- #1

friend

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I'm told that a product of distributions is undefined. See,

http://en.wikipedia.org/wiki/Distribution_(mathematics)#Problem_of_multiplication

where the Dirac delta function is considered a distribution.

Now the Dirac delta function is defined such that,

[tex]\[

\int_{ - \infty }^{ + \infty } {{\rm{f(x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{f(x}}_0 )

\]

[/tex]

for all continuous compactly supported functions ƒ. See,

http://en.wikipedia.org/wiki/Dirac_delta_function

But the question is can we make [tex]\[

{\rm{f(x}}_1 ) = {\rm{\delta (x - x}}_1 )

\]

[/tex], in order to get,

[tex]\[

\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{\delta (x - x}}_0 )

\]

[/tex]

which is a very convenient recursion relation?

But then we are faced with the product of distributions inside the integral. So does the recursion relation actually exist?

We are told that the delta function is not everywhere continuous so it is not allowed to be [tex]\[

{\rm{f(x}}_1 )

\]

[/tex].

Nevertheless, it seems obvious if we consider the limits of the delta function individually, then of course the recursion relation is allowed. For if we use the gaussian form of the delta function, we have,

[tex]\[

{\rm{\delta (x - x}}_1 ) = \mathop {\lim }\limits_{\Delta _1 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_1 )^2 /\Delta _1 ^2 }

\]

[/tex]

and

[tex]\[

{\rm{\delta (x}}_1 {\rm{ - x}}_0 ) = \mathop {\lim }\limits_{\Delta _0 \to 0} \frac{1}{{(\pi \Delta _0 ^2 )^{1/2} }}e^{ - (x_1 - x_0 )^2 /\Delta _0 ^2 }

\]

[/tex]

Then,

[tex]\[

\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = \int_{ - \infty }^{ + \infty } {\left( {\mathop {\lim }\limits_{\Delta _1 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_1 )^2 /\Delta _1 ^2 } } \right)\left( {\mathop {\lim }\limits_{\Delta _0 \to 0} \frac{1}{{(\pi \Delta _0 ^2 )^{1/2} }}e^{ - (x_1 - x_0 )^2 /\Delta _0 ^2 } } \right){\rm{dx}}_1 }

\]

[/tex]

[tex]\[

= \mathop {\lim }\limits_{\Delta _1 \to 0} \int_{ - \infty }^{ + \infty } {\left( {\frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_1 )^2 /\Delta _1 ^2 } } \right){\rm{\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = \mathop {\lim }\limits_{\Delta _1 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_0 )^2 /\Delta _1 ^2 } = {\rm{\delta (x - x}}_0 )

\]

[/tex]

For if we let [tex]\[

{\Delta _1 }

\]

[/tex] remain a fixed non-zero number until after the integration then the exponential delta function is a continuous compactly supported function and qualifies to be [tex]\[

{\rm{f(x}}_1 )

\]

[/tex]. Or

[tex]\[

= \mathop {\lim }\limits_{\Delta _0 \to 0} \int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 )\left( {\frac{1}{{(\pi \Delta _0 ^2 )^{1/2} }}e^{ - (x_1 - x_0 )^2 /\Delta _0 ^2 } } \right){\rm{dx}}_1 } = \mathop {\lim }\limits_{\Delta _0 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_0 )^2 /\Delta _0 ^2 } = {\rm{\delta (x - x}}_0 )

\]

[/tex]

if we let [tex]\[

{\Delta _0 }

\]

[/tex] remain a fixed non-zero number until after the integration so that [tex]\[

{\rm{f(x}}_1 )

\]

[/tex] becomes a continuous compactly supported function as before.

Since the result is [tex]\[

{\rm{\delta (x - x}}_0 )

\]

[/tex] for any order in which we take the limits. Does this prove that the limit is valid and the recursion relation holds? Thank you.

http://en.wikipedia.org/wiki/Distribution_(mathematics)#Problem_of_multiplication

where the Dirac delta function is considered a distribution.

Now the Dirac delta function is defined such that,

[tex]\[

\int_{ - \infty }^{ + \infty } {{\rm{f(x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{f(x}}_0 )

\]

[/tex]

for all continuous compactly supported functions ƒ. See,

http://en.wikipedia.org/wiki/Dirac_delta_function

But the question is can we make [tex]\[

{\rm{f(x}}_1 ) = {\rm{\delta (x - x}}_1 )

\]

[/tex], in order to get,

[tex]\[

\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = {\rm{\delta (x - x}}_0 )

\]

[/tex]

which is a very convenient recursion relation?

But then we are faced with the product of distributions inside the integral. So does the recursion relation actually exist?

We are told that the delta function is not everywhere continuous so it is not allowed to be [tex]\[

{\rm{f(x}}_1 )

\]

[/tex].

Nevertheless, it seems obvious if we consider the limits of the delta function individually, then of course the recursion relation is allowed. For if we use the gaussian form of the delta function, we have,

[tex]\[

{\rm{\delta (x - x}}_1 ) = \mathop {\lim }\limits_{\Delta _1 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_1 )^2 /\Delta _1 ^2 }

\]

[/tex]

and

[tex]\[

{\rm{\delta (x}}_1 {\rm{ - x}}_0 ) = \mathop {\lim }\limits_{\Delta _0 \to 0} \frac{1}{{(\pi \Delta _0 ^2 )^{1/2} }}e^{ - (x_1 - x_0 )^2 /\Delta _0 ^2 }

\]

[/tex]

Then,

[tex]\[

\int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 {\rm{)\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = \int_{ - \infty }^{ + \infty } {\left( {\mathop {\lim }\limits_{\Delta _1 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_1 )^2 /\Delta _1 ^2 } } \right)\left( {\mathop {\lim }\limits_{\Delta _0 \to 0} \frac{1}{{(\pi \Delta _0 ^2 )^{1/2} }}e^{ - (x_1 - x_0 )^2 /\Delta _0 ^2 } } \right){\rm{dx}}_1 }

\]

[/tex]

[tex]\[

= \mathop {\lim }\limits_{\Delta _1 \to 0} \int_{ - \infty }^{ + \infty } {\left( {\frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_1 )^2 /\Delta _1 ^2 } } \right){\rm{\delta (x}}_1 {\rm{ - x}}_0 ){\rm{dx}}_1 } = \mathop {\lim }\limits_{\Delta _1 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_0 )^2 /\Delta _1 ^2 } = {\rm{\delta (x - x}}_0 )

\]

[/tex]

For if we let [tex]\[

{\Delta _1 }

\]

[/tex] remain a fixed non-zero number until after the integration then the exponential delta function is a continuous compactly supported function and qualifies to be [tex]\[

{\rm{f(x}}_1 )

\]

[/tex]. Or

[tex]\[

= \mathop {\lim }\limits_{\Delta _0 \to 0} \int_{ - \infty }^{ + \infty } {{\rm{\delta (x - x}}_1 )\left( {\frac{1}{{(\pi \Delta _0 ^2 )^{1/2} }}e^{ - (x_1 - x_0 )^2 /\Delta _0 ^2 } } \right){\rm{dx}}_1 } = \mathop {\lim }\limits_{\Delta _0 \to 0} \frac{1}{{(\pi \Delta _1 ^2 )^{1/2} }}e^{ - (x - x_0 )^2 /\Delta _0 ^2 } = {\rm{\delta (x - x}}_0 )

\]

[/tex]

if we let [tex]\[

{\Delta _0 }

\]

[/tex] remain a fixed non-zero number until after the integration so that [tex]\[

{\rm{f(x}}_1 )

\]

[/tex] becomes a continuous compactly supported function as before.

Since the result is [tex]\[

{\rm{\delta (x - x}}_0 )

\]

[/tex] for any order in which we take the limits. Does this prove that the limit is valid and the recursion relation holds? Thank you.

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