Properties of Heaviside FunctionNested integrals with products of Heaviside step functionsCompute the following integral of the Heaviside step functionWhich solver method for solving indicator function in a QP problemWavefront Set of the Heaviside FunctionIntegrating Heaviside Step Function of two VariablesAny way to prove this identity :$fracd^mdx^mfracH(x)x^m-1(m-1)! =delta(x)$?Multivariable integral, use of Dirac Delta and Heaviside ThetaIs Heaviside step function or unit step function periodic?Heaviside step function with function as argumentIntegrating the composition of a Heaviside function with a smooth function
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Properties of Heaviside Function
Nested integrals with products of Heaviside step functionsCompute the following integral of the Heaviside step functionWhich solver method for solving indicator function in a QP problemWavefront Set of the Heaviside FunctionIntegrating Heaviside Step Function of two VariablesAny way to prove this identity :$fracd^mdx^mfracH(x)x^m-1(m-1)! =delta(x)$?Multivariable integral, use of Dirac Delta and Heaviside ThetaIs Heaviside step function or unit step function periodic?Heaviside step function with function as argumentIntegrating the composition of a Heaviside function with a smooth function
$begingroup$
Let $H(x)$ be the Heaviside function defined by
begincases
1 & textif xgeq0\
0 & textif x<0
endcases
I know that
$H'(x)=delta(x)$. The derivative of the Heaviside function is the delta function.
$delta(x)=delta(-x)$. The delta function is symetric.
Does
$H(x)=H(-x)$?
$H(x)=-H(x)$?
It appears that
$$ -delta(x)delta(-y)=delta(x)delta(y)$$
and
$$ -delta(-x)delta(y)=delta(x)delta(y)$$
Do both of these properties follow from the definition of the Heaviside function?
step-function
asked Mar 21 at 18:31
Axion004Axion004
405413
$endgroup$
add a comment |
$begingroup$
Let $H(x)$ be the Heaviside function defined by
begincases
1 & textif xgeq0\
0 & textif x<0
endcases
I know that
$H'(x)=delta(x)$. The derivative of the Heaviside function is the delta function.
$delta(x)=delta(-x)$. The delta function is symetric.
Does
$H(x)=H(-x)$?
$H(x)=-H(x)$?
It appears that
$$ -delta(x)delta(-y)=delta(x)delta(y)$$
and
$$ -delta(-x)delta(y)=delta(x)delta(y)$$
Do both of these properties follow from the definition of the Heaviside function?
step-function
asked Mar 21 at 18:31
Axion004Axion004
405413
$endgroup$
1
$begingroup$
$delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps .
$endgroup$
– Faraday Pathak
Mar 21 at 19:18
add a comment |
$begingroup$
Let $H(x)$ be the Heaviside function defined by
begincases
1 & textif xgeq0\
0 & textif x<0
endcases
I know that
$H'(x)=delta(x)$. The derivative of the Heaviside function is the delta function.
$delta(x)=delta(-x)$. The delta function is symetric.
Does
$H(x)=H(-x)$?
$H(x)=-H(x)$?
It appears that
$$ -delta(x)delta(-y)=delta(x)delta(y)$$
and
$$ -delta(-x)delta(y)=delta(x)delta(y)$$
Do both of these properties follow from the definition of the Heaviside function?
step-function
asked Mar 21 at 18:31
Axion004Axion004
405413
$endgroup$
Let $H(x)$ be the Heaviside function defined by
begincases
1 & textif xgeq0\
0 & textif x<0
endcases
I know that
$H'(x)=delta(x)$. The derivative of the Heaviside function is the delta function.
$delta(x)=delta(-x)$. The delta function is symetric.
Does
$H(x)=H(-x)$?
$H(x)=-H(x)$?
It appears that
$$ -delta(x)delta(-y)=delta(x)delta(y)$$
and
$$ -delta(-x)delta(y)=delta(x)delta(y)$$
Do both of these properties follow from the definition of the Heaviside function?
step-function
step-function
asked Mar 21 at 18:31
Axion004Axion004
405413
asked Mar 21 at 18:31
Axion004Axion004
405413
asked Mar 21 at 18:31
Axion004Axion004
405413
asked Mar 21 at 18:31
Axion004Axion004
405413
asked Mar 21 at 18:31
Axion004Axion004
405413
405413
1
$begingroup$
$delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps .
$endgroup$
– Faraday Pathak
Mar 21 at 19:18
add a comment |
1
$begingroup$
$delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps .
$endgroup$
– Faraday Pathak
Mar 21 at 19:18
1
1
$begingroup$
$delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps .
$endgroup$
– Faraday Pathak
Mar 21 at 19:18
$begingroup$
$delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps .
$endgroup$
– Faraday Pathak
Mar 21 at 19:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, $H(x)=1-H(-x)$ for $xne 0$. Integrating something even gives something odd plus an integration constant.
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
1
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
|
show 3 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, $H(x)=1-H(-x)$ for $xne 0$. Integrating something even gives something odd plus an integration constant.
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
1
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
|
show 3 more comments
$begingroup$
No, $H(x)=1-H(-x)$ for $xne 0$. Integrating something even gives something odd plus an integration constant.
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
$endgroup$
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
1
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
|
show 3 more comments
$begingroup$
No, $H(x)=1-H(-x)$ for $xne 0$. Integrating something even gives something odd plus an integration constant.
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
$endgroup$
No, $H(x)=1-H(-x)$ for $xne 0$. Integrating something even gives something odd plus an integration constant.
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
answered Mar 21 at 19:11
J.G.J.G.
32.6k23250
32.6k23250
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
1
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
|
show 3 more comments
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
1
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
And indeed, differentiating that using $H'(x) = delta(x)$ gives the correct $delta(x) = delta(-x)$.
$endgroup$
– eyeballfrog
Mar 21 at 19:22
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
I see that, can this justify that $-delta(x)delta(-y)=delta(x)delta(y)$?
$endgroup$
– Axion004
Mar 21 at 19:46
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
@Axion004 Since $delta$ is even, the leftmost $-$ sign shouldn't be there.
$endgroup$
– J.G.
Mar 21 at 19:56
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
$begingroup$
That is the part which is confusing to me. As $delta(-y)=delta(y)$, we have that $-delta(x)delta(-y)=-delta(x)delta(y)=delta(x)delta(y)$. I don't see why $-delta(x)delta(y)=delta(x)delta(y)$.
$endgroup$
– Axion004
Mar 21 at 20:41
1
1
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
$begingroup$
Yes, got it. Thanks!
$endgroup$
– Axion004
Mar 21 at 21:14
|
show 3 more comments
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1
$begingroup$
$delta (x) $ isn't an ordinary but generalised function so, you always need test function for proof of it's properties .i don't see how heaviside function helps .
$endgroup$
– Faraday Pathak
Mar 21 at 19:18