Convolution of a Gaussian with itself from the definition

Click For Summary
SUMMARY

The convolution of the Gaussian function g(x) = e^{-πx^{2}} with itself is calculated using the definition of convolution, specifically integrating from -∞ to ∞. The process involves integration by parts, which is essential for solving the integral correctly. A common mistake is concluding that the convolution equals zero; however, the correct approach reveals that the convolution results in another Gaussian function, specifically g(x) = e^{-πx^{2}/2}. This outcome is derived from completing the square in the exponent during the integration process.

PREREQUISITES
  • Understanding of convolution integrals
  • Proficiency in integration by parts
  • Knowledge of Gaussian functions
  • Familiarity with completing the square in algebra
NEXT STEPS
  • Study the properties of convolution for Gaussian functions
  • Practice integration by parts with various functions
  • Explore the implications of convolution in signal processing
  • Learn about the Central Limit Theorem and its relation to Gaussian distributions
USEFUL FOR

Students in advanced calculus, mathematicians studying convolution operations, and anyone interested in the properties of Gaussian functions in mathematical analysis.

GuiltySparks
Messages
11
Reaction score
0

Homework Statement



Find the convolution of g(x) = e^{-πx^{2}} with itself from -∞ to ∞ using the definition of convolution, not the Fourier Transform.



The Attempt at a Solution



See my attachment. My professor said that you have to use integration by parts, but I keep getting stuck. No matter what I do, I reach the conclusion that the convolution is 0. Is there something that I'm missing here?
 

Attachments

  • gauss.jpg
    gauss.jpg
    11.8 KB · Views: 658
Physics news on Phys.org
GuiltySparks said:

Homework Statement



Find the convolution of g(x) = e^{-πx^{2}} with itself from -∞ to ∞ using the definition of convolution, not the Fourier Transform.

The Attempt at a Solution



See my attachment. My professor said that you have to use integration by parts, but I keep getting stuck. No matter what I do, I reach the conclusion that the convolution is 0. Is there something that I'm missing here?

Complete the square.

\displaystyle -2\pi\left( y^2-xy\right)-\pi x^2
\displaystyle =<br /> -2\pi\left( y^2-xy+\frac{x^2}{4}-\frac{x^2}{4}\right)-\pi x^2

\displaystyle =-2\pi\left( y-\frac{x}{2}\right)^2+2\pi\frac{x^2}{4}-\pi x^2

\displaystyle =-2\pi\left( y-\frac{x}{2}\right)^2-\pi\frac{x^2}{2}​
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 19 ·
Replies
19
Views
30K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
1
Views
2K
  • · Replies 11 ·
Replies
11
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K