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

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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 $(y',-x')=-{\frac {\lambda }{{\sqrt {x'^{2}+y'^{2}}}^{3}}}(x''(x'^{2}+y'^{2})-x'x'x''-x'y''y',y''(x'^{2}+y'^{2})-y'x'x''-y'y''y')$ which gets simplified to $-y'/x'={\frac {x''y'^{2}-x'y'y''}{y''x'^{2}-y'x'x''}}$, 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 $y''x'=y'x''$. 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

$y'=-{\frac {\lambda }{{\sqrt {x'^{2}+y'^{2}}}^{3}}}(x''y'^{2}-x'y''y').$

Cancel y' and divide by -λ to get

$-{\frac {1}{\lambda }}={\frac {x''y'-x'y''}{{\sqrt {x'^{2}+y'^{2}}}^{3}}},$

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#Several functions of single variable with single derivative but without the "=0". --RDBury (talk) 07:40, 16 December 2019 (UTC)

De Morgan's laws#Formal notation

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)

@CiaPan:, 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)