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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0let \(P(n)\) denote the population after \(n\) years. we're given that \(P(0)=10000\) and the recurrence relation $$P(n)=0.04P(n1)0.03P(n1)+500\\P(n)=0.01P(n1)+500$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$P(n)0.01P(n1)=500$$ignoring our initial condition, a constant particular solution \(P(n)=k\) gives \(k0.01k=500\implies 0.99k=500\implies k=500/0.99=\frac{50000}{99}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Ahh that looks much better than what im doing.. i ended up with a geometric series after writing out the first few terms

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now consider the homogeneous recurrence relation $$P(n)0.01P(n1)=0$$suppose that we had a geometric sequence \(P(n)=r^n\) which gives $$r^n0.01r^{n1}=0\\r^{n1}\left(r0.01\right)=0\implies r=0.01$$ so our homogeneous solution looks like \(P(n)=C(0.01)^n\) and thus the general solution to our original recurrence is \(P(n)=C(0.01)^n+\frac{50000}{99}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we want to solve for \(C\) using our initial condition: $$P(0)=10000\\C(0.01)^0+\frac{50000}{99}=10000\\C=10000(15/99)\approx 9494.95$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, I messed up the recurrence  it should've actually read

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(P(n)=C(1.01)^n+\frac{50000}{99}\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$P(n)=P(n1)+0.04P(n1)0.03P(n1)+500\\P(n)=1.01P(n1)+500$$ in which case we'd get \(r=1.01\) and our particular solution would be different as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$k=1.01k+500\\0.01k=500\\C=50000$$ so the general solution is \(P(n)=C(1.01)^n50000\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now applying our initial condition: $$10000=C(1.01)^050000\\60000=C$$ so our model is $$P(n)=60000(1.01)^n50000$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we plug in \(n=30\) so $$P(30)\approx 30870.93$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in general, the theory of linear recurrence relations with constant coefficients $$x_n+c_1x_{n1}+c_2x_{n2}+\dots+c_{n1}=g(n)$$ 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 \(x_n=r^n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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 :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.2wolfram 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, @dan815 your approach will overshoot since this is essentially the same as dealing with continuous compounding

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 we pick a constant because we want \(P(n)1.01P(n1)\) to be a constant; if there were nonconstant terms then they would not be entirely annihilated  consider \(P(n)=An+B\) which gives \(An+B1.01(A(n1)+B)=0.01An+(1.01A0.01B)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we'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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, 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

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0yeah i did solve it by expanding like that, i think oldrin's method looks more elegant and general..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it'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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$P(n)=500+1.01P(n1)$$is an arithmeticogeometric sequence which permits a very easy trick to solve it: $$P(n)=500+1.01(500+P(n2))\\\quad\quad=500(1+1.01+\dots+1.01^{k1})+1.01^kP(nk)$$ so $$P(n)=500\cdot\frac{1.01^n1}{1.011}+1.01^nP(0)$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then you just \(\dfrac{1.01^n1}{1.011}=100(1.01^n1)=100\cdot1.01^n100\) so $$P(n)=50000\cdot1.01^n50000+10000\cdot1.01^n=60000\cdot1.01^n50000$$
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