Wikipedia:Reference desk/Archives/Mathematics/2007 February 23

= February 23 =

How to start?
How do you start to solve the following integral? $$\int\frac{dx}{\sqrt{16+4x-2x^2}}$$

Thanks --MentalPower 01:43, 23 February 2007 (UTC)


 * You should recognize that it resembles the integral for arcsin:
 * $$\int \frac{1}{\sqrt{1-x^2}}\,dx$$
 * The next step is to transform the integral until it looks exactly like the known one. I would proceed by completing the square inside the square root and then performing linear changes of variables. Fredrik Johansson 02:31, 23 February 2007 (UTC)


 * Try a substitution of variables - I quickly tried t = 16+4x-2x^2 but didn't get too far. The square root factorises nicely into (-2)(x+2)(x-4), so that could possibly be a clue to something. --h2g2bob 16:30, 24 February 2007 (UTC)


 * Fredrik is right. Complete the square first, then use a change of variables. –King Bee (T • C) 17:05, 24 February 2007 (UTC)

Check math for time dilation
I was very sleepy when I did this last night so theres a good chance it is wrong. I was trying to figure out much time would go by for a person going at 99% of c from here to the Andromeda galaxy and back. Using the equations found in time dilation, I figured that about 721,520 years would go by for them (more than I expected), while about 5,114,716 years would go by for a person back on Earth. Could one of you mathies check my math real quick? Thanks much in advance. Imaninjapiratetalk to me 02:08, 23 February 2007 (UTC)


 * Well, the Andromeda galaxy is 2.5Mly away; γ for 0.99c is 7.09. The trip as viewed from Earth is not affected by relativity: it's just $$\frac{2(2.5\text{Mly})}{0.99c}=5.05\text{My}$$.  The person travelling sees the distance as reduced by γ, so they count $$\frac{2(2.5\text{Mly})}{7.09(0.99c)}=712\text{ky}$$.  Remember that for any actual such trip, there must be acceleration at the destination to come back: see twin paradox.  --Tardis 17:08, 23 February 2007 (UTC)
 * Re: more than I expected. It really is depressing just how far away everything is, even at ludicrous speed. It's like a cosmic caste system where you're destined to die where you were born, give or take 100 light years.  :(  --TotoBaggins 20:29, 23 February 2007 (UTC)
 * Exactly what I thought. Hopefully we can eventually find a way to break that barrier that is the speed of light. Right now that notion sounds ludicrous to science, but so would have most of our modern technology and science to someone only a few hundred years ago, so I suppose it could be possible sometime in the far future. Thats supposing we don't kill ourselves first. :-/ Imaninjapiratetalk to me 02:22, 24 February 2007 (UTC)

270 degree triangles
I was wondering, if it is possible to make a triangle in conjunction with Non-Euclidean geometry that is more than 270 degrees? —Preceding unsigned comment added by 124.191.241.62 (talk • contribs)


 * Sure it is. On a sphere draw the equator, then draw two lines of constant longitude from the North pole down to the equator. These lines will each meet the equator at an angle of 90 degrees, so you have two right angles in your triangle straight away. If these longitude lines meet at an obtuse angle at the North pole (say they are 0 degrees longitude and 135 degrees longitude, for example) then you have a triangle with an angle sum that exceeds 270 degrees. Gandalf61 11:33, 23 February 2007 (UTC)


 * Well, you can get even more... Draw a small triangle on the sphere, and then declare its "interior" being "outside" it. Now you have a triangle with a sum of angles arbitrarily close to (3&times;360°)–180°=900°. --CiaPan 12:34, 23 February 2007 (UTC)

"Blackboard bold 1" with subscript
In my probability courses I've come across (and used) the notation 1{A} (where the 1 is meant to be "Blackboard Bold") to respresent a function that takes a value 1 when the event A is true and 0 when the event A is false.

I've just used it in a different course to avoid the joy of writing 0r when dealing with the case r=0. Instead, I used 1{r=0}, i.e. 1 if r=0 or 0 otherwise.

However, I'm aware this may not be standard notation, so my question is:

What would you call this function? Only thing I could think of was the identity function, and I know that's not right.

Many thanks. Rawling 4851 12:12, 23 February 2007 (UTC)


 * You could call it the indicator function of the set. &#x2013; b_jonas 13:12, 23 February 2007 (UTC)
 * Indeed, what you are describing would be the indicator function of the singleton zero. Also, I've seen some (logic) textbooks call a (usually discrete) function taking the value 1 at zero and 0 everywhere else a 0-test function. Not very original, but I guess it works. Phil s 16:29, 23 February 2007 (UTC)
 * Also read Kronecker delta and Iverson bracket. --mglg(talk) 17:14, 23 February 2007 (UTC)
 * Outside probability, it is also called the Characteristic function of the set. In probability characteristic function means something totally different of course. Algebraist 17:48, 23 February 2007 (UTC)
 * Our indicator function article provides an assortment of names, notations, and applications for variations of this. --KSmrqT 18:25, 23 February 2007 (UTC)


 * thanks to all, guys :D Rawling 4851 14:15, 24 February 2007 (UTC)