Can this be turned into a differential equation? Recursive diffy Q?

[DrummingAtom]

Let's say you have a loss of 3% of the current energy on a device for every time is gets turned on. It starts at a 100% energy.

The only way I could think of it is a recursive function. Here's what I came up:

E(t) = E_t - .03*E_t-1

E(0) = 1

E(1) = 1 - .03*E₀

E(2) = E₁ - .03*E₁

...

Where E = Energy and t = each time turned on.

I feel like the solution should be something close to E^-.03t but it's not. I know the recursive function describes it exactly but I have to wonder if it's even possible to turn this into a differential equation?

Thanks.

 

[Skrew]

I don't think so considering you are look at specifc steps(each time you turn it on and off) and not a continuous interval..

 

[Office_Shredder]

Let me propose an alternative recursive formula (it should be obvious why this is the same)
E₁ = .97 E₀=.97
E₂ = .97 E₁=.97²

 

[TylerH]

I don't know anything about them, but this looks like a delay differential equation. http://en.wikipedia.org/wiki/Delay_differential_equation

 

[blue_raver22]

Yes, this can be turned into a differential equation. A recursive function is essentially a discrete version of a differential equation, where the variable is changing over discrete time steps instead of continuously.

To turn this into a differential equation, we can use the definition of a derivative to represent the change in energy over time as a continuous function. Let's call E(t) the energy at time t, and let's also define a function f(t) as the rate of change of energy over time, which in this case is -3% of the current energy:

f(t) = -0.03*E(t)

Now, using the definition of a derivative, we can write:

E'(t) = lim(h->0) (E(t+h) - E(t))/h

This represents the change in energy over a small time interval (h) as it approaches 0. We can rewrite this using our function f(t):

E'(t) = lim(h->0) (E(t+h) - E(t))/h = lim(h->0) (-0.03*E(t+h))/h

Now, as h approaches 0, we can approximate E(t+h) as E(t), so we end up with:

E'(t) = lim(h->0) (-0.03*E(t))/h = -0.03*E(t)

This is the differential equation that represents the loss of energy in the device over time. It is a first-order linear differential equation, and its solution is indeed E(t) = E0*e^(-0.03*t), where E0 is the initial energy at t=0.

So, in summary, yes, this can be turned into a differential equation and its solution is E(t) = E0*e^(-0.03*t). I hope this helps!