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Thank you in advance.

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- #1

- 254

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Thank you in advance.

- #2

blue_leaf77

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- #3

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- #4

Ssnow

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1)is fixed before ...

2)doesn't appear in the equation as variable ...

3) there is an implicit dependence ##f(y(x))## so ##f## is a functional ...

- #5

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##f \left(x, y\right) = A \left(\frac{x}{x_0}\right) \left(\frac{y_0}{y}\right)^{\frac{1}{2}}##

So, it appears as a variable in the equation, and it doesn't seem to be a functional. The variables ##A##, ##x_0## and ##y_0## are constants.

- #6

Ssnow

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- #7

Mark44

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Thank you in advance.

This seems to me like the writer was being sloppy. The notation f(x, y) indicates a function of two variables. If x is held fixed at ##x_0##, then you have ##f(x_0, y)##, which is effectively a function of y alone. Clearer would be to say that ##f(x_0, y) = g(y)##, so as to not reuse f now appearing as a function of a single variable.

- #8

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This seems to me like the writer was being sloppy.

I think so too... Anyway, does everyone agree that the function ##f\left( y\right)## is a function with ##x = x_0##?

- #9

Samy_A

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It's impossible to be sure without knowing the full context.I think so too... Anyway, does everyone agree that the function ##f\left( y\right)## is a function with ##x = x_0##?

It is a reasonable hypothesis, but x could also be some other fixed number in the calculation where the author uses f(y) instead of f(x,y).

- #10

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Sorry, let me re-phrase:

Is the function ##f\left( y\right)## has its ##x## value fixed?

Is the function ##f\left( y\right)## has its ##x## value fixed?

- #11

Samy_A

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##f(y)## has no dependence on ##x##, so if this ##f## refers to the same ##f## used in the definition of ##f(x,y)##, then yes, it most probably means that ##x## is considered fixed in the calculation. If not, it is very sloppy indeed.Sorry, let me re-phrase:

Is the function ##f\left( y\right)## has its ##x## value fixed?

- #12

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Thank you very much! Your help is highly appreciated.

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