What is the Cross-Correlation Theorem for Continuous Fourier Transform?

  • Thread starter T.Engineer
  • Start date
  • Tags
    Integral
In summary, the conversation discusses finding the result for a particular equation involving the integral of two functions, p(t) and p(t-d). The conversation also mentions using integration by parts and recurrence relations to find a general formula for the result. Maxima, a free math software, is suggested as a tool to compute the nth case. The conversation also briefly touches on the topic of autocorrelation function for Hermite polynomials.
  • #1
T.Engineer
87
0
can anybody find the result for the following equation:

F(d)= [tex]\int^{T_f}_{0}p(t)p(t-d) dt [/tex]

where
916; = d but it doesn't appears very well
and
p(t) = (-1)^n * e^(t^2) * d/dt * e^(-t^2)

thanks alot!
 
Last edited:
Physics news on Phys.org
  • #2
The way you've written, it appears that

[tex] p(t)=-2 (-1)^n t [/tex]

I believe that in this case the integral would be a formality.
 
  • #3
I am sorry I didnt write it correctly

p(t) = (-1)^n * e^(t^2) * d^n/dt^n * e^(-t^2)

where n= 1,2,...,N

I just want a general formula for the result
 
  • #4
Have you attempted at an answer? What technique(s) have you tried? Integration by parts comes to mind...
 
  • #5
my question is how to find a general formula for the following

F(d)= [tex]\int^{T_f}_{0}p(t)p(t-d) dt [/tex]

where
p(t) = (-1)^n * e^(t^2) * d/dt * e^(-t^2)

and
n=1,2,...,N
Thanks alot!
 
  • #7
And my response is 'why do you keep putting n's in and then taking them out?' State the question precisely. I presume you're just supposed to do it for n an integer, rather than for n=1,2,3,...
 
  • #8
That p_{n} is, up to a normalization constant, a Hermite polynomial of degree "n". So try looking that integral in mathematical tables of integrals or ask Mathematica software about the result.
 
  • #9
"There are many functions - called special functions - which fail to
have an anti-derivative expressible as a finite combination of
elementary functions. The so-called elliptic functions, the error
function, and the gamma function are a few examples. The error
function, which is extremely useful in both physics and statistics, is
defined as:

erf(x) = (2/sqrt(pi))integral from 0 to x of e^(-t^2)dt

Extensive tables of the error function would not exist if the
anti-derivative of e^(-t^2) were expressible as a finite combination
of elementary functions."
http://mathforum.org/library/drmath/view/53554.html
 
  • #10
EnumaElish said:
"There are many functions - called special functions - which fail to
have an anti-derivative expressible as a finite combination of
elementary functions. The so-called elliptic functions, the error
function, and the gamma function are a few examples. The error
function, which is extremely useful in both physics and statistics, is
defined as:

erf(x) = (2/sqrt(pi))integral from 0 to x of e^(-t^2)dt

Extensive tables of the error function would not exist if the
anti-derivative of e^(-t^2) were expressible as a finite combination
of elementary functions."
http://mathforum.org/library/drmath/view/53554.html

p(t) is not an exponential. It's a polynomial as dextercioby has already pointed out.
 
  • #11
I have realized the same because, as dextercioby had pointed out, there is a closed solution for n=1.

Here's how I'd approach the problem (and I communicated this to T.Engineer at least once before, under Statistics & Probability). I'd start with n=1 and calculate the closed solution, which is easy. Then move on to n=2, 3, ..., and see if there is a pattern.
 
  • #12
EnumaElish said:
I have realized the same because, as dextercioby had pointed out, there is a closed solution for n=1.

Here's how I'd approach the problem (and I communicated this to T.Engineer at least once before, under Statistics & Probability). I'd start with n=1 and calculate the closed solution, which is easy. Then move on to n=2, 3, ..., and see if there is a pattern.

There are also recurrence relations for the Hermite polynomials which could be exploited without rediscovering them empirically. It's clear there is no exponential in the final integral since exp(-t^2) comes out from the differentiation intact and cancels with the exp(t^2) without even knowing there are closed form solutions for particular n.
 
  • #13
Dick said:
There are also recurrence relations for the Hermite polynomials which could be exploited without rediscovering them empirically. It's clear there is no exponential in the final integral since exp(-t^2) comes out from the differentiation intact and cancels with the exp(t^2) without even knowing there are closed form solutions for particular n.
Right; the fact that a closed solution exists is sufficient but not necessary to see that it is not the error function, or any other special function.
 
  • #14
If anybody is interested in pursuing this, here's a Maxima (free math software) program to compute the nth case. It should be pretty self explanatory.

p(t,n):=expand(exp(t^2)*(-1)^n*diff(exp(-t^2),t,n));
integ(n):=p(t,n)*subst((t-d),t,p(t,n));
final(n):=integrate(integ(n),t,0,Tf);

E.g. typing 'final(7);' after this is entered computes the n=7 case.
 
Last edited:
  • #15
Dick said:
If anybody is interested in pursuing this, here's a Maxima (free math software) program to compute the nth case.

Please, can you tell me from where can I download this program.
Thanks a lot!
 
  • #17
T.Engineer said:
Please, can you tell me from where can I download this program.
Thanks a lot!

Try http://maxima.sourceforge.net/. If you are running a debian flavor linux it might be as simple doing 'apt-get install xmaxima'.
 
  • #18
Dick said:
Try http://maxima.sourceforge.net/. If you are running a debian flavor linux it might be as simple doing 'apt-get install xmaxima'.

I am runing windows Xp.
 
  • #19
T.Engineer said:
I am running windows Xp.

Then get, for example, maxima-5.12.0a.exe from the downloads section.
 
  • #20
Dick said:
If anybody is interested in pursuing this, here's a Maxima (free math software) program to compute the nth case. It should be pretty self explanatory.


can you help to find a general formula for the autocorrelation function Hermite polynomials.
Thanks a lot!
 

1. What is an integral?

An integral is a mathematical concept that represents the area under a curve in a graph. It is used to calculate a quantity or value that changes continuously over a given interval.

2. Why is it important to find an integral?

Finding an integral is important in many fields of science and engineering. It is used to solve problems related to motion, heat transfer, probability, and many other areas. It helps us understand and predict how different systems behave.

3. How do you find an integral?

To find an integral, you need to use integration techniques such as substitution, integration by parts, or partial fractions. These techniques help to simplify the integral and solve it using basic integration rules.

4. What is the purpose of "Find the following integral" in scientific calculations?

The purpose of "Find the following integral" is to help scientists and engineers solve real-world problems by finding the area under a curve. This area represents a quantity or value that is changing continuously, and finding the integral allows us to understand and analyze this change.

5. Can you give an example of "Find the following integral" in real-life applications?

One example of "Find the following integral" in real-life applications is calculating the amount of water flowing through a pipe over a given time. By finding the integral of the flow rate over time, we can determine the total volume of water that has passed through the pipe, which is useful in engineering and environmental studies.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
160
  • Calculus and Beyond Homework Help
Replies
1
Views
537
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
3
Views
571
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
Replies
0
Views
459
  • Calculus and Beyond Homework Help
Replies
3
Views
763
  • Calculus and Beyond Homework Help
Replies
5
Views
799
  • Calculus and Beyond Homework Help
Replies
12
Views
992
  • Calculus and Beyond Homework Help
Replies
3
Views
418
Back
Top