Convolution product between x² and x³

In summary, the conversation is discussing the convolution product between two functions, x² and x³. The first attempt to find the convolution product using the integral ##\int_{- \infty}^{+ \infty} f(u) g(x-u) du## resulted in an indeterminate form. However, the second attempt using the definition ##\int_{0}^{x} f(u) g(x-u) du## resulted in the answer of 1/60 x6. The conversation then moves on to discussing a possible symmetric algebraic definition of convolution, but the suggested expression ##\int_{- \infty}^{+ \infty} f(\frac{1}{2}x+u) g(\
  • #1
Jhenrique
685
4
I tryied make the convolution product between x² and x³ by ##\int_{- \infty}^{+ \infty} f(u) g(x-u) du## and the result is an indeterminate form, however, by defintion ##\int_{0}^{x} f(u) g(x-u) du##, the result is 1/60 x6. So, [tex]\int_{- \infty}^{+ \infty} f(u) g(x-u) du \overset{?}{=} \frac{1}{60} x^6[/tex] ?
 
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  • #2
Jhenrique said:
I tryied make the convolution product between x² and x³ by ##\int_{- \infty}^{+ \infty} f(u) g(x-u) du## and the result is an indeterminate form, however, by defintion ##\int_{0}^{x} f(u) g(x-u) du##, the result is 1/60 x6. So, [tex]\int_{- \infty}^{+ \infty} f(u) g(x-u) du \overset{?}{=} \frac{1}{60} x^6[/tex] ?
Show us what you did, please.
 
  • #3
Mark44 said:
Show us what you did, please.
imagem.png


BTW, exist a symmetric algebraic definition for convolution where the graphic of the function f(x) is moved oppositely to g(x) and vice-versa (different of the standard definition, where a function is fixed and the another is moved over the first)? I tryied ##\int_{- \infty}^{+ \infty} f(\frac{1}{2}x+u) g(\frac{1}{2}x-u) du##, but this expression isn't equal to the convolution of f and g...
 
  • #4
Jhenrique said:
I tryied make the convolution product between x² and x³ by ##\int_{- \infty}^{+ \infty} f(u) g(x-u) du## and the result is an indeterminate form, however, by defintion ##\int_{0}^{x} f(u) g(x-u) du##, the result is 1/60 x6.

OK, so far.

According to you, how does this follow exactly:

So, [tex]\int_{- \infty}^{+ \infty} f(u) g(x-u) du \overset{?}{=} \frac{1}{60} x^6[/tex] ?
 
  • #5
micromass said:
Jhenrique said:
[tex]\int_{- \infty}^{+ \infty} f(u) g(x-u) du \overset{?}{=} \frac{1}{60} x^6[/tex]
According to you, how does this follow exactly:

This equation isn't an affirmation, is a question. In acctually, I'm asking if ##\int_{- \infty}^{+\infty} f(u)g(x−u)du## is equal to ##\int_{0}^{x} f(u)g(x−u)du##.
 
  • #6
Jhenrique said:
This equation isn't an affirmation, is a question. In acctually, I'm asking if ##\int_{- \infty}^{+\infty} f(u)g(x−u)du## is equal to ##\int_{0}^{x} f(u)g(x−u)du##.

I know it's a question. But you're not going to learn much if we solve the question for you. So again, why do you think it's equal? Try to give the proof you're thinking of.
 

1. What is the convolution product between x² and x³?

The convolution product between x² and x³ is a mathematical operation that combines two functions to create a new function. It is expressed as (x² * x³) and is also known as the Cauchy product. The result of this operation is a function that represents the area under the curve of the two original functions.

2. How is the convolution product different from regular multiplication?

The convolution product differs from regular multiplication in that it not only multiplies the two functions, but also integrates their product over the entire range of values. This results in a new function that represents the combined influence of the two original functions.

3. What is the significance of the convolution product in science?

The convolution product is significant in science as it allows for the analysis of complex systems by breaking them down into simpler components. It is commonly used in signal processing, image processing, and probability theory. It also has applications in physics, biology, and engineering.

4. How is the convolution product calculated?

The convolution product is calculated by integrating the product of two functions over the entire range of values. This can be done using the convolution integral formula or by graphically representing the functions and finding the area under the curve of their product. In some cases, the convolution product can also be calculated using simpler mathematical operations, such as polynomial multiplication.

5. Can the order of the functions be changed in a convolution product?

Yes, the order of the functions can be changed in a convolution product without changing the result. This is because the convolution operation is commutative, meaning that the order of the functions does not affect the final outcome. However, the integration limits may change, so it is important to take this into consideration when swapping the order of the functions.

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