Wikipedia:Reference desk/Archives/Mathematics/2019 December 16

= December 16 =

Proving the isoperimetric inequality with Lagrange multipliers applied to calculus of variations
I'm having a little bit of a conundrum here.

I am formulating the problem as maximizing $$\int_a^b -xy'\ dt$$ subject to $$\int_a^b \sqrt{x'^2 + y'^2} dt = K$$. If I take the functional derivative of the inside of each integral, multiply the right hand side's functional derivative by a Lagrange multiplier, equate components, and divide the equations, I end up with a tautology. More explicitly, I have $$(y', -x') = -\frac{\lambda}{\sqrt{x'^2 + y'^2}^3}( x(x'^2 + y'^2) - x'x'x - x'yy', y(x'^2 + y'^2) -y'x'x - y'yy')$$ which gets simplified to $$-y'/x' = \frac{xy'^2 - x'y'y}{yx'^2 - y'x'x}$$, which is satisfied for all x and y such that the numerator and denominator aren't zero. This leads me to think that one must instead look at where they are zero, which occurs when $$yx' = y'x$$. This is satisfied only in the case $$\ln(y') = \ln(x') + C$$ which, however, does not exclude an ellipse.--Jasper Deng (talk) 03:49, 16 December 2019 (UTC)


 * I think the problem is you eliminated lambda. You get two equations but they are redundant, so just look at the first equation
 * $$y' = -\frac{\lambda}{\sqrt{x'^2 + y'^2}^3}( xy'^2 - x'yy').$$
 * Cancel y' and divide by -λ to get
 * $$-\frac{1}{\lambda} = \frac{xy' - x'y}{\sqrt{x'^2 + y'^2}^3},$$
 * in other words curvature is constant, i.e. the curve is a circle. Btw, for those following along at home, the functional derivative in question is given in Euler–Lagrange equation but without the "=0". --RDBury (talk) 07:40, 16 December 2019 (UTC)

De Morgan's laws
For this,
 * $$\neg(P \land Q) \vdash (\neg P \lor \neg Q)$$

why its reverse direction,
 * $$(\neg P \lor \neg Q) \vdash \neg(P \land Q)$$

is not included? --Ans (talk) 14:02, 16 December 2019 (UTC)


 * Possibly it's because that is not a separate law, but rather a result obtained from the second law:
 * $$(\neg P \lor \neg Q) = \neg\big(\neg((\neg P) \lor (\neg Q))\big)$$
 * And by the law of negation of disjunction applied to the part in outermost brackets:
 * $$\vdash\ \ \neg\big(\neg(\neg P) \land \neg(\neg Q))\big) = \neg(P \land Q)$$
 * CiaPan (talk) 14:13, 16 December 2019 (UTC)


 * In your third step, Are you are using $$A \vdash B$$ to conclude $$\neg A \vdash \neg B$$? RoxAsb (talk) 15:31, 16 December 2019 (UTC)
 * , your last step can imply $$\neg B$$ from $$\neg A$$ if $$A = B$$, but the law of negation of disjunction in sequent notation (in the article) is in the form, $$A \vdash B$$, not the form, $$A = B$$. $$A \vdash B$$ is not sufficient to imply $$\neg B$$ from $$\neg A$$ --Ans (talk) 05:25, 18 December 2019 (UTC)

If no any other opposed comments, I will add the reverse rules in the article, then. --Ans (talk) 05:49, 18 December 2019 (UTC)