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[Solved] A problem with Golden ratio in Wolfram|Alpha ?

Posted 1 month ago
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I just tried to find with Wolfram|Alpha a possible closed form of this number : 0.61803398875 .

( https://www.wolframalpha.com/input/?i=0.61803398875 )

The answer for Wolfram is :

1/Φ - 1≈0.618033988749894

Φ≈0.618033988749894

So I would like to know what is Φ for Wolfram.

Thank you.

2 Replies

Ok, it is just the golden ratio conjugate.

Posted 13 days ago

Just as a side note, I find this terminology rather confusing. Why call Φ the "conjugate" golden ratio instead of the reciprocal ratio? Since ϕ is a quadratic irrational, the conjugate of such a number is generally well understood as the other root of the quadratic it satisfies (x^2 - x - 1 = 0). This would be -1/ϕ = -Φ.

The conjugate irrational in this case happens to have the same absolute value as the reciprocal ratio Φ, but the negation makes these different ratios, whereas ϕ and Φ are essentially the same ratio, as they are just reciprocals of one another. That's like referring to the "difference" between numbers, where positive and negative difference are essentially the same, and typically taken to mean the absolute (positive) difference. But for ratios, it's the reciprocals that are equivalent in this sense, and not negations. The term "conjugate" is ambiguous if used for this type of equivalence, and also unnecessary since the term "reciprocal" clearly means this already for ratios.

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