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While reading an article online, I encountered this expression.

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I was wondering if anyone knows what does the square brackets in this expression do?

Thank you.

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  • $\begingroup$ It is only a way of grouping the expression to be evaluated at "point" $s=L$. $\endgroup$ Commented Feb 16, 2018 at 9:40

2 Answers 2

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If $f:A \to B$ is a function and if $a \in A$ then $[f(x)]_{x=a}$ simply means $f(a)$.

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  • $\begingroup$ If that is the case, why couldn't the equation just be written exactly the way it is, except with $s = L$? I mean, it would be much simper for me to write $f(a) = b$ rather than $\left[f(x)\right]_{x=a} = b$. $\endgroup$
    – Mr Pie
    Commented Feb 16, 2018 at 11:52
  • $\begingroup$ Because the above expression involves derivatives. What this notation implies that you need to take the derivative with respect to $s$ before you plug in $s=L$ $\endgroup$
    – Andrei
    Commented Feb 16, 2018 at 15:03
  • $\begingroup$ But the semantics of the symbol f(x) is tantamount to the content of the brackets, isn't it? $\endgroup$
    – von spotz
    Commented Jul 14, 2022 at 10:00
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What if the bracket is inside an integral? Or with a cipher or varible at the top and bottom see https://en.wikipedia.org/wiki/Integration_by_parts. But what else can it mean in calculus? Can it also mean a derivation?

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  • $\begingroup$ How does this answer the question? $\endgroup$ Commented Jul 14, 2022 at 10:18

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