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Differentiation is commonly written simply with a prime mark and an equation, as $(x^2)' = 2x$. Although I find this confusing and think it'd better be written $D(x\mapsto x^2) = x\mapsto 2x$, as $x^2$ itself isn't a function. But this notation doesn't allow for specific values, so I was wondering if there was a common way of writing it?

Would this be acceptable: $D(x\mapsto x^2)|_{x=a} = 2a$?

And how would one write integration in similar terms? $I(x\mapsto x^2) = x\mapsto \frac{1}{3}x^3+C$ ? And how would one write definite integration? $I(x\mapsto x^2)_{x_1=a}^{x_2=b} = \frac{1}{3}b^3-\frac{1}{3}a^3$?

Edit: What I'd like to know is if there is a more rigorous notation for differentiation and integration that doesn't hide the fact that you're dealing with functions and not equations. Simple and shortened notation is useful when doing calculations, but not for getting an understanding of what you're doing. I want a better understanding.

Thanks in advance

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    $\begingroup$ Neither is $x\mapsto x^2$ a function, since you haven't told us the domain. Completeness in notation can be the death of clarity. $\endgroup$ – Thomas Andrews Jan 22 '15 at 15:55
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    $\begingroup$ How would you write the differential equation $y''=y'$? $\endgroup$ – Julián Aguirre Jan 22 '15 at 15:56
  • $\begingroup$ @ThomasAndrews Lets say it's $\mathbb{R}$ the the sake of simplicity. $\endgroup$ – Frank Vel Jan 22 '15 at 15:56
  • $\begingroup$ @ThomasAndrews can I borrow your last phrase (for signatures on forum posts, etc)? $\endgroup$ – TZakrevskiy Jan 22 '15 at 15:57
  • $\begingroup$ My point wasn't to ask a question, but to point out that you still haven't defined a function, even though you seem to think $x\mapsto x^2$ is somehow more a function definition than just writing $x^2$. The fact that you have to step out and say $\mathbb R$ separate from the notation is my point. $\endgroup$ – Thomas Andrews Jan 22 '15 at 15:58
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You could say $D(x\mapsto x^2)(a) = 2a$. It's incorrect to say that is $a \to 2a$, because evaluation of the function $D(x \mapsto x^2) = x \mapsto 2x$ at $x=a$ should give a number, not a function.

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  • $\begingroup$ I fixed the $\mapsto$-part of my question. This answer does not fully answer my question about integration. $\endgroup$ – Frank Vel Jan 22 '15 at 16:11
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When the dash notation is employed, the respective variable is usually implicit from the context -- typically it's $x$, or the first argument of a function.   So we generally know what $y', {(x^2+h(x))}', f'(x), g'(u)$ all mean.

If confusion can arise, you can always make the respective variable explicit by placing it in a subscript. $${\big(x^2+f\circ g(x)+y\big)}'_x = 2x + g(x)\cdot f'\circ g(x)+y'_x$$

Another way to express a derivative is to use the linear operator, $\mathcal D_x\;$, to mean $\frac{\mathrm d\;}{\mathrm d x}$.

$$\begin{align} \dfrac{\mathrm d\;}{\mathrm d x} x^2 & = \mathcal D_x \,x^2 \\ & = {(x^2)}'_x \\ & = 2x \end{align}$$

$$\begin{align} \dfrac{\mathrm d\;}{\mathrm d x} f(x) & = \mathcal D_x \,f(x) \\ & = f'(x) \end{align}$$

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