Strange Calculus Question

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Eastwinn
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Strange Calculus Question

Postby Eastwinn » Sun Dec 19, 2010 9:07 pm UTC

I have no idea how to search this. Let's say I have two continuous functions on (0, c), f(t, x) and g(t, x), for which the following have no known closed forms:

[math]\int_{0}^{c}f(t,x)dt[/math]
[math]\int_{0}^{c}g(t,x)dt[/math]

If I write this:

[math]\int_{0}^{c}f(t,x)dt\int_{0}^{c}g(t,x)dt = \int_{0}^{c}h(t,x)dt[/math]

What can I say about h(t, x)? Can I define it in terms of f(t,x) and g(t,x)? Thanks :)
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z4lis
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Re: Strange Calculus Question

Postby z4lis » Mon Dec 20, 2010 12:11 am UTC

Hm. Smells a bit like it could be homework, so I'll mention that under sufficient smoothness assumptions (what you have there should be good enough, but someone check my sloppy thinking :P ) on f and g, you could apply the fundamental theorem of calculus and get an expression for h valid almost everywhere. But that still leaves you with an expression containing the integrals that have no closed form.

Are you looking to define h with that property from f and g, or just say what you can about h, given that it satisfies that property?
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Eastwinn
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Re: Strange Calculus Question

Postby Eastwinn » Mon Dec 20, 2010 12:40 am UTC

This is question I've been curious about for ages. If h can be defined in terms of f and g, sweet.
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Nitrodon
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Re: Strange Calculus Question

Postby Nitrodon » Mon Dec 20, 2010 1:41 am UTC

The convolution of f and g will give a function on (-c,c) with the desired property. If you want h to be defined on (0,c), you can rescale it fairly easily.

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Sizik
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Re: Strange Calculus Question

Postby Sizik » Mon Dec 20, 2010 4:35 am UTC

If you differentiate both sides, you get
[math]g(c,x)\int_{0}^{c}f(t,x)dt + f(c,x)\int_{0}^{c}g(t,x)dt = h(c,x)[/math]
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Re: Strange Calculus Question

Postby Sagekilla » Mon Dec 20, 2010 5:14 am UTC

Sizik wrote:If you differentiate both sides, you get
[math]g(c,x)\int_{0}^{c}f(t,x)dt + f(c,x)\int_{0}^{c}g(t,x)dt = h(c,x)[/math]



Which you can just as easily re-write as:

[math]h(t, x) = g(t, x)\int_{0}^{t}f(t',x)dt' + f(t, x)\int_{0}^{t}g(t', x)dt'[/math]
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z4lis
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Re: Strange Calculus Question

Postby z4lis » Mon Dec 20, 2010 6:19 am UTC

Nitrodon wrote:The convolution of f and g will give a function on (-c,c) with the desired property. If you want h to be defined on (0,c), you can rescale it fairly easily.


Hm. So integration turns convolutions into products. That nugget will probably come in handy, one day.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

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Eastwinn
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Re: Strange Calculus Question

Postby Eastwinn » Tue Dec 21, 2010 6:46 pm UTC

Interesting, thanks for the responses. :D
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gorcee
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Re: Strange Calculus Question

Postby gorcee » Tue Dec 21, 2010 8:02 pm UTC

z4lis wrote:
Nitrodon wrote:The convolution of f and g will give a function on (-c,c) with the desired property. If you want h to be defined on (0,c), you can rescale it fairly easily.


Hm. So integration turns convolutions into products. That nugget will probably come in handy, one day.


It's super-handy in signal processing.


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