- #1

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for eg. < [tex]\frac{df}{dt}[/tex] > = [tex]\frac{d<f>}{dt}[/tex]

Here < > is temporal averaging. If the differentiation is w.r.t. a spatial coordinate it makes sense, but could someone help me with the above equation?

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- Thread starter plasmoid
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- #1

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for eg. < [tex]\frac{df}{dt}[/tex] > = [tex]\frac{d<f>}{dt}[/tex]

Here < > is temporal averaging. If the differentiation is w.r.t. a spatial coordinate it makes sense, but could someone help me with the above equation?

- #2

Stephen Tashi

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Is it as simple as moving differentiation past the integral sign? Physicists do that all the time.

[tex] \int \frac{d}{dt}(f(t)) [/tex]

[tex] = \frac{d}{dt} \int f(t) dt [/tex]

- #3

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Is it as simple as moving differentiation past the integral sign? Physicists do that all the time.

[tex] \int \frac{d}{dt}(f(t)) [/tex]

[tex] = \frac{d}{dt} \int f(t) dt [/tex]

That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involve

Edit - I just cant get the latex to work :( - you're missing a dt after f(t) in the L.H.S. of your equation.

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- #4

lurflurf

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- #5

Stephen Tashi

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That's it actually. But all the math texts I saw move differentiation past the integral sign only when the two involveindependentvariables.

That's a good point. What I wrote suggests an idea, but it doesn't really make sense to write

[tex] \frac{d}{dt} \int f(t) dt [/tex]

since the integralion shouldn't leave a function that still depends on t. So you should double check what you said about <f> being a temporal average. Perhaps it is a spatial average.

Is there a rigorous justification for doing it when both the differentiation and integration involve thesamevariable?

Yes, there are conditions that say when one may do this. I can look them up. They aren't something that most people remember since functions describing physical phenomena are usually "well behaved".

Edit - I just cant get the latex to work :( - you're missing a dt after f(t) in the L.H.S. of your equation.

LaTex on the forum looks crazy because of a problem with the pages showing a stale cache of LaTex expressions. (There are threads discussing this.) What you must do is to "preview" your page and then uses the browsers refresh or reload function after that.

- #6

lurflurf

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It would help if the definition given were given here. The usual temporal average does not remove temporal dependence. Often an overbar is used instead of a bracket to avoid confusion. For example

[tex]\int_{-\infty}^{\infty} f(x,s)w(s-t) ds[/tex]

or

[tex]\frac{1}{2h}\int_{t-h}^{t+h} f(x,s) ds[/tex]

[tex]\int_{-\infty}^{\infty} f(x,s)w(s-t) ds[/tex]

or

[tex]\frac{1}{2h}\int_{t-h}^{t+h} f(x,s) ds[/tex]

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- #7

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<f(t,x)> = [tex]\stackrel{lim}{T\rightarrow\infty}[/tex] [tex]\frac{1}{T}[/tex] [tex]\int_{t_{0}}^{t_{0}+T} f(t,x) dt [/tex]

And it's all making even less sense to me now. With this definition, <f(t,x)> is independent of t (as it should if the "average" is to make sense.).

So is

[tex]\frac{\partial<f>}{\partial t}[/tex] = < [tex]\frac{\partial f}{\partial t} [/tex] > ?

Shouldn't the L.H.S. be zero? I'm begiining to think there's something wrong in the problem I'm working on ....

- #8

lurflurf

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It would be more usuual and less nonsensical if it were

[tex]\frac{1}{T}\int_{t}^{t+T} f(s,x) ds [/tex]

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