Operator for the local average of a growing oscillating function

In summary, the conversation discusses two examples of functions where simple integration cannot accurately retrieve a bias value. The solution is using a windowing method and integrating over the window. The question then asks if there is a more rigorous way to recover the bias value without explicitly knowing the envelope function. This leads to a basic question about the uniqueness of the average value for a function that can be represented in two different ways. All functions in the examples are assumed to be differentiable with positive derivatives.
  • #1
Swamp Thing
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TL;DR Summary
How can we get the average value around a point of an oscillating function that has an increasing envelope?
First some background, then the actual question...
Background:
(a) Very simple example: if we take ##Asin(x+ϕ)+0.1##, the average is obviously 0.1, which we can express as the integral over one period of the sine function. (assume that we know the period, but don't know the phase or other parameters of the function).

(b) More complicated example: Take ##x^2 sin(x+\phi) + 0.1##
1574221983920.png
Here simple integration won't help, because the integral depends on where in the cycle we start from. If we integrate a positive half-cycle first and then a negative half-cycle, we get a negative result, and vice versa. To avoid favoring either the positive or negative half cycles, we can multiply our function by a window function:
1574222148151.png
And now if we integrate over the window we can retrieve the bias (average) of 0.1.

Question:
Can we have a more rigorous way of recovering the "0.1" that gives the same result as the windowing method? Something that involves a smart way of integrating over one cycle (or a whole number of cycles) in different ways, then combining them and calibrating out the growth (envelope) function? -- But without actually knowing the envelope function explicitly?
 
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  • #2
To have hope of finding an average value, we'd have to know that the definition of such a value makes it unique.

So a basic question (which I cannot answer) is:

Let f(x) be a function such that f(x) = A(x) p(x) + k where A(x) is a non-periodic function, p(x) is periodic function and k is a constant. For some f(x), can there exist a different representation of f(x) as f(x) = B(x) q(x) + m where B(x) is a non-periodic function, q(x) is a periodic function and m is a constant different than k ?

For your examples, we can assume all the above functions are differentiable. The examples are also cases where A'(x) > 0, B'(x) > 0.
 
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  • #3
Useful perspective, thanks!
 

1. What is an operator for the local average of a growing oscillating function?

An operator for the local average of a growing oscillating function is a mathematical tool used to calculate the average value of a function over a specific interval. It takes into account the oscillating behavior of the function and provides a more accurate average compared to a simple arithmetic mean.

2. How is the operator for the local average of a growing oscillating function calculated?

The operator is calculated by first dividing the function into smaller intervals and calculating the average value of each interval. Then, these average values are combined using a weighted average formula to give the final result. The weights are determined by the amplitude of the oscillations in each interval.

3. What is the purpose of using this operator?

The purpose of using this operator is to obtain a more accurate representation of the average value of a function that exhibits oscillating behavior. It is especially useful in situations where the function is continuously growing and the oscillations make it difficult to determine the average using traditional methods.

4. Can this operator be applied to any type of function?

Yes, this operator can be applied to any type of function as long as it exhibits oscillating behavior. However, it is most commonly used for functions that are continuously growing and have a periodic or repetitive nature.

5. Are there any limitations to using the operator for the local average of a growing oscillating function?

One limitation is that the operator may not provide an accurate result if the function has rapidly changing oscillations or if the intervals chosen are too large. In these cases, other methods may be more suitable for calculating the average value of the function.

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