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Knowing that: $y(0)=100, y(1)=90$. And:
$y(k+1)-y(k)=-0.128(y(k)-22)$ for every $k=0,1,2,...$
How to draw a graph of $y(k)$ as a function of $k$ for every $k=0,1,2,...,100$.

I'm really stuck, I know for sure the purpose isn't to start calculating each $y(k)$ one by one, I've tried to find a pattern to find a function $y(k)$ that I can draw or estimate, but things got too complicated and I cannot see a pattern.
How do I deal with this question?
Thanks in advance for any help.

Pwaol
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  • Are you able to compute the value of $y(k)$ for $k\to\infty$? – PC1 Nov 01 '21 at 17:01
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    Hint: subtract 22 from both sides and add $y(k)$ to both sides, to see that $y(k+1) - 22$ is some constant times $y(k) - 22$. – Joe Nov 01 '21 at 17:11
  • @Joe Thanks for the hint, I'm trying to continue from there, but it's also complicated to find a pattern, am I supposed to find a pattern here by calculating $y(1),y(2),y(3)...$ until I could write $y(k)$? – Pwaol Nov 01 '21 at 17:28
  • Once you find that constant, you can plot by knowing that, starting from $y(0)-22 = 78$. I mean, if you actually have to plot the precise location of all 101 points by hand, it's still going to be tedious, if not impossible. But if you just need a rough sketch, it should be clear from the equation in my previous hint. – Joe Nov 01 '21 at 17:52
  • @Joe My work is aimed to find $y(k)$ as a function like $y(k)=3^k$, and then I would plug values $k=0,10,20,...,100$ and draw a sketch, but I'm having a hard time finding $y(k)$, I've found that $y(k+1)-22=0.872(y(k)-22)$, I'm trying to start from $y(1)$, and work my way up to find a pattern for $y(k)$ but it's just getting complicated and I'm not finding a pattern. – Pwaol Nov 01 '21 at 18:01
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    Hint 2: $y(1) - 22 = 0.872(y(0) - 22)$. What does $y(2) - 22$ equal? Can you express that in terms of $(y(0) - 22)$? What does $y(3) - 22$ equal? Can you express that in terms of $(y(0) - 22)$? ... – Joe Nov 01 '21 at 18:04
  • @Joe Thanks alot! I was just substituting $y(1)-22$, and then subtracting $22$ again for some reason, I complicated myself. But I really appreciate the help and this has opened my mind to these kinds of tricks! – Pwaol Nov 01 '21 at 18:16
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    You're welcome. FYI - I don't want to give you the impression that math is about a bunch of tricks. There are some general techniques in dynamical systems, like finding equilibrium points: by solving for $y(k+1)=y(k)$. Then you can often express the recursion equation in terms of distance to an equilibrium point (that's what I did). So hopefully your book/professor is discussing these concepts. – Joe Nov 01 '21 at 18:18
  • @Joe I haven't heard of that actually, my professor didn't really "solve" anything like this before, all the examples were just easy to see like $y(0)=10$, $y(k+1)=1.05y(k)$, but if you could recommend any book/online material/What to search for.. that could give me a deeper understanding of these concepts, I would appreciate that. Thanks alot for the information – Pwaol Nov 01 '21 at 18:28
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    Well, I'm not sure the context in which you are studying this, but this is a "discrete dynamical system". Unfortunately, I'm not familiar with any specific references, but perhaps you could search that. This is a single variable DDS, but if you get into systems of recursion equations, that's multi-variable, and you would study them with matrix equations, and using concepts like eigenvalues. This post is a linear DDS, of the form $y(k+1) = a y(k) + b$. If $a\ne 1$, the equilibrium value is $\frac{b}{1-a}$. If $|a|<1$, the equilibrium is attracting. There are also nonlinear DDS. – Joe Nov 01 '21 at 18:37

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