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Equation 87 · Part 1 · No Particle Without a Cosigner

Symbol u^ast

u∗≈0.25363u^\ast \approx 0.25363
u∗u^\ast

What this part means

uau^ast is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

Its job in the formula

uau^ast is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

The passage around this formula

…u}(1-π\pi u)=1 , the same class of transcendental equation that fixes the peak of a Planck spectrum weighted by an extra power of frequency. Solved by Newton’s method to five-figure precision, the nontrivial root is u∗u^\ast ≈\approx 0.25363 , at which u∗u^\ast g(u∗u^\ast) ≈\approx 0.3937 . The leading-order sensitivity is therefore

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the article section

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