- #1

Apashanka

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Now if the average ##f^2(x,y,z,t)## is given 0 over a volume,then ##f(x,y,z,t)## has to be necessarily 0 in the volume domain??

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In summary, if the average of a function over a volume is defined as ##\frac{\int_v f(x,y,z,t) dv}{\int_v dv}## and the average ##f^2(x,y,z,t)## is given 0 over the volume, then the function ##f(x,y,z,t)## must be necessarily 0 in the volume domain. This is true if the function is real-valued, both positive and negative valued, and is continuous. Additionally, if a function is non-negative and has a zero integral over a domain, it is 0 almost everywhere and if it is continuous, it is 0 everywhere in the domain.

- #1

Apashanka

- 429

- 15

Now if the average ##f^2(x,y,z,t)## is given 0 over a volume,then ##f(x,y,z,t)## has to be necessarily 0 in the volume domain??

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- #2

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Can you list all the relevant assumptions on ##f## that would make that true?

- #3

Apashanka

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I am asking ,is it will be??if so then what ##f## had to satisfy??PeroK said:Can you list all the relevant assumptions on ##f## that would make that true?

- #4

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Apashanka said:I am asking ,is it will be??

Surely it's better if you learn to think for yourself.

- #5

Apashanka

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Yes it is,but can't you give any hints or ideas that would help to think better...PeroK said:Surely it's better if you learn to think for yourself.

- #6

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Apashanka said:Yes it is,but can't you give any hints or ideas that would help to think better...

Is ##f## real- or complex-valued?

- #7

Apashanka

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Real,and both positive and negative valuedPeroK said:Is ##f## real- or complex-valued?

- #8

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Apashanka said:Real,and both positive and negative valued

What about continuity?

- #9

Apashanka

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Its is well defined for every point in the volume.PeroK said:What about continuity?

- #10

Mark44

Mentor

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PeroK said:What about continuity?

@PeroK asked if the function was continuous. That's not the same as being well-defined.Apashanka said:Its is well defined for every point in the volume.

- #11

Apashanka

- 429

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Yes it is continousMark44 said:@PeroK asked if the function was continuous. That's not the same as being well-defined.

- #12

mathman

Science Advisor

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- 572

The Spatial Average of squared functions is a mathematical concept used in statistics and signal processing to calculate the average value of a function over a specific spatial domain. It is calculated by squaring the function, integrating it over the domain, and then dividing by the size of the domain.

Spatial Average of squared functions is commonly used in data analysis to measure the average value of a function over a specific area or region. It is particularly useful in analyzing spatial data, such as images or maps, to understand the overall pattern or trend within the data.

While both concepts involve squaring a function and calculating an average, Spatial Average of squared functions is used to measure the average value of a function over a specific spatial domain, while Mean Squared Error is used to measure the average difference between two functions.

No, Spatial Average of squared functions cannot be negative. Since the function is squared before calculating the average, all values will be positive. If the function has negative values, they will be squared to become positive before being averaged.

In practice, Spatial Average of squared functions is calculated by dividing the sum of the squared values of the function over the domain by the size of the domain. This can be done manually or using mathematical software or programming languages such as MATLAB or Python.

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