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July 2

Convexity

Hi, here's a problem I was given by a friend :)

"Is that true that for every two smooth functions f, g, such that f is convex, and g is concave, there exists a constant c, such that c*f+g is convex? If it's true, find this constant." 31.154.81.30 (talk) 05:33, 2 July 2016 (UTC)[reply]

Not true. The function f(x)=0 is convex. Bo Jacoby (talk) 06:15, 2 July 2016 (UTC).[reply]

Thank you! 31.154.81.30 (talk) —Preceding undated comment added 06:40, 2 July 2016 (UTC)[reply]

It's more interesting if you require strict convexity, but I think it's still false (e.g. f(x) = x², g(x) = −e^x). -- BenRG (talk) 07:25, 2 July 2016 (UTC)[reply]

No, it's false. As a counterexample take f(x) = x² and g(x) = –x⁴; there is no such c, which makes –x⁴ + cx² convex. --CiaPan (talk) 10:57, 6 July 2016 (UTC)[reply]

((((x+3)*2)+x)/3)-x returning 2

This return 2 for every x, odd or even integer. Why?

Is there a name for this trick or type of tricks? Wannabe psyches seem to enjoy using it to read minds. --Hofhof (talk) 13:00, 2 July 2016 (UTC)[reply]

Multiply out the inner parentheses, do the addition, and do the division and then the subtraction and you'll see that it equals 2. This is called simplifying an algebraic expression. Loraof (talk) 14:30, 2 July 2016 (UTC)[reply]

{\frac {2(x+3)+x}{3}}-x={\frac {2x+6+x}{3}}-x={\frac {3x+6}{3}}-x=x+2-x=2

— crh ₂₃ (Talk) 16:08, 2 July 2016 (UTC)[reply]

Justification of abuse of notation in the Dirac equation

It is indeed quite tempting to formally take a square root of a derivative operator, or square it, but in what sense does applying a differential operator twice lead to energy squared? The differential operator is of course the functional square root of applying the operator twice, but in what sense does the numerical equation $E^{2}=(mc^{2})^{2}+(pc)^{2}$ imply the functional equation saying that applying the energy operator twice obeys this identity with momentum applied twice on the right? I'm unfamiliar with the derivation of the energy and momentum operators, but I had expected this to be cleared up in the article itself.--Jasper Deng (talk) 17:40, 2 July 2016 (UTC)[reply]

The Dirac and Klein-Gordon equations are ultimately justified by the fact that their solutions are linear combinations of plane waves of the form

\exp \mathbf {i} (\mathbf {k} \cdot \mathbf {x} -\omega t)

where

\omega ^{2}-\mathbf {k} ^{2}=\mu ^{2}

. When you quantize you also get a de Broglie relationship between

(\omega ,\mathbf {k} )

and

(E,\mathbf {p} )

which implies

E^{2}-\mathbf {p} ^{2}=m^{2}

. I don't think that the less fundamental relation

E^{2}-\mathbf {p} ^{2}=m^{2}

implies the more fundamental wave physics deductively. It's just a hint as to what's really going on. -- BenRG (talk) 19:07, 2 July 2016 (UTC)[reply]

To directly answer "in what sense does applying a differential operator twice lead to energy squared", it's because

(\partial /\partial t)^{2}e^{i\omega t}=-\omega ^{2}e^{i\omega t}

. -- BenRG (talk) 04:08, 3 July 2016 (UTC)[reply]

Thanks. I hadn't thought to think of it as an eigenvalue equation. I do think this clarification should be included in the article.--Jasper Deng (talk) 06:28, 3 July 2016 (UTC)[reply]

The justification of all the relativistic equations appearing in quantum physics is really that their solutions transform under some representation of the Lorentz group, and the transformed solutions satisfy the same equation one started with. (This is quite precisely what is meant by relativistic covariance, and is a direct demand of special relativity.) In particular, every component of a solution must satisfy the Klein-Gordon equation, which is the quantized version of the energy-momentum relation. These equations can all be deduced in some manner using representation theory. (The level of logical rigor that goes into such derivations is up to someone else to judge.) In particular, the free Dirac equation results if the representation is taken to be

(.mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠1/2⁠, 0) ⊕ (0, ⁠1/2⁠)

. With

(1, 1)

(presumably applicable to gravitons) you'd get the source-free Einstein equations. Other equations, like the Maxwell equations, the Rarita–Schwinger equations, and more generally the Bargmann–Wigner equations appear in the same way. To see how this can be done in quantum field theory, one might check out

Weinberg, S. (2002), The Quantum Theory of Fields, vol. 1, Cambridge University Press, ISBN 0-521-55001-7, Chapter 5.

This freely available reference,

Bekaert, X.; Boulanger, N. (2006). "The unitary representations of the Poincare group in any spacetime dimension". arXiv:hep-th/0611263. Expanded version of the lectures presented at the second Modave summer school in mathematical physics (Belgium, August 2006),

(outlining the Bargmann-Wigner programme) is also good. YohanN7 (talk) 11:52, 6 July 2016 (UTC)[reply]