Let's discuss Mathematics

[Souron]

Cubic feet may be less ambiguous than feet cubed, but that doesn't save you from the same problem with higher exponents. The size of a tesseract may be 10 feet⁴, which can only be stated as 10 feet to the forth.

 

[peter grimes]

Ok, so there's this Wired article talking about the new paper on Twin Primes that's coming out. It mentions an extremely large pair:
3,756,801,695,685 x 2^666,669 +/- 1

My brain short circuits when presented with numbers like this. If it were in base10 I *might* have a shot at comprehension, but in binary I'm lost. So how many digits is this number? In an order-of-magnitude estimation is this number 666,669+12 digits long?




EDIT: the Wired bit:
http://www.wired.com/wiredscience/2013/05/twin-primes/

 

[sanabas]

2^3.32 = 10, roughly. So it's about 666,669/3.32 + 12 digits, a little under 201,000.

 

[PlutonianEmpire]

When I went to an online conversion site, I looked to see how many square meters an acre was, which was about four thousand. But, when I switched to square kilometers, it said an acre was 0.004.... square km, rather than 4.0.... as I would expect, considering 1000 meters is one kilometer. Is that a bug or miscalculation, or is there really that big of a difference? If so, how or why?

 

[Atticus]

It's correct. There are 1000 meters in a kilometer, but million squaremeters in a squarekilometer. If you draw a square whose sides are 1000 meters long, you can fit in 1000*1000 squares with one meter side.

 

[nc-1701]

Does anybody know anything about the actuary exams?

I was thinking bout trying to pass the first one as a backup/resume builder, but searching around the internet people seem to be claiming they are incredibly hard etc. And perhaps it isn't a good idea to do that, of course it's hard to tell whether people saying it's hard have more/less mathematical proficiency than I do. Does anybody have any experience with them?

btw I have a B.S. in math and am 1 year into a pure math grad program, but am considering getting out with just a masters and heading into industry to do something financial, where passing an actuary exam would really help.

 

[Atticus]

I don't know how they are in the US and A, but here in Finland they're not that hard from all that I've gathered.

However, I at least find most of the material extremely boring and have had difficulties to go into it. This is of course very subjective thing though.

What kind of maths have you done this far? If it's stochastics, there should be no problem.

 

[peter grimes]

Does anybody know anything about the actuary exams?

I was thinking bout trying to pass the first one as a backup/resume builder, but searching around the internet people seem to be claiming they are incredibly hard etc. And perhaps it isn't a good idea to do that, of course it's hard to tell whether people saying it's hard have more/less mathematical proficiency than I do. Does anybody have any experience with them?

btw I have a B.S. in math and am 1 year into a pure math grad program, but am considering getting out with just a masters and heading into industry to do something financial, where passing an actuary exam would really help.

Yes, they are tough. My old roommate became an actuary, but it took her a few years. She could have done it sooner if her heart was in it. But at that time in her life she wasn't committed to the idea of being shackled to a corporate career.

She got a job at Aetna in Hartford, CT. She could only stay there for a couple of years before she would have been forced to leave if she chose not to pursue Actuarial stuff. So she bit the bullet and went for it - but I think it took her a few tries (not positive - I fell out of touch with her around that time, and this was all very much pre-facebook).

She had a very solid general science background - BS in something biological. But it's not like she had a Mathematics degree or anything, so you may not find it as challenging as she did. But don't assume that they won't require preparation and focus!

 

[PlutonianEmpire]

I've notice that astronomers say something is x milliarcseconds (mas) wide or y mas apart, and from there say it is x kilometers wide or y AU apart. So, how do I convert mas into km or au or other units? (And/or sky coordinates into lengths or distances in general, or something?)

 

[Blue Emu]

I've notice that astronomers say something is x milliarcseconds (mas) wide or y mas apart, and from there say it is x kilometers wide or y AU apart. So, how do I convert mas into km or au or other units? (And/or sky coordinates into lengths or distances in general, or something?)

For very small angles (such as the apparent width of objects in the sky), sine theta approaches theta very closely... so you can use the Small Angle Formula to estimate the actual linear size of the object:

D = X · d / 206,265

... where D is the actual width of the object, X is the apparent width in arc-seconds, and d is the distance to the object in whatever units you wish to use (au, light-years, km, etc).

The constant 206,265 just represents the number of arc-seconds in a circle divided by (2 x pi).

 

[PlutonianEmpire]

Is it possible for someone to explain the equations presented in this paper in layman's terms to me? Or, alternatively, can they be simplified somewhat for easier calculating?

 

[Petek]

I came up with the following puzzle the other day. It's likely too easy for the members of this forum, but I thought I'd post it since I can't recall ever having seen it.

A standard deck of 52 playing cards is dealt to four players in a game of bridge. All four players pass. The hand is re-dealt in the following unusual manner: Player #1 places his cards face down on the table. Player #2 places his cards on top of player #1's cards and players #3 and #4 do the same. The cards are now dealt in the usual way but without shuffling. What is the probability that at least one of the hands contains a void? (It is assumed that in the initial deal each player sorted his cards by suit.)

 

[sanabas]

Spoiler :

Assuming void = 0 cards of a suit, and the usual way = 1 card at a time around the table, rather than something like 3-2-3-2-3, then the answer is zero. A suit has 13 cards, 4 players with 3 each can't happen, therefore at least one player has 4 cards of that suit in the initial deal. When putting the cards back, not shuffling, and dealing, those 4 cards will end up as 1 for each player. That applies to all 4 suits, therefore a void is impossible.

 

[Petek]

Correct!

 

[PlutonianEmpire]

I did a quick search of this and the science thread, but zilch.

How do I calculate parts per million for a particular gas in an atmosphere of whatever distant planet I have in mind, for a given surface pressure and/or temperature?

 

[Gigaz]

You can assume that all the different gases behave like the other ones are not there. Each of them obeys the ideal gas equation of state.

 

[Robert FIN]

Ultimate maths science from my maths school book: Why is 2*(-5)E+2=50.. Why its no -50?

 

[Arakhor]

If you're asking why doubling the square of -5 is +50 and not -50, if you multiply two negative numbers together, you get a positive number.

 

[Robert FIN]

If you're asking why doubling the square of -5 is +50 and not -50, if you multiply two negative numbers together, you get a positive number.

Yeah, that's about it. Thanks I got it now.

 

[Samson]

How do I calculate the Jacobian matrix of y = log base a (x) + b?

If that is hard, what function would approximate a growth curve from age 10 to 20? Preferably with 2 unknown parameters.

I need to curve fit a large number of growth curves, and I could use the example here but I need to provide a Jacobian matrix function, in function jacobian. For the funtion y = a * x^2 + b * x + c the Jacobian matrix is calculated with:

       double[][] jacobian = new double[x.size()][3];
        for (int i = 0; i < jacobian.length; ++i) {
            jacobian[i][0] = x.get(i) * x.get(i);
            jacobian[i][1] = x.get(i);
            jacobian[i][2] = 1.0;
        }

I cannot work out how to do this for y = log base a (x) + b