Some Simple Mathematical Models - SACEMA
Some simple mathematical modelsSome simple mathematical modelsJuly 1, 2011Some simple mathematical models
Some simple mathematical modelsThe birth of modern science Philosophy is written in this grand book the universe, which standscontinually open to our gaze. But the book cannot be understoodunless one rst learns to comprehend the language and to read thealphabet in which it is composed. It is written in the language ofmathematics, and its characters are triangles, circles and othergeometric gures, without which it is humanly impossible tounderstand a single word of it; without these, one wanders about ina dark labyrinth. Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneGalileo's assumption/hypothesis When . I observe a stone initially at rest falling from an elevatedposition and continually acquiring new increments of speed, whyshould I not believe that such increases take place in a mannerwhich is exceedingly simple? .A motion is said to be uniformly accelerated when starting fromrest, it acquires, during equal increments of time, equal incrementsof speed. That is the concept of accelerated motion that is mostsimple and easy, and it is con rmed to be that actually occurring inNature, by exact correspondence with experimental results basedupon it. Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneGalileo's deduction/prediction:Distances are to each other as the squares of the times.Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneGalileo's experiment:The ball and the inclined plane .timetdistancey123451491625Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneGalileo's other assumption: In order to handle this matter in a scienti c way, it is necessary tocut loose from such irksome di culties [as friction or airresistance], otherwise it is not possible to give any exactdescription. Once we have discovered our theorems about thestone, assuming there is no resistance, we can correct them orapply them with limitations as experience will teach. I can show byexperiments that these external and incidental resistances arescarcely observable for bodies travelling over short distances. Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneOur modern version of Galileo's simple mathematical model:Accelerationso velocityhence distanced 2y g,dt 2 Zdy g dt gt ,v (t ) dtZa(t ) y (t ) 1gt dt gt 2 .2Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneBall rolling on an inclined plane:N Fmg sin θ mg cos θθSome simple mathematical models
Some simple mathematical modelsGalileo's falling stoneFalling stone with air resistance:Taking into account the air resistance kmvshould improve ourreal-life modelling.mass accel. force,sohenceand therefored 2ydy mg km ,2d tdtdvma m mg kmv ,dtdva (t ) g kv .dtma mSome simple mathematical models
Some simple mathematical modelsGalileo's falling stoneSeparating variables and integrating:dvg kvZhence 1kZ dt ,g kv ) t c1 ,ln(g kv ) kt c2 .Taking exponentials,g kv c3 e kt ,gsov (t ) c4 e kt ,kg as t .kso thatln(This equation tells us that in the long run the velocity will approacha constant limiting valueg /k , called the terminal velocity.But wecould have found the terminal velocity much more easily. How?Some simple mathematical models
Some simple mathematical modelsGalileo's falling stoneHow can we test our model?What were our assumptions?When might it be necessary to improve our model?How could we do this?Some simple mathematical models
Some simple mathematical modelsGalileo's and Huygens' simple pendulumGalileo's rst observation One must observe that each pendulum has its own time ofvibration so de nite and determinate that it is not possible to makeit move with any other period than that which nature has given it.For let any one take in his hand the cord to which the weight isattached and try, as much as he pleases, to increase or decrease thefrequency of its vibrations; it will be time wasted. Some simple mathematical models
Some simple mathematical modelsGalileo's and Huygens' simple pendulumGalileo's assumptions:That the angle of swing is small, and that friction/resistance maybe ignored.Galileo's experimental conclusion: As to the times of vibration of bodies suspended by threads ofdi erent lengths, they bear to each other the same proportion asthe square roots of the lengths of the thread; or one might say thatthe lengths are to each other as the squares of the times. Some simple mathematical models
Some simple mathematical modelsGalileo's and Huygens' simple pendulumθLMFOur modern modelling of the pendulum:From the geometry,Force towards centreysSmgy arc MS Lθ.mgF mg sin θ mg θ yLSome simple mathematical models
Some simple mathematical modelsGalileo's and Huygens' simple pendulumForceso that givingGeneral solution:where mass acceleration,mgd 2yy F m 2,Ldt2gd y y.dt 2Ly (t ) A sin qgLt α ,A is the amplitude , α is the phase ,The period of this function is 2πqLgqgLis the frequency ., in harmony with Galileo'sexperimental conclusion.Some simple mathematical models
Some simple mathematical modelsGalileo's and Huygens' simple pendulumHow can we test our model?What assumptions have we made in this model?When might it fail to model vibrations accurately?How could we improve it?Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsLiber Abbaci (Book of Counting), 1202, by Leonardo of Pisa, alsocalled Leonardo Fibonacci son of Bonacci.Problem: A certain person places one pair of rabbits in a certainplace surrounded on all sides by a wall. We want to know howmany pairs can be bred from that pair in one year, assuming that itis in their nature that each month they give birth to another pair,and in the second month after birth each new pair can also breed.1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, · · ·Fibonacci's answer is 377 pairs of rabbits, and his sequence ofnumbers is now known as theFibonacci sequence.Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsLeonardo's solution: After the rst month there will be two pairs, after the second,three. In the third month, two pairs will produce, so at the end ofthat month there will be ve pairs. In the fourth month, three pairswill produce, so there will be eight pairs. Continuing thus, in thesixth month there will be ve plus eight equals thirteen, in theseventh month, eight plus thirteen equals twenty-one, etc. Therewill be 377 pairs at the end of the twelfth month. For the sequenceof numbers is as follows, where each is the sum of the twopredecessors, and thus you can do it in order for an in nite numberof months. Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbits( B , Y , A)TotalB(1, 0, 0)1Y(0, 1, 0)1?A -B(1, 0, 1)2?RA -B Y(1, 1, 1)3?R RA -B Y A - B(2, 1, 2)5(3, 2, 3)8?Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsAfternmonths, let there be(Bn , Yn , An )pairs of babies, youngrabbits and adults, respectively. Thenso thatAn 1Bn 1Yn 1An 1 An Yn An Yn Bn Bn 1 Yn 2 .Let the total number of (pairs of ) rabbits atnmonths be given byFn An Bn Yn .Then we deduce thatFn 2 Fn 1 Fn ,forn 0,F0 F1 1.Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbits1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, · · ·The number of pairs of rabbits after Fibonacci's 12th month isas he starts withbabies. We haveF1 , not F0 , with a pair of young rabbits, notF13 377. In the next month after that,F13 ,F11 233 pairs will produce;F98 pairs will produce.and in the 100th month,Leonardo got there without much fuss. But our notation andtechnique is very powerful and can be applied to a host of otherproblems.Is there a quick way of arriving at this numberFibonacci numbers likeF100 ?F13 , or higherSome simple mathematical models
Some simple mathematical modelsFibonacci's rabbits Fn 2Fn 1 F2F1 1110 Fn 1Fn ,forn 0.Hence F3F2 F13F12 1110111011101110 F1F0 1 ,1 2 11 12 11 , 377233 .Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsThis same model for population growth can be applied to othersituations. Assuming that each branch of a tree gives rise to a newbranch, but only after skipping a season's maturation period, weobtain a visually very plausible model:85321Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsIn certain species of bee, the forebears follow a Fibonacci law.Female bees are born from the mating of male and female, malebees are born asexually to single females.MExercise:FMFMFMProve that the numberBn3 great grandparents2 grandparents1 parent1 male beeof bee ancestors in thenthgeneration backward, satis es the Fibonacci generating equation.(LetBn Mn Fn , the sum of the number of males and females,Mn , Fn , Mn 1 , Fn 1 .)and write down the equations connectingSome simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsFibonacci numbers also arise from the following problem. Why?Problem: Taking either one or two steps at a time, how manydi erent ways are there to climb n steps?Some simple mathematical models
Some simple mathematical modelsFibonacci's rabbitsHow accurate is the Fibonacci model for the biologicalsituations we are modelling?What were the assumptions we made?What could be done to improve the model?Some simple mathematical models
Some simple mathematical modelsRed blood cellsThese give your blood its red colour.Their function is to carry oxygen around the body.When there are too few in your body you su er fromanaemia.They are manufactured in your bone marrow, and each cell lives forabout 4 months before dying, probably in your kidney or spleen.You should normally have about 2 1013red blood cells in yourbody at any moment, unless you live in Lesotho. Why?Your bone marrow produces about 28second, or 1728 10 per day. 106red blood cells perSome simple mathematical models
Some simple mathematical modelsRed blood cellsthat a xednumber b of cells are produced every day, andproportion m of existing cells die each day.Problem 1:What percentage of cells normally survive each day inAssume that a xedthe equilibrium situation?Problem 2:Is this equilibrium stable? On our assumptions above,what will happen if the red blood cell count is suddenly di erentfrom normal?Some simple mathematical models
Some simple mathematical modelsRed blood cellsCn be the number of red blood cells present at day n.m be the proportion of red blood cells that die each day.Let b be the number of red blood cells produced each day in theLetLetbone marrow.ThenCn 1 (1 m)Cn b.For equilibrium,we haveCn C , say, for all n.C (1 m)C b,m b C17282Substituting in the equationhence 108 1013mC b, 0.00864.Thus, the percentage of red blood cells that die each day is0.864%, so that 99.136% of them survive each day.Some simple mathematical models
Some simple mathematical modelsRed blood cellsNext, supposeCn 6 C , and let Dn Cn C , so Dnis thedeviation from the normal. ThenHenceDn 1 Cn 1 C (1 m)Cn b C (1 m)Cn mC C (1 m)(Cn C ) (1 m)Dn .Dn (1 m)n D0 0, so Cn C .The equilibrium situation is astable one:after a small deviationthe system will return to the equilibrium.Some simple mathematical models
Some simple mathematical modelsRed blood cellsHow can we test our model?What assumptions have we made in this model?When might it fail to model blood cell count accurately?How could we improve it?Some simple mathematical models
Some simple mathematical modelsCrop yields and bacteria culturesIn the Chinese classic Jiu zhang suan shu (The Nine Chapters ofthe Mathematical Art), written over 2000 years ago, the followingproblem is solved on the Chinese counting board:Problem:Now there are 3 classes of paddy: top, medium andlow grade. Given 3 bundles of top grade paddy, 2 bundles ofmedium grade paddy and 1 bundle of low grade paddy, the yield is39dou.For 2 bundles of top grade, 3 bundles of medium grade and1 bundle of low grade, the yield is 34dou.And for 1 bundle of topgrade, 2 bundles of medium grade and 3 bundles of low grade, theyield is 26Answer:dou.How much does one bundle of each grade yield?dou perdou; and low gradeTop grade paddy yields nine and a quarterbundle; medium grade paddy four and a quarterpaddy two and three quartersdou.Some simple mathematical models
Some simple mathematical modelsCrop yields and bacteria cultureshigh grade123medium grade232low gradeyieldExercise:3112634393x 2y z 39(1)2x 3y z 34(2)x 2y 3z 26(3)Solve the Chinese problem by doing row operations onthe matrix of coe cients.Some simple mathematical models
Some simple mathematical modelsCrop yields and bacteria culturesIn the 21st century, three species of bacteriaB1 , B2 , B3are mixedtogether in the same culture in a laboratory. It is known that eachrequires three types of food,B1B2B3F1 , F2 , F3 , and:F1 , F2 , F3consumes F1 , F2 , F3consumes F1 , F2 , F3consumesat the ratesat the ratesat the rates(1, 2, 2);(3, 2, 1);(8, 4, 1).It is found experimentally that when 2750 units ofF1 , and 500 units of F3F1 , 1500 units ofare supplied per day, a constant situation isreached in which all food is consumed.Problem 1:Problem 2:Find the amounts of bacteria present.Which set of constant states is possible, and which isnot, regardless of the food supply?Some simple mathematical models
Some simple mathematical modelsCrop yields and bacteria culturesSolution:Suppose that there are amountsX1 , X2 , X3 ,respectively, of the three bacteria present. Then 138 224211SinceX1 X2 X3 2750 1500 ,AX F .500det (A) 0, so there are many possible solutions:(X1 , X2 , X3 ) (t 250,1000Notice that for positive values of the250or t 10003 .Xi 3t , t ), t R.we must haveSome simple mathematical models
Some simple mathematical modelsCrop yields and bacteria culturesFor the second question, consider the mapf : R3 R3 ,f (X ) AX .The rank is easily found to be 2, so the kernel of f has dimension 1de ned by(it's a line) and the range is the 2-dimensional plane through theorigin, spanned by the two image vectors: f 1 1 0 2 ,02 f 0 3 1 2 .01Taking the vector product of these two, we get a vector in direction( 2, 5, 4)perpendicular to the required plane, hence the equationof the plane is2x 5y 4z 0.Some simple mathematical models
Some simple mathematical modelsCrop yields and bacteria culturesThe range offis the plane2x 5y 4z 0.One point on this plane is that experimentally discoveredequilibrium state(2750,1500, 500).But all possible equilibrium states must lie on this plane!Some simple mathematical models
Some simple mathematical modelsAge-strati ed populations & a butter y's life cycleConsider a population of individuals each of which gives birth, livesfor a while, then dies. Assume there is a maximum life span, anddivide the population intokage-groups, numberinga1 (n), a2 (n), . . . , ak (n) individuals at time n, measured in years orbreeding seasons. Suppose thatk1everyone in stage2the probability that an individual in classyear/season is3dies at the end of the year/season;pj ;the probability that an individual in classexactly one individual isfjjsurvives anotherjgives birth to(the fertility rate).Some simple mathematical models
Some simple mathematical modelsAge-strati ed populations & a butter y's life cyclek 4.Leslie matrix:For example, takeby theThe numbers in the next season are givena1 (n )a1 (n 1 )f1 f2 f3 f4 a2 (n 1) p1 0 0 0 a2 (n) a3 (n 1) 0 p2 0 0 a3 (n) .a4 (n )a4 (n 1 )00p3 0 AfterN years we will have N a1 (N )f1 f2 f3 f4a1 (0) a2 (N ) p1 0 0 0 a2 (0) a3 (N ) 0 p2 0 0 a3 (0) .a4 (N )00p3 0a4 (0) Some simple mathematical models
Some simple mathematical modelsA butter y's life cycleConsider the four stages of a butter y's life cycle: egg, caterpillar,pupa (or chrysalis) and adult butter y.fEGGp BUTTERFLYCATERPILLARqro/PUPA(CHRYSALIS)Clearlyf1 f2 f3 0.Assume that a butter y succeeds inf , and then always lays NThen f4 must be replaced by fN . Let the respectivesurviving to lay eggs with probabilityeggs.probabilities of an egg, caterpillar and pupa surviving to move intothe next class bep, q, r .Some simple mathematical models
Some simple mathematical modelsA butter y's life cycleLetEn , Cn , Pn , Bnbe the numbers of individual eggs, caterpillars,pupa, and butter ies, in a population after thenth season.Thenthe numbers for the next season will be0 0 0 fNEnfNBnEn 1 Cn 1 pEn p 0 0 0 Cn Pn 1 qCn 0 q 0 0 Pn .rPn0 0 r0BnBn 1 Some simple mathematical models
Some simple mathematical modelsA butter y's life cycleThe parametersf , p, q, rcould be estimated from experimentaldata. The long term fate of the population will depend on theirvalues, and that of 00 p 0 0 q00N.To see how, we look at powers of the matrix:0fN0000r 4 1000 fNpqr 0100001000010En 4 Cn 4 Pn 4 fNpqr Bn 4 EnCn Pn Bn Some simple mathematical models
Some simple mathematical modelsA butter y's life cycleIt's easy to check that matrix equation, but how do we arrive at itin the rst place? When powers of a matrix are in view, it is naturalto nd the eigenvalues.The Cayley-Hamilton theorem says that the matrix will satisfy itscharacteristic equation, and the eigenvalues of this Leslie matrix areeasily seen to be the solutions of the cubicλ3 fNpqrSome simple mathematical models
Some simple mathematical modelsA butter y's life cycleThus we conclude:ifififfNpqr fNpqr fNpqr 1,ultimate extinction occurs;1,there is unstable equilibrium;1,there is unsustainable expansion.How realistic is this model?What were our assumptions?How can we test the model?How can we improve it?What can we learn from it anyway?Some simple mathematical models
Some simple mathematical models Some simple mathematical models July 1, 2011 Some simple mathematical models. Some simple mathematical models The birth of modern science Philosophy is written in this grand book the universe, which stands . Our modern modelling of the pendulum: F mg
mathematical metaphysics for a human individual or society. 1 What Mathematical Metaphysics Is Quite simply, the position of mathematical metaphysics is that an object exists if and only if it is an element of some mathematical structure. To be is to be a mathematical o
. mathematical models based on the stochastic evolution laws . mathematical models based on statistical regression theory . mathematical models resulting from the particularization of similitude and dimensional analysis.When the grouping criterion is given by the mathematical complexity of the process model (models), we can distinguish:
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