# Inverse integral of this integration

1. Apr 18, 2014

Hi
I am facing a mathematical problem in my research. I am not a maths magor and i need to do this to move on with my research. Please check the picture for the equation http://i.stack.imgur.com/jQroR.jpg

Mod note: Image was too large, so deleted it, and replaced it with LaTeX. Left the link, though.
If g(x) is defined as
$$g(x) = \int_{x - h/2}^{x + h/2} y(x)dx$$
then y(x) = ?

Last edited by a moderator: Apr 19, 2014
2. Apr 18, 2014

### micromass

OK, so this equation bothers me.

First, what is $h$? Is it a fixed known number?

Second, you have written $x$ both in the limits of the integral and as a dummy variable. This is an abuse of notation. Did you mean

$$g(x) = \int_{x-h/2}^{x+h/2} y(t)dt$$

Anyway, do you know what happens if you differentiate $g$?

3. Apr 18, 2014

No I meant x as it is , a variable
Its like convoluting with a rectangular function (I think )
And h is a constant btw

4. Apr 18, 2014

### pwsnafu

The convolution integrates with respect to a dummy variable.
Say $u(t) = 1$ for $-\frac{h}{2} \leq t \leq \frac{h}{2}$ and $u(t) = 0$ elsewhere.
Then the convolution is
$\int_{-\infty}^\infty y(t) \, u(x-t) \, dt = \int_{x-h/2}^{x+h/2} y(t) \, dt$.

5. Apr 18, 2014

### jbunniii

If we assume that the functions are nice enough, then $g = x*y$ can be Fourier transformed to $\hat{g} = \hat{x}\hat{y}$. If $\hat{x}$ is nonzero everywhere, then this can be solved: $\hat{y} = \hat{g} / \hat{x}$ and in some cases you may be able to invert the Fourier transform to find $y$.

Certainly this will NOT be possible if $\hat{x}$ is zero at too many points, for example on an open set where $\hat{y}$ is nonzero - in that case, information is irretrievably lost. If $\hat{x}$ is never zero or is only zero on a set of measure zero, then you may be able to recover $y$ in principle as long as $\hat{g} / \hat{x}$ has an inverse Fourier transform. (E.g., if $\hat{g} / \hat{x}$ is an $L^2$ function.)

6. Apr 19, 2014

Yes Ok that makes sense , So how do i get y(x)???
Note that I am expecting it to be of a Gaussian nature but I cant know for sure

7. Apr 19, 2014

I am not following you on the not zero part but its a Measured signal (from lab experiment) and I think it should be Gaussian so I think it has zeros on its extremes
What do I do?

8. Apr 19, 2014

### Staff: Mentor

You missed micromass's point. The variable of integration (not the variable in the limits) is a dummy variable. It's bad form to have the same variable as one or both limits of integration, and as the dummy variable (which appears in the integrand).,

9. Apr 19, 2014

Yes I got it thanks.
I still cant solve it though, can you help :) :)

10. Apr 19, 2014

### jbunniii

Can you be more specific about what you mean by "it should be Gaussian"?

Note that this problem is not going to have a unique solution. If $y(x)$ is a particular solution such that
$$g(x) = \int_{x-h/2}^{x+h/2} y(t) dt$$
then, for example, $y(x) + a \cos(2\pi n x / h + \phi)$ is another solution, where $n$ is any nonzero integer and $\phi$ is any real number.

More generally, if $p$ is any integrable periodic function with $p(x) = p(x+h)$ which satisfies $\int_{0}^{h}p(x) dx = 0$, then $y(x) + p(x)$ is also a solution.

11. Apr 20, 2014

Its a physical quantity of a Gaussian nature (optical intensity = E*exp(x^2/a^2)) where a is a constant
However due to the nature of the experiment we are trying to PROVE that its Gaussian to test our equipment
If it is Gaussian we want to measure the constant a

12. Apr 20, 2014

### disregardthat

It's an interesting problem, and basically boils down to the problem of finding a set of solutions of locally integrable functions f(x) such that

$f(x+h)-f(x) = g^{\prime}(x)$.

Given such an f(x), a solution to the original equation is recovered by putting $y(x) = f(x+\frac{h}{2})+\frac{\int^h_0f(t)dt-g(0)}{h}$.

As jbunniii points out, any periodic locally integrable function of period h may be added to f(x) do get a new solution to the equation above. This can be considered a homogeneous solution, and a particular solution added with such an homogeneous solution will exhaust the solution space.

Do you know anything about the function g(x)?

Some examples:
If $g^{\prime}(x) = x$, then $f(x) = \frac{1}{2h}x^2-\frac{1}{2}x$ is a particular solution. In fact, one may find a polynomial solution f(x) for any polynomial $g^{\prime}(x)$.

If $g^{\prime}(x) = e^x$, then $f(x) = \frac{1}{e^h-1}e^x$ is a particular solution.

If $g^{\prime}(x) = xe^x$, then $f(x) = \frac{x-\frac{h}{1-e^{-h}}}{e^h-1}e^x$ is a particular solution. This might be generalized for a solution for any polynomial multiplied with $e^x$ as $g^{\prime}(x)$.

I don't know if one can find a solution for f(x) in terms of a general g(x), but it would certainly be interesting to see.

Last edited: Apr 20, 2014
13. Apr 20, 2014

### jbunniii

So is $y(x)$ a deterministic function of the form $E\exp(-x^2/a^2)$, or is $y(x)$ a stochastic process whose sample values are random variables with Gaussian distribution?

14. Apr 20, 2014

Deterministic

15. Apr 20, 2014

The y(x) is Gaussian as i pointed out and the measured g(x)looks Gaussian on the osciliscope by visual inspection but I don't know for sure
Can the methods you did at the end of your post apply for y(x)=E*exp(x^2/a^2) ?
If so what would it be ?

16. Apr 20, 2014

### disregardthat

If $f(x) = \sum^{\infty}_{n = 0}-g^{\prime}(x+hn)$ converges for all x, then this is a solution.

EDIT: This above sum does converge if $g(x) = E e^{-\frac{x^2}{a^2}}$, but I don't know if this is what you mean.

Last edited: Apr 20, 2014
17. Apr 20, 2014

### disregardthat

Isn't the point to find y(x), knowing g(x)? So don't you mean g(x) = E*exp(x^2/a^2) ?

If you already know what y(x) is, then what is the problem exactly?

18. Apr 20, 2014

### jbunniii

OK, so $y$ looks like the normal bell curve, where $a$ is the standard deviation controlling the "width." Note that $y(a)/y(0) = e^{-1} \approx 0.3679$, so finding the value of $x$ for which $y(x)/y(0)$ equals this value is equivalent to finding $a$.

Now the convolution is a nuisance to deal with, but if we Fourier transform the problem, the convolution becomes multiplication, which isn't so annoying. Thus $\hat{g} = \hat{y}\hat{x}$ as I mentioned before. Now it's well known that if $y(x) = E\exp(-x^2/a^2)$ then $\hat{y}(\omega)$ is of the form $EK\exp(-a^2 \omega^2)$ for some scale factor $K$. There might also be a $2\pi$ or something scaling the argument to the $\exp$ function - I'll let you check the details. So essentially a gaussian function transforms to another gaussian function with the reciprocal standard deviation.

So if not for the multiplication by $\hat{x}$, we could find $1/a$ by looking for the value of $\omega$ where $\hat{y}(\omega)/\hat{y}(0)$ has the right ratio.

But fortunately you know $\hat{x}$: it is the Fourier transform of a rectangular pulse-shaped function, so the result is something of the form $\hat{x}(\omega) = a\sin(c\omega)/\omega$. (I'll let you work out $a$ and $c$.) Now this function is nonzero in the "main lobe", i.e. where $|c\omega| < \pi$. If your $1/a$ happens to fall in this range, then you are in business, because the multiplication by $\hat{x}$ can be inverted (just take its reciprocal). Whether $1/a$ falls into this range depends on how big $h$ is compared with $a$.

Thus, calculate $\hat{y}(\omega) = \hat{g}(\omega) / \hat{x}(\omega)$ and then look for the value of $\omega$ that gives the desired ratio for $\hat{y}(\omega)/\hat{y}(0)$.

I doubt you'll get a closed form answer, but you should be able to solve it numerically.

Last edited: Apr 20, 2014
19. Apr 20, 2014

Yes you are right :) I want to know y(x).
I thought you meant what I expected it to be.
The thing about g(x) is that its a measured array of points that looks Gaussian but could be "not exactly". One cannot depend on visual inspection to call out the type of a function.
Fitting it with a gaussian curve does produce results but it is not a "good fit"

20. Apr 20, 2014

Thanks that looks like its worth a shot
I will let you know when I try it :D

21. Apr 20, 2014

### disregardthat

Maybe it's me, but I'm just not understanding the situation. Could you explain clearly what you do know for sure about g(x) and y(x), and what exactly it is that you want to prove or find?

22. Apr 20, 2014

### jbunniii

You might consider formulating it as a least-squares problem, where you hypothesize that the samples of $g(x)$ are of the form $u*y + n$, where $n$ is some noise and $y$ is a "signal" of the form $E \exp(-x^2/a^2)$ with $E$ and $a$ unknown. The goal would then be to solve for the values of $E$ and $a$ which minimize the error $\sum_{k=1}^{N} (g(x_k) - [u*y](x_k))^2$. This can often be reduced to a linear algebra problem involving the calculation of the Moore-Penrose pseudoinverse.

On the other hand, if you are not sure whether the "signal" really is of the form $E \exp(-x^2 / a^2)$, you might try fitting it to a more general class of curve and see how closely that fit matches a Gaussian function.

23. Apr 20, 2014