Natural Logarithm of Convolution

[tramar]

If I have a convolution of two variables, say x * y, and I take the natural logarithm of this operation, ln(x*y), do the same properties of logarithms apply?

So, does ln(x*y) = ln(x)+ln(y) ?

 

[mathman]

In your notation, what are x, y, x*y and ln? Usually ln(x) is the natural log of a number (x).

 

[LCKurtz]

If I have a convolution of two variables, say x * y, and I take the natural logarithm of this operation, ln(x*y), do the same properties of logarithms apply?

So, does ln(x*y) = ln(x)+ln(y) ?

I'm afraid your question doesn't make any sense as stated. The convolution operation is an operation on two functions, not two variables, and it gives another function. And if you ask the same question for functions, the answer is no.

 

[tramar]

I realize that now since a convolution is an integral... just wishful thinking on my part I guess.

 

[CellCoree]

I can confirm that the properties of logarithms do indeed apply to the natural logarithm of a convolution. This means that ln(x*y) is equivalent to ln(x) + ln(y). This is because the natural logarithm of a product is equal to the sum of the natural logarithms of the individual factors. Therefore, the same rules and properties that apply to normal logarithms also apply to the natural logarithm of a convolution. This property can be useful in simplifying calculations and understanding the behavior of convolutions in mathematical models.