- anonymous

anyone able to help me with difference equations?

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

What specifically? I know a bit...

- 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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- 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

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Correct. Do you have Mathematica?

- anonymous

Mathematica, no. is it software?

- anonymous

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

- 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

Yup. np.

- anonymous

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

- 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

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

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

- anonymous

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

- 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

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.