Q., The Sclr Triple Product Given tht c c sin θ nˆ where nˆ is the unit vector perpendiculr to oth nd c, nd.n ˆ nˆ cos ϕ with nˆ. c (.n ˆ ) c sinθ c sinθ cosϕ.nˆ c ( ) The Sclr Triple Product in component form. c If,, c c, c c c c c c c c,. c. c c c c c c ( c c ) + ( c c ) + ( c c ) c c c Volume of the prllelepiped OAEBCGFD where the three sets of prllel sides re given y OA, OB, OC c is given y G F V.( c) A E C D c O B If the volume is then the four points lie on the sme plne. Volume of the tetrhedron OABC where the three sides re given y OA, OB, OC c is given y V.( c) 6 Distnce etween two skew lines If l is defined y point nd direction n nd l y point nd direction n then the shortest distnce is given y ( ).( n n ) d n n If the shortest distnce is then the lines intersect. FP; Further Applictions of Advnced Mthemtics Version B: pge 5 Competence sttements v8, v9, v, v, v C c O B A E.g. Given tht,, c, find. c.. c + + ( 4) + ( 5) + () 6 Show tht the points A(,, ), B(, 4, 7), C(,, ) nd D(,, 4) re coplnr. AB, AC, AD 6 AB AC 8 AD.AB AC. 4 + + 4 8 E.g. You re given the equtions of two lines: x y z l : x y z k l : (i) Find the distnce etween the lines when k 9. (ii) Find the vlue of k if the two lines intersect., k k n, n n n, 5 n n 9 + + 5 5. 5k k 5 5k d 5 (i) When k 9, d 5 (ii) When d, 5k k ( ) ( n n )
Summry FP Option : Multivrile Clculus - Chpter Pges 4-5 Exmple. Pge 45 Exercise A Q., 7 Exmple. Pge 49 Exercise B Q., Chpter Pges 5-5 Exmple.4 Pge 5 Exercise C Q. (i), (iii), 5 A function of three vriles Just s the function y f(x) represents curve in two dimensions, the function z f(x, y) represents surfce. If the x- nd y-xes re horizontl nd the z-xis is verticl then for ny coordinte pir (x, y) vlue of z cn e found. All the points where z is equl is known s contour. A verticl plne cuts the surfce in wht is clled section. Prtil differentition This is the process of differentiting the function z f(x, y) with respect to x keeping y constnt nd differentiting z f(x, y) with respect to y, keeping x constnt. The nottion is nd E.g. z x + y E.g. Determine the section z of the ove function. z gives x + y In the plne this is prol x y Chpter Pges 55-59 Exmple.5 Pge 58 Exercise D Q. (i), (ii) Differentiility- The Tngent Plne At point on continuous surfce the plne which touches the surfce t point is sid to e the tngent plne t tht point. It contins the tngent of the section of the surfce prllel to the x-xis t tht point nd lso the tngent of the section of the surfce prllel to the y- xis t tht point. The directions of these lines re given y nd. z c mens tht from the point, the chnge in y is zero (ecuse we keep it constnt!) nd for chnge of in x there is chnge of c in z. Alterntive nottion If z f( x, y) then differentiting with respect to x gives. f This cn lso e written or f x ( x, y). FP; Further Applictions of Advnced Mthemtics Version B: pge 6 Competence sttements c, c, c, c7 z x + x y E.g. Find nd when x +, E.g. Find the eqution of the tngent plne to the z x xy+ x surfce + 4 t the point (,, 9). x+ y+ 4 giving the direction x giving the direction n So the plne cn e written x+ y z d nd the eqution is stisfied y the point (,, 9) d + 4 9 5 x+ y z 5
Summry FP Option : Multivrile Clculus - Chpter Pges 6-6 Exercise E Q., Chpter Pges 64-66 Exmple.7 Pge 65 Exercise F Q., Chpter Pges 69-7 Exmple.9 Pge 7 Exercise G Q., 4 Chpter Pges 74-76 Exmple. Pge 75 Exercise H Q., 5 Chpter Pges 77-79 Exmple. Pge 78 Exercise I Q., 4 Directionl derivtives In ny -D representtion, the horizontl direction of line my e denoted y the unit cosα vector u ˆ, where the line mkes n sinα ngle of α with the x-xis. Then uˆi uˆigrd f cosα + sin α is the directionl derivtive. Sttionry points A sttionry point on surfce is defined s point where the tngent plne is prllel to the x-y plne. This is point which is locl mximum or minimum of z. This occurs when. The nture of sttionry point my e determined y considering sections or the vlue of z for smll chnges in x nd y. Smll chnges f f For z f( x, y), δ z δx+ δ y This formul is pplicle for ny numer of vriles. f f f For z f( xyw,, ), δ z δx+ δy+ δw w The pproximtion cn e used to estimte the effects of errors in clcultion. The Directionl Derivtive If w g( x, y, z) then the directionl derivtive is w w ˆ u.grd g, where grd g w The surfce g(x,y,z) k. For the point A with position vector on the surfce g(x,y,z) k, the tngent plne is (r ).grd g, where grd g is evluted t point A. The norml line is r + λgrd g. E.g. Find grd f when f. x+ xy Find the grdient on this surfce t (,, ) in the.6 direction uˆ..8 f f + y, xy + y grd f. At A, grd f xy.6 Grdient..8.8 +.6.4 E.g. Investigte the sttionry points on the curve z x 8xy + y + 4x. z x 8xy y 4x + + x 8y+ 4, 8x+ 4y x 8y 4 8x+ 4y x, y, z 7 E.g. in the exmple ove, the sttionry point is (,, 7). z x 8xy y 4x + + When y, z x x 8 + x t x + Also, when x, z y 8y 5 4y 8 t y z z At this point > nd > i.e. the sttionry vlue is minimum. FP; Further Applictions of Advnced Mthemtics Version B: pge 7 Competence sttements c4, c5, c6, c8
Summry FP Option : Differentile Geometry - Chpter Pges 85-9 Exmple. Pge 89 Exercise A Q.,, Chpter Pge 9-97 Exmple.4 Pge 96 Exercise B Q.,, 6 Chpter Pge 99 Chpter Pges - Exmple.6 Pge Exercise C Q. (i), 5 Envelopes The fmily of lines oeying rule is the set of equtions f(x,y,p). The eqution of the envelope is given y the two equtions f( x, y, p), f ( x, y, p) p If p cn e eliminted then the Crtesin eqution results. Alterntively rerrnge to give prmetric equtions x g(p), y h(p). Arc length Crtesin coordintes: y f( x) x dy dy + s dx dx dx + dx x Polr coordintes: r f( θ ) θ dr dr r + s r dθ dθ dθ + dθ θ Prmetric coordintes: x f( θ ), y f( θ ) dx dy + dθ dθ dθ θ dx dy s + dθ dθ θ dθ Volume of solid of revolution The volume of the solid swept out when the curve y f( x) is rotted through π out x the x-xis is given y V π y d x. x The volume of the solid swept out when the curve x f( y) is rotted through π out y the y-xis is given y V π x d y. y Surfce re of solid of revolution The surfce re of the solid swept out when the curve y f( x) is rotted through π out x the x-xis is given y S π yd s. x In crtesin coordintes this ecomes x S π y dx dx x The surfce re of the solid swept out when the curve x f( y) is rotted through π out the y-xis is given y S π xd s. In cresin coordintes this ecomes y S π x dy dy y y y E.g. Find the circumference of circle.. Crtesin coordintes: dy dy x. x + y x + y d x d x y Length of positive qudrnt ( x to x ) x x dy x s dx + + dx dx y x x x x x y + x dx dx dx y x x x x x dx Let x sin θ : cos θ, x cos θ dθ π When x, θ ; when x, θ θ π π s dθ θ π So for whole circle, c 4 π. Prmetric coordintes: x cos θ, y sinθ [ θ] [ θ] θ π dr r r s π θ FP; Further Applictions of Advnced Mthemtics Version B: pge 8 Competence sttements g, g, g dx dy dx dy sin θ, cos θ; + dθ dθ dθ dθ dθ sin θ + cos θ dθ π s π. Polr coordintes + dθ dθ π E.g. To find the surfce re of sphere. x x + y etween x nd x dθ Rotte circle through 6 out the -xis. dy S π y d x where + dx dx dx dy dy x + + d d x y x y x x y dy + dx x x + y + y y y S y. dx dx x dx y π π π y S 4π [ ]
Summry FP Option : Differentile Geometry - Chpter Pges 5-8 Exmple.9 Pge 8 Exercise D Q., Chpter Pge 9- Exmple. Pge Exercise E Q., 6 Chpter Pge 5-7 Exmple. Pge 6 Exercise F Q., Chpter Pge 8-9 Exmple. Pge 8 Exercise G Q., 4 Intrinsic Equtions An lterntive wy to descrie curve is in terms of the rc length, s, with the ngle ψ, which its tngent mkes with fixed direction. We determine the eqution uniquely we lso need the point of the curve where s, the direction where ψ nd lso sense of direction. (ψ is usully mesured in rdins nticlockwise.) dy dx dy tn ψ, cos ψ, sinψ dx Curvture The curvture of curve t point P is the rte of chnge of ψ with s t P. dψ κ If κ is positive, then ψ increses with s nd the curve curves to the left. For curve given in intrinsic form the formul ove cn e used. If the eqution is given in Crtesin coordintes, y f(x) then d y κ dx dy + dx Centre of curvture The circle of curvture of curve t point P is the circle with centre on the norml t P. The rdius is ρ κ dψ We define unit vectors in the direction of the positive tngent nd positive norml to e tˆ cos sin nd ˆ where ˆ ψ ψ n t nd nˆ sinψ cosψ Then, if the vector for P is r nd the vector for the centre, C, is c then c r +ρnˆ The Evolute of Curve As the point P moves long curve, the centre of curvture, C, lso moves. The locus of C is clled the evolute of the curve. As shown, the centre of curvture cn e found for specific point. If, insted, the prmetric point is used then the form of c will e in prmetric form, which will e the eqution of the evolute. dc dρ An lterntive form is nˆ. E.g. The curve with intrinsic eqution s 4 ( cos ψ ) hs sttionry point t the origin. Find the prmtric equtions for the curve. dx dy cos ψ, sin ψ. From the curve 4 sinψ dψ dx dx cosψ 4sinψ sinψ dψ dψ x k cos ψ; x when ψ k x ( cos ψ ) dy dy sinψ 4sinψ ( cos ψ) dψ dψ y k + ψ sin ψ; y when ψ k y (ψ sin ψ) Writing ψ θ gives prmetric equtions x ( cos θ), y ( θ sin θ ) y x 4 is positive constnt), find (i) the rdius of curvture, (ii) the coordintes of the centre of curvture. E.g. For the point P, on the curve 4 (where dy x d y 6x (i) At P,, dx 4 4 dx 4 dy + + dx 4 ρ d y dx 5 4 5 5 64 96 (ii) Norml vector is nˆ 4 5 4 5 Centre of curvture is + 96 5 4 4 75 96 7 i.e. which is, 4 + 4 96 FP; Further Applictions of Advnced Mthemtics Version B: pge 9 Competence sttements g4, g5, g6, g7, g8
Summry FP Option 4: Groups - Chpter 4 Pges - Chpter 4 Pge Exercise 4A Q., Chpter 4 Pge 5-4 Exercise 4B Q., Exmple 4. Pge 4 Exercise 4C Q.,, Sets nd opertions A set is collection of items hving common property. A inry opertion is n opertion comining two items of set to form third item. The result of inry opertion is often referred to s the product (though most people restrict this word to the result of the inry opertion multiply.) The opertion is closed with respect to set if, for ll elements x, y of the set, the product x*y lies in the set. The opertion is commuttive if, for ll x, y є S, x*y y*x. The opertion is ssocitive if, for ll x, y, z є S x*(y*z) (x*y)*z An identity element, e є S, is n element such tht e*x x * e x for ll x є S. The inverse x, of n element x is n element such tht x * x x * x e Modulr rithmetic Within the set of integers, two numers re sid to e congruent modulo m if the difference etween them is multiple of m. E.g. 7 (mod ) ecuse 7. In modulo rithmetic ll integers cn e reduced to the numers, or. Groups A Group (S,*) is non-empty set S with inry opertion * such tht * is closed in S - i.e. for ll x, y є S, x*y є S * is ssocitive in S i.e. for ll x, y, z є S, x*(y*z) (x*y)*z There is n identity element, e є S such tht e*x x * e x for ll x є S For every element of the set, x, there exists n inverse element x є S such tht x * x x * x e An element tht is its own inverse is sid to e self-inverse. If, in ddition, the opertion is commuttive, then the group is sid to e Aelin. The tle showing the comintion of elements is clled the Cyley Tle. In ech row nd column ech element will occur once nd once only. The order of Group The order of finite group is the numer of elements in the group. The order of n element, x, is the smllest positive integer n such tht x n e. Properties of Group The identity element is unique. Ech element hs n unique inverse. If x*y x*z then y z (known s the cncelltion lw) The eqution x hs the unique solution x. E.g. The inry opertion dd is closed with respect to the set of positive numers ecuse the ddition of ny two positive numers is positive. The inry opertion sutrct however is not closed. For exmple 4 5 is not positive numer. E.g. The inry opertion dd is commuttive ecuse the ddition of ny two positive numers is sme whichever wy round you comine the numers. I.e. 4 + 5 5 + 4. The inry opertion sutrct however is not commuttive. For exmple 4 5 5 4. E.g. The inry opertion dd is ssocitive: E.g. 6 + (5 + 4) 6 + 9 5 nd (6 + 5) + 4 + 4 5 The inry opertion sutrct however is not ssocitive: E.g. 6 (5 4) 6 5 nd (6 5) 4 4 E.g. Consider the set G nd the inry opertion of multipliction modulo, where G {,, 7, 9,,, 7, 9} Show tht G is group under this opertion. The comintion tle is 7 9 7 9 7 9 7 9 9 7 9 7 7 7 9 7 9 9 9 7 9 7 7 9 7 9 9 7 9 7 7 7 9 7 9 9 9 7 9 7 (i) The set is closed under the opertion (ii) Multipliction is ssocitive (iii) The identity element is (iv) There is n inverse for ech element (i.e. ppers in ech row nd ech column). E.g. Stte the order of G nd find the order of ech element. The order of the group is the numer of elements i.e. 8. The order of ech element, x, is the smllest integer, n, such tht x n For the order is. For 9,, 9, the order is (These re the elements where is in the leding digonl) For, 7,, 7, the order is 4. (i.e. 9, 9 7, 7 ) FP; Further Applictions of Advnced Mthemtics Version B: pge Competence sttements,, 4
Summry FP Option 4: Groups - Chpter 4 Pge 46-49 Exercise 4D Q., 8 Chpter 4 Pges 5-5 Exmple 4.4 Pge 5 Exercise 4E Q., 6 Chpter 4 Pges 54-56 Exercise 4F Q., 4 Chpter 4 Pges 59-6 Exmple 4.5 Pge 6 Exercise 4G Q., 5 Isomorphism Consider two groups with the sme order. If the mpping of the elements of one group to the other preserves the structure then the two groups re sid to e isomorphic. Sugroups A sugroup of group (S, *) is non-empty suset of S which forms group under the opertion *. Every group hs trivil su-group {e}. As with fctors of numer, you my lso consider the set s sugroup of itself. A proper sugroup is sugroup tht is not one of the ove. Lgrnge s Theorem The order of ny su-group is fctor of the order of the group. For instnce, group of order 4 cn hve sugroups only of order, or 4, ut not. Cyclic groups If memer of group is x then x, x, etc re lso memers of the group. There must e smllest numer, m, such tht x m e. m must e less thn or equl to n, the order of the group. If m n, then ech memer of the group is power of x. x is sid to generte the group nd the group is sid to e cyclic. The group {e,,, } with 4 e is cyclic. Groups with order prime numer must e cyclic. This is ecuse the order of ech element is fctor of p, the order of the group (Lgrnge s Theorem). Since e is the only element with order, ll others much hve order p nd so must generte the group. FP; Further Applictions of Advnced Mthemtics Version B: pge Competence sttements, 5, 6, 7, 8 E.g. List ll the su-groups of G in exmple on previous pge. The identity element {} lwys forms su-group or order. Other proper sugroups re found y scrutinising the comintion tle. Any element with order will, with e form proper sugroup of order if e is in the leding digonl position for the element. So {,9}, {,} {,9}re proper sugroups of order. By Lgrnge s Theorem there cnnot e ny sugroups of order. Scrutiny of the comintion tle will revel tht the top left lock of 4 elements contins only those 4 elements. Therefore {,,7,9}is proper sugroups of order 4. There re two others: {, 9,, 7}, {, 9,, 9} Note the comintion tle for these sugroups which re n extrction of the comintion tle of G. 9 7 9 9 9 7 9 9 9 9 7 9 9 9 7 9 9 9 7 7 9 9 9 9 The set G itself is sugroup of itself, of order 8. E.g. List the sugroups of G tht re isomorphic to one nother. Sugroups of order will lwys e isomorphic to one nother. I.e. {, 9}, {, } nd {, 9} Likewise, sugroups of order 4 will e isomorphic to ech other providing the identity element is in the sme plce within their tles. This is so for {,, 7, 9} nd {, 9,, 7}. [ The sugroup {, 9,, 9} hs the identity element in every plce of the leding digonl.] E.g. Prove tht G is not cyclic. For ll elements in G, there is lest vlue of n for which x n. We hve seen ove tht the vlues of n for the elements re, nd 4. In order for G to e cyclic there must e t lest one element, x, for which x 8 with 8 the smllest such vlue.
Summry FP Option 5: Mrkov Chins - Chpter 5 Pges 7-75 Exercise 5A Q. (i), 5 Chpter 5 Pges 77-79 Exmple 5. Pge 77 Exercise 5B Q. (i), Chpter 5 Pges 8-87 Exercise 5C Q., 6 Terminology A sequence of events where the proility of n outcome t one stge depen only on the outcome t the previous stge is known s Mrkov Chin. The conditionl proilities of pssing from one stge to the next re clled trnsition proilities. They re most usefully rrnged in squre trnsition mtrix, P. Ech column of P is proility vector. It follows tht the sum of elements of ech column is. If the column vector p represents the proilities t one stge nd P is the trnsition mtrix then Pp represents the proilities t the next stge. E.g. if there re, t ny stge, two outcomes then P is mtrix. If the two outcomes re A nd B, then P is given y P(A A) P(A Β) P P(B A) P(B Β) Trnsition Mtrices If t ny stge there re three sttes, then the trnsition mtrix P will e mtrix. There will e 9 trnsition proilities. The clcultion of the product of these mtrices (nd those which re lrger!) cn e tedious so it is importnt tht you re le to use your clcultor effectively. Usully (ut not lwys) the trnsition mtrix from stge to stge is the sme s tht from stge to stge. The trnsition mtrix from stge to stge is therefore P. You my therefore e required to clculte P n for ny integer vlue, n. Equilirium proilities If, for some given strting stte, successive sttes converge to fixed proilities then those vlues re clled equilirium proilities. This mens tht t limiting stge which gives proility vector p then Pp p. This my occur in two situtions: (i) Whtever the initil column proility this stge is eventully reched. (ii) If the initil column proility is the equilirium proilities then this stte will e constnt t ll stges. This column proility vector cn e found s follows p P p Pp p If nd then gives c d p p + p p cp + dp p These cn e solved simultneously to find p nd p..5.8 E.g. P find P..5..5.8.5.8.5 +.4.4 +.6 P.5..5..5 +..4 +.4.65.56.5.44 A wether forecster clssifies the wether s dry or wet. If it is wet on one dy then the proility tht it is wet the next dy is.7. If it is dry one dy then the proility tht it is dry the next is.8. (i) Form the trnsition mtrix. (ii) If it is dry one Mondy wht is the proility tht it will e wet on Wednesdy? (i) wet dry wet.7. P. dry..8 (ii) The stte for Mondy is M The stte for Tuesdy is T PM The stte for Wednesdy is W PT P M.7..7..55. W..8..8.45.7..7 so the proility tht it will e wet on Wednesdy is.. E.g. Find the Equilirium proilities for the mtrix ove..7. x x Solve..8 y y.7x+.y x.y.x y x Also x+ y x+ x x y 5 5 FP; Further Applictions of Advnced Mthemtics Version B: pge Competence sttements m, m, m, m4, m6
Summry FP Option 5: Mrkov Chins - Chpter 5 Pges 9-9 Exmple 5. Pge 9 Exercise 5D Q. (i),(iii), Chpter 5 Pges 96- Exmple 5. Pge 98 Exmple 5.4 Pge 99 Exercise 5E Q., Run lengths in Mrkov Chins The run length is the numer of no chnge trnsitions. This is one less thn the numer of times the stte is repeted. So for ny stte A, if p is the proility tht the system remins in tht stte t the next stge. Hence A the proility tht it chnges from stte A to stte is p. Let X represent the numer of further consecutive stges in which the stte of the system is A, given tht it is initilly in stte A then P( X r) p r ( p) for r,,,, 4,... The expected run length is given y E( X) [ r P( X r)] p( p) + p ( p) + p ( p) +... ( ) p( p) + p+ p +... p( p) ( p) Clssifying Mrkov Chins Regulr chins A trnsition mtrix is regulr if some power of the mtrix hs only positive entries. A Mrkov Chin is regulr if its trnsition mtrix is regulr. In regulr Mrkov chin it is possile to pss from ny stte to ny other stte nd there is unique limiting proility vector. Rndom Wlks This is n expression tht descries process of moving etween ordered sttes. Periodic chins A periodic Mrkov chin is one where successive powers of P form pttern where there is vlue of k such tht P k P. The period of the Mrkov chin is k where k is the smllest vlue for which this is true. Reflecting rriers A Mrkov chin hs reflecting rrier if following one prticulr stte, the next stte is inevitle. In the corresponding column of the trnsition mtrix there is in ech position including position (i,i) except which hs the entry. Asoring sttes A Mrkov chin hs n soring stte if the system is unle to leve tht stte once it hs reched it. In the corresponding column of the trnsition mtrix there is n entry of in position (i,i) nd elsewhere. FP; Further Applictions of Advnced Mthemtics Version B: pge Competence sttements m5, m7, m8, m9, m p p E.g. Expected run length of wet dys for exmple on previous pge. p.7 Length p.7 E.g. The following mtrix represents the trnsition mtrix for the purchse y customers of three rn of commodity, AB, nd C. A B C A 4 B 8 6 C 8 6 (i. e. If cutomer uys rnd A then the chnce of him uying it gin is ; otherwise there is n equl 4 chnce of uying B or C.) (i) Find the equilirium proilities. (i.e. the long-run proportion of purchses.) (ii) A customer uys rnd A.Find the expected numer of consecutive further occsions on which this customer purchsed rnd A. (i) PX X x+ y+ z x 4 x+ y+ z y 8 6 4 x ( y+ z) with x+ y+ z 4 x, y z 7 4 p (ii) Expected run of purches of A where p p is the proility tht A is purchsed gin fter A 4 4 E.g. The following mtrix hs n soring stte. A B C.5.8 P.... If the strt is t A then it will remin there. On further steps, B will trnsfer to A with proility.5, B with proility. nd C with proility.. These proilities will reduce to, so A is n soring stte.