anonymous
  • anonymous
anyone able to help me with difference equations?
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
What specifically? I know a bit...
anonymous
  • anonymous
@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...
anonymous
  • anonymous
\[d _{n}=-2p_{n}+3 \] and \[s _{n+1}=p^{2}_{n}+1\] need a difference equation that relates \[p_{n+1}\] to \[p_{n}\]

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

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