Oct 4, 2015 · Yesterday a friend asked me what a Dirac delta function really is. I tried to explain it but eventually confused myself. It seems that a Dirac delta is defined as a function that satisfies these

Sep 12, 2024 · Explore related questions convolution dirac-delta See similar questions with these tags.

Sep 11, 2020 · The Dirac Delta is simply NOT a function (it is a Generalized Function). And use of an integral operator symbol to represent the functional $\langle \delta,\phi\rangle$ is abuse of notation.

Aug 25, 2019 · On Wikipedia, the definition of the dirac delta function is given as: Suppose I have a function where at two points, the function goes to infinity. Given that the distance between the two …

45 I am given to understand that the Dirac delta function is strictly not a function in the conventional sense and it is a "functional or a distribution".

Using this definition and the fact that the $\delta$-distribution is half of the second derivative of the absolute value function, one can give a rigorous proof of the formula in the query.

Mar 9, 2023 · The Dirac delta is defined as a distribution by $$ \langle \delta_0,\varphi\rangle = \varphi (0). $$ The compact support is actually not needed for $\varphi$ because the value here just …

Feb 20, 2023 · This should hold by definition of Dirac delta: limit of some sequence of function with property that $ \int_ {-\infty}^\infty \delta (x)\cdot g (x)dx = g (0)$. Checking this definition property …

Dec 27, 2021 · So, if we accept, that the above is the real meaning of \begin {equation*} \int_ {-\infty}^\infty \delta (x) f (x) dx =f (0) \end {equation*} then, we have proved the f (0) property, for real …

Aug 30, 2018 · The answer to your question becomes quite easy if you are able to build the correct mathematical framework. Note that I try to build an answer adapted to the OP background, whence it …