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I have the following expression (which I find in the article, Eq. (136).)

$$\delta=\prod_{\mu, \nu =0}^{1}\left[\left(a_{1} + \left(-1 \right)^\mu a_{2} + \left(-1\right)^{\nu}a_{3}\right)^2 - 1\right], \tag{1}$$

where, If I understand correctly, the $\prod_{\mu, \nu =0}^{1}$ implies a double productory. Then, my question is: how to expand the above expression?

In an attempt to answer my own question I try the following:

$$\delta =\prod_{\mu, \nu =0}^{1}\left[\left(a_{1} + \left(-1 \right)^\mu a_{2} + \left(-1\right)^{\nu}a_{3}\right)^2 - 1\right] \\ = \prod_{\mu =0}^{1} \prod_{\nu =0}^{1}\left[\left(a_{1} + \left(-1 \right)^\mu a_{2} + \left(-1\right)^{\nu}a_{3}\right)^2 - 1\right]. \tag{2}$$

Then, I expand the productory in $\mu$, hence I obtain

$$\delta=\prod_{\nu =0}^{1} \left[\left(a_{1} + \left(-1 \right)^0 a_{2} + \left(-1\right)^{\nu}a_{3}\right)^2 - 1\right] \left[\left(a_{1} + \left(-1 \right)^1 a_{2} + \left(-1\right)^{\nu}a_{3}\right)^2 - 1\right]. \tag{3}$$

Now I expand the productory in $\nu$, obtaining

$$\delta=\left[\left(a_{1} + \left(-1 \right)^0 a_{2} + \left(-1\right)^{0}a_{3}\right)^2 - 1\right] \left[\left(a_{1} + \left(-1 \right)^1 a_{2} + \left(-1\right)^{0}a_{3}\right)^2 - 1\right] \\\times \left[\left(a_{1} + \left(-1 \right)^0 a_{2} + \left(-1\right)^{1}a_{3}\right)^2 - 1\right] \left[\left(a_{1} + \left(-1 \right)^1 a_{2} + \left(-1\right)^{1}a_{3}\right)^2 - 1\right].\tag{4}$$

Is my above procedure correct?

Thanks in advance.

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    $\begingroup$ Looks good to me. $\endgroup$ Commented Apr 7, 2021 at 21:47

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