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I thought I had pretty much figured out the difference between $\equiv$ and $=$. Then I came across this while reading about partial derivatives (in Colley's Vector Calculus): $$ \frac{\partial^2f}{\partial z^2} = \frac{\partial}{\partial z} \left(\frac{\partial f}{\partial z}\right) = \frac{\partial}{\partial z} (y^2) \equiv 0 $$ when $f(x,y,z)=x^2y+y^2z$. Why do they use $\equiv$ in stead of $=$ here?

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In this case, it means "identically $0$" (that is, the value is always $0$), rather than "find a value where it is equal to $0$". – Arturo Magidin Mar 28 '11 at 15:20
I think you can safely ignore it, just imagine it is = instead. The author may be using it in some weird way that normally isn't done. – quanta Mar 28 '11 at 17:23
Could you please provide a reference for the above excerpt. – Gone Mar 28 '11 at 20:26
@Bill: I'm very sorry for the late response. I found this in Susan Colley's Vector Calculus. – please delete me Apr 22 '11 at 15:43

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$\equiv$ is often used (between functions) to mean they are identical (instead of being equal at some point). in your example this means identically $0$. if someone write something like $f(z)=g(z)$ it might be thought these functions are equal at some point $z$ instead of every point $z$, so you could write $f\equiv g$ or $f(z)\equiv g(z)$ to mean they are equal everywhere.

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So $\equiv$ is stronger than $=$? – please delete me Mar 28 '11 at 15:25
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@Eivind: It's not "stronger" or "weaker", it's meant to emphasize that this is an equality of functions rather than an equality of values of the function. When one writes, for example, "$x^2+x = x^3+x^2$", it may be unclear if one is talking about the values of $x$ will make the two expressions equal, or if one is talking about an equality of functions (which holds, for example, if you consider the two polynomials as functions on the field of two elements). Use of $\equiv$ emphasizes that you are talking about equality of functions, not values. But both are statements about equality. – Arturo Magidin Mar 28 '11 at 15:34
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@Arturo: OK, that makes it clear. Thank you. – please delete me Mar 28 '11 at 18:39
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You could think about it like this: $f(x) \equiv g(x) \iff f(x)=g(x) \forall x$ – Fernando Martin Apr 22 '11 at 19:08

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