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What specifically? I know a bit...

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

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.

Yup. np.

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

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

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.