One of my friend sent me to proof this identity :$\frac{d^m}{dx^m}\frac{H(x)x^{m-1}}{(m-1)!} =\delta(x)$ Where $\delta$ is the dirac delta function and $H$ is heaviside step function , I knwo only that derivative of the first order of heaviside step function present dirac function, But i never saw the titled formula with $m$ th derivative , Then does the title identity a well known formula ? and how i can show it ?

  • $\begingroup$ what definition of $\delta$ and derivative are you using? $\endgroup$ – GFauxPas Aug 9 '18 at 0:19
  • $\begingroup$ delta is dirac delta function $\endgroup$ – zeraoulia rafik Aug 9 '18 at 0:19
  • $\begingroup$ look this :mathworld.wolfram.com/DeltaFunction.html $\endgroup$ – zeraoulia rafik Aug 9 '18 at 0:21
  • $\begingroup$ Are you sure of this formula ? It's well known that $ \frac{d}{dx} H(x) = \delta(0)$, doesn't work for $m=1$ ? $\endgroup$ – Phoenix Aug 9 '18 at 0:25
  • $\begingroup$ From there, you see the definition of $\delta$ is $\langle \delta, f \rangle = f(0)$, so what are you using to define the derivative of that? $\endgroup$ – GFauxPas Aug 9 '18 at 0:25

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