- #1

- 109

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter mathnerd15
- Start date

- #1

- 109

- 0

- #2

- 70

- 0

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

- #3

- 109

- 0

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?

Last edited:

- #4

- 20,105

- 10,843

[tex]\delta(k x)=\frac{1}{|k|} \delta(k x).[/tex]

Of course you have to assume that [itex]k \neq 0[/itex]. Otherwise the equation doesn't make any sense to begin with.

Then you have to just do the substitution [itex]y=k x[/itex] in the integral with the distribution times an arbitrary test function to prove this formula. For [itex]k>0[/itex] you find

[tex]\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).[/tex]

For [itex]k<0[/itex] you have

[tex]\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).[/tex]

On the other hand the distribution

[itex]\frac{1}{|k|} \delta(x)[/itex]

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 [itex]y(x)[/itex] which have only single-order zeros, i.e., [itex]y(x_k)=0[/itex] but [itex]y'(x_k) \neq 0[/itex] for [itex]k \in \{1,\ldots,n\}[/itex]. Then you can prove in pretty much the same way as the above example

[tex]\delta[y(x)]=\sum_{k=1}^{n} \frac{1}{|y'(x_k)|} \delta(x-x_k).[/tex]

- #5

- 109

- 0

[tex]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.[/tex] 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.

Here is a vast generalization of the fundamental theorem of algebra

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.

Here is a vast generalization of the fundamental theorem of algebra

Last edited:

- #6

- 109

- 0

[tex](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}[/tex]

Last edited:

Share: