How to prove that convolution is associative and distributive with “plus”
up vote
1
down vote
favorite
The convolution is defined as:
$$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
$$f*(g+h)=f*g+f*h$$
I expand them but I find that it seems difficult to process especially how to change the order of integration.
Thank you a lot!
calculus integration
asked Jul 1 '14 at 7:30
zgzhen
1013
add a comment |
up vote
1
down vote
favorite
The convolution is defined as:
$$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
$$f*(g+h)=f*g+f*h$$
I expand them but I find that it seems difficult to process especially how to change the order of integration.
Thank you a lot!
calculus integration
asked Jul 1 '14 at 7:30
zgzhen
1013
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The convolution is defined as:
$$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
$$f*(g+h)=f*g+f*h$$
I expand them but I find that it seems difficult to process especially how to change the order of integration.
Thank you a lot!
calculus integration
asked Jul 1 '14 at 7:30
zgzhen
1013
The convolution is defined as:
$$int_{-infty}^{+infty}f(x-t)g(t)dt=f*g(x)$$
I want to prove the associativity and distributivity of it:$$f*(g*h)=(f*g)*h$$
$$f*(g+h)=f*g+f*h$$
I expand them but I find that it seems difficult to process especially how to change the order of integration.
Thank you a lot!
calculus integration
calculus integration
asked Jul 1 '14 at 7:30
zgzhen
1013
asked Jul 1 '14 at 7:30
zgzhen
1013
asked Jul 1 '14 at 7:30
zgzhen
1013
asked Jul 1 '14 at 7:30
zgzhen
1013
asked Jul 1 '14 at 7:30
zgzhen
1013
1013
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Hint:
For the first try writing out the integral then using a simple substitution like $u=x-t$
The second one follows from the fact that the integral is a linear operator. So,
begin{align*}
f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
&=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
end{align*}
edited Nov 20 at 7:40
Kei
297
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hint:
For the first try writing out the integral then using a simple substitution like $u=x-t$
The second one follows from the fact that the integral is a linear operator. So,
begin{align*}
f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
&=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
end{align*}
edited Nov 20 at 7:40
Kei
297
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
add a comment |
up vote
4
down vote
accepted
Hint:
For the first try writing out the integral then using a simple substitution like $u=x-t$
The second one follows from the fact that the integral is a linear operator. So,
begin{align*}
f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
&=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
end{align*}
edited Nov 20 at 7:40
Kei
297
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hint:
For the first try writing out the integral then using a simple substitution like $u=x-t$
The second one follows from the fact that the integral is a linear operator. So,
begin{align*}
f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
&=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
end{align*}
edited Nov 20 at 7:40
Kei
297
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
Hint:
For the first try writing out the integral then using a simple substitution like $u=x-t$
The second one follows from the fact that the integral is a linear operator. So,
begin{align*}
f*(g+h) &= int_{-infty}^infty f(x-t)(g(t)+h(t)),dt\
&=int_{-infty}^infty f(x-t) g(t) , dt +int_{-infty}^infty f(x-t)h(t) ,dt
end{align*}
edited Nov 20 at 7:40
Kei
297
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
edited Nov 20 at 7:40
Kei
297
edited Nov 20 at 7:40
Kei
297
edited Nov 20 at 7:40
Kei
297
297
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
answered Jul 1 '14 at 7:39
Millardo Peacecraft
9101619
9101619
add a comment |
add a comment |
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