Natural Logarithm of Convolution

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Discussion Overview

The discussion revolves around the properties of logarithms in relation to the convolution of two functions. Participants explore whether the logarithmic identity ln(x*y) = ln(x) + ln(y) holds when applied to convolutions, which are operations on functions rather than simple variables.

Discussion Character

  • Conceptual clarification, Debate/contested

Main Points Raised

  • One participant questions if the properties of logarithms apply to the convolution of two variables, specifically asking if ln(x*y) equals ln(x) + ln(y).
  • Another participant seeks clarification on the notation used, noting that ln(x) typically represents the natural logarithm of a number.
  • A third participant asserts that the initial question is flawed because convolution operates on functions, not variables, and states that the logarithmic property does not hold in this context.
  • A later reply acknowledges the misunderstanding regarding convolution as an integral operation, indicating a realization of the initial error.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as there are competing views regarding the application of logarithmic properties to convolutions and the nature of the convolution operation itself.

Contextual Notes

The discussion highlights the potential confusion between variables and functions in the context of convolution and logarithmic properties, which remains unresolved.

tramar
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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) ?
 
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In your notation, what are x, y, x*y and ln? Usually ln(x) is the natural log of a number (x).
 
tramar said:
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.
 
Last edited:
I realize that now since a convolution is an integral... just wishful thinking on my part I guess.
 

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