Are these statements valid ... ?

[trfrm]

Om Swastyastu.

Excuse me ... .

We can define the signature of a real number $x$ as

$\textrm{sgn}\,{x}:=\begin{cases}1&\textrm{if}~x>0\\0&\textrm{if}~x=0\\-1&\textrm{if}~x<0\end{cases}$ ,

then its absolut value is $|x|=x\,\textrm{sgn}\,x$ ... .

The unit-step-function is $u(x):=(\textrm{sgn}\,x+1)/2$ ... .

The delta Dirac $\delta(x)$ is defined such that $\int_{-\infty}^\infty\delta(x-y)\,f(x)\,dx=f(y)$ where $f$ is any integrable-real-function, and $y$ is any real number ... .

The derivative of unit-step-function is delta Dirac ... . $d\,u(x)/dx=\delta(x)$ ... .

Since $\textrm{sgn}\,x=2u(x)-1$, then $d\,\textrm{sgn}\,x/dx=2\,\delta(x)$ ... .

$d\,|x|/dx=d\,(x\,\textrm{sgn}\,x)/dx=\textrm{sgn}\,x+2x\,\delta(x)$ ... .

$\int\textrm{sgn}\,x\,dx=|x|+\textrm{constant}$ ... .

$\int\,u(x)\,dx=(|x|+x)/2+\textrm{constant}$ ... .

$\int\,|x|\,dx=\frac{1}{2}x|x|+\textrm{constant}$ ... .

Are those statements valid for any real number $x$ ... ?

Thank you and see you ... .

Alhamdulillah hirobbil alamin.

[trfrm]

Om santi santi om.

O.K. ... . Thank you for the inspection ... .

But ... , since $\displaystyle\frac{d}{dx}(f(x)\,g(x))=\left(\frac{d}{dx}f(x)\right)g(x)+f(x)\left(\frac{d}{dx}g(x)\right)$ and $\displaystyle\frac{d}{dx}\textrm{sgn}\,x=2\,\delta(x)$ , then

$\displaystyle\frac{d}{dx}|x|=\frac{d}{dx}(x\,\textrm{sgn}\,x)=\frac{dx}{dx}\,\textrm{sgn}\,x+x\frac{d}{dx}\textrm{sgn}\,x=\textrm{sgn}\,x+2x\,\delta(x)$ ... .

Wal bi Taufiq wal Hidayah.

[trfrm]

Sanctus, Sanctus, Dominus Deus Sabaoth.

The signum function is the derivative of the absolute value function (up to the indeterminacy at zero). Note, the resultant power of $x$ is $0$, similar to the ordinary derivative of $x$. The numbers cancel and all we are left with is the sign of x.
$\displaystyle\frac{d}{dx}|x|=\textrm{sgn}\,x$ for $x\neq0$ ... .

The formula in this quote is valid, since $\delta(x)=0$ for $x\neq0$ ... . However, for all $x\in\mathbb{R}$, this formula will become

$\displaystyle\frac{d}{dx}|x|=\textrm{sgn}\,x+2x\,\delta(x)$ ... .

Testing for $x\to0^+$ ... ,

$\displaystyle\lim_{x\to0^+}\left(\frac{d}{dx}|x|\right)=\lim_{x\to0^+}\textrm{sgn}\,x+2\lim_{x\to0^+}(x\,\delta(x))=1+0=1$ ... .

Testing for $x\to0^-$ ... ,

$\displaystyle\lim_{x\to0^-}\left(\frac{d}{dx}|x|\right)=\lim_{x\to0^-}\textrm{sgn}\,x+2\lim_{x\to0^-}(x\,\delta(x))=-1+0=-1$ ... .

Clearly, that $|x|$ has no derivative at $x=0$ ... .

The Dirac delta accomodates this derivative that is not continuous at $x=0$ ... .

http://www.thescienceforum.com/mathematics/34571-these-statements-valid.html#post406815

Om Swastyastu.