# Dirac's Delta as a derivative of the unit step

1. Sep 17, 2006

### Ali 2

The Dirac detla or unit impulse function is defined as :

$\delta (t) = \left \{ \begin {matrix} \infty \quad \ t = 0 \\ 0 \quad : \ t \neq 0 \end{matrix}$

and the unit step function :

$u(t) = \left \{ \begin {matrix} 1 \quad \ t \geqslant 0 \\ 0 \quad : \ t < 0 \end{matrix}$

It is said that the
$$\frac d {dt} u(t) = \delta (t) [/itex] .. but .. if we came to the definition of the derivative , we should find the limit from left and right .. : [tex] { \displaysytle u'_+ (0) = \lim { h \to 0^+ } \frac { u (0 + h ) - u (0) } h = \lim { h \to 0^+ } \frac {1-1} h = 0 }$$

${ \displaystyle u'_- (0) = \lim { h \to 0^- } \frac { u (0 + h ) - u (0) } h = \lim { h \to 0^- } \frac {0-1 }h = \infty}$

We see that the derivative is infinity from left only .. and it is not equal to the right limit , so the derivative doesn't exist .. and it is not an imuplse ..

But IF we define u (0) to be between 0 and 1 , For instance 0.5 , then :

$u'_+ (0) = \lim { h \to 0^+ } \frac { u (0 + h ) - u (0) } h = \lim { h \to 0^+ } \frac {1-0.5} h = \infty$

Which makes the derivative from both sides infinity , which gives us the impulse ..

As a result , we should chane the definition of the unit step function ..

Do you agree with me ?

Last edited: Sep 17, 2006
2. Sep 17, 2006

### arildno

No, because you cannot, strictly speaking, take the derivative of a discontinuous function in the first place.

3. Sep 17, 2006

### arildno

Besides, what you have given as a "definition" of the Dirac delta "function" isn't any valid definition, nor is the Dirac delta "function" a function, it's a distribution.

4. Sep 17, 2006

### lokofer

- There're even more "bizarre" approach for example the n-th derivative of the delta function is equal to (Fourier transform):

$$2\pi D^{n} \delta (x) =\int_{-\infty}^{\infty}du(iu)^{n}e^{iux}$$

By the way...why can't you differentiate a "discontinous" function?..for example you ca get the formula for the derivative of prime number function:

$$\frac{d\pi (x)}{dx}= \sum_{p}\delta (x-p)$$ and PNT theorems stablishes the asymptotic equality:

$$\frac{d\pi (x)}{dx} \sim 1/log(x)$$

"ramanujan gave the formula"...(Mathworld)

$$\frac{d\pi (x)}{dx}= \sum_{n>1}\frac{ \mu(n)}{n}x^{1/n}(log(x))^{-1}$$

and the derivative of any function that has a "jump" at a real number c the derivative can be interpreted as $$f'(x)\delta (x-c)$$ that you can check is oo..if we express the derivative in the sense of "distributions" you can define it even with discontinuities.

5. Sep 17, 2006

### arildno

"if we express the derivative in the sense of "distributions" you can define it even with discontinuities."
Evidently, but then you've changed the sense of the term "derivative".

6. Sep 17, 2006

### HallsofIvy

Well, no, that isn't the way the Dirac Delta function is defined. The "Dirac Delta function" is defined as the linear functional that, to every function f in its domain, assigns the value f(0). It can then be shown that the Dirac Delta "function" satisfies those properties. Yes, it is true,that the Dirac Delta function is the derivative of the Heavyside Step Function under the definition of "derivative" for "distributions" or "generalized functions".

7. Sep 28, 2006

### jpr0

Isn't the step function already defined as taking the value 1/2 when it's argument is zero anyway?

8. Oct 7, 2006

### quinn

ad absurdum

I would answer by trying to intergrate the Dirac Delta.

The Dirac delta would give a step increase, but of infinite size and I am fairly certain that there would be no good way to determine at what point the step would fall.

That is to say the derivative of the unit step, u(t), or the unit step function times some constant, a*u(t), would yield the Dirac Delta for both:

u'(t) = delta = a*u'(t)

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