How to Extract Data from an Integral

In summary, the original problem of finding B(x) from the given equation is not possible as there are infinitely many solutions for B(x). One possible approach is to look for the "minimal" solution which implies B(x) being a scalar multiple of C(x), and this can be found by calculating the scalar product of C(x) with itself and using it to calculate the minimal value for B(x). However, this does not give a unique solution for B(x).
  • #1
ecastro
254
8
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?

Thank you in advance.
 
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  • #2
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
Given any B(x) there are an infinite number of different functions, C(x), that will give the same A.
 
  • #4
ecastro said:
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?

Thank you in advance.

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
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
Ssnow said:
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?

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

the_wolfman said:
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.

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.

zinq said:
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.

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
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.

Possibly if you have more information about B(x) and C(x), something more intelligent can be said.
 
  • #8
zinq said:
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.
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
In Hilbert space terminology, your problem can be written as [itex]A=<B, C> [/itex], where < , > is the scalar product of that particular space (in this case the scalar product is the integral). This means that, for example, [itex] <B+H, C>=<B, C>+<H, C>[/itex]. From this simple fact follows that whenever [itex]<H, C>=0 [/itex], [itex]A = <B, C>+<H, C> = <B+H, C>[/itex]. Therefore, there can be no single solution for [itex]B [/itex].

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 [itex] A=<\lambda\cdot C, C>=\lambda<C, C>[/itex]. Since you know C(x), you can calculate [itex]<C, C>=\int_{a}^{b}C^{2}(x)dx [/itex] and then finding λ is trivial.
 
  • #10
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.
 

1. How do I extract data from an integral?

To extract data from an integral, you can use the fundamental theorem of calculus. This theorem states that the definite integral of a function can be evaluated by finding its antiderivative and evaluating it at the upper and lower limits of integration.

2. Can I use numerical methods to extract data from an integral?

Yes, you can use numerical methods such as the trapezoidal rule or Simpson's rule to approximate the value of an integral. These methods use a series of smaller intervals to estimate the area under the curve represented by the integral.

3. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration, meaning that it will give a numerical value as a result. An indefinite integral, on the other hand, does not have specified limits and will give a function as a result. To extract data from an indefinite integral, you will need to evaluate it at specific points.

4. Can I use calculus to extract data from a non-numerical integral?

Yes, you can use techniques such as integration by substitution or integration by parts to extract data from non-numerical integrals. These methods involve manipulating the integral using algebraic techniques to make it easier to solve.

5. Are there any limitations to extracting data from an integral?

Yes, there are some integrals that cannot be solved analytically and require the use of numerical methods to extract data. Additionally, some integrals may have complex or infinite solutions, making it difficult to extract meaningful data from them.

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