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Jul 9, 2023 at 20:56 vote accept onamoonlessnight
Jul 9, 2023 at 18:47 comment added LSpice Re, the answer.
Jul 9, 2023 at 18:21 history edited LSpice CC BY-SA 4.0
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Jul 9, 2023 at 16:25 comment added Carlo Beenakker if I'm not mistaken, there is a simple closed-form answer: $$\max_{j_3} \sum_{m_3} \left(C^{j_3 m_3}_{j_1 m_1 j_2 (m_3 - m_1)} \right)^2=\frac{1+2j_1+2j_2}{(1+2j_1)(1+2j_2)},$$ independent of $m_1\in[-j_1,j_1]$. This simple result may well be in the literature, I arrived at it via Mathematica (see the answer box).
Jul 9, 2023 at 14:42 answer added Carlo Beenakker timeline score: 9
Jul 8, 2023 at 21:27 history edited onamoonlessnight CC BY-SA 4.0
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Jul 8, 2023 at 19:53 history asked onamoonlessnight CC BY-SA 4.0