## TedG 2 years ago anyone able to help me with difference equations?

1. vf321

What specifically? I know a bit...

2. TedG

@vf321 sorry was watching the rugby :D. how do you take two recurrance equations ( i think thats correct term) and create a difference equation that relates the two. More specifically...

3. TedG

\[d _{n}=-2p_{n}+3 \] and \[s _{n+1}=p^{2}_{n}+1\] need a difference equation that relates \[p_{n+1}\] to \[p_{n}\]

4. vf321

Well hang on-what is the relation between \(d_n\) and \(s_n\)?

5. TedG

you may need the information that P_{n} represents price per unit in period n. s_{n}, d_{n} represents supply and demand respectively. the question says asumming market price is price at which supply equals demand... i think that is answering what you have just asked.

6. vf321

OK then. Well, then what's the problem?\[s_{n+1} = d_{n+1} = -2 p_{n+1}+3\]

7. TedG

but d_{n} is given, not d_{n+1}

8. vf321

Yes, but that's the point of difference equations - they're true for all integral \(n > 0\)

9. vf321

Or, depending on your definitions, \(n = 0\) may be valid too.

10. TedG

so, d_{n+1} is simply -2p_{n+1}+3

11. vf321

yeah. Given that, can you replace \(s_{n+1}\) in the other equation?

12. vf321

BTW, for inline latex surround your formula in \.(formula\.) ( without the periods)

13. TedG

so the difference equation would be -2p_{n+1}+3=p^{2}_{n}+1, and rearranged?

14. vf321

Correct. Do you have Mathematica?

15. TedG

Mathematica, no. is it software?

16. vf321

Yes, it would let you solve the difference equation if you needed to.

17. TedG

oh... I use maple. pretty sure that will help solve it. also thanks for the tip on the inline latex. been wandering how to do that for ages.

18. vf321

Yup. np.

19. TedG

20. vf321

Hmm. If I had to guess (so, no, I haven't really done them before), then I would say that steady states are situations where the price has asymptotic behavior, so as \(n\) goes up, \(p_n\) continually gets closer to a certain value.

21. TedG

hmm ok. Well I think the steady states are a diferrent way of saying stationary points. as we were taught them with an example using dx/dt=f(x*) and the steady states were solutions of f(x*)=0.

22. TedG

not sure if that will change the way you think about it.

23. TedG

oh ok. so do you have an idea on how you would approach finding steady states of a difference equation?

24. vf321

Well, (and again I dont know for sure), being in a steady state for something discrete like a difference equation probably means solving \(p_{n+1}-p_{n} = 0\), because the difference between the two steps is 0 means it's in a steady state.

25. vf321

And you already have a formula for \(p_{n+1}\), so I would just plug it in and solve.

26. TedG

oh ok. that makes sense. well anyway i guess that is everything. thanks for the help.