# oen

 Dec 3 comment Poker, number of three of a kind, multiple formulaes@CodeKingPlusPlus: Glad to help. Cheers! Dec 3 comment Poker, number of three of a kind, multiple formulaes@CodeKingPlusPlus: Notice that $\binom{13}{1}\binom{4}{3}\binom{49}{2}$ counts hands consisting of three of a kind and any two other cards. The shufflings of the three of a kind have been accounted for by the factor $\binom{4}{3}$, but if the fourth card creates four of a kind we must correct for the fact that A1 A2 A3 A4 and A1 A2 A4 A3 are treated as distinct hands. Nov 23 comment What is the intuition behind the Wirtinger derivatives?@laovultai: If $f(x,y)=x^2-y^2+i(y^2-x^2)$, then $f(z,\bar z) = \frac{1}{2}(1-i)(z^2+\bar z^2)$. Thus, $f$ cannot be expressed in terms of $z$ only. In fact, $\partial f/\partial \bar z = (1-i)\bar z = (1-i)(x-i y)$. This is the result you should find using the Wirtinger derivative: $$\begin{eqnarray*} \partial_{\bar z}f &=& \frac{1}{2}(\partial_x + i \partial y)(x^2-y^2+i(y^2-x^2)) \\ &=& \frac{1}{2}(\partial_x + i \partial y)((1-i)x^2-(1-i)y^2) \\ &=& \frac{1}{2}((1-i)2x-i(1-i)2y) \\ &=& (1-i)(x-iy) \\ &=& (1-i)\bar z. \end{eqnarray*}$$ Nov 21 comment What is the intuition behind the Wirtinger derivatives?@laovultai: If you calculate carefully you will find that $\partial f/\partial \bar z = (1-i)\bar z = (1-i)(x-i y)$. You will find this result whether you use the Wirtinger derivative or instead making the transformation $(x,y)\to(z,\bar z)$ in $f$. Nov 16 comment What is the intuition behind the Wirtinger derivatives?@laovultai: As described above, if you solve $\partial_x=\partial_z+\partial_{\bar z}$ and $\partial_y = i(\partial_z-\partial_{\bar z})$ for $\partial_z$ and $\partial_{\bar{z}}$ you will find the usual expressions for the Wirtinger derivatives, $\partial_z = \frac{1}{2}(\partial_x-i\partial_y)$ and $\partial_{\bar z} = \frac{1}{2}(\partial_x+i\partial_y)$. Nov 14 awarded Necromancer Nov 10 comment What are some examples of notation that really improved mathematics?This is a great example of an advance in notation. Reducing clutter like the summation symbol is not trivial. A less cluttered notation allows one to think more clearly about mathematical ideas. It also saves chalk. Mathematicians avoid going to components but for some calculations it is preferable or even necessary. That is when Einstein's notation shows its power. (+1) Nov 10 awarded Nice Answer Nov 7 awarded Nice Answer Oct 27 answered Dirac Delta Constraint Question Oct 27 answered Indices Contraction in Minkowski Spacetime Oct 24 awarded integration Oct 1 comment Where do variables outside of a square root go?I also prefer the numerical part on the left. Another option is to "close" the root. (+1) Aug 26 revised Complex isosceles triangle...added 516 characters in body Aug 26 answered Complex isosceles triangle... Aug 25 revised Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$added 963 characters in body Aug 23 revised Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$added 825 characters in body Aug 23 comment Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$@JonasMeyer: Perhaps I misunderstand you, but since the original equation we wish to solve is $az+b\color{red}{\bar z}+c=0$, isn't the restriction $y=\bar x$ part of the original problem statement? Aug 23 comment Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$@Hurkyl: I appreciate your comments. While it is true that the system $$\begin{eqnarray*} az+bw+c&=& 0\\ \bar a w+\bar b z+\bar c &=& 0 \end{eqnarray*}$$ can have degenerate solutions not of the form $(z,\bar z)^T$ (I only claim the unique solutions are of this form) these solutions have nothing to do with the original problem (to solve $az+b\bar z+c=0$) unless $w=\bar z$. Aug 22 comment Working with $z$, $\overline{z}$ instead of $\operatorname{Re}(z)$, $\operatorname{Im}(z)$There is no mystery here. The second equation, being the conjugate of first, forces the solution to be of the form $(z,\bar z)^T$. Then linear algebra tells us that a unique solution exists if and only if $\mathrm{det}(A)\ne 0$.