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At moment t I have:

x(t) = w(t) / h^2

e(t) = 10w(t) + 6.25h - 5g + 5

f(t) = a*(x(t))^2 + b*(x(t)) + c

And at moment t+1 I have:

x(t+1) = x(t) - [e(t) * f(t)] / (k*h^2)

a, b, c, g, h and k are constants. t is time starting at 0.

I need to find how much time does it take to change, for example, x=200 to x=100.

I have been trying to create a generic function x(t) and integrate it.

But I am having problems with it ...

Update

To better clarify the algorithm given a x(t) to obtain x(t+1) I do: 1 - Calculate current energy level, e(t), using x(t); 2 - Calculate factor e(t) using x(t); 3 - Calculate x(t+1) using x(t), e(t) e f(t)

This said x(t+1) as a function of only x(t) = x and constants is:

x(t+1) = [-10ah*x^3 - (6.25a+10bh)*x^2 -(6.25b+10ch+hk)*x - 6.25chk] / (hk)

So it is a third degree polynomial ...

But how to find how much time does it take decrease x from 200 to 100?

How to plot such a function? I mean, can I define x(t) as not recursive?

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  • $\begingroup$ You also need a formula for $w(t)$, then there is hope. At least you could plug it into a spreadsheet and copy down $\endgroup$ Jun 23, 2017 at 20:11
  • $\begingroup$ @Miguel Moura : Try the following....: 1) Solve for $x(t+1)$ in terms of solely x(t) and the constants. 2) Evaluate the expression at x(1) to get it in terms of x(0). 3) From there, you can do a change of variable with $u=t+1$ from t=1 onwards, and then integrate x in terms of u. From there, you should be good. Let us know how it goes. $\endgroup$
    – Johnq
    Jun 23, 2017 at 20:30
  • $\begingroup$ @Johnq I just added an update with x(t+1) in terms of x(t) which gives a Polynomial of third degree. How to calculate the time and plot a function? $\endgroup$ Jun 23, 2017 at 21:27
  • $\begingroup$ MiguelMoura : Backtracking right now, I don't think so. @Ross Millikan kind of hit the nail on the head. Right now you have three unknowns ($x(t),e(t),and f(t)$) but only two equations. In order to get something that isn't implicit, you would need another equation. $\endgroup$
    – Johnq
    Jun 23, 2017 at 22:55

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