Re: a rabbits problem

On 25 Nov., 02:18, r...@xxxxxxxxxxxxxx (Rob Johnson) wrote:
In article <18535560.1195820890508.JavaMail.jaka...@xxxxxxxxxxxxxxxxxxxxxx>,

tommy1729 <tommy1...@xxxxxxxxx> wrote:
fibonacci like

m(0) = 17
f(0) = 23

m(i) = 5 m(i-1) + 4 m(i-2) - m(i-3) + 3 f(i) + 2 f(i-1) + 7 f(i-2)

f(i) = 7 f(i-1) + 5 f(i-2) - f(i-3) + 8 m(i) + 3 m(i-1) + m(i-2)

I assume you are looking for a general solution for m and f.

There is not enough information to compute m and f. We need at least
three prior values. Once we have those, we can compute a recursive
relation for m and f as follows:

The relation above can be rewritten as

[ m(i-1) ]
[ m(i-2) ]
[ 1 -3 ] [ m(i) ] [ 5 4 -1 3 2 7 ] [ m(i-3) ]
[ ] [ ] = [ ] [ f(i-1) ]
[ -8 1 ] [ f(i) ] [ 8 3 1 7 5 -1 ] [ f(i-2) ]
[ f(i-3) ]

Multiplying both sides by [[1 3][8 1]], yields

[ m(i-1) ]
[ m(i-2) ]
[ m(i) ] [ 1 3 ] [ 5 4 -1 3 2 7 ] [ m(i-3) ]
-23 [ ] = [ ] [ ] [ f(i-1) ]
[ f(i) ] [ 8 1 ] [ 8 3 1 7 5 -1 ] [ f(i-2) ]
[ f(i-3) ]

[ m(i-1) ]
[ m(i-2) ]
[ m(i) ] [ 29 13 2 24 17 4 ] [ m(i-3) ]
-23 [ ] = [ ] [ f(i-1) ]
[ f(i) ] [ 48 35 -7 31 21 55 ] [ f(i-2) ]
[ f(i-3) ]

Fill in some rows and divide by -23 to get the recursive relation:

[ m(i) ] [ 29 13 2 24 17 4 ] [ m(i-1) ]
[ m(i-1) ] [ -23 0 0 0 0 0 ] [ m(i-2) ]
[ m(i-2) ] -1 [ 0 -23 0 0 0 0 ] [ m(i-3) ]
[ f(i) ] = -- [ 48 35 -7 31 21 55 ] [ f(i-1) ]
[ f(i-1) ] 23 [ 0 0 0 -23 0 0 ] [ f(i-2) ]
[ f(i-2) ] [ 0 0 0 0 -23 0 ] [ f(i-3) ]

The eigenvectors and eigenvalues of the matrix are

v_1 = -0.674390073391

[ -14319.1830892 ]
[ 21232.7904193 ]
e_1 = [ -31484.4349837 ]
[ -2623.61224799 ]
[ 3890.34825912 ]
[ -5768.69146062 ]

v_2 = -2.24451505133

[ 140790969.554 ]
[ -62726676.513 ]
e_2 = [ 27946649.9793 ]
[ 229082081.822 ]
[ -102063063.327 ]
[ 45472211.588 ]

and the following two with their conjugates:

v_3 = 0.411048660425 cis(-2.72298498109)
= r_3 cis(a_3).

[ (-861530.534436,-306752.72497) ]
[ (2218311.30552,-170140.579448) ]
e_3 = [ (-4762484.17701,2571881.61324) ]
[ (-251879.20628,-92736.9275557) ]
[ (651571.062158,-42954.5983813) ]
[ (-1405797.34077,739819.993421) ]

v_4 = 1.00995407163 cis(-1.01757409525)
= r_4 cis(a_4)

[ (18963808.3492,11706096.5468) ]
[ (4174.55573831,22066192.4636) ]
e_4 = [ (-18587493.5951,11483517.6395) ]
[ (-29410474.3981,3440216.05182) ]
[ (-18199095.073,-22987073.858) ]
[ (9897329.89246,-27290920.9539) ]

Write the initial conditions as linear combinations of the eigenvectors

c e + c e + c Re(e ) + d Im(e ) + c Re(e ) + d Im(e )
1 1 2 2 3 3 3 3 4 4 4 4

Then the general solution is

n n
c e v + c e v
1 1 1 2 2 2

n
+ c (Re(e ) cos(na ) - Im(e ) sin(na )) r
3 3 3 3 3 3

n
+ d (Im(e ) cos(na ) + Re(e ) sin(na )) r
3 3 3 3 3 3

n
+ c (Re(e ) cos(na ) - Im(e ) sin(na )) r
4 4 4 4 4 4

n
+ d (Im(e ) cos(na ) + Re(e ) sin(na )) r
4 4 4 4 4 4

Note that as time progresses, the solution is dominated by e_2 v_2^n,
since v_2 has the greatest absolute value of any of the eigenvalues.
However, this implies that both m and f attain negative values, which
seems bad if this is modelling populations.

In other words, the problem statement implicitly gives us c_2 = 0.
Also, since r_4 is only slightly bigger than 1, we might get some
conditions on c_1 so that no negative values occur while the v_2 term
is still big enough.
If the original problem is well-posed, such reasoning should
exclude all but one initial state. But I'm too lazy to try that out
(and also too little confident that the well-posedness assumption
holds).

hagman



Rob Johnson <r...@xxxxxxxxxxxxxx>
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Relevant Pages

  • Re: a rabbits problem
    ... I assume you are looking for a general solution for m and f. ... Fill in some rows and divide by -23 to get the recursive relation: ... Write the initial conditions as linear combinations of the eigenvectors ... since v_2 has the greatest absolute value of any of the eigenvalues. ...
    (sci.math)
  • Re: a rabbits problem
    ... I assume you are looking for a general solution for m and f. ... Write the initial conditions as linear combinations of the eigenvectors ... (and also too little confident that the well-posedness assumption ... The same reasoning implies that c_4 and d_4 must also be 0. ...
    (sci.math)
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