Summation to a Continuous Function

[aryajur]

The problem is defined as follows

if the value of a function at time t = Y

then the value of the function at time t+Δt = A(10 - Y)/(A+B)

so I want to find the expression for Y as a function of t as Δt -> 0

How can this be done?? Any help will be appreciated.

 

[steve10]

It can be done as follows:

Let f(t) be the function. Then, according to your assumption, when f(t)=Y(t), we have f(t+Δt)=A(10-Y(t+Δt))/(A+B).
Therefore, f(t+Δt)-f(t)=A(10-Y(t+Δt))/(A+B) - Y(t)
=(10A-A*Y(t+Δt)-A*Y(t)-B*Y(t))/(A+B).

Now, we need some continuity for both f(t) and Y(t). Let Δt->0, the left hand side -> 0 according to the continuity. Then, the right hand side should also tend to zero, which means,
10A-A*Y(t)-A*Y(t)-B*Y(t)=0

=> Y(t)=10A/(2A+B).

 

[aryajur]

See but this way you get Y(t) a constant ! Which is not the case if you look at the original problem where it is defined recursively

 

[steve10]

(1) "Constant" is also a kind of function, isn't it? You get a constant function mainly because A and B are constants. If they are not, neither is Y(t) (generally).
(2) "Recursive" doesn't mean the function can't be a "constant".

Those are not keys. The key is the "uniqueness" of your solution. Now, look at these, for any t, f(t) = Y(t), and f(t+Δt)=A(10 - Y(t+Δt))/(A+B), where Δt can be arbitrarily small, That essentiallly means Y(t)= A(10 - Y(t))/(A+B) as long as the continuity is assumed for both f(t) and Y(t).

I'd be surprised if you can get any solution that is NOT a constant.

 

[aryajur]

Well actually the function is not a constant if you evaluate it using the initial definition. You got the initial definition wrong when you assumed f(t+Δt)=A(10-Y(t+Δt))/(A+B)
actually f(t+Δt)=A(10-Y(t))/(A+B) = Y(t+Δt) so you can see the value of Y changes in the Δt interval so there is no way this is a constant, this is a discrete funtion with a sample every Δt time interval.

 

[steve10]

Well, using your definition, it's even easier to reach a conclusion. Look what you wrote f(t+Δt)=A(10-Y(t))/(A+B), for any Δt. Then what is it if it's not a constant. f(t+Δt) doesn't change whatever Δt is. In this case, We don't even need a continuity.

 

[aryajur]

I think you are getting confused, lets say A = B = 1 for a simple case now since
A(10-Y(t))/(A+B) = Y(t+Δt) it means Y(t+Δt) = (10-Y(t))/2 so lets say if Y(t) = 2
so Y(t+Δt) = 4 now as time again increases by Δt we have Y(t+2Δt) = (10-4)/2 = 3

so you see...
Y(t) = 2
Y(t+Δt) = 4
Y(t+2Δt) = 3
and so on...
this is not a constant, it is changing with time !!! I hope you understand the problem now.

 

[steve10]

Oh, well, I thought Y(t) is what you want to find according to your earlier post. Now you "lets say if Y(t) = 2 ...". You turn it over to an assumption? Then, have you tried Y(t)=10/3, and "let's what we have then ...".

 

[aryajur]

Yes when it is 10/3 then it will come out to be a constant. I guess I should have mentioned that take Y(t) to be 5 initially. But as you see it is not a constant in a general case. So any more ideas?

 

[steve10]

If you allow Δt vary, then the intial value has been decided and you are not allowed to specify it. In fact, from what you wrote, A(10-Y(t))/(A+B) = Y(t+Δt). You let Δt->0, and you immediately reach a controdiction. A reasonal choice is that Δt is fixed and so are A and B (they should depend on Δt). As such, you will end up with an expression of Y(t+mΔt) in terms of Y(t) (m is an integer) and you can then specify an initial value Y(t). Good night.

 

[aryajur]

Let me explain better, Δt was a fixed sampling value. A and B are fixed. So Y(t) is a discrete function that increases in steps, but it is a function of t. I guess I should not have said Δt -> 0, actually what will the Y(t) function look like if we see the waveform for say a million Δt's then it does look like a continuous function which is a function of t. I want to find the expression for that function. I hope this explains it better.
So still looking.... any more help will be appreciated.

 

[steve10]

Set h=A/(A+B). Then,
Y(t+Δt)=h(10-Y(t)).
Since A and B do not depend on t and Δt, we have (for any integer m),
Y(t+mΔt)=h(10-Y(t+(m-1)Δt)),
which means
Y(t+mΔt)=h(10-Y(t+(m-1)Δt)) (=10h-Y(t+(m-1)Δt)h)
=h(10-h(10-Y(t+(m-2)Δt))) (=10h-10h^2+Y(t+(m-2)Δt)h^2)
=h(10-h(10-h(10-Y(t+(m-3)Δt)))) (=10h-10h^2+10h^3-Y(t+(m-3)Δt)h^3)
=...
=h(10-h(10-h(10-...Y(t)...))) (=10h-10h^2+10h^3-...-10(-h)^m+Y(t)(-h)^m)
=10h(1+(-h)+(-h)^2+...+(-h)^(m-1))+Y(t)(-h)^m
=10h*(1-(-h)^m)/(1+h)+Y(t)(-h)^m.

You may try Z-transform also, which is born to solve this kind of problem.

 

[Tahar]

For me this problems sounds strange at many regards:


1) what is i)mplied in your mathematical function is that there is a discrete index, so in fact your problem is suppose to define a sequence.

But any sequence is defined by two things:

a) its initial value
b) a recursive relation or induction relation that express the value of function at index n+1 function of the value at index n.

2) IN the framework of sequence problems we generally dont care about the values at some small or initial value of n, (recall in my way of explaining t=n.Δt ), we are rather interested in the value of the function of sequence for infinite value of n.

3)Finally, if your step Δt is so small, it becomes maybe non relevant to use a sequences as a model and therefore stating that Δt becomes the infinitesimal δt or dt seems to be a better model for your problem.

in this case, you initial assumption can lead to a differential equation that involve Y(t+dt)-Y(t), (er are close to get a derivation here )