What does this Change in Representing a Function mean?

  • #1
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I have encountered this form of a function: ##f \left(x, y\right)##. Later throughout the discussion, the function was represented as ##f \left(y \right)##. Does this change in representation imply anything? I have been looking through the discussion if ##x## was considered constant, but it doesn't say anything.

Thank you in advance.
 

Answers and Replies

  • #2
blue_leaf77
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What about the derivation that follows? Is the ##x## still omitted? Otherwise, I can't say too much without knowing the major part of the derivation.
 
  • #3
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There was no derivation. The function ##f\left(y \right)## simply appeared as a term on an equation after it was defined as ##f \left(x, y\right)##.
 
  • #4
Ssnow
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It is possible that the dependence of ##f## by ##x## was omitted because

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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It was defined as this:

##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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Ok, it is a function in two variables ##x,y##. In this case I think ##x## is fixed or not considered as variable if they write ##f(y)## ...
 
  • #7
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I have encountered this form of a function: ##f \left(x, y\right)##. Later throughout the discussion, the function was represented as ##f \left(y \right)##. Does this change in representation imply anything? I have been looking through the discussion if ##x## was considered constant, but it doesn't say anything.

Thank you in advance.

There was no derivation. The function ##f\left(y \right)## simply appeared as a term on an equation after it was defined as ##f \left(x, y\right)##.
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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I think so too... Anyway, does everyone agree that the function ##f\left( y\right)## is a function with ##x = x_0##?
It's impossible to be sure without knowing the full context.
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?
 
  • #11
Samy_A
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Sorry, let me re-phrase:

Is the function ##f\left( y\right)## has its ##x## value fixed?
##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.
 
  • #12
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Thank you very much! Your help is highly appreciated.
 

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