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TedG

  • one year ago

anyone able to help me with difference equations?

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  1. vf321
    • one year ago
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    What specifically? I know a bit...

  2. TedG
    • one year ago
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    @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
    • one year ago
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    \[d _{n}=-2p_{n}+3 \] and \[s _{n+1}=p^{2}_{n}+1\] need a difference equation that relates \[p_{n+1}\] to \[p_{n}\]

  4. vf321
    • one year ago
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    Well hang on-what is the relation between \(d_n\) and \(s_n\)?

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

  7. TedG
    • one year ago
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    but d_{n} is given, not d_{n+1}

  8. vf321
    • one year ago
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    Yes, but that's the point of difference equations - they're true for all integral \(n > 0\)

  9. vf321
    • one year ago
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    Or, depending on your definitions, \(n = 0\) may be valid too.

  10. TedG
    • one year ago
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    so, d_{n+1} is simply -2p_{n+1}+3

  11. vf321
    • one year ago
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    yeah. Given that, can you replace \(s_{n+1}\) in the other equation?

  12. vf321
    • one year ago
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    BTW, for inline latex surround your formula in \.(formula\.) ( without the periods)

  13. TedG
    • one year ago
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    so the difference equation would be -2p_{n+1}+3=p^{2}_{n}+1, and rearranged?

  14. vf321
    • one year ago
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    Correct. Do you have Mathematica?

  15. TedG
    • one year ago
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    Mathematica, no. is it software?

  16. vf321
    • one year ago
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    Yes, it would let you solve the difference equation if you needed to.

  17. TedG
    • one year ago
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    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
    • one year ago
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    Yup. np.

  19. TedG
    • one year ago
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    do you know about steady states? because thats what i am asked to find.

  20. vf321
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    not sure if that will change the way you think about it.

  23. TedG
    • one year ago
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    oh ok. so do you have an idea on how you would approach finding steady states of a difference equation?

  24. vf321
    • one year ago
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    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
    • one year ago
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    And you already have a formula for \(p_{n+1}\), so I would just plug it in and solve.

  26. TedG
    • one year ago
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    oh ok. that makes sense. well anyway i guess that is everything. thanks for the help.

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