# Bug: "\fbox" is not correctly displayed in my browser if there is a "\tag" inside "\fbox"

#### Problem

Some formulas I write in \fbox are not displayed strictly for me. The box around the equation has shifted to the left. For example, the equation \fbox{ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right] \tag{4}\\ \end{align*} } becomes this / looks like this: (this formula is from my answer here)

It's obvious that this shouldn't be the case... This always occurs when I write \tag inside \fbox. Other examples are e.g.: \begin{align*} \fbox{ a + b = c \tag{1}}\\ \fbox{ i^{2} = -1 \tag{2} }\\ \fbox{ i \in \mathbb{C} \tag{3} }\\ \end{align*} \begin{align*} \fbox{ \mathscr{B}\left( H \right)\\ q = w + x \cdot i + y \cdot j + z \cdot k \in \mathbb{H}\\ i^{2} = j^{2} = k^{2} = i \cdot j \cdot k = -1 \tag{5} }\\ \end{align*} #### Browser

I'm using "Google Chrome" with "Version $$112.0.5615.138$$ (Offizieller Build) ($$64$$-Bit)" aka Google Chrome: version $$112.0.5615.138$$ (official build) ($$64$$-bit). The only extensions I use is an adblocker and an anti-virus program and the problem isn't with these, because it still shows up erroneously when I disable both.

• This is fixed in version 3 (and version 4) of MathJax. Apr 27 at 11:08

Taking the tag out of the fbox works for me: \fbox{ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{align*} }\tag{4}
\fbox{ \begin{align*} \operatorname{D_{x}^{\alpha}}\left[ \frac{1}{1 - x} \right] = \sum\limits_{k = 1}^{\infty}\left[ \frac{\Gamma\left( k + 1 \right)}{\Gamma\left( k - \alpha + 1 \right)} \cdot x^{k - \alpha} \right]\\ \end{align*} }\tag{4}
\begin{align*} \fbox{ \mathscr{B}\left( H \right)\\ q = w + x \cdot i + y \cdot j + z \cdot k \in \mathbb{H}\\ i^{2} = j^{2} = k^{2} = i \cdot j \cdot k = -1}\\ \end{align*}\tag{5}
\begin{align*}
\fbox{$\mathscr{B}\left( H \right)\\ q = w + x \cdot i + y \cdot j + z \cdot k \in \mathbb{H}\\ i^{2} = j^{2} = k^{2} = i \cdot j \cdot k = -1$}\\
\end{align*}\tag{5}