Symbol used to indicate having done a substitution

Question

When I have an equation like:

$$f(x,y)\tag1$$

And I use a subsitution $y=6+a$ and $x=9-q$ I get the following equation:

$$f(9-q,6+a)\tag2$$

Question: how do I write that mathematically, to go from the first equation to the second?

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I think that I should use (according to the given answer):

$$f(x,y)\space\space\space\Longleftrightarrow\space\space\space f(9-q,6+a)\tag3$$

Or is using the arrow wrong? The question is how do I write $(3)$.

Answer 1

The problem with your writing in (1) and (2) is that these expressions are not equations, as you claim. Using an equivalence would make sense in the following context:

Consider the equation $$f(x,y) = 0. \tag1$$ Subsituting $9-q$ for $x$ and $6+a$ for $y$, one gets the following equivalent equation: $$f(9-q,6+a) = 0\tag2$$

If you insist to use an equivalence, which I would not recommend in this case, you could write:

Setting $x = 9-q$ and $y=6+a$, one gets the following equivalence $$f(x)=0 \iff f(9-q,6+a)=0$$

but (3) as you write it does not make much sense. And once again, I would simply avoid any equivalence symbol in your case.

Answer 2

So one should not use arrows to replace sentences, unless one is aware of the precise signification of the arrows. The symbols of the logical connectors $\Rightarrow$, $\Leftrightarrow$, $\vee$ (denoting the logical "or"), $\wedge$ (denoting the logical "and") are used for closed propositions (sentences that are true or false), or predicates (propositions depending on a variable) but NOT to explain something, not to indicate what you are doing. To do that, it is better to use the appropriate words. And a lot of words in mathematics have really precise meaning actually.

The answer from JE Pin seems good for me. And you can replace $\Leftrightarrow$ by $\Rightarrow$, this is actually not a problem (I do not see the problem of Rebellos about that ?). The $\rightarrow$ is used to denote a limit.

Answer 3

Personally, I would do this:

Suppose $f(x)=y$. Let $x\equiv g(s)$. Then $f(g(s))=y$. The use of the triple equals sign "$\equiv$" denotes that $x$ is equal to $g(s)$ as a definition, not just by construction. Here is an example from my own work:

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Which can be simplified to $$-mc_{R} f^{\prime }_{tra}( -c_{R} t) =T\Bigl[\frac{2}{c_{L}} f_{inc}( -c_{L} t) -\left(\frac{1}{c_{R}} +\frac{1}{c_{L}}\right) f_{tra}( -c_{R} t)\Bigr]$$ We can make a change of variable $\displaystyle p\equiv -c_{R} t$ and rearrange to obtain:$$mc_{R} f^{\prime }_{tra}( p) -T\left(\frac{1}{c_{R}} +\frac{1}{c_{L}}\right) f_{tra}( p) =-\frac{2T}{c_{L}} f_{inc}\left(\frac{c_{L}}{c_{R}} p\right)$$ This is a first order linear const-coeff ODE for $f_{tra}$, the solution of which, although generally quite complicated (due to $f_{inc}$ not being a function simply of $p$), is usually at least of closed-form.