# Is $2\delta(x) \neq \delta(x)$?

$2\delta(x) \neq \delta(x)$ since, by definition,

Can this been seen graphically though?

If so, how?

If not, why is it that they are mathematically different but graphically the same?

Btw, I don't understand any precise definitions here so please dumb down your answers to the level of someone who knows basic Real Analysis and Partial Differential Equations, and just pretend I don't really know what the Dirac Delta function is. Thanks.

http://en.wikipedia.org/wiki/Dirac_delta_function#Definitions

Context: I thought the derivative of the sgn function was the Dirac Delta Function(al)/Distribution when it is really twice of it (http://en.wikipedia.org/wiki/Signum_function#cite_ref-2). I was thinking they would turn out to to be equal. Is it really wrong to say that the derivative of the sgn function is the Dirac Delta Function(al)/Distribution? I just can't visualize this.

• Think of it as a functional defined as follows $\langle \delta, f\rangle:=f(0)$. Since $\delta$ is not a function, but a "distribution" I don't think you will gain much by trying to compare the graphs of $\delta$ and $2\delta$. It might be helpful to compare the graphs of functions whose "limits" give $\delta$ and $2\delta$. – EPS May 17 '14 at 15:16
• @Sam Ah you mean like the ones here? upload.wikimedia.org/wikipedia/commons/b/b4/… So if I multiply each by 2, they would be different...Thanks. Sorry I forgot to add the context: I thought the derivative of the sgn function was the Dirac Delta Function(al)/Distribution when it is really twice of it (en.wikipedia.org/wiki/Signum_function#cite_ref-2). I was thinking they would turn out to to be equal. Is it really wrong to say that the derivative of the sgn function is the Dirac Delta Function(al)/Distribution? I just can't visualize this. – BCLC May 17 '14 at 15:24
• The sign makes a jump of size 2 at 0 (jumping from -1 to +1) hence it is only natural that the derivative of the sign function in the distributional sense of the term derivative should be twice the Dirac distribution at 0, not the Dirac distribution at 0. – Did May 17 '14 at 15:38
• If one plots a delta function, then its graph would be zero everywhere except the origin where it is infinite. Since $2\cdot\infty=\infty$, the graph of twice a delta function looks the same as that of a single delta function although they are different objects in the sense of distributions. – Riemann1337 May 17 '14 at 15:41
• The usual symbol for $2\,\delta(x)$ is a vertical vector of length $2$. This convention was chosen since integration will give $2\,H(x)+C$ with $H$ the Heaviside step function : we go two units higher at $x=0$ since $\int_{-\epsilon}^{\epsilon}2\,\delta(x)\,dx=2$. Note that for $C=-1$ you get the $\operatorname{sgn}$ function. – Raymond Manzoni May 17 '14 at 23:12