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TedG
anyone able to help me with difference equations?
What specifically? I know a bit...
@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...
\[d _{n}=-2p_{n}+3 \] and \[s _{n+1}=p^{2}_{n}+1\] need a difference equation that relates \[p_{n+1}\] to \[p_{n}\]
Well hang on-what is the relation between \(d_n\) and \(s_n\)?
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.
OK then. Well, then what's the problem?\[s_{n+1} = d_{n+1} = -2 p_{n+1}+3\]
but d_{n} is given, not d_{n+1}
Yes, but that's the point of difference equations - they're true for all integral \(n > 0\)
Or, depending on your definitions, \(n = 0\) may be valid too.
so, d_{n+1} is simply -2p_{n+1}+3
yeah. Given that, can you replace \(s_{n+1}\) in the other equation?
BTW, for inline latex surround your formula in \.(formula\.) ( without the periods)
so the difference equation would be -2p_{n+1}+3=p^{2}_{n}+1, and rearranged?
Correct. Do you have Mathematica?
Mathematica, no. is it software?
Yes, it would let you solve the difference equation if you needed to.
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.
do you know about steady states? because thats what i am asked to find.
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.
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.
not sure if that will change the way you think about it.
oh ok. so do you have an idea on how you would approach finding steady states of a difference equation?
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.
And you already have a formula for \(p_{n+1}\), so I would just plug it in and solve.
oh ok. that makes sense. well anyway i guess that is everything. thanks for the help.