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Delta function question

  1. Aug 31, 2013 #1
    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
     

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  3. Aug 31, 2013 #2
    ques. not visible
     
  4. Aug 31, 2013 #3
    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: Aug 31, 2013
  5. Aug 31, 2013 #4

    vanhees71

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    It's just an example for pretty unclear notation. I guess the formula is an attempt to prove the equation
    [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]
     
  6. Aug 31, 2013 #5
    [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
     

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    Last edited: Aug 31, 2013
  7. Sep 1, 2013 #6
    [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: Sep 1, 2013
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