Wikipedia:Reference desk/Archives/Mathematics/2010 December 11

= December 11 =

Formulas for finding sums
hi all. is there a general way to find formulas like $$\sum_{x=1}^{n}{x}=\frac{(n+1)(n)}{2}$$ for sigma sums? I am at the moment only interested in polynomials but might be interested in other functions in the future. Memorizing all of them is a pain. Thanks 24.92.94.65 (talk) 01:16, 11 December 2010 (UTC)
 * Faulhaber's formula gives a general formula for powers of x. This can be extended to all polynomials. The more general Euler–Maclaurin formula can be used to find sums of other functions.--RDBury (talk) 04:08, 11 December 2010 (UTC)


 * There's a nice inductive way to do it, which I worked out as a student. I'll explain a very simple example to show the idea. Consider a square number, say n2. You can get from 0 to n2 be travelling from 0 to 12, then from 12 to 22, then from 22 to 32, &hellip;, then from (n–1)2 to n2. The total distance travelled is the sum of the differences between each square number and the next. The difference between any two successive square numbers is k2 – (k–1)2 = 2k – 1. This tells you that
 * $$ n^2 = \sum_{k=1}^n (2k - 1) = 2\left(\sum_{k=1}^n k \right) - n \implies n^2 - n = 2\left(\sum_{k=1}^n k \right) \implies \sum_{k=1}^n k = \frac{1}{2}n(n+1) . $$
 * This means that considering the the difference between squared numbers (power 2) tells you the sum of integers (power 1). If we consider a cubed number, say n3. You can get from 0 to n3 be travelling from 0 to 13, then from 13 to 23, then from 23 to 33, &hellip;, then from (n–1)3 to n3. Now, the difference between two two cube numbers is k3 – (k–1)3 = 3k2 – 3k + 1. Using our knowledge of the sum of integers tells us that
 * $$ n^3 = \sum_{k=1}^n (3k^2 - 3k + 1) = 3\left(\sum_{k=1}^n k^2 \right) -\frac{3}{2}n(n+1) + n.$$
 * If we simplify and factorise, we get
 * $$ \sum_{k=1}^n k^2 = \frac{1}{6}n(n+1)(2n+1) . $$
 * For the next step, you consider the differences in fourth powers. Follow the same method, use your knowledge of the sum of integers and squares to give the sum of cubes. — Fly by Night  ( talk )  01:14, 12 December 2010 (UTC)
 * For degree-n polynomial, just compute n+1 of the partial sums, then fit the Lagrange interpolation polynomial through them. 67.117.130.143 (talk) 06:56, 12 December 2010 (UTC)

a^x+b^x≠c^x when x>2
Who discovered this and is there a proof somewhere? — Preceding unsigned comment added by Foljiny (talk • contribs) 03:10, 11 December 2010 (UTC)
 * Fermat discovered this, which became known as Fermat's Last Theorem, albeit he did not publish a proof, and most likely did not have one, in spite of believing he did. A proof was not known until Andrew Wiles published one in 1995. I would not advise that you attempt to read his proof, however, as it is a tad tricky to follow. --COVIZAPIBETEFOKY (talk) 03:47, 11 December 2010 (UTC)
 * Another proof, which some say is easier to understand, was done in 2006 by proving the Serre conjecture. This proof is also impossible to read by any nonspecialist. Staecker (talk) 13:25, 11 December 2010 (UTC)

Proper Use of Heuristic Involving PNT?
I have a new coincidence to report. I know it's rather large, but I'm wondering if anybody can think of a clear reason why my use of standard heuristics using the Prime Number Theorem is making it seem larger than it actually is. The prime number which first translates 6 times in succession as a prime when taking base-4 representations as though they were base 10 is 81875178313, and is the 222nd that translates 5 times. The crude--but I'm sure conservative--calculation I did was to take this number, divide it by (log104)5, take the natural log of this, divide 2 by it, subtract this from 1, and raise the result to the 221st power. It seems I'm doing this roughly well enough to get an upper bound on a pseudo-probability for the likelihood of the result, around 6*10-8, but perhaps there's some reasonable argument for why a number that translates in this way 5 times has a lower likelihood of translating again than the typical number of its size with final digit 1 or 3. Much appreciated if there's something I need to be set straight on.Julzes (talk) 14:56, 11 December 2010 (UTC)


 * As noted below by PrimeHunter, this calculation is totally off in a couple of ways, and the proper equivalent calculation yields no coincidence at all without using non-conservative realistic figures.Julzes (talk) 01:33, 13 December 2010 (UTC)


 * The result obtained using the actual values and the probability estimate below was about 0.16.Julzes (talk) 02:49, 18 December 2010 (UTC)

Hold on a second. I need to substitute 2.5 for 2 in the calculation, but the question is going to end up the same.Julzes (talk) 15:05, 11 December 2010 (UTC)

Yeah, of course this makes it even smaller. 8*10-10. Incidentally, I kind of regret not writing all the details. I'm going to need to run it again to nail this down the way I'd like, but first I'm hoping someone will have something to add to things to make this either seem more reasonable or confirm just how peculiar it is.Julzes (talk) 15:14, 11 December 2010 (UTC)

In attempting to resolve this myself, it occurs to me that maybe the conditional probability of divisibility by 3, 7, and/or other small primes vis a vis the 5-fold translation is enhanced over the non-conditional. I can test this hypothesis without very much effort.Julzes (talk) 15:34, 11 December 2010 (UTC)

It just dawned on me that a prime fails the base-10 divisibility-by-3 test when written in base 4, but this is just going in the wrong direction.Julzes (talk) 15:51, 11 December 2010 (UTC)

Divisibilty by 11 is going to be ruled out as well, for the great majority, considering a base-4 test for 5. Random walk theory might even be usefully employed (Is the absolute alternating sum at least 11?), which is a little cute.Julzes (talk) 16:22, 11 December 2010 (UTC)

Well, that last thing was wrong. I wasn't considering how large the numbers are. There should be a substantial effect on divisibility by 11, but an alternating digital sum of a multiple of 11 other than 0 is going to be far from rare. I obtained empirical data that indicates no effect at all on the other primes through 31, but at the present I have no reason not to think that an opposite kind of effect on some collection of larger primes might be the root cause of this large coincidence. I can't immediately think of how or why that could be so, but I don't really know right now.Julzes (talk) 03:55, 12 December 2010 (UTC)

I haven't gotten a handle on it yet, but something I hadn't even noticed--that the first to make it through 5 iterations as a prime is the 121st to make it through 4--has me pretty convinced there is a theoretical reason for so many consecutive composites. The next step is to find the small factors of the 5th iterates of these 120. Maybe a disproportionate number are divisible by certain primes. Incidentally, the 2nd to make it through 5 iterations almost makes it through 6, giving 43 times a prime.Julzes (talk) 18:49, 12 December 2010 (UTC)

I'm not even getting a hint from the small factors. The closest I've gotten just appear to be more coincidences even less likely to have a theoretical basis. None of the composites have a prime factor between 233 and 2^20, and none have more than 2 distinct small prime factors (only 9 having that many). There is no prime that's overly represented to a degree I consider suspicious. I just don't get it. There aren't many options other than just calling it a dumb coincidence--that I can see anyway. I'll do precise calculation of this part of it once I've tested primes between 2^20 and 2^26.Julzes (talk) 20:28, 12 December 2010 (UTC)


 * If we ignore divisibility by odd primes then your initial computation has two errors. First, you say (log104)5 instead of (log104)6 which has the right exponent for 6 consecutive base conversions. Second, instead of dividing the initial prime by (log104)6 you must raise it to the exponent 1/(log104)6 in order to get a number with approximately the size of the final prime (it has 224 digits but your original computation assumes it only has 13 digits). Then we get probability 0.43 and no coincidence. PrimeHunter (talk) 20:48, 12 December 2010 (UTC)


 * Ah, sorry I didn't even notice this remark before I posted below. As I say there, I had a suspicion I had at least one error there.  This does not come as a surprise to me.  A bit fuzzy these days, and having to do my internet by cellphone is no help.  Thanks for straightening that out.  Anyway, I'll nail down a better estimate in a day or two.Julzes (talk) 22:14, 12 December 2010 (UTC)


 * Yeah, off the top of my head it looks like the original coincidence was next to nothing and the actual coincidence lies in the thing I didn't even notice at first. A bit boneheaded of me.  I do expect to get a smaller probability than you give, though, after I get rid of conservative assumptions and replace them with a nearly accurate calculation, as below.Julzes (talk) 22:27, 12 December 2010 (UTC)

Okay, the search for primes did not even pick up a single new one by expanding to 2^26. I'm finding this now to be the strange thing for the 120 that translate 4 times. I think my original probability calculation above likely had a serious problem with it, as I'm only getting a probability of 0.0001977013 for this collection by a non-conservative computation (each 5th iterate is assigned probability (495/128)/log(n) of being prime, where the fraction in the numerator is given by an appropriate factor for each of the first three primes and 11 is handled by assuming about 1/3 of all multiples of 11 are excluded--2*(3/2)*(5/4)*(33/32)). I'll see how it turns out differently from what I said earlier once the list is fully regenerated. Right now I'm having an awful time understanding how the current list of composites lacks a single prime factor between 233 and 2^26.Julzes (talk) 21:57, 12 December 2010 (UTC)

Okay, this has become stranger still. Removing all prime factors up through 233 doesn't leave a single one of the 120 numbers prime. No doubt, my sense of things is slightly off, but this still has to be quite a coincidence. Not only are there no primes between 233 and 2^26, but none of this is even accounted for with failure to factor further.Julzes (talk) 23:40, 12 December 2010 (UTC)

For the last several hours I've had a program running trying to isolate the next prime factor in the collection. Nothing so far, which leads me to guess 2^26 can be replaced by at least 2^32. I may have to approach the question in a more refined manner if nothing comes out. At any rate, I'm going to have to assess this coincidence at some point also, whenever I get or give up on finding that next factor.Julzes (talk) 05:16, 13 December 2010 (UTC)

I did something wrong in my programming after the point where I entered the 120 numbers from paper. The probability is 0.052 and the stuff about what the prime factors are is totally incorrect. In fact I would have known this if I hadsimplyconsidered the size of the 5th iterate of 5.Julzes (talk) 18:32, 13 December 2010 (UTC)

Integral
I'm doing a treasure hunt with a mathematical clue. I need to work out Integral of e^(-2x+9).sin(x) dx. Any suggestions? -- SGBailey (talk)
 * You could do it using integration by parts. Algebraist 22:56, 11 December 2010 (UTC)


 * You need to repeatedly apply integration by parts. The formula says that
 * $$ \int \! u \, \text{d}v = uv - \int \! v \, \text{d}u. $$
 * Start by labelling your integral, and then start to apply integration by parts:
 * $$ I := \int e^{9-2x}\sin x \, \text{d}x, $$
 * We can set u := e9–2x and dv := sin&thinsp;x&thinsp;dx, and so du = –2e9–2x&thinsp;dx and v = –cos&thinsp;x. The first application gives
 * $$ I = -e^{9-2x}\cos x - 2\int e^{9-2x}\cos x \, \text{d}x . $$
 * Now we apply integration by parts to this new integral, let's put u := e9–2x and dv := cos&thinsp;x&thinsp;dx; meaning that du = –2e9–2x&thinsp;dx and v = sin&thinsp;x. This gives
 * $$ I = -e^{9-2x}\cos x - 2\left( e^{9-2x}\sin x + 2\int e^{9-2x}\sin x \, \text{d}x \right) . $$
 * Making a quick expansion, and remembering what we defined I to be, gives
 * $$ I = -e^{9-2x}\cos x - 2e^{9-2x}\sin x - 4I . \, $$
 * Solving this last expression in terms of I tells us that
 * $$ I = -\frac{1}{5}e^{9-2x}(\cos x + 2\sin x) . $$
 * I hope this helps. If any of the steps aren't clear then just ask. — Fly by Night  ( talk )  23:21, 11 December 2010 (UTC)


 * ... plus a constant. -- Bk314159 (Talk to me and find out what I've done) 03:54, 12 December 2010 (UTC)
 * Quite right: plus a constant! «blush» — Fly by Night  ( talk )  15:25, 12 December 2010 (UTC)

Thanks -- SGBailey (talk) 23:31, 11 December 2010 (UTC)