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If I say $x$ is defined as $y+z$ then I can say $x := y+z$. If I want to say $x$ is defined as proportional to $y+z$, then how can I say that? Would I say $x :\propto y+z$?

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    $\begingroup$ I would say that strictly speaking there is no such symbol because, e.g., saying "$x$ is proportional to $y + z$" doesn't pin $x$ down---rather, it describes a property of $x$ shared by many other quantities. Depending on context, you could still frame this relationship as a definition by writing something like, "For a fixed parameter $a$ (to be determined later), set $x := a (y + z)$. $\endgroup$ – Travis Feb 18 at 23:30
  • $\begingroup$ While it is perhaps a meaningless distinction, I would say "being defined as" is more a statement of identity while "being proportionate to" is more descriptive, that is it assigns a predicate to the variable (thinking in terms of first-order logic), as opposed to just a value. In my writing, I generally would add a statement about $x$'s identity before declaring any properties it might have. For example, I may say "Let $x$ be a number such that $x\propto y+x$". $\endgroup$ – Nico Feb 18 at 23:32
  • $\begingroup$ Any two number are proportional. What the context here? $\endgroup$ – Somos Feb 19 at 0:01
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    $\begingroup$ Context: "The MSE cost function is defined as $C(w, b) \propto \sum_{x}{||y(x)-a(x)||^2}$" $\endgroup$ – Shrey Joshi Feb 19 at 0:27
  • $\begingroup$ I would not recommend using a special symbol for this. I think the quote in your comment is clear enough, given the context where "cost functions" like that arise. $\endgroup$ – Mark S. Feb 19 at 12:29
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The symbol $\propto$ is indeed meaning that. It means that there exists $k$ a constant in a $\mathbb{K}$ field so that $x=k\cdot (y+z)$.

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  • $\begingroup$ I mean to ask, how do I say it is defined as that. I can always say that $x=y+z$ but it is better to say $x := y+z$. Or does no such notation exist? $\endgroup$ – Shrey Joshi Feb 18 at 23:25
  • $\begingroup$ Prop to already fullfil this task. For example, you can write $PV \propto T$ $\endgroup$ – PackSciences Feb 18 at 23:25
  • $\begingroup$ But your example does not say "$PV$ is defined to be proportional to $T$". In fact, $PV$ is defined to be the product of $P$ and $V$. I think you have not understood what @Shrey is asking for, which is a symbol for "defined to be the following function, up to a multiplicative constant" $\endgroup$ – Mark S. Feb 19 at 12:28

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