Fourier transformation of distribution
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How to prove this Fourier transformation please? What does it mean $x^0_+$?
FT
fourier-transform
asked Nov 23 at 18:01
Elisabeth
1034
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up vote
0
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How to prove this Fourier transformation please? What does it mean $x^0_+$?
FT
fourier-transform
asked Nov 23 at 18:01
Elisabeth
1034
I guess that $x_+ = x , H(x),$ where $H$ is the Heaviside step function. Then $x_+^lambda = (x , H(x))^lambda,$ which equals $x^lambda$ for $x>0$ and $0$ for $x leq 0.$ When $lambda=0$ it should equal $H(x).$
– md2perpe
Nov 23 at 18:38
Thank you very much
– Elisabeth
Nov 24 at 17:47
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
How to prove this Fourier transformation please? What does it mean $x^0_+$?
FT
fourier-transform
asked Nov 23 at 18:01
Elisabeth
1034
How to prove this Fourier transformation please? What does it mean $x^0_+$?
FT
fourier-transform
fourier-transform
asked Nov 23 at 18:01
Elisabeth
1034
asked Nov 23 at 18:01
Elisabeth
1034
edited Nov 23 at 18:08
asked Nov 23 at 18:01
Elisabeth
1034
asked Nov 23 at 18:01
Elisabeth
1034
asked Nov 23 at 18:01
Elisabeth
1034
1034
I guess that $x_+ = x , H(x),$ where $H$ is the Heaviside step function. Then $x_+^lambda = (x , H(x))^lambda,$ which equals $x^lambda$ for $x>0$ and $0$ for $x leq 0.$ When $lambda=0$ it should equal $H(x).$
– md2perpe
Nov 23 at 18:38
Thank you very much
– Elisabeth
Nov 24 at 17:47
add a comment |
I guess that $x_+ = x , H(x),$ where $H$ is the Heaviside step function. Then $x_+^lambda = (x , H(x))^lambda,$ which equals $x^lambda$ for $x>0$ and $0$ for $x leq 0.$ When $lambda=0$ it should equal $H(x).$
– md2perpe
Nov 23 at 18:38
Thank you very much
– Elisabeth
Nov 24 at 17:47
I guess that $x_+ = x , H(x),$ where $H$ is the Heaviside step function. Then $x_+^lambda = (x , H(x))^lambda,$ which equals $x^lambda$ for $x>0$ and $0$ for $x leq 0.$ When $lambda=0$ it should equal $H(x).$
– md2perpe
Nov 23 at 18:38
I guess that $x_+ = x , H(x),$ where $H$ is the Heaviside step function. Then $x_+^lambda = (x , H(x))^lambda,$ which equals $x^lambda$ for $x>0$ and $0$ for $x leq 0.$ When $lambda=0$ it should equal $H(x).$
– md2perpe
Nov 23 at 18:38
Thank you very much
– Elisabeth
Nov 24 at 17:47
Thank you very much
– Elisabeth
Nov 24 at 17:47
add a comment |
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I guess that $x_+ = x , H(x),$ where $H$ is the Heaviside step function. Then $x_+^lambda = (x , H(x))^lambda,$ which equals $x^lambda$ for $x>0$ and $0$ for $x leq 0.$ When $lambda=0$ it should equal $H(x).$
– md2perpe
Nov 23 at 18:38
Thank you very much
– Elisabeth
Nov 24 at 17:47