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A similar question has been asked but the answer states that it should be obvious from context, which is not satisfactory for my situation. I have a function $l_x(\theta) = (x-1)\log(1-\theta)+\log \theta$ and wish to check that the second derivative at $\theta = 1/x$ is negative. \begin{align} \frac{d}{d\theta} \frac{dl_x(\theta)}{d\theta}\left(\frac{1}{x}\right) &= \left(\frac{1-x}{(1-\theta)^2}-\frac{1}{\theta^2}\right)\left(\frac{1}{x}\right) \\\\ &= \frac{x^2}{1-x}-x^2 < 0 \iff x \notin [0, 1] \end{align} However, the second step does not clearly differentiate between evaluation at the point $\theta = 1/x$ and multiplication by $1/x$. In fact, if I saw just that equation I would be convinced we were talking about multiplication. The only reason this can be seen is because the entire derivation of the second derivative has been omitted so that we can clearly see that the previous step was evaluating a derivative. Looking just at the line, one would think that the correct continuation would be \begin{align} \frac{d}{d\theta} \frac{dl_x(\theta)}{d\theta}\left(\frac{1}{x}\right) &= \left(\frac{1-x}{(1-\theta)^2}-\frac{1}{\theta^2}\right)\left(\frac{1}{x}\right) \\\\ &= \frac{1-x}{x(1-\theta)^2}-\frac{1}{x\theta^2} < 0 \end{align} which is entirely wrong.

In order to make the difference clear, I have introduced in my work the notation $@$ to mean specifically evaluation at \begin{align} \frac{d}{d\theta} \frac{dl_x(\theta)}{d\theta}@\left(\frac{1}{x}\right) &= \left(\frac{1-x}{(1-\theta)^2}-\frac{1}{\theta^2}\right)@\left(\frac{1}{x}\right) \\\\ &= \frac{x^2}{1-x}-x^2 < 0 \iff x \notin [0, 1] \end{align} However, I fear that this symbol might already have a different meaning in mathematics. (for one, in the Python programming language, @ means matrix multiplication).

Is the notation $@$ for function evaluation unambiguous and are there better alternatives avaiable?

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  • $\begingroup$ One notation that is in use is $\frac {d^2I_x(\theta)}{d\theta^2}|_{\theta=1/x}$. Your proposal is clear, and clearly communicated, and has some advantages too. I can't think of any prior use for the ampersand symbol in higher math. $\endgroup$ Nov 7, 2021 at 10:49
  • $\begingroup$ @Servaes . Riiiiight.............. $\endgroup$ Nov 8, 2021 at 16:31

2 Answers 2

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Why not write something like $$\left.\frac{d}{d\theta} \frac{dl_x(\theta)}{d\theta}\right\vert_{\theta=\frac1x}$$

which I think is quite common?

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The most common notation is the one you mention first, which is $$\frac{d}{d\theta} \frac{dl_x(\theta)}{d\theta}\left(\frac{1}{x}\right).$$ If it is not clear from the context that you mean to evaluate the derivative at the point $\tfrac1x$, then the best solution is to provide more context. For example, you could lead this expression with a sentence such as

  • "The second derivative of $l_x$ at $\tfrac1x$ satisfies...".
  • "Evaluating the second derivative of $l_x$ at $\tfrac1x$ shows that...".

As for alternate notation; the second derivative is often denoted $\frac{d^2}{d\theta^2}$, so you could instead write $$\frac{d^2l_x(\theta)}{d\theta^2}\left(\frac1x\right).$$ Moreover, because $l_x$ is a function of a single variable, you may further reduce this to $$\frac{d^2l_x}{d\theta^2}\left(\frac1x\right),$$ without any risk of confusion. Finally, some common notation that cannot be confused for multiplication: $$\left.\frac{d^2l_x}{d\theta^2}\right\vert_{\theta=\frac1x}\qquad\text{ and }\qquad\left[\frac{d^2l_x}{d\theta^2}\right]_{\theta=\frac1x}.$$ I would be hesitant to introduce new notation for something so commonly encountered, not in the least because commonly used notation already exists. There seems to be no need to deviate from it.

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