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August 21

Has this formula ever been used before in some way or form?

$v={C_{m}+({(C_{e}-C_{m})\cdot {D_{g} \over {C_{m}\div 4}})} \over 2}\div t$

• $v$ = velocity (i.e. rate of expansion)
• $C_{m}$ = circumference at the meridian (i.e. where the spheroid's cross-section is most oblate)
• $C_{e}$ = circumference at the equator (i.e. where the spheroid's cross-section is least oblate)
• $D_{g}$ = geodesic distance from a given surface point to the nearest of either two surface points closest to the spheroid's center
• $t$ = time taken for a circular phenomenon to encase the entire spheroid

-- MrHumanPersonGuy (talk) 10:13, 21 August 2019 (UTC)

And what is this formula supposed to produce? Ruslik_Zero 20:09, 21 August 2019 (UTC)

To put it this way; if a gray-goo type mass spawns somewhere on a planet and it was known when it would encase the entire planet, then this formula could be used to figure out the average speed of the mass's outward expansion across the planet's surface. -- MrHumanPersonGuy (talk) 22:37, 21 August 2019 (UTC)

One might naïvely expect (a) that the antipodal point was the last to be affected and (b) that for an oblate spheroid, the shortest route between antipodal points was along a meridian. In which case

$v={\frac {C_{m}}{2t}}$

which is similar, but not the same. catslash (talk) 22:41, 21 August 2019 (UTC)

Though (a) could be incorrect. A pair of points, just (roughly) east and west of the antipodal point might be the last affected (supposing that the wavefronts converging on the antipodal point from the north and south were convex when they met there). catslash (talk) 22:51, 21 August 2019 (UTC)

August 25

Multiple integration tricks

If I wanted to integrate something like cos(ax+by+cz) or exp(ax+by+cz) over the unit ball, I could simply apply a coordinate rotation to make the vector (a,b,c) point along one of the axes, and then it'd become easy. Is there a trick to make this work for doing something like x cos(ax+by+cz) or x exp(ax+by+cz) over the same region? Normally in 1D you'd use integration by parts for products xe^ax, but in 3D I can't rotate to make the argument look nice without messing up the other factor. Double sharp (talk) 10:31, 25 August 2019 (UTC)

But you should "mess up" the other factor if it simplifies what's in the exponential! I assume you want something more than just splitting the exponential into factors.--Jasper Deng (talk) 10:37, 25 August 2019 (UTC)

Your integrals can be re-written as:

$\oint \limits _{|\mathbf {r} |=1}(\mathbf {re} _{x})\exp(\mathbf {ar} )dS,$ and

$\oint \limits _{|\mathbf {r} |=1}(\mathbf {re} _{x})\cos(\mathbf {ar} )dS,$

which means averaging over the unit sphere. You can try to calculate them using the spherical coordinates. Ruslik_Zero 12:34, 25 August 2019 (UTC)

@Ruslik0: Could you walk me through the first few steps of this? If I try to simplify that dot product by coordinate rotation before converting to spherical coordinates (which is the trick I saw for the analogous easy integral without that r·e_x factor, I end up with my original problem of messing up the e_x component outside; is this the way I should go, or should I try to convert to spherical coordinates first? (I've been trying to teach myself some basic multivariable calculus over the last few days and got through the usual basic stuff about changing coordinates for multiple integrals that lets you do the analogous integrals without those x factors easily, so if the problem is that I'm asking something that becomes clear with some more advanced stuff please do tell me what I should be looking at.) Double sharp (talk) 16:24, 26 August 2019 (UTC)

August 26

Bell curve formula with skew

I am using the following formula for a bell curve (please feel free to format this so it looks good): y = (1/(s*sqrt(2*p)))*(e^(-((x-a)^2)/((2*s)^2)), where a=mean, s=standard deviation, p=pi, and e=euler. In case I typo'd, this is the actual code:

ls = 1/(s*sqrt(2*p));
dn = pow(2*s,2);
y = ls*pow(e, -(pow(x-a,2)/dn));

What I want to do is add a skew to it. For example, if I had a skew from -1 to 1, I could say that -1 places the mean pretty much all the way to the left side. A skew of 1 will place it almost all the way at the right side. A skew of 0 will be a normal bell curve. Is there a commonly accepted formula that includes this type of skew that I can add? 199.164.8.1 (talk) 13:58, 26 August 2019 (UTC)

A bit off-topic: the parentheses don't seem to balance. Possibly you meant: y = (1/(s*sqrt(2*p)))*e^(-((x-a)^2)/((2*s)^2)).

BTW, wouldn't it be easier to say y = (1/(s*sqrt(2*p)))*exp(-((x-a)/(2*s))^2) ...? --CiaPan (talk) 15:01, 26 August 2019 (UTC)

That is not the optimized code, so your simplification will work. In the optimized code, dn = 1/dn. Then, in the equation, I have *dn instead of /dn. I have to loop through about four million values of x to get the value of y. The optimization replaces millions of floating point divisions with floating point multiplication for a noticeable reduction in clock cycles. What isn't shown is that I'm editing a Fortran procedure, so optimization is manual. 199.164.8.1 (talk) 15:46, 26 August 2019 (UTC)

You may want the skew normal distribution, although it's not necessarily the only possible generalization. Add: yeah, a bit more is mentioned at that article; see also the exponentially modified Gaussian distribution. Different options are going to have different properties, and which you'd want probably depends on what you're trying to do. –Deacon Vorbis (carbon • videos) 15:08, 26 August 2019 (UTC)

I read through that, but didn't see anything that I could use. I will look again. For example, I saw that I could multiply y by dy(x)/dy. But, "dy(x)" is not a appliable formula for computer programming (and neither is dx). I have to differentiate the equation first. That made me wonder if there was a standard solution instead of differentiating the equation. 199.164.8.1 (talk) 15:48, 26 August 2019 (UTC)