I How to Extract Data from an Integral

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1. Mar 2, 2016

ecastro

Consider this form:

$A = \int B\left(x\right) C\left(x\right) dx$

I have the values for $A$ and $C\left(x\right)$ (a value of $C$ per value of $x$), is there a way that I can extract $B\left(x\right)$ numerically or analytically?

2. Mar 2, 2016

Ssnow

A question, is $A$ a number or a function $A=A(x)$, in other words the integration is definite on an interval or is indefinite?

3. Mar 2, 2016

HallsofIvy

Given any B(x) there are an infinite number of different functions, C(x), that will give the same A.

4. Mar 2, 2016

the_wolfman

Yes, equations of this for are called integral equations. There are numerous methods that have been developed to solve integral equations. However, it's not always possible to find a unique solution.

5. Mar 2, 2016

zinq

I will assume that the integral in the original post is a definite integral, or else the questions won't make sense.

This is analogous to the discrete case, where you have number Bj and Cj, for 1 ≤ j ≤ n and a number A such that

A = ∑ Bj Cj

where the sum is over j = 1, 2, 3, ..., n.

Consider the simplest non-trivial case, where n = 2:

A = B1 C1 + B2 C2,​

where you know A and the Cj, and are trying to determine the B1 and the B2.

This gives you one linear equation in two unknowns. As we know from algebra, we need two linear equations (and independent ones at that) to determine two unknowns. It does, of course, present a constraint on the B1 and B2 — but that is not enough to solve for them.

This not only gives an idea of why the original integral equation cannot be solved for B(x), but this can even be translated into a corresponding integral equation, with the C(x) function defined as a step function, constant on each of two consecutive unit-length intervals. This discrete case shows that even if B(x) were likewise required to be a step function constant on each of those unit intervals, you still wouldn't be able to solve for it.

6. Mar 2, 2016

ecastro

The variable $A$ is a number for this case. I apologise if I haven't mentioned that the integral is definite.

I'm trying to search these methods with Google, but I always find sites on how to solve a particular integral equation, not to extract a function from it.

Is Singular Value Decomposition method valid for this case? I just happen to read on this technique and I do not know if it is applicable for this problem.

In point of fact, I have encountered this form of equation in image filters, where $C\left(x\right)$ is the filter and $x$ is the wavelength. I have a filtered image, which is $A$, and I want to extract the unfiltered image which is $B\left(x\right)$. Do you have any thoughts on how to solve this kind of problem?

7. Mar 2, 2016

zinq

I'm no expert on image reconstruction, but I can say that the math problem as you have described it cannot be solved uniquely for the function B(x) — not even close.

8. Mar 3, 2016

Samy_A

Yes, and "not even close" is an understatement.
If the equation you are trying to solve is $$A=\int_{a}^{b} B(x)C(x) dx \ \ (1)$$
any function $D$ satisfying $\displaystyle \int_{a}^{b} D(x)C(x) dx \neq 0$ is a solution up to a constant, as $$B(x)=\frac{A}{\int_{a}^{b} D(x)C(x) dx} D(x)$$ will be a solution to $(1)$.

9. Mar 3, 2016

Svein

In Hilbert space terminology, your problem can be written as $A=<B, C>$, where < , > is the scalar product of that particular space (in this case the scalar product is the integral). This means that, for example, $<B+H, C>=<B, C>+<H, C>$. From this simple fact follows that whenever $<H, C>=0$, $A = <B, C>+<H, C> = <B+H, C>$. Therefore, there can be no single solution for $B$.

An interesting question might be: find the "minimal" solution for B. This implies that B is some scalar multiple of C, say λ⋅C. This again means that $A=<\lambda\cdot C, C>=\lambda<C, C>$. Since you know C(x), you can calculate $<C, C>=\int_{a}^{b}C^{2}(x)dx$ and then finding λ is trivial.

10. Mar 11, 2016

OrangeDog

If you assume it is a type of function, such as a polynomial or exponential, then maybe it is possible with enough equations for the unknowns.