Wikipedia:Reference desk/Archives/Mathematics/2007 February 7

= February 7 =

Finding the center and radius of a circle
How do I find the center and radius of a circle when given only the coordinates of three points on the circle? Thanks, anon.


 * Plug the three points into the circle equation:

\left( x - a \right)^2 + \left( y - b \right)^2=r^2 $$.
 * and solve for a, b, and r as a system of equations. Splintercellguy 00:59, 7 February 2007 (UTC)


 * Or get the perpendicular bisectors of two of the lines joining the points, these will meet at the centre. The radius is the distance from this point to any of the original ones.81.154.107.174 01:10, 7 February 2007 (UTC)


 * Are you sure such a circle exists? If the three points are on a common line, it does not. When doing numeric computations, trouble is to be expected in situations even just approximating this, so do be careful.
 * Every circle in the xy plane satisfies a quadratic polynomial which is a weighted sum of
 * $$ -1, -2x, -2y, x^2+y^2 . \,\!$$
 * Call the weights a, b, c, d. Then we have a system of three homogeneous linear equations:
 * $$\begin{align}

0 &{}= -a - 2 b x_1 - 2 c y_1 + d (x_1^2 + y_1^2) \\ 0 &{}= -a - 2 b x_2 - 2 c y_2 + d (x_2^2 + y_2^2) \\ 0 &{}= -a - 2 b x_3 - 2 c y_3 + d (x_3^2 + y_3^2) \end{align}$$
 * Of course, if (a,b,c,d) is a solution, so is (&lambda;a,&lambda;b,&lambda;c,&lambda;d). We can remove this freedom by requiring d to be 1, and rearrange to obtain
 * $$\begin{align}

a + 2 b x_1 + 2 c y_1 &{}= x_1^2 + y_1^2 \\ a + 2 b x_2 + 2 c y_2 &{}= x_2^2 + y_2^2 \\ a + 2 b x_3 + 2 c y_3 &{}= x_3^2 + y_3^2 \end{align}$$
 * This is simpler than it looks, because each xi and yj will be replaced by a given numeric coordinate; we are only solving for a, b, and c. For readers familiar with matrices, we have
 * $$\begin{bmatrix}

1 & 2 x_1 & 2 y_1 \\ 1 & 2 x_2 & 2 y_2 \\ 1 & 2 x_3 & 2 y_3 \end{bmatrix} \begin{bmatrix}a\\b\\c\end{bmatrix} = \begin{bmatrix}x_1^2 + y_1^2 \\ x_2^2 + y_2^2 \\ x_3^2 + y_3^2\end{bmatrix}. $$
 * Example:
 * Given
 * $$ (17,21), \quad (9,25), \quad (-23,1) \,\!$$
 * Solve
 * $$\begin{align}

a + 34 b + 42 c &{}= 17^2 + 21^2 &&{}= 730 \\ a + 18 b + 50 c &{}= 9^2 + 25^2 &&{}= 706 \\ a + (-46) b + 2 c &{}= (-23)^2 + 1^2 &&{}= 530 \end{align}$$
 * Solution
 * $$ a = 620, \quad b = 2, \quad c = 1 \,\!$$
 * $$ x^2 + y^2 - 4x - 2y = 620 \,\!$$
 * $$\begin{align}

\mathrm{center} &{}= (b,c) &&{}= (2,1) \\ \mathrm{radius} &{}= \sqrt{a+b^2+c^2} &&{}= \sqrt{625} = 25 \end{align}$$
 * The center coordinates appear directly in the solution, with the radius little extra trouble. Although this method requires that we know how to solve a system of linear equations, any method will require comparable work. This one has the advantage that it generalizes to a wide variety of curves, and to least-squares approximations. Collinearity reveals itself in lack of independence among the three linear equations (equivalently, in singularity of the matrix). --KSmrqT 20:47, 7 February 2007 (UTC)

Airy function coefficients
Solving the Airy differential equation with a power series gives


 * $$y(x) = \sum_{k=0}^\infty c_k x^k$$

where


 * $$c_2 = 0$$
 * $$c_k = \frac{c_{k-3}}{k (k-1)}$$

but no values for $$c_0$$ and $$c_1$$ pop out of the equation. Setting ($$c_0$$ and $$c_1$$) to (1, &minus;1) or (1, 1) gives two solutions, but the actual Airy functions Ai and Bi use specific transcendental constants. Where do they come from? Fredrik Johansson 17:16, 7 February 2007 (UTC)


 * Often we choose particular constants to make other conditions fit better. Perhaps it has something to do with the Wronskian, quoted as being 1/&pi;, which may have other implications.


 * As you can see in the article, function Ai is defined by a specific definite integral, which fixes all coefficients of the power series. For example,
 * $$c_0 = \frac{1}{\pi} \int_0^\infty \cos\left(\frac{t^3}{3}\right)\, dt.$$
 * This fixes the numeric value at about 0.355. --Lambiam Talk  00:27, 10 February 2007 (UTC)

Vandalism
I just removed some (what I assumed to be minor) vandalism from Maillard reaction See http://en.wikipedia.org/w/index.php?title=Maillard_reaction&diff=105518582&oldid=105452096

Quote:

Diagram of the Maillard Reaction (English fairly obviously not the $$n^{th}$$ language of the website creator $$\forall\,n\in\mathbb{Z}^+\ll\,17$$)

Can anyone tell me what it meant - or is it just gibberish? Thanks.83.100.250.165 19:35, 7 February 2007 (UTC)


 * It is a rude and inappropriate insult of the English language skills exhibited on the linked web site. The meaning is that, for the web site creator, English is not the native language, nor second language, nor third language, nor any language acquired significantly less than seventeenth. Ironically, anyone who could learn seventeen languages, and who can diagram and explain the Maillard reaction as chemistry, is surely far more intelligent and sophisticated than the vandal who added the insult.
 * Incidentally, this method of producing flavor and aroma is well-known to trained chefs; even I deliberately take advantage of it almost every time I cook. Once we used to hear that when making a beef stew, say, we should sear (brown) the chunks of meat on all sides to "seal in juices". This is scientific nonsense, as it does no such thing. Its real benefit is producing lovely rich flavor through the Maillard reaction. Vegetarians might like to try roasting onions, celery, and carrots in the oven for a comparable experience. --KSmrqT 21:13, 7 February 2007 (UTC)
 * A very odd form of vandalism by all accounts...And a very long way round to give an insult.83.100.250.165 21:27, 7 February 2007 (UTC)