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November 2

Consilience in combined estimates

How does one adjust for consilience or lack thereof when combining multiple estimates (i.e. determine how much smaller the combined error margin should be when the input estimates all agree)? Neon Merlin 04:54, 2 November 2017 (UTC)[reply]

Depending on what exactly you had in mind, articles such as Bayesian inference or Standard deviation#Estimation may answer your question, or give you the vocabulary to ask a relevant follow-up. Tigraan^{Click here to contact me} 08:12, 2 November 2017 (UTC)[reply]

Maybe Meta-analysis will help. Loraof (talk) 19:35, 2 November 2017 (UTC)[reply]

Differential equations

I had trouble with two problems:

1. Find a homogeneous second-order Cauchy-Euler equation with real coefficients if the given number is a root of its auxiliary equation.

m 1 = i

2. Use systematic elimination to solve the given system.

.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}⁠dx/dt⁠ = 2x + y + t − 2⁠dy/dt⁠ = 3x + 4y − 4t

(x (t), y (t)) =

For 1, I believe there is something easy that I am missing to generate the equation. For 2, I attempted to use differential operators to solve the two equations, but I might have made an arithmetic error somewhere. link

Any help would be appreciated. 147.126.10.148 (talk) 08:23, 2 November 2017 (UTC)[reply]

When you apply a differential operation to an equation, you should be sure to apply it to both sides (at least, if you hope to preserve equality). --JBL (talk) 18:35, 2 November 2017 (UTC)[reply]

I apologize, but I don't understand how that helps me solve the equation, unless I didn't do this somewhere (a likely possibility). 147.126.10.129 (talk) 22:55, 2 November 2017 (UTC)[reply]

Because the Cauchy-Euler equation should have real coefficients and is of the second order its auxiliary equation is of the second order as well and has two roots that are complex conjugates of each other, which means that

m_{1}=i

and

m_{2}=-i

. To find the coefficients of the auxiliary equation you can use Vieta's formulas, which leads to

\lambda ^{2}+1=0

. The final diff-equation is

x^{2}{\frac {d^{2}y}{dx^{2}}}+x{\frac {dy}{dx}}+y=0

. Ruslik_Zero 20:14, 2 November 2017 (UTC)[reply]

Thank you so much for the help. 147.126.10.129 (talk) 22:55, 2 November 2017 (UTC)[reply]

As to the linear system you can solve it using standard methods. Ruslik_Zero 18:09, 3 November 2017 (UTC)[reply]

Miniumum distance between two points and a line

Give two points, A & B, and a line L on a plane, with A and B not on the line, I'm trying to find a way of constructing a point C on the line such that the distance AC + CB is a minimum. I suspect the angles ACL and BCL are equal, but cannot prove it. Any help would be appreciated! Note this isn't homework (I left school more years ago than I care to remember) but is for a brain-teaser app I'm trying to work through. Optimist on the run (talk) 13:07, 2 November 2017 (UTC)[reply]

That's the derivation of the law of reflection from Fermat's principle. See e.g. wikibooks for how it's done. --Wrongfilter (talk) 13:25, 2 November 2017 (UTC)[reply]

Is it a variant of the Distance between two straight lines problem? After all, you can define a line AB, and then find the shortest distance from a point on that line equidistant from both points; the perpendicular distance from that point to the other line should be the shortest from the two points??? --Jayron 32 14:45, 2 November 2017 (UTC)[reply]

In this particular case, the distance AC "costs" as much as the distance CB (in other cases à la Snell's law, you may want to minimize AC + 2*CB, for instance). This allows a neat trick to solve without calculus: let B' be the symmetrical point of B with respect to L. Then, wherever C is, CB'=CB. The question is then to find the shortest path A -> B' which crosses L, which is obviously the straight line. (I am assuming A and B' are on different sides, which ensures AB' crosses L; if not, then A and B are on different sides to start with, and the starting problem is trivial). Tigraan^{Click here to contact me} 14:56, 2 November 2017 (UTC)[reply]

@Tigraan: Mirror trick gave me the hint I needed - I constructed the mirror of B in the line, and the rest was simple. Thanks

Optimist on the run (talk) 17:01, 2 November 2017 (UTC)[reply]