Let's discuss Mathematics

[Mise]

even google... https://www.google.co.uk/#q=sqrt(1-0.4^2)

 

[PlutonianEmpire]

Why use windows calculator? Why not use for example Wolfram Alpha?

Because a) my internet access is NOT 24/7, and b) wolfram requires you to pay to use some features, and I am VEHEMENTLY opposed to paywalls.

 

[PlutonianEmpire]

There is no part of Classical Mechanics I hated studying more than elliptical orbits and their geometry. I am pretty sure it is the most vile thing ever invented. Too many possible lines to draw which have too many names and too many relations between them. I can feel my blood boiling already.

You'll have to take it up with God then.

 

[Tahuti]

Does someone here know of a good (re)introductionary text on mathematics? While I used to study Computer Science, I've changed studies since, and my math skills have grown rusty somewhat.

 

[dutchfire]

Any specific subjects you want to learn about?

 

[peter grimes]

I'm not a math guy, but I've been reading this:
Mathematics for the Non-Mathematician by Morris Kline
http://books.google.com/books/about/Mathematics_for_the_Nonmathematician.html?id=f-e0bro-0FUC

I've found it to be an excellent guide for a second time around. I never pursued math beyond pre-calc (big regrets), and this book is a helpful way to re-learn or just re-acquaint myself with the fundamentals.

I'd love to hear some titles from people more knowledgeable.

 

[sanabas]

http://mathworld.wolfram.com/ is a website, not a textbook, but is still a pretty good resource.

http://www.amazon.com/Calculus-Complete-Course-Robert-Adams/dp/0201828235 is a good basic calculus text.

 

[Tahuti]

http://www.amazon.com/Calculus-Complete-Course-Robert-Adams/dp/0201828235 is a good basic calculus text.

Interestingly, I went to my university's mathematics department not too long ago, and read exactly the book you suggested in the quote, except I wasn't aware of the fact you had.

Thanks anyway!

 

[sanabas]

You weren't a fan of it?

 

[Jouzou]

Does someone here know of a good (re)introductionary text on mathematics? While I used to study Computer Science, I've changed studies since, and my math skills have grown rusty somewhat.

I recently discovered Khan Academy. It's video lessons instead of text, but it's all free. I have watched some of them and I think they're pretty good.

 

[PlutonianEmpire]

I was reading up on Roche Lobes on Wiki, and when I looked at the formulas, I got confused by the syntax/format, so had my father bring me up to speed, and while helping me remember the important stuff, he noticed an apparent peculiarity, but didn't know if it was a typo or something else.

The mass ratio is indicated by this graphic:

(0.3 and 20)

but the graphic above it shows different numbers:

(0.38 and 0.2)

Were they supposed to be the same numbers, or were we misunderstanding the formulas?

 

[Mise]

No, you're misunderstanding the formulas. The r1/A = ... formula gives an approximation for the radius of the sphere when the mass ratio is between 0.3 and 20. That's what it says in the wikipedia article. There are other approximations listed in the wikipedia article, which you can use for mass ratios outside of this range.

 

[PlutonianEmpire]

No, you're misunderstanding the formulas. The r1/A = ... formula gives an approximation for the radius of the sphere when the mass ratio is between 0.3 and 20. That's what it says in the wikipedia article. There are other approximations listed in the wikipedia article, which you can use for mass ratios outside of this range.

Ok, that makes a little bit more sense, although I am unsure as to what the 0.3 and 20 represent? The mass of the bodies given a specified unit (Msun, Mearth, etc)?

 

[dutchfire]

It is a mass ratio. Basically it means that this formula is a good approximation as long as object two is maximally 20 times as heavy as object one, and minimally 0.3 times as heavy.

 

[PlutonianEmpire]

Ok, yet another question from moi. :*

This one is a triangulation/observer issue. In space.

Ok, you know the center of the galaxy, Sgr A? On the celestial sphere, I have it as
RA: 17h 45m 37s
Dec: -28° 56' 10"
Distance: 27720 ly
on my space program (see sig), which, as we all know, is the location of the black hole as seen from Earth. However, for a Sci-Fa space opera (again, see sig), I'm intending to build an inter-stellar version of the "equal-but-opposite-earth-on-opposite-side-of-sun" thingy.

But, it's not with our solar system, but rather, this happens to a nearby star system, which is where the triangulation/observer issue comes in. It happens to this system. As observed in my space program (which sadly doesn't account for light delay at interstellar distances, nor does it have stellar drift), Sgr A, when using Earth's celestial sphere, is shown to be located at
RA: 17h 45m 23.999s-ish
Dec: -28° 57' 14.47"-ish
Distance: 27745-ish ly
as observed from the system I linked to above.

The problem? The program forces the user to use an Earth-based observer point when defining the locations of stars, whether they like it or not.

So, when defining the equal-but-opposite twin of Delta Tri on the direct opposite side of Sgr A to Delta Tri*, how do I figure out where it might be located in the sky when observed from Earth?

*Also, what's the proper grammer and syntax for that part of the sentence?

 

[Jouzou]

I've been refreshing my math skills lately and after watching a tutorial on combinations and probabilities, I made up this exercise for my own amusement and thought I'd share it:

You are playing Civ 4 BtS and choose a random leader and 6 random AI rivals. There are 34 civilizations in the game. What is the probability that the Zulus (and therefore, Shaka) end up being one of your rivals*?

Here's my solution (spoilered in case you want to figure it out yourself). Please do point out if I made any mistakes:

Spoiler :

There are 34!/(27!*7!) unique ways to pick 7 civs out of 34.

If the Zulus take up one slot, 6 civs out of 33 have to fill the rest. That gives 33!/(6!*27!) unique line-ups that include Shaka.

Let
34!/(27!*7!) = a
and
33!/(6!*27!) = b

The probability that Shaka is in the game at all is b/a. However, you will play as him, on average, every seventh time this happens. Therefore, the final probability of having Shaka as a rival is (b/a)*(6/7) which is approximately 0.176 or 17.6%.

*The real answer is, of course, "Too damn high".

 

[Mise]

Yeah I got the same answer.

P(zulu rival) = P(you are not a zulu) * P(one of the other 6 out or 33 are zulu)
P(you are not a zulu) = 33/34
P(one out of the other 6 are zulus) = 1 - P(no zulus)
P(no zulus) = 32/33 * 31/32 * 30/31 * 29/30 * 28/29 * 27/28 = 0.81818181...
P(one out of the other 6 are zulus) = 0.18181818...
P(zulu rival) = 33/32 * 0.181818... = 0.176470588 = 17.6%

 

[dutchfire]

You don't need to take yourself into account. It is a classical vase problem (apparently, this word isn't used on the internet ). You have a vase with 34 marbles, 1 of which is red (Shaka). You pick 6 marbles (enemies) at random from this vase. The chance to find the red marble in this set is:
p = 1 combination 1 * 33 combination 5 / 34 combination 6 = 3/17 = 0.176

The chance to have both Ragnar and Shaka against you is:
p = 2 combination 2 * 32 combination 4 / 34 combination 6 = 5/187 = 0.0267

 

[Petek]

New largest known prime number discovered. From the article:

On January 25th at 23:30:26 UTC, the largest known prime number, 2^57,885,161-1, was discovered on Great Internet Mersenne Prime Search (GIMPS) volunteer Curtis Cooper's computer. The new prime number, 2 multiplied by itself 57,885,161 times, less one, has 17,425,170 digits. With 360,000 CPUs peaking at 150 trillion calculations per second, 17th-year GIMPS is the longest continuously-running global "grassroots supercomputing" project in Internet history.

 

[nerdfighter13]

This is cool and all, but why? There will always be a higher prime number.