Why can k in the Delta Function equation be both positive and negative?

[mathnerd15]

in this equation why does k take on both positive and negative values? isn't k a fixed constant that can only be positive or negative

 

[namanjain]

in this equation why does k take on both positive and negative values? isn't k a fixed constant that can only be positive or negative

ques. not visible

 

[mathnerd15]

oh you have to click on the attachment and then the picture to see the equation. so the domain of the function is kx so if k ranges from -infinity to infinity then you need a +- before the integral?

 

[vanhees71]

It's just an example for pretty unclear notation. I guess the formula is an attempt to prove the equation
$$\delta(k x)=\frac{1}{|k|} \delta(k x).$$
Of course you have to assume that $k \neq 0$. Otherwise the equation doesn't make any sense to begin with.

Then you have to just do the substitution $y=k x$ in the integral with the distribution times an arbitrary test function to prove this formula. For $k>0$ you find
$$\int_{\mathbb{R}} \mathrm{d} x f(x) \delta(k x)=\int_{\mathbb{R}} \mathrm{d}y \frac{1}{k} f(y/k) \delta(y)=\frac{1}{k} f(0).$$
For $k<0$ you have
$$\int_{\mathbb{R}} \mathrm{d} x f(x) \delta(k x)=\int_{\mathbb{R}} \mathrm{d} y \left (-\frac{1}{k} \right ) f(y/k) \delta (y)=-\frac{1}{k} f(0).$$
On the other hand the distribution
$\frac{1}{|k|} \delta(x)$
has the same outcome under an integral with an arbitrary test function, which proves the above statement about the Dirac distribution.

You can generalize this for arbitrary function $y(x)$ which have only single-order zeros, i.e., $y(x_k)=0$ but $y'(x_k) \neq 0$ for $k \in \{1,\ldots,n\}$. Then you can prove in pretty much the same way as the above example
$$\delta[y(x)]=\sum_{k=1}^{n} \frac{1}{|y'(x_k)|} \delta(x-x_k).$$

 

[mathnerd15]

$$f(x)=\frac{1}{2\pi}\int_{R}e^{ipx}\left ( \int_{R}f(\alpha)d\alpha \right )dp=\frac{1}{2\pi}\int_{R}\left (\int_{R}e^{ipx}e^{-ip\alpha}\right)f(\alpha)d\alpha= \int_{R}\delta (x-\alpha)f(\alpha)d\alpha.$$ where,[itex]\delta(x-a)=\int_{R}e^{ip(x-\alpha)}\left dp. [/tex]as you know Cauchy expressed Fourier's integral as exponentials and the delta distribution can be expressed in this way (he also pointed out that the integrals are non-commutative in some circumstances). In modern times there is L Schwartz's theory of distributions. Dirac called it the delta function because he used it as a continuous analogue of the discrete Kronecker delta.<br /> Here is a vast generalization of the fundamental theorem of algebra[/itex]

 

[mathnerd15]

$$(x+y)^{N+1}=_{0}^{N+1}\textrm{C}x^{N+1}+\sum_{k=1}^{N}(_{k}^{N}\textrm{C}+_{k-1}^{N}\textrm{C})x^{N+1-k}y^{k}+_{N+1}^{N+1}\textrm{C}y^{N+1}=_{0}^{N+1}\textrm{C}x^{N+1}+\sum_{k=1}^{N}_{k}^{N+1}\textrm{C}x^{(N+1)-k}y^{k}+_{N+1}^{N+1}\textrm{C}y^{N+1}=\sum_{k=0}^{N+1}_{k}^{N+1}{C}x^{(N+1)-k}y^{k}$$