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B What does this Change in Representing a Function mean?

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  1. Apr 7, 2016 #1
    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.
     
  2. jcsd
  3. Apr 7, 2016 #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.
     
  4. Apr 7, 2016 #3
    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)##.
     
  5. Apr 7, 2016 #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 ...
     
  6. Apr 7, 2016 #5
    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.
     
  7. Apr 7, 2016 #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)## ...
     
  8. Apr 7, 2016 #7

    Mark44

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    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.
     
  9. Apr 8, 2016 #8
    I think so too... Anyway, does everyone agree that the function ##f\left( y\right)## is a function with ##x = x_0##?
     
  10. Apr 8, 2016 #9

    Samy_A

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    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).
     
  11. Apr 8, 2016 #10
    Sorry, let me re-phrase:

    Is the function ##f\left( y\right)## has its ##x## value fixed?
     
  12. Apr 8, 2016 #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.
     
  13. Apr 8, 2016 #12
    Thank you very much! Your help is highly appreciated.
     
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