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ganeshie8

  • one year ago

The starting population of a certain kind of migratory birds in a forest is \(10000\). \(3\%\) of the population die each year. Reproduction rate is \(4\%\) per year and \(500\) birds immigrate to the forest each year. Estimate the population of birds at the end of \(30\) th year

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  1. anonymous
    • one year ago
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    let \(P(n)\) denote the population after \(n\) years. we're given that \(P(0)=10000\) and the recurrence relation $$P(n)=0.04P(n-1)-0.03P(n-1)+500\\P(n)=0.01P(n-1)+500$$

  2. anonymous
    • one year ago
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    so $$P(n)-0.01P(n-1)=500$$ignoring our initial condition, a constant particular solution \(P(n)=k\) gives \(k-0.01k=500\implies 0.99k=500\implies k=500/0.99=\frac{50000}{99}\)

  3. ganeshie8
    • one year ago
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    Ahh that looks much better than what im doing.. i ended up with a geometric series after writing out the first few terms

  4. anonymous
    • one year ago
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    now consider the homogeneous recurrence relation $$P(n)-0.01P(n-1)=0$$suppose that we had a geometric sequence \(P(n)=r^n\) which gives $$r^n-0.01r^{n-1}=0\\r^{n-1}\left(r-0.01\right)=0\implies r=0.01$$ so our homogeneous solution looks like \(P(n)=C(0.01)^n\) and thus the general solution to our original recurrence is \(P(n)=C(0.01)^n+\frac{50000}{99}\)

  5. anonymous
    • one year ago
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    now we want to solve for \(C\) using our initial condition: $$P(0)=10000\\C(0.01)^0+\frac{50000}{99}=10000\\C=10000(1-5/99)\approx 9494.95$$

  6. anonymous
    • one year ago
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    oops, I messed up the recurrence -- it should've actually read

  7. ganeshie8
    • one year ago
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    \(P(n)=C(1.01)^n+\frac{50000}{99}\) ?

  8. anonymous
    • one year ago
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    $$P(n)=P(n-1)+0.04P(n-1)-0.03P(n-1)+500\\P(n)=1.01P(n-1)+500$$ in which case we'd get \(r=1.01\) and our particular solution would be different as well

  9. ganeshie8
    • one year ago
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    Ohk gotcha!

  10. anonymous
    • one year ago
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    $$k=1.01k+500\\-0.01k=500\\C=-50000$$ so the general solution is \(P(n)=C(1.01)^n-50000\)

  11. dan815
    • one year ago
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    |dw:1437282863276:dw|

  12. dan815
    • one year ago
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    would that work?

  13. anonymous
    • one year ago
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    now applying our initial condition: $$10000=C(1.01)^0-50000\\60000=C$$ so our model is $$P(n)=60000(1.01)^n-50000$$

  14. anonymous
    • one year ago
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    now we plug in \(n=30\) so $$P(30)\approx 30870.93$$

  15. anonymous
    • one year ago
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    in general, the theory of linear recurrence relations with constant coefficients $$x_n+c_1x_{n-1}+c_2x_{n-2}+\dots+c_{n-1}=g(n)$$ mirrors that of hte analogous differential equations in that we can first find a particular solution and also investigate the general homogeneous solution of the homogeneous case \(g=0\) using the ansatz \(x_n=r^n\)

  16. anonymous
    • one year ago
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    the method of undetermined coefficients works here, too, only with slightly different 'rules' since we're no longer dealing with differentiation, and even variation of parameters carries over :-)

  17. dan815
    • one year ago
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    wolfram gives 30,991.5, with DE method http://www.wolframalpha.com/input/?i=da%2Fdt%3D0.01*a%2B500%2C+a%280%29%3D10000%2C+t%3D30

  18. anonymous
    • one year ago
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    yep, @dan815 your approach will overshoot since this is essentially the same as dealing with continuous compounding

  19. ganeshie8
    • one year ago
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    I see what you're doing to get a particular solution, but I'm not so sure why making the function a constant, \(P(n)=k\) is working

  20. dan815
    • one year ago
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    yeah i think

  21. anonymous
    • one year ago
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    @ganeshie8 we pick a constant because we want \(P(n)-1.01P(n-1)\) to be a constant; if there were non-constant terms then they would not be entirely annihilated -- consider \(P(n)=An+B\) which gives \(An+B-1.01(A(n-1)+B)=-0.01An+(1.01A-0.01B)\)

  22. anonymous
    • one year ago
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    we'd want \(-0.01An\) to be identically zero, which would make \(A=0\) so this reduces to a constant solution \(P(n)=B\). it's the same logic that is used in the method of undetermined coefficients with linear differential equations

  23. ganeshie8
    • one year ago
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    so a partiuclar solution need not be a function of independent variable (\(n\)) ? oh nvm i think i get it, we just want "some" particular solution and the constant funciton is good enough i tihnk xD

  24. dan815
    • one year ago
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    |dw:1437283645284:dw|

  25. anonymous
    • one year ago
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    yeah, the particular solution is just any \(P_p\) that satisfies the recurrence regardless of initial conditions. its purpose is that we can then consider solutions of the equation in \(P_p+P_c\); since our equation is linear, this reduces to: $$(P_p+P_c)(n)-1.01(P_p+P_c)(n)=500\\(P_p(n)-1.01P_p(n))+(P_c(n)-1.01P_c(n))=500\\500+(P_c(n)-1.01P_c(n))=500\\\quad\implies P_c(n)-1.01P_c(n)=0$$so finding *any* particular solution reduces the problem to solving the homogeneous case

  26. ganeshie8
    • one year ago
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    yeah i did solve it by expanding like that, i think oldrin's method looks more elegant and general..

  27. dan815
    • one year ago
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    |dw:1437283962537:dw|

  28. anonymous
    • one year ago
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    it's the same thing in linear algebra with \(Ax=b\) -- you find a particular solution \(x_p\) so you can then look at \(x=x_p+x_c\) so \(Ax=A(x_p+x_c)=Ax_p+Ax_c=b+Ax_c\) and thus \(b+Ax_c=b\implies Ax_c=0\); a particular solution is the origin vector that 'steps' us into the affine space of solutions

  29. dan815
    • one year ago
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    ah ok

  30. anonymous
    • one year ago
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    $$P(n)=500+1.01P(n-1)$$is an arithmetico-geometric sequence which permits a very easy trick to solve it: $$P(n)=500+1.01(500+P(n-2))\\\quad\quad=500(1+1.01+\dots+1.01^{k-1})+1.01^kP(n-k)$$ so $$P(n)=500\cdot\frac{1.01^n-1}{1.01-1}+1.01^nP(0)$$

  31. anonymous
    • one year ago
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    so then you just \(\dfrac{1.01^n-1}{1.01-1}=100(1.01^n-1)=100\cdot1.01^n-100\) so $$P(n)=50000\cdot1.01^n-50000+10000\cdot1.01^n=60000\cdot1.01^n-50000$$

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