# B What does this Change in Representing a Function mean?

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1. Apr 7, 2016

### ecastro

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.

2. Apr 7, 2016

### blue_leaf77

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. Apr 7, 2016

### ecastro

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. Apr 7, 2016

### Ssnow

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. Apr 7, 2016

### ecastro

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. Apr 7, 2016

### Ssnow

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. Apr 7, 2016

### Staff: Mentor

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. Apr 8, 2016

### ecastro

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

9. Apr 8, 2016

### Samy_A

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. Apr 8, 2016

### ecastro

Sorry, let me re-phrase:

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

11. Apr 8, 2016

### Samy_A

$f(y)$ 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. Apr 8, 2016

### ecastro

Thank you very much! Your help is highly appreciated.