> #OK to post
> #Timothy Nasralla, HW5, 9/20/21
> #Question 0: Put the given 6*a(n-1) + a(n+3) + 5*a(n+1) = 0 into canonical
> form, then use RecToSeq to find a(1000)
> #Canonical form: Let n = n-3, a(n) = 0*a(n-1) - 5*a(n-2) + 0*a(n-3) - 6*a(n-4)
> read "/Users/tan88/OneDrive - Rutgers University/mw51.txt"
> Help5()
           RecToSeq(INI,REC,N), GrowthC(INI,REC,K) , GrowthCe(REC)

           LeslieMod(SUR,FER): e.g. LeslieMod([9/10,9/10],[0,1,1]);

           LeslieMat(SUR,FER); e.g. LeslieMat([9/10,9/10],[0,1,1]);

> INI := [1, 2, 4, 11]

> REC := [0, -5, 0, -6]

> RecToSeq(INI, REC, 1000)[1000];
181801458979349684211926335397716595590116925130008115201730179162903000957919\

  4774209925134910767767993350034005595962441714858161276739646642515466061813\

  1176283941650552170945484194399749328351304786759734718454695940190410974568\

  4403540309


> #Question 1: Find the growth constant of the given summation in two different
> ways #GrowthC and GrowthCe
> INI := [a, a, a, a, a, a, a, a, a, a]

> REC := [1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

> GrowthC(INI, REC, 1000)
                                 1.999018633

> GrowthCe(REC)
                                 1.999018633

> #Approximately the same.
> #Question 2: Find the growth rate of the given situation with two different
> techniques.
> #Technique 1 FER := [1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2, 1/2,
> 1/2, 1/2, 1/2, 1/2, 1/4, 1/4, 1/4, 1/4, 1/4, 1/4, 1/4, 1/4, 1/4, 1/4, 1/4,
> 1/4, 1/4, 1/4, 1/4]

> SUR := [.99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99,
> .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99, .99]

> LeslieMod(SUR, FER)
  [1                                                                        
  [-, 0.4950000000, 0.4900500000, 0.4851495000, 0.4802980050, 0.4754950250, 
  [2                                                                        

    0.4707400747, 0.4660326740, 0.4613723472, 0.4567586238, 0.4521910375, 

    0.4476691271, 0.4431924358, 0.4387605115, 0.4343729064, 0.2150145887, 

    0.2128644428, 0.2107357984, 0.2086284404, 0.2065421560, 0.2044767344, 

    0.2024319671, 0.2004076474, 0.1984035710, 0.1964195352, 0.1944553399, 

                                                          ]
    0.1925107865, 0.1905856786, 0.1886798218, 0.1867930236]
                                                          ]


> GrowthCe(%)
                                 1.489452967

> #Technique 2 LeslieMat(SUR, FER)

> Eigenvalues(%)

> #After comparing each eigenvalue, it is clear that the first, 1.4894... + 0i
> is the greatest one after applying absolute value, which means that it is the
> growth constant (and it is equal to what was found using the first method)



> #Question 3, find the growth rate of the given equation.
> REC := [0, 0, 0, .16, .41]

> GrowthCe(REC)
                                 0.8879729192




> #Question 4: Using the given information, write a procedure that inputs the
> relevant botanical parameters and outputs the first K terms
> PlantGseq := proc(alpha, beta, gamma, sigma, INI, K) local a,b,i,L,k,newguy,
> REC: a := alpha*sigma*gamma: b := beta*sigma*sigma*gamma*(1 - alpha): REC :=
> [a, b]; if not (type(INI,list) and type(REC,list) and nops(INI)=nops(REC) and
> type(K,integer) and K>=nops(INI)) then print("Yikers!"): RETURN(FAIL): fi: k:=
> nops(INI): L:= INI: while nops(L)<K do newguy:=add(REC[i]*L[-i],i=1..k):
> L:=[op(L),newguy]: od: L: end:
> INI := [100, 80]

> PlantGseq(.5, .25, 2.0, 0.8, INI, 21)
[100, 80, 80.000000, 76.80000000, 74.24000000, 71.68000000, 69.22240000, 

  66.84672000, 64.55296000, 62.33784320, 60.19874816, 58.13305344, 56.13824246, 

  54.21188252, 52.35162481, 50.55520105, 48.82042081, 47.14516882, 45.52740239, 

  43.96514892, 42.45650352]



> INI := [100, 96]

> PlantGseq(.6, .3, 2, 0.8, INI, 21)
[100, 96, 107.5200, 117.964800, 129.7612800, 142.6902221, 156.9139458, 

  172.5546061, 189.7544040, 208.6686153, 229.4681472, 252.3409206, 277.4935912, 

  305.1534130, 335.5702921, 369.0190446, 405.8018797, 446.2511298, 490.7322533, 

  539.6471367, 593.4377253]


> #These values are the same as the ones found in the table.
> 
> #Question 5: Write a function that calculates the growth constant of the same
> table. PlantGseq := proc(alpha,beta,gamma,sigma) local a,b,i,ii: a :=
> alpha*sigma*gamma; b := beta*sigma*sigma*gamma*(1 - alpha); i := 1/2*(a -
> (a*a+4*b)^(1/2)); ii := 1/2*(a + (a*a+4*b)^(1/2)); if (abs(ii)>abs(i)) then
> RETURN(abs(ii)): else RETURN(abs(i)): fi: end:
> PlantGseq(2,2,4,.8)
                                 5.462741700

> #Raising the gamma value allowed for the entire function to increase
> exponentially, so long as you have a positive alpha & beta
> PlantGseq(2,2,.4,.01)
                                0.008944271910

> #As sigma approaches 0, the growth constant shows extinction.
> PlantGseq(.6, .3, 2, .8)
                                 1.099677336

> #Around this level, as well as the given parameters in the original question,
> offered a relatively stable population growth.