Question

anyone able to help me with difference equations?

Answer 1

What specifically? I know a bit...

Answer 2

@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...

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Correct. Do you have Mathematica?

Answer 15

Mathematica, no. is it software?

Answer 16

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

Answer 17

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.

Answer 18

Yup. np.

Answer 19

do you know about steady states? because thats what i am asked to find.

Answer 20

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.

Answer 21

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.

Answer 22

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

Answer 23

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

Answer 24

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.

Answer 25

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

Answer 26

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