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[edit] March 14

[edit] Messy integral

I was working on some probability stuff and reduced the problem to the integral

\int_{0}^{\infty} \frac{\exp(-x^2)}{c+\exp(x)} dx

The c is a constant. Any ideas or thoughts on how to proceed? I tried mathematica's online integrator, and it finds no closed form indefinite integral. It looks convergent. Thanks in advance, --TeaDrinker 01:42, 14 March 2007 (UTC)

Are you supposed to find out whether it is convergent or divergent, or are you supposed to find a value?
because if you want to merely find out if it is convergent of divergent, you can use the comparison theorem and prove that a larger function is convergent, or a smaller function is divergent. In this case, exp(x) is always positive for all real numbers x (I'm assuming we are working in the reals here). So we can eliminate that from the denominator, and we have a larger function (which needs to be proven to be convergent):
\int_{0}^{\infty} \frac{\exp(-x^2)}{c} dx
You can then take the 1/c factor outside of the integral:
\frac{1}{c} \int_{0}^{\infty} \exp(-x^2)dx
Now you need to prove that exp(-x^2) is convergent. You can either use the maclaurin series and integrate term by term, or you can use another application of the comparison theorem and find a bigger function. --Ķĩřβȳ♥ŤįɱéØ 03:51, 14 March 2007 (UTC)
If you need to find the exact value, then you can use integration by parts, using the fact that
\int \exp(-x^2) dx = \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x) + C
where erf is the Error function. --Ķĩřβȳ♥ŤįɱéØ 03:59, 14 March 2007 (UTC)
Thanks for the replies; I was actually trying to find the exact value... Worst case is I can do a numerical integration, since as you point out, it is bounded by exp(-x^2) which convergest fairly quickly. The integration by parts seems like a great way to try, I'll give it a go. Thanks in any event. --TeaDrinker 20:15, 14 March 2007 (UTC)

I love math!

\int_{0}^{\infty} \frac{\exp(-x^2)}{c+\exp(x)} dx = \lim_{t \to \infty} \int_{0}^{t} \frac{\exp(-x^2)}{c+\exp(x)} dx

  • [u = \frac{1}{c + \exp(x)}]
  • [du = -(c + \exp(x))^{-2}(\exp(x))dx] \!
  • [dv = \exp(-x^2) dx] \!
  • [v = \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x)]

And of course, since: \int u\,dv = u v - \int v\,du

= \lim_{t \to \infty} \int_{0}^{t} \frac{1}{c + \exp(x)}\,\exp(-x^2) dx
= \lim_{t \to \infty} (\frac{1}{c + \exp(x)} \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x))|_0^t - \lim_{t \to \infty} \int_{0}^{t} \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x)\,-(c + \exp(x))^{-2}(\exp(x))dx

Simplifying:


= 0 - 0 - \lim_{t \to \infty} \int_{0}^{t} \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x)dx + \,\lim_{t \to \infty} \int_{0}^{t}(c + \exp(x))^{-2}(\exp(x))dx

The first integral can be simplified to:

-\lim_{t \to \infty}\frac{1}{2} \sqrt{\pi} \int_{0}^{t} \operatorname{erf}(x)dx

Noting that

\int \operatorname{erf}(x) dx = x*\operatorname{erf}(x) + \frac{exp(-x^2)}{\sqrt{\pi}}

...to be continued. (Wow, I love this problem)

--Ķĩřβȳ♥ŤįɱéØ 21:55, 14 March 2007 (UTC)

So now we have:

= -\lim_{t \to \infty}(\frac{1}{2} \sqrt{\pi} (x*\operatorname{erf}(x) + \frac{exp(-x^2)}{\sqrt{\pi}})|_0^t)

Evaluated at t and 0:

= -\lim_{t \to \infty}(\frac{1}{2} \sqrt{\pi} (t*\operatorname{erf}(t) + \frac{exp(-t^2)}{\sqrt{\pi}}) - (0*\operatorname{erf}(0) + \frac{exp(-0^2)}{\sqrt{\pi}}))

Simplify the right side:

= -\lim_{t \to \infty}(\frac{1}{2} \sqrt{\pi} (t*\operatorname{erf}(t) + \frac{exp(-t^2)}{\sqrt{\pi}}) - \frac{1}{\sqrt{\pi}})

Further simplification:

= -\lim_{t \to \infty}(\frac{\sqrt{\pi}t*\operatorname{erf}(t)}{2} + \frac{exp(-t^2)}{2} - \frac{1}{\sqrt{\pi}})

Limit laws:

= -\lim_{t \to \infty}(\frac{\sqrt{\pi}t*\operatorname{erf}(t)}{2}) -\lim_{t \to \infty}(\frac{exp(-t^2)}{2}) + \lim_{t \to \infty}(\frac{1}{\sqrt{\pi}})

Simplify the last 2 terms:

= -\lim_{t \to \infty}(\frac{\sqrt{\pi}t*\operatorname{erf}(t)}{2}) - 0 + \frac{1}{\sqrt{\pi}}

Oh boy, I think I am confused now, as the first time seems to evaluate to infinity... someone else help? --Ķĩřβȳ♥ŤįɱéØ 22:31, 14 March 2007 (UTC)

Wow, impressive work. I think the mistake is on the line
= \lim_{t \to \infty} (\frac{1}{c + \exp(x)} \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x))|_0^t - \lim_{t \to \infty} \int_{0}^{t} \frac{1}{2} \sqrt{\pi} \operatorname{erf}(x)\,-(c + \exp(x))^{-2}(\exp(x))dx
The \int v du should be the product of the terms; I think a negative sign may have slipped in as a subtraction. I think the integral should be
\lim_{t\to \infty} \int_0^t \frac{\frac{1}{2}\sqrt{\pi}\operatorname{erf}(z)\exp(z)}{(c+\exp z)^2}dz
I was trying a substitution on this, but did not get anywhere. Thanks again, --TeaDrinker 23:24, 14 March 2007 (UTC)

Wow, all that work, for nothing, because of a mistake. I'm sorry about that. In any case, I think another application of parts is needed, with \frac{d}{dz} \operatorname{erf}(z)\exp(z) being evaluated with the product rule. Denominator as dv. I have to eat lunch now, and then I have tutor someone in Calculus. Cheers! --Ķĩřβȳ♥ŤįɱéØ 23:40, 14 March 2007 (UTC)

[edit] what is the limit of this integral ?

f(x)=x^n now presum the arc lengh of f(x) on the interval(0,1) arclenght=integral{sqr(1+n^2x^2(n-1)}limits of integral is from 0to 1 what is, lim(n→∞)integral{sqr(1+n^2x^2(n-1)}?is it=2or what? 80.255.40.168 12:23, 14 March 2007 (UTC)ARTHER

Yes it is 2.  --LambiamTalk 15:12, 14 March 2007 (UTC)

[edit] is this right?

IF,integral(f(x)dx)=integral(dx),does this mean f(x)=1? 80.255.40.168 12:29, 14 March 2007 (UTC)NEIL

It depends on the precise definition of integral used. If you use the Riemann integral, the answer is simply yes. For the Lebesgue integral, f can be any function that is 1 almost everywhere, which means that exceptions are allowed on a set of Lebesgue measure 0. For example, f(x) = 0 if x is an integer, and f(x) = 1 otherwise.  --LambiamTalk 15:22, 14 March 2007 (UTC)
Umm - maybe I am missing something here, but isn't the function "f(x) = 0 if x is an integer, and f(x) = 1 otherwise" bounded and continuous almost everywhere, and so Riemann integrable ? (Unlike the Dirichlet function, which is nowhere continuous and not Riemann integrable). If so, it's still a good counterexample to the questioner's proposition, but Riemann integrability is not a sufficient condition to the make the proposition true. Gandalf61 15:34, 14 March 2007 (UTC)
I think you're right (meaning I was wrong).  --LambiamTalk 16:40, 14 March 2007 (UTC)
What we need is that f is continuous. Given that, the fundamental theorem of the calculus and the mean value theorem imply that f is everywhere 1. Algebraist 22:25, 14 March 2007 (UTC)

[edit] Sudoku grids

How many possible standard sudoku grids are there? (Reflections and rotations counting as distinct). Obviously cell (1,1) can be any of 9 values, cell (1,2) any of 8 etc to cell (1,9) which only has one option left. On the next row, cell (2,1) can be any of 6, so we get a number of options per cell grid as:

987|654|321
654|???|321
321|321|321
-----------
6??|???|321
5??|???|321
4??|???|321
-----------
333|333|321
222|222|2?1
111|111|111

But I can't work out what the options are for the ? cells. The final number of grids is presumably the product of the values in each cell. -- SGBailey 13:59, 14 March 2007 (UTC)

This problem is quite difficult. As stated on Sudoku, the answer is 6,670,903,752,021,072,936,960. You'll have to read the original paper for the proof. Algebraist 14:03, 14 March 2007 (UTC)
Note that the factorization of that number is 2^{20}\cdot3^8\cdot5\cdot7\cdot27704267971, so it is not a product solely of small integers. --Tardis 15:39, 14 March 2007 (UTC)
BECAUSE for cell 4,2, there are a few amounts of possible numbers - if 4,1 = 1,2, 5,1 = 2,2, 6,1 = 3,2, then there are 6. If those are 6 different values, it's 3. It'd require you to use some sort of tree, I think. ST47Talk 20:17, 14 March 2007 (UTC)
The paper referenced above used computers to give an answer through brute force. It uses some damn good optimization techniques to do that. Using a rather naive approach on my computer suggests I'll have an answer in 211 billion years :D I did run it for a 4x4 sized grid rather than a 9x9 (it ran instantly), and there are 288 distinct combinations for that. --h2g2bob 15:59, 17 March 2007 (UTC)

[edit] Any Mathematics Tests?

I have learned advanced mathematics and I am applying for univeristy , I wanna ask is there any test for me to impress the people at the admission department? —The preceding unsigned comment was added by 76.199.98.111 (talk) 16:29, 14 March 2007 (UTC).

If you really want to test yourself you can try a past paper from the International Mathematical Olympiad. A lot will depend on the university, some will make offers purely on exam grades, other like Oxford/Cambridge (UK) set a special maths paper for applicants. Others will use an interview, I don't remember getting any maths quiz during my interviews. The situation in the US may be different. --Salix alba (talk) 17:48, 14 March 2007 (UTC)
The IMO would certainly impress admissions departments, but it doesn't test "advanced mathematics" so much as a deeper knowledge and understanding of elementary and somewhat offbeat mathematics. For a more helpful response, you should tell us at least what country you're in, what qualifications/tests you have/will have already taken, what extra mathematics you have learned, and whether you're applying for a degree in mathematics or in some other scientific subject. —Blotwell 18:26, 14 March 2007 (UTC)
Somewhat personal interjection: If you have a propensity for math, are in high school in the states, and want to go to college, see if your school of choice accepts Advanced Placement credit, and take one of the AP exams for calculus. (their website is here: [1]) They are dirt cheap compared to college courses, I took 3 for $70 each, and saved myself a few thousand dollars by getting credit at my college for entry level courses. I'm not an admissions worker, but demonstrating that you can handle college level material (the AP exams) doesn't hurt. Atropos235 00:10, 15 March 2007 (UTC)
Check out our list of mathematics competitions. − Twas Now ( talk • contribs • e-mail ) 10:28, 17 March 2007 (UTC)