I just wanted to wish you all a Merry Christmas who celebrate it, even if a little late for some of us (it's already past midnight where I live).

Or for those celebrating old style (January 7), good luck with preparations.

Totally off-topic thread

Posted: 26 Dec 2017 09:40

by Mosfet

List compiled from five million leaked credentials

The list was put together by SplashData, a company that provides various password management utilities such as TeamsID and Gpass. The company said it compiled the list by analyzing over five million user records leaked online in 2017 and that also contained password information.

There's a link to a PDF with top 100 worst passwords of 2017. If one of your passwords is in there, or it's similar, well, it's really time to change it.

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Posted: 31 Dec 2017 16:01

by midtskogen

Happy New Year to everybody!

(Midnight from our porch)

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Posted: 05 Jan 2018 23:39

by Betelgeuze

Earth was an Oceania planet (Aquaria in 0.9.9.0) at Vaalbara supercontinent stage because there only small island and rest is full ocean

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Posted: 12 Jan 2018 21:41

by Watsisname

Earlier in the science Q&A thread I introduced the function y=sin(1/x)/x, which has some interesting properties, like a singularity where the function not only gets infinitely large, but also oscillates infinitely many times, infinitely faster. In technical terms, "it gets REALLY squiggly!"

But today I stumbled across another interesting property of it. What do you think is the area under the curve (from -infinity to +infinity)? That is, if you shaded in all the area below the curve and above the x-axis, and subtracted all the area above the curve and below the x-axis, how much area will it be?

I suspected the answer would be finite, and I was right, but the actual value was a cool surprise.

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Posted: 13 Jan 2018 01:01

by midtskogen

Pure guessing would lead me to 0, -1 or 1, but since you were surprised, I suspect that the answer is more exciting? Then I thought -π or π (for no particular reason apart from the surprise and π being related to the sine), and I typed in "integral of sin(1/x)/x from -infinity to infinity" at wolframalpha.com, and sure enough:

x.gif (2.06 KiB) Viewed 3849 times

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Posted: 13 Jan 2018 02:13

by ngx

[quote="Mosfet"][quote]List compiled from five million leaked credentials

The list was put together by SplashData, a company that provides various password management utilities such as TeamsID and Gpass. The company said it compiled the list by analyzing over five million user records leaked online in 2017 and that also contained password information.[/quote]

There's a link to a PDF with top 100 worst passwords of 2017. If one of your passwords is in there, or it's similar, well, it's really time to change it.[/quote] just quote test

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Posted: 13 Jan 2018 02:17

by ngx

please somebody help me, I cannot quote. When I submit the post, it has quote tags shown instead of quote frames. What am I doing wrong?

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Posted: 13 Jan 2018 10:15

by JackDole

ngx wrote:

Source of the post please somebody help me, I cannot quote. When I submit the post, it has quote tags shown instead of quote frames. What am I doing wrong?

You have to do a few more posts.

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Posted: 13 Jan 2018 15:06

by ngx

[quote="JackDole"] You have to do a few more posts.[/quote]

By the way, for perhaps less obvious reasons, the answer for cos(1/x)/x is not finite.

Totally off-topic thread

Posted: 14 Jan 2018 02:22

by Watsisname

midtskogen wrote:

Source of the post By the way, for perhaps less obvious reasons, the answer for cos(1/x)/x is not finite.

Yes. Cos(1/x)/x is an odd function (symmetric about the origin), much like the function y=1/x. Because of this particular symmetry, one might naively think that all of the area on the right side of the y-axis should exactly cancel the area on the left side, leaving the answer to be exactly zero. However, that does not work rigorously, because of the problem of taking limits. The integrals blow up or don't converge.

When I first learned that, I hated it. :p It flies in the face of the geometric intuition of the areas cancelling by symmetry. But it does turn out that there is a way to redefine the integral (Cauchy Principle Value) to deal with these scenarios. In this case, returning the expected value of zero, as Wolfram reports with the PV Integral.

The case for the function sin(1/x)/x is a bit more complicated. It is an even function (symmetric about y-axis), so areas generally do not cancel. (But one could imagine it still being zero by some fluke of cancellation of area above and below the x axis.) I basically guessed that this was not the case and started wondering if it would be finite or infinite.

As x goes to zero in this function, the amplitude blows up very fast, but so does the frequency. If we zoom in, it looks like an infinite number of very tall, very thin waves. It's not obvious how much area these contribute. Imagine a 1x1 square (area 1), and then squeeze it in half horizontally while stretching by two vertically. Now it's a 0.5x2 rectangle, still with area 1. One may imagine repeating this process -- 0.25x4, 0.125x8, etc, and taking the limit that the width goes to zero and the height goes to infinity. This still has an area of 1! (In fact this is the definition of the Dirac Delta Function, which is useful in many areas of math and physics).

It isn't obvious that these infinite oscillations in sin(1/x)/x do not act in a similar way, each one continuing to add enough area so that the total area blows up to infinity. I again sort of guessed that it didn't -- that the widths would collapse faster than the heights would grow, or some other magic would happen, to make the area finite.

Evidently, some magic does happen as the answer works out to be exactly pi, but geometric insight as to why eludes me completely.

Totally off-topic thread

Posted: 14 Jan 2018 05:03

by midtskogen

The sine function needs to be carefully aligned to get a solution. Only sin(1/x)/x and sin(-1/x)/x have solutions (π and -π respectively) - and similarly cos(1/x - π/2)/x and cos(1/x + π/2)/x, of course. So clearly the key is symmetry around origo, not simply around the y axis.

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Posted: 16 Jan 2018 03:37

by Watsisname

Symmetry about the origin (odd functions) won't work -- this will always make the integral either exactly zero, or not converge. Every bit of area to the right of the y-axis (if not infinite) will be cancelled by an equivalent but negative area on the left side, since f(-x) = -f(x) by this symmetry.

Examples: integral over x/(e^x^2) is zero. Integral over x, 1/x and 1/x^3 don't converge.

So a function must be even (symmetric about y-axis), or neither even nor odd, to possibly give a finite and non-zero integral. If even, then the area from -infinity to infinity is double the area from 0 to infinity, since f(-x) = f(x). Examples:

Integral over cos(x) doesn't converge, since it continues to oscillate as x-->infinity. (Area is incomputable). 1/x^2 doesn't converge because it blows up too much at origin. (Area is infinite). 1/(e^x^2) yields the square root of pi. (Cousin to the normal distribution. Area is finite since it decays very fast.) Our friend sin(1/x)/x is even and gives pi. sin(x)/x also gives pi, but in this case it's a bit easier to visually see how since there is no nasty singularity.

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Posted: 21 Jan 2018 21:47

by Aborygen

Hello I dont know if this is good place to write about that but, I just wanted to ask about become a beta tester of SE. Is this still valid?