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Dirac function

  • Thread starter Aroma2010
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  • #1
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Dirac function :(

Hello everyone...

I have some triple with my PDEs course especially with the Dirac function.
How can I prove it is discontinuous function?
I do not know where can I start...
Could somebody help me, please.
 

Answers and Replies

  • #2
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Well, for a start, the Dirac "function" is not even considered a function. It doesn't qualify for it. There are two approaches two define it. You can considere it as a limit of several continuous sequences of functions, for example:

[tex]\delta_{n}=\frac{1}{n\sqrt{\pi}} e^{-x^{2}/n^{2}} , n\rightarrow 0 [/tex]

is one such sequence. The limit, however, is not defined as a function. The right name is distribution. See http://en.wikipedia.org/wiki/Distribution_(mathematics)
 
  • #3
Redbelly98
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(Moderator's note: thread moved from "Differential Equations")

Hello everyone...

I have some triple with my PDEs course especially with the Dirac function.
How can I prove it is discontinuous function?
I do not know where can I start...
Could somebody help me, please.
Start with the definition of a continuous function. At least one of the conditions is not satisfied, at at least one value of x, for a function to be discontinuous.
 
  • #4
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Aroma2010 said:
I have some triple with my PDEs course especially with the Dirac function.
How can I prove it is discontinuous function?
I do not know where can I start...
Could somebody help me, please.
It's pretty simple really. Just prove that the left and right limits at 0 are 0 (pretty easy), so in order to be continuous, f(0) must be 0, but it isn't, QED.
 

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