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September 18

Turing thesis

Does Turing thesis entail that every algorithm must have a "starting point", i.e. a first step starting that algorithm? 84.229.81.123 (talk) 06:40, 18 September 2013 (UTC)[reply]

Please don't cross post the same Q on multiple desks. Your post at the Computer Ref Desk already has a reply. StuRat (talk) 06:59, 18 September 2013 (UTC)[reply]

Resolved

Inverse of gross tonnage

I'd like to modify Template:GT to include a parenthetical figure in cubic meters, at least as an option.

The calculation of gross tonnage is kind of crazy: you take the volume V times a fudge factor K.

K = 0.20 + 0.02 * log10 (V) . So reputedly the range is around 0.20 to 0.32 - 0.28 for a ship of 10,000 cubic meters.

After playing around with V^V type expressions for a bit, I should admit, I don't think I ever learned how to deal with this kind of expression, or if I did, I forgot! Is there a closed form inverse for the formula?

Of course, I realize it can be done easily with a simple method of successive approximation in a Lua module, but I risk getting grief over "original research" as it is, and using those is only going to make it more problematic. Besides, I'm just curious how you do this. Wnt (talk) 23:08, 18 September 2013 (UTC)[reply]

The inverse of x^x is

{\frac {\ln x}{W(\ln x)}}

, where W is the Lambert W function. Not sure if this counts as "closed form", but it's probably as good as it's going to get. Staecker (talk) 11:43, 19 September 2013 (UTC)[reply]

Sigh. I'm still chasing my tail with stuff like "10^GT = ((V ^ V) ^ 0.02) * (N ^ V)"... What is not encouraging is that a search for "gross tonnage" and "lambert function" yielded zero results. Wnt (talk) 20:27, 19 September 2013 (UTC)[reply]

I'm pretty sure you're best off using Newton's method or creating a Taylor series for the inverse.--Jasper Deng (talk) 04:03, 20 September 2013 (UTC)[reply]

I know this is an archived discussion but @Wnt: I did succeed in deriving a solution to this in terms of the Lambert W function.--Jasper Deng (talk) 06:53, 16 April 2015 (UTC)[reply]

September 19

e Day & Euler Day

September, both historically and etymologically, has been the seventh month of the Roman calendar. On the other hand, ⁠19/7⁠ is a very good approximation for Euler's famous constant e, much like ⁠22/7⁠ is for pi. To top it all, Euler died on September 18, 1783. Since a huge part of his life's work is relevant to the fields of statistics and probability, I guess my question would be: What are the odds of such a remarkable coincidence ? (Or is the coincidence perhaps un-remarkable ?) — 79.113.221.167 (talk) 06:05, 19 September 2013 (UTC)[reply]

OK, before all the arguments about "odds" — I don't even get what the coincidence is supposed to be. 19/7 is about e, and Euler died (in the dd/mm reckoning) on 18/7, if you take September to be 7? How is that a coincidence? Or does the 1783 figure into it somehow? --Trovatore (talk) 03:02, 20 September 2013 (UTC)[reply]

Nope. :-) That's pretty much it. (And thank you for indirectly answering my question). — 79.113.214.51 (talk) 05:47, 20 September 2013 (UTC)[reply]

The odds of those specific things are extremely extremely low (supposing the notion of odds makes sense), I'd wager- the odds that there is some seeming coincidences, however, is absurdly high. Most everything has a huge number of "weird" relationships with most anything else- such relationships may be interesting if you are beginning with a relationship and looking at what satisfies it, but it is not interesting to start with things and highlight just the weird things they satisfy (in this context). The first case may reveal something about the relationship (or that the relationship is arbitrary/pointless), the latter does nothing since it's just singling out data without reason, it doesn't elucidate. For related notions, check out: Texas sharpshooter fallacy, Law of truly large numbers, Littlewood's law, and Junkyard tornado. :-)Phoenixia1177 (talk) 06:35, 19 September 2013 (UTC)[reply]

I didn't know there was a name for "Texas sharpshooter". I've often told a friend "you're shooting an arrow and then drawing a bull's eye around it." Bubba73 ^{You talkin' to me?} 02:53, 20 September 2013 (UTC)[reply]

In my `defense`, so to say, I wasn't exactly `trying to connect` anything... :-) The three facts which I've mentioned are related to completely different times, areas, and aspects of my life. The first is religious in nature, and I've known it since my early teens. It is also related to matters of language: the Latin septem gives the Romanian sapte, so it's hardly an `obscure` observation. The second is mathematical, and was discovered by accident quite a number of years back, by noticing that what e lacks to become a whole number is twice that which pi needs to `let go`. The third is rather recent, and is related to matters of history. — 79.113.221.167 (talk) 08:12, 19 September 2013 (UTC)[reply]

Not to be rude, but I don't really see what it is you're defending or how that's a defense anyways. If you said you flipped a coin 100 times and got all heads, it would be no less remarkable than any other sequence, except to people because we put stock in neat little things like that. That you only flipped the coins 100 times, as opposed to flipping until you got that sequence, doesn't make it anymore remarkable- in other words, that you didn't dig up weird relations, just happened upon them, does not make them anymore interesting, at best it makes them more interesting to us. I guess the distinction I'm trying to draw is this: the things you point out may be neat to various people who care about Euler and e, but there is no significant connection among them. I'm not trying to be a dick, by the way, it just happens that the coincidence is not remarkable at all and, if the things you mentioned weren't all true, you could have easily replicated this post with other coincidences instead.Phoenixia1177 (talk) 09:01, 19 September 2013 (UTC)[reply]

In your previous response, you wrote:you are beginning with a relationship and looking at what satisfies it. I was simply pointing out that this was not the case. That is all.

that you didn't dig up weird relations, just happened upon them, does not make them anymore interesting — True... but at least it shows that my question was asked in good faith: That was all I was trying to say. :-) — 79.113.221.167 (talk) 09:29, 19 September 2013 (UTC)[reply]

I don't think you were asking in anything but good faith:-) I was speaking from the general- and that's not what that line was saying, it was saying that if you start with a given set of relations, then specific instances may be of interest; but starting with objects and highlighting relations, in this context, is not of interest.Phoenixia1177 (talk) 09:42, 19 September 2013 (UTC)[reply]

Would you say, for instance, that the same observations would apply to this question as well ? — 79.113.221.167 (talk) 10:34, 19 September 2013 (UTC)[reply]

As far as I can see, yes. Someone even mentions the Texas Fallacy there too. At best, it could be a neat pattern of the numbers you mention, but it doesn't say much about irrationals in general- it almost reduces to: these numbers are close to sevenths, how close is close? As close as these are! Not that I'm saying you should quit considering the question, or that that observation can't lead to something interesting, but that it, by itself, doesn't answer any questions nor pose any new ones. I had to write this quick, if poor, sorry:-)Phoenixia1177 (talk) 11:15, 19 September 2013 (UTC)[reply]

No, your answer is not poor. And I wasn't obviously trying to make a statement "about irrationals in general" either. Thank you for taking the time to answer my questions. — 79.113.221.167 (talk) 11:53, 19 September 2013 (UTC)[reply]

The claim of "All Basic Irrationals" is of course highly dependent on what you include. So five numbers are close. Many others at Mathematical constant#Table of selected mathematical constants aren't, for example the Golden ratio. PrimeHunter (talk) 11:33, 19 September 2013 (UTC)[reply]

The most famous and/or most used ones, with the notable exception of the golden mean. They each represent a different thing (circle, square, triangle, combinatorics, etc), so the fact that they exhibited a commonality was surprising. But whether this was merely coincidental or -on the contrary- systemic, I could not have known before asking the question. — 79.113.221.167 (talk) 11:53, 19 September 2013 (UTC)[reply]

Actually there's one more constant there in that list close to a multiple of ⁠17⁠, namely Ω, whose value, just like that of γ, is close to ⁠47⁠ ... And another two not on that list, namely Gelfond's constant, e^π ~= 23⁠17⁠, and ⁠eπ⁠ ~= ⁠67⁠. — 79.113.221.167 (talk) 13:11, 19 September 2013 (UTC)[reply]

e and π are both transcendental numbers, and therefore are more readily approximated by fractions than are algebraic numbers such as (say) √2 (apologies if that's already been mentioned - I haven't read every word above). --catslash (talk) 14:49, 19 September 2013 (UTC)[reply]

Phoenixia - I think you meant to say "If you said you flipped a coin 100 times and got all heads, it would be no more remarkable than any other sequence, except to people because we put stock in neat little things like that." And, that's completely wrong. A sequence of all heads from a fair coin is astonishingly remarkable. It was supposedly generated by a process with 100 bits of entropy, but has Kolmogorov complexity of only a few bits. That's virtually impossible by the laws of probability, and you don't need to be a human to recognize that. In fact, this is so remarkable that if observed in practice, the only logical conclusion is that the coin is not fair at all but rather is heavily biased towards heads, maybe even having heads on both sides.

A slightly trickier case is if the sequence has equal portions of all 2-grams, yet still has very low Kolmogorov complexity. Then you don't have the bias "out", and must conclude that the universe works very differently than what we thought.

Of course, if the universe behaves more or less as we expect, we will not encounter this in practice. When we toss a fair coin 100 times, we'll obtain a sequence with complexity of about 100 bits. -- Meni Rosenfeld (talk) 18:03, 19 September 2013 (UTC)[reply]

Getting 100 heads in a row is as likely as any other particular sequence of heads and tails. Bubba73 ^{You talkin' to me?} 02:48, 20 September 2013 (UTC)[reply]

Probably in real life, but definitely not in mathematics. :-) In theory, the overall mean or distribution should be about 50% for each. — 79.113.214.51 (talk) 05:47, 20 September 2013 (UTC)[reply]

No, there are two issues here. The probability of getting 100 heads out of 100 is a lot less than getting about 50. But any particular sequence is equally likely. Take 10 flips: HHHHHHHHHH has the same probability as HTTTHHTHHT. Both are 1 in 1024. Bubba73 ^{You talkin' to me?} 16:17, 20 September 2013 (UTC)[reply]

Way to completely miss the point of my comment.

Yes, obviously 100 heads in a row is as likely as any particular sequence, and Phoenixia said it before you.

And yet, if you were to actually encounter such a sequence in real life, you wouldn't just shrug it off, you'll recognize that something very weird is going on, a recognition that a sequence such as

HTTTTTTHTTHHTHTHHHHHHTTHHHTHHHTHHHHTTTTHHTTHTTHHTHTTHTHTTTHTHTHTTHTTHHHTHTHTTHTTTHTTHTTTHHTTTHHTTTHT

wouldn't evoke.

And what I was trying to explain is that your intuition isn't playing tricks on you. There are very real reasons why a sequence of all heads is remarkable, and would force you to update your beliefs about the coin or about the world, in a way most sequences wouldn't. -- Meni Rosenfeld (talk) 19:39, 21 September 2013 (UTC)[reply]

Doesn't the Kolmogorov Complexity depend on the description language though? So, say, the class of strings {"H", "HH", "HHH",...} might tend to have low complexity, no specific element does for all languages. Or, more to the point, wouldn't it be that no matter what language we pick, doing trials of sequences of flips should tend to have complexity equal to the number of flips? Ultimately, for any given sequence of flips, each outcome sequence is going to be equally likely, so I don't see how you'd get around needing to look at multiple sequences to determine if something goofy was going on. --This is half counterargument, half question; the less Abstract nonsense something becomes the goofier my brain's interaction with it becomes, so, most likely, I'm missing a very obvious point:-)Phoenixia1177 (talk) 07:05, 20 September 2013 (UTC)[reply]

It does indeed depend on the language, but only up to an additive constant - which will be fairly small if the languages are not contrived. So we can pick some reasonable-seeming language and stick to it. It may be that for a short sequence like 10 heads there would not be a very short description, but for 100 we should see a real difference.

Yes, for a given description language and a given large number of random, independent fair flips, the resulting complexity will be virtually always very close to the number of flips. Which means that if it isn't, this is strong evidence that the flips aren't "random, independent and fair".

"Multiple sequences" is the same as one long sequence.

A similar problem is if we know the flips are random and independent, but we just don't know the bias of the coin. If we observe a sequence of 100 heads - which is astonishingly unlikely if the coin is fair - we have strong evidence that the coin is heavily biased towards heads, because the number of heads (100) is very different from what we expected (50). -- Meni Rosenfeld (talk) 19:50, 21 September 2013 (UTC)[reply]

That's extremely fascinating, I definitely need to read more about the subject, thank you for the response:-) It's somewhat humorous to me that the intuitive case seems to be unintuitive; if that makes sense (tone = light). Thank you:-)Phoenixia1177 (talk) 21:57, 21 September 2013 (UTC)[reply]

You might like mathematical coincidence. Staecker (talk) 15:23, 19 September 2013 (UTC)[reply]

I am well familiar with the article, as well as with that about almost integers. :-) — 79.113.221.167 (talk) 15:47, 19 September 2013 (UTC)[reply]

9 isn't a prime number, the next prime is 11. So, by including factors of 7 in denominators (note that when adding fractions, prime factors in denominators will be preserved, at most they can get cancelled but you won't get new prime numbers) you can get an approximation that on first sight may look better than it should be. 18:14, 19 September 2013 (UTC)

To the general discussion about 7ths. Every number is within 0.07ish of a 7th, so what you're calling close makes all the difference- 1 in 8 numbers are going to be within 0.009ish of a seventh, 1 in 64 numbers will be closer than your approx. for pi. Ultimately, that doesn't seem like that big of a deal- especially since there is no clear reason why anything is a constant except that it get's used a lot. Do you have some related hypothesis that this observation adds support too? That could make it notable, but without some framework it is helping to support, I don't see that it connects up with anything else in a meaningful way.Phoenixia1177 (talk) 07:19, 20 September 2013 (UTC)[reply]

there is no clear reason why anything is a constant except that it get's used a lot. — Well... the reason that "it get's used a lot" is usually because it "appears" all over the place, and the reason for that is usually "structural" in nature. Eg, if you define pi as the ratio between the circumference and its diameter, the next thing you'll notice is that the ratio between the area of a disc and the square of its radius is the same. Then, when going from circles to ellipses, you'll notice that their areas and circumferencess are also dependent upon it; etc. Then, when you'll start rotating graphics around their axes of coordinates, it will appear there as well; etc. — 79.113.214.51 (talk) 07:46, 20 September 2013 (UTC)[reply]

The human minds divides more complicated structures into simpler ones, and that which is more complex is ultimately expressible into that which is simpler. For instance, every polygon can be divided into n-2 triangles, so if you understand the triangle, you understand all polygons. But there are things which are not composed of line segments, such as curves: the simplest of which is the circle. Then taking these realities into 3D, we have polyhedrons being divided into tetrahedrons; and then we also have spheres and other round (as opposed to flat) surfaces. — 79.113.214.51 (talk) 07:59, 20 September 2013 (UTC)[reply]

You're oversimplifying it, though. Those numbers, specifically, come up because of what is obvious to us, because of what geometry out universe approximates, etc. If we lived in a 50 dimensional world and had a different visual system, we might have constants derived from the really simple case of weird irregular 32 dimensional areas bounded by odd curves. By the same token, there are theories of computation for limit ordinals not omega, if we lived in a universe governed by weird uncountables, we'd have a different theory of computation and number theory, most likely. Human experience, situation, and thought selected the constants that we have, none of those are a priori required. Nonetheless, what you've demonstrated isn't even particularly interesting for those specific constants, your approx for pi works for more than 1 in 64 numbers, that's not astounding.Phoenixia1177 (talk) 08:15, 20 September 2013 (UTC)[reply]

If we lived in a 50 dimensional world and had a different visual system, we might have constants derived from the really simple case of weird irregular 32 dimensional areas bounded by odd curves — Actually, no, we wouldn't. :-) Because even in our humble 3-dimensional universe, pi was dicovered by studying the circle first, not the sphere.... So even if these things would exist, they would exist as expressions containing 1D or 2D constants. — 79.113.214.51 (talk) 08:20, 20 September 2013 (UTC)[reply]

if we lived in a universe governed by weird uncountables — Perhaps they are called 'uncountables' for a reason... such as the fact that they don't really count ;-) — 79.113.214.51 (talk) 08:30, 20 September 2013 (UTC)[reply]

...and even if they would count... that would not mean that their number (of the uncountables that count) is itself uncountable, OR that -even "worse"- these uncountables are not related to each other in some way; ie, even pi can be written as Γ(⁠12⁠)² because it describes the geometric shape of equation x² + y² = 1. Generalizing, all curves belonging to the family of equations of the form xⁿ + yⁿ = 1 are described by constants of the form ⁠Γ(⁠1n⁠)²Γ(⁠2n⁠)⁠ . — 79.113.214.51 (talk) 08:41, 20 September 2013 (UTC)[reply]

(edit conflict) Yes, but low dimensional cases are of interest because our experience only goes up to 3- we can toss out euclidean too and make it some really weird space. But moreover, why is the circle simpler? The circle is simple because it is convenient and simple to us in the geometry that is natural to us, if we lived in some weird space based off of the 3-adics, I doubt we'd have the same idea, circles would be the unusual object. Uncountables count as much as anything else, unless you reject the axiom of choice; it's like suggesting the imaginary numbers don't exist because of the name, they can be used to base a theory as much as omega. My ultimate point is this: you've found a not that unlikely way to approx. a small collection of numbers that people call special; it is neat, but unless you are positing something more, it's just a random factoid. Do you have some connection this is supposed to make? — Preceding unsigned comment added by Phoenixia1177 (talk • contribs) 08:44, 20 September 2013 (UTC)[reply]

(to what you just posted) I don't think we are talking about the same uncountables, or I don't know what you're trying to say if we are. As for what you've highlighted, that's all well and good, but why how does that make pi universally special? And if it isn't mean to, what has it to do with the interest of pi being roughly 22/7 and euler's number being related to 7ths as well? I don't see the connection.Phoenixia1177 (talk) 08:47, 20 September 2013 (UTC)[reply]

Generally speaking, people are of two kinds: practical people, and intellectuals. What both these usually-antithetical groups have in common is their appreciation for simplicity. On one hand, simplicity delights the minds of the latter; on the other hand, it proves useful to the interests of the former. But not all minds belong into these two broad categories. — 79.113.214.51 (talk) 09:02, 20 September 2013 (UTC)[reply]

how does that make pi universally special? — probably because the Universe itself is made out of countless circular/disc-shaped galaxies, filled by spherical stars, surrounded by spherical planets, around which revolve equally-spherical moons or satellites, all of which are characterized by the same constant ... :-) And as far as imaginary numbers are concerned, these too have several fundamental applications in location or coordinates, as well engineering, electrical or otherwise. — 79.113.214.51 (talk) 09:02, 20 September 2013 (UTC)[reply]

we can toss out euclidean too and make it some really weird space — Actually, no, we kinda can't, since we are not God(s). :-) — 79.113.214.51 (talk) 09:02, 20 September 2013 (UTC)[reply]

What? That I'm not God means I can't imagine different geometries? You had no objection to a geometry with 50 dimensions, why can we pretend one and not the other? By the way, "universal" need not mean "pertaining to the physical universe", especially when the context is mathematical interest. What does your artificial splitting of people have to do with anything? What isn't simple about 3-adics anyways? It's ultrametric, that's simplifies a lot of things, actually. As for imaginaries having uses, I'm well aware of that, hence my point: using the name "uncountable" to somehow single out omega as the "right" ordinal is like using the name "imaginary" to single out the reals as the "right" numbers- in other words, bullshit.Phoenixia1177 (talk) 09:10, 20 September 2013 (UTC)[reply]

Some things cannot be explained, they can only be understood. :-) (BTW, I do not necessarily "disagree" with you... I was just trying to explain my perspective, so that you might understand me as well, that's all). :-) — 79.113.214.51 (talk) 09:19, 20 September 2013 (UTC)[reply]

The imaginary numbers you keep mentioning are constructed from real numbers... There is a logical order inherent to things: We did not just discover the imaginary unit before the actual unit 1, whose opposite or negative image we've called -1. There's a structure inherent to everything... — 79.113.214.51 (talk) 09:29, 20 September 2013 (UTC)[reply]

Likewise, generally speaking, the avegage human mind, for some random, mysterious reason, places great emphasis on the difference between reality and imagination... I'm not saying that this is "right" or "wrong", all I'm saying is that this is how things are, or happen to be... The reason behind this dichotomy lies probably in the fact that we are not disembodied minds. — 79.113.214.51 (talk) 09:29, 20 September 2013 (UTC)[reply]

Now, as to the simplicity of the circle: Think of it this way: First degree equations describe lines, segments, and polygons. Second degree equations describe the first curves: parabolae [0], circles [+], and hyperbolae [-]. There are other curves out there, but 2 comes before 3 and other numbers. — 79.113.214.51 (talk) 09:36, 20 September 2013 (UTC)[reply]

But we're talking about these constants and your relation and mathematical interest. That a handful of numbers useful to humans are related to 1/7 in the same way a massive chunk of other numbers are is not interesting. The problem is that your relation to these numbers does not "fall out" of 1/7, the only reason those numbers show up is because we, mankind, find them neat- in other words, that it is the constants and not some other handful of numbers instead means nothing-- in the same way that the naturals inside of the theory of the reals is just a countable set, it is not special. As for the rest of it, you can start with the complex numbers and get the reals; to be honest, the complex numbers seem more foundational to me than the reals do (both in physics and mathematics). I still submit that your argument for circles being simple is a rather pointless endeavour unless we mean "simple for humans", which is my point anyways- but that's beside the point, which is "your observation is not related to pi anymore that it is related to the other 1/64 numbers that satisfy the same thing." In short, you selected as an example of a fairly common relation a number that people regard as special, then concluded that there is something special; but there isn't, pi is just another number as far as this is concerned.Phoenixia1177 (talk) 09:55, 20 September 2013 (UTC)[reply]

It's a question of logic and logical sequence; of structure and construction. You can't get to i before passing through -1, nor to -1 before getting to 1 first. — 79.113.214.51 (talk) 10:15, 20 September 2013 (UTC)[reply]

Or perhaps you meant to say that you could get to certain complex numbers without having to construct ALL reals first... which is an obvious statement, with which I perfectly agree, of course. — 79.113.214.51 (talk) 10:34, 20 September 2013 (UTC)[reply]

As for e, e^x is the neutral element for derivation and integration, which are as basic to mathematical analysis as the four operations are to arithmetic. It's not merely useful, it's fundamental, and has a logical precedence to calculus equivalent to that of 0 and 1 in number theory. — 79.113.214.51 (talk) 10:15, 20 September 2013 (UTC)[reply]

MChesterMC makes the point better than I've been below, but I'll answer anyways. First, what does any of that have to do with 1/7 and what I pointed out, mainly that pi isn't special in this context (or in general). Second, that's all neat, but all sorts of magical stuff holds in all sorts of other spaces, why do I need to pretend the reals/naturals are the root of all thought? After Cantor, especially after Grothendieck, I don't think Kronecker's "God made natural numbers..." is given. I'm not disputing that these numbers are neat because of humans, but they aren't "important" in and of themselves- there's ton's of crazy spaces out there, pi isn't universally potent, it's a real number associated with some human baubles of thought, that's it. As for the complex numbers, why do you need all of that? Here: let C be the metric completion of the algebraic closure of the p-adics (pick a p), C is isomorphic as a field to the complex numbers. It is not a question of "logic and logical sequence", there is no sequence to things. For example, how should I get the reals? Dedekind cuts, something else? What about: Let R be the complete ordered field? Why do I need to start where you say I must, I can start all over the place. I'm sure if you really wanted to you could do some category theory fun and end up with "reals" that do what you need. The same applies to just about everything mentioned here.Phoenixia1177 (talk) 10:36, 20 September 2013 (UTC)[reply]

Maths is about numbers and logic, not "spaces". And I think I've shown that Pi would still hold meaning even in MC's universe. — 79.113.214.51 (talk) 10:43, 20 September 2013 (UTC)[reply]

Again, what? That's just absolute nonsense. Mathematics is all about spaces: you can found most everything starting with categories and that, essentially, is nothing but a generalization of spaces and morphisms. What is it you think mathematicians study?Phoenixia1177 (talk) 10:48, 20 September 2013 (UTC)[reply]

there is no sequence to things. Let R be the complete ordered field — If there is no sequence, then there is no order.

What is it you think mathematicians study? — Numbers, my dear. Numbers and shapes. "Spaces" are for astronauts — a respectable profession, BTW. — 79.113.214.51 (talk) 10:53, 20 September 2013 (UTC)[reply]

You have to be joking, that's about the tenth time you've made giant obvious equivocations. At any rate, I'm done.Phoenixia1177 (talk) 11:01, 20 September 2013 (UTC)[reply]

"Fields" and "spaces" are for cattle-herds and astronauts... Maths deals with numerical fields, and numerical spaces, and numerical functions. You can't have maths without numbers. (Logic, yes. Not mathematics). And numbers start at 0, since nothing logically precedes something. And the first something in logical order is 1. — 79.113.214.51 (talk) 11:08, 20 September 2013 (UTC)[reply]

Wait, what? You definiely CAN have maths without numbers, unless Euclidean geometry is some form of interpretive dance... MChesterMC (talk) 15:56, 20 September 2013 (UTC)[reply]

The time for waiting is over ! :-) Shapes devoid of numerical relationships are called art, not geometry. — 79.113.239.23 (talk) 17:19, 20 September 2013 (UTC)[reply]

(edit conflict)I think the difference here is that the IP is very much grounded in this universe (not a bad thing, it's handy for a lot of intuitive understandings), while Phoenixia is no so much (also not a bad thing, it allows some degree of inuitive undesanding of things that are well divorced from examples that can be experienced). 79... Consider this. Our mathematics starts from Euclidean geometry, because that is (to a decent approximation) the kind of geometry we live in. Consider Flatland, a 2D world, and consider what would happen if that world were projected on a sphere rather than a flat plane. Pi would be next to meaningless (aside from some esoteric thories about a supposed "third dimension"), since the ratio of the size of a circle to it's area would be different (it's not one I know offhand, but someone can probably provide it. Pi will appear, but with a lot of factors involved). The actual constant used might be some multiple of pi, and it might not be as well appoximated by a number of the form n/7. It's pure chance that the choice we made (3.14159....) is close to 22/7, but even in eucllidean geometry we could just have easily have chosen tau (=2pi=6.28319...), which is not as close to 44/7. Or we could have focussed on semicircles, and chosen pi/2, and we'd consider 11/7 to be an even more fantastic approximation. and any of these would seem even more wondrous if we had seven fingers and counted in base 7, making each of them n/10 MChesterMC (talk) 10:06, 20 September 2013 (UTC)[reply]

Actually, the ratio would no longer be constant, but variable (ranging randomly inbetween 2 and Pi, depending on the position of that circle on the spherical "plane"). As far as rational multiples of Pi are concerned, they may have different values (than Pi itself), but not a different nature. — 79.113.214.51 (talk) 10:23, 20 September 2013 (UTC)[reply]

syllogism test

Hi,

I would like to know whether there is a flaw in the following syllogism.

Major premise: "If 9/11 were an inside job, there would be a credible leak, since it would weigh on every person's conscience who knew about it and someone would leak."
Minor premise: "Someone that has a specific conspiracy theory fits the definition of a paranoid schizophrenic"
Second minor premise: "A paranoid schizophrenic is not credible."
Minor conclusion: there can be no credible specific leak about 9/11 being an inside job."
Conclusion: 9/11 cannot be an inside job.

Specifically, I seem to agree that it's out of the question that nobody would leak it if 9/11 were in any way an inside job: if it were in any way an inside job, it would have a leak. Then I notice that Susan Lindauer fits the definition of a leaker - but, she is very clearly a paranoid schizophrenic by virtue of having this one, specific story: http://en.wikipedia.org/wiki/Susan_Lindauer . Specifically, she was incarcerated in a mental hospital but reported as es." Ms.Lindauer also spent many hours at the prison law library studying up on her case. In its monthly reports, Carswell wrote: "Good physical health. Socializes well. Good intellectual functioning." And "no behavioral problems." ". So, the major reason that she is obviously a paranoid schizophrenic is due to her single theory. I find it convincing as an argument that she is therefore clearly not a credible witness.

More generally, this would apply to any credible witness. If that witness had a delusional history of inventing a specific story like this, this seems to me on its face to disqualify them.

So it seems to me that I've mathematically proved - if I bleieve 1-5, which I do - that 9/11 could not have been in any way an inside job.

Could you comment on the strength of my reasoning? I suppose you could move this to the Humanities desk, but I am really interested in the Logical aspect of it, which is a part of Mathematics as far as I understand. Thank you. Sylloguy (talk) 19:13, 19 September 2013 (UTC)[reply]

As far as I can see your logic is correct -- your premises lead to your conclusions. But your second premise ("Someone that has a specific conspiracy theory fits the definition of a paranoid schizophrenic") is invalid. It might be valid to have the premise "Someone that has a specific conspiracy theory without any evidence for it fits the definition of a paranoid schizophrenic"; but then the conclusions no longer follow. Duoduoduo (talk) 19:44, 19 September 2013 (UTC)[reply]

You're saying that someone with inside information would be a credible leak. You're also saying that someone with a conspiracy theory ... would not be a credible leak. It's not entirely clear that inside information must equal a conspiracy theory, but at least as read your premises contradict each other, and you're coming up with "false proves anything". Wnt (talk) 20:31, 19 September 2013 (UTC)[reply]

I don't see how 4 leads to 5. "We cannot know that something is true" =/= "Something cannot be true". Ah, wait, I see how i comes from 1, but if anything, 2-4 disprove that formulation of 1. Something more along the lines of:

1. If 9/11 were an inside job, there would be a ~~credible~~ leak, since it would weigh on every person's conscience who knew about it and someone would leak

2-4 as above

5. Therefore a credible leak about 9-11 is impossible

6. Therefore we cannot know that 9-11 is an inside job (NB: "that" not "whether". This says nothing about the possibility of confirming that it was not an inside job, though there's probably a similar argument)

Note that this does not assume that someone in a position to leak is in a position to give sufficient evidence for it to be credible, unlike the original (since that is refuted by 2-4 under the assumption that 9-11 being an inside job is possible) MChesterMC (talk) 09:44, 20 September 2013 (UTC)[reply]

I don't think it was an inside job, but I don't think 1 holds. The people who did 9/11 didn't leak it before hand because their conscience. If 9/11 wasn't an inside job, then the hijackers had conviction to another set of ideas, this allowed them to not have moral issues- by the same token, if it was an inside job, I'd imagine it wasn't for the hell of it. It's not impossible that it was an inside job by people with different knowledge and values, those things keeping them from leaking it. More to the point, unless somebody has actual evidence, there's no reason to believe it, this feels as if you're trying to prove a negative- it can always be the case that it was an inside job, but a long string of improbable things prevented it from getting out. I don't think you can disprove that, but that doesn't mean you need to accept it was an inside thing; nor does it mean that you can't tentatively dismiss it, much as with anything else.Phoenixia1177 (talk) 13:00, 20 September 2013 (UTC)[reply]

To simplify slightly: you don't believe hypothesis

\scriptstyle {H}

(efficacy of ESP, telepathy, crystal healing etc.), meaning you assign to it an extremely low prior probability. Along comes evidence (data)

\scriptstyle {D}

, supporting

\scriptstyle {H}

, but because your prior for

\scriptstyle {H}

is much lower than your prior for

\scriptstyle {D}

being flawed, your analysis is that the evidence is almost certainly unreliable, and your assessment of the plausibility of

\scriptstyle {H}

, is virtually unchanged. This is entirely logical and rational, despite being little affected by the nature of

\scriptstyle {D}

.

Conversely an individual who has no opinion concerning

\scriptstyle {H}

, and assigns it a moderate prior, will be swayed by

\scriptstyle {D}

to consider

\scriptstyle {H}

more plausible. This is also entirely rational, though it means that the same evidence has caused his/her assessment of

\scriptstyle {H}

to diverge from yours (albeit that the plausibility

\scriptstyle {H}

has increased for both of you).

Given conflicting evidence (partly supporting and partly contradicting

\scriptstyle {H}

), it is entirely possible that it increases one person's belief in

\scriptstyle {H}

while decreasing another person's belief - even though both people apply a rational analysis. This is Bayesian belief polarization and is not covered by WP as far as I am aware. It is however discussed by Jaynes Probability Theory, The Logic of Science pp. 128 - 131. --catslash (talk) 15:35, 22 September 2013 (UTC)[reply]

September 20

Linear Programming

Maximize Z=4x+9y, subject to teh constraints x>=0,y>=0 s+5y<=200 2x+3y<=134 — Preceding unsigned comment added by Bikramckc (talk • contribs) 04:07, 20 September 2013 (UTC)[reply]

Hello. The reference desk will not do your homework for you, although we can help you if you're stuck - only if you have made an attempt to solve the problem.--Jasper Deng (talk) 04:12, 20 September 2013 (UTC)[reply]

I assume that you meant x+5y<=200. You don't say whether you are looking for integer values or any real number as the solution. One way to get a solution is by drawing graphs of the constraints to identify the area of the graph that satisfies all the constraints, then draw a line such as 4x + 9y = 36 (just taking Z=36 as a convenient value) and look at lines parallel to this that just cross the constraint area. You may have been taught other methods, so you should use whatever method you have been taught. Dbfirs 14:55, 20 September 2013 (UTC)[reply]

Modelling Dissipative Process

I want to model the dissipative process $d\theta =-a\sin \theta dt+\sigma dW(t)$ where $dW(t)=N(t)dt$ is a normally distributed noise process. I would like to model the evolution of $\cos \theta$ , using the runge-kutta method.

Given the stochastic nature of the process is some sort of Ito's lemma style correction neccesarry when using runge-kutta models?

In other words, for the purposes of phrasing the problem as ${\frac {dy}{dt}}=f(y,t)$ where $y=\cos \theta$ , so that RK methods can be used, is it correct to simply say

${\frac {dy}{dt}}={\frac {dy}{d\theta }}{\frac {d\theta }{dt}}=a\sin ^{2}\theta -\sigma \sin \theta N(t)=a(1-y^{2})-\sigma {\sqrt {1-y^{2}}}N(t)$

where, when modelling, the value of $N(t)$ is drawn from a unit normal distribution at each time step.

Or is it instead correct to say, following Ito, that this is drift diffusion process with $\mu _{t}=-a\sin \theta$ and ${\sigma }_{t}=\sigma$

${\frac {dy}{dt}}=-a\sin \theta {\frac {dy}{d\theta }}+{\frac {\sigma ^{2}}{2}}{\frac {d^{2}y}{d\theta ^{2}}}+\sigma {\frac {dy}{d\theta }}N(s)=a\sin ^{2}\theta -{\frac {\sigma ^{2}}{2}}\cos \theta -\sigma \sin \theta N(t)=a(1-y^{2})+{\frac {\sigma ^{2}}{2}}y-\sigma {\sqrt {1-y^{2}}}N(t)$

I feel like it should be the latter, as the former clearly has a fixed point at y=1, whereas the starting equation does not have a corresponding one at θ=0, however I am not convinced I full understand what are the circumstances under which Ito's correction is valid. — Preceding unsigned comment added by 144.82.172.150 (talk) 17:33, 20 September 2013 (UTC)[reply]

The Ito correction should be included, unless the time steps are so small that they fall well within the correlation time of the noise. But if N(t) is purely white noise (it has arbitrary high frequency components), then N(t) and N(t+delta t) are totally uncorrelated for arbitrary small delta t. Count Iblis (talk) 18:04, 20 September 2013 (UTC)[reply]

The article Runge–Kutta method (SDE) has a Markov chain approximation method that looks suitable to your problem. --Mark viking (talk) 18:05, 20 September 2013 (UTC)[reply]

Thanks guys. For the record (if anyone ever looks back over this) I found this paper useful [1]. — Preceding unsigned comment added by 109.144.154.56 (talk) 10:33, 22 September 2013 (UTC)[reply]

September 21

annuities, compound interest and simple interest

Is there a website that shows you how to do a simple interest calculations, compound interests and annuities in an easy way and and how you can tell if the question is talking about either a future value or present value or other things? — Preceding unsigned comment added by 174.89.40.56 (talk) 16:09, 21 September 2013 (UTC)[reply]

Start with this link describing Interest Cite error: There are <ref> tags on this page without content in them (see the help page). Describes multiple types of interest and their calculations.

This page Cite error: There are <ref> tags on this page without content in them (see the help page). walks you through a simple interest calculation, eventually getting you to the answer. It is set up in a way that it is easy to understand and the data fields are self-explanatory.

There are many places to find help with compound interest and their calculations. Try this link Cite error: There are <ref> tags on this page without content in them (see the help page). It has data fields where you can enter all the information needed in order to find the amount of interest to be charged or paid. Ice24710 (talk) 20:05, 21 September 2013 (UTC)[reply]

automobile speeds at passing

how fast do I have to drive to pass a vehicle doing 90kph? Is there a graph for this? It always seems like they speed up when I pass, or is that just an illusion? — Preceding unsigned comment added by 209.121.225.128 (talk) 18:57, 21 September 2013 (UTC)[reply]

Suppose that you pass from 10 meters behind and you want to move 10 meters past it before moving in front of the vehicle. The time needed for that will then be 20 meters/(relative speed). So, to be able to pass the car in 15 seconds the relative speed needs to be 4/3 m/s = 4.8 km/h. If you travel at 94.8 km/h you will travel a distance of 395 meters in that 15 seconds.

I think the distance needed affects the perception of time, obviously if you need to pass in short distance, you will need to travel at a large speed. E.g. to pass in 50 meters you need to travel at 150 km/h, which may be a lot faster than what you might intuitively expect. Count Iblis (talk) 19:13, 21 September 2013 (UTC)[reply]

A couple additional points:

1) The length of the vehicle you are passing also may be significant, especially in the case of long cargo trucks. This distance needs to be added to the 20m in the above example.

2) The vehicle you are trying to pass may indeed speed up. The drivers aren't just being jerks, though. Subconsciously, we judge our own speed based on the speed of other vehicles on the road. So, if we are passing every other vehicle on the road, we think we are going too fast, and slow down, while if every other car on the road is passing us, we think we are going too slow, and speed up. In the case of only two vehicles, that makes each driver adjust their speed to be closer to the other, which increases passing time. The obvious solution to this problem is cruise control. StuRat (talk) 00:20, 22 September 2013 (UTC)[reply]

Random walk problem

Hi, I've got a problem I can't figure out. A player flips a weighted coin where he wins $1 with probability p or loses $1 with probability 1-p, where p>0.5. The game ends when he reaches some positive amount x. What is the average amount of flips it takes to end the game? I'm thinking it has something to do with Stirling's numbers but have had no success thus far 78.0.223.206 (talk) 20:59, 21 September 2013 (UTC)[reply]

For x=1, the probability is p that 1 flip is required, p²(1-p) that 3 flips are required, 2p³(1-p)² that 5 flips are required and in general C_npⁿ⁺¹(1-p)ⁿ that 2n+1 flips are required, where C_n is the nth Catalan number. From this I get the expectation is 1+2p²(1-p)C'(p(1-p)) where C(x) is the generating function for C_n. (Note, C(p(1-p))=1/p for p>.5.) Plugging in the formula for C(x) (see the article) and simplifying, I get the expectation is 1/(2p-1) (p>.5). For x>1 you'd probably utilize Bertrand's ballot theorem so get the probabilities, but I think the calculations are a bit too hairy for me to tackle at the moment. --RDBury (talk) 22:26, 21 September 2013 (UTC)[reply]

Once you know the answer for 1, can't you just use linearity of expectation to argue the general answer is x/(2p-1)? Our random variables are v_1 through v_x, where v_i is the number of flips from the first moment we are up by i-1 dollars until the first moment we are up by i dollars. Clearly E[v_i] = E[v_1], which you've just argued is 1/(2p-1). We want E[v_1 + ... + v_x] = E[v_1] + ... + E[v_x] = x/(2p-1).--80.109.106.49 (talk) 22:44, 21 September 2013 (UTC)[reply]

This sort of problem is called (in some disciplines) a first hitting time problem for a Bernoulli process. The number of flips is distributed according to a negative binomial distribution. --Mark viking (talk) 23:10, 21 September 2013 (UTC)[reply]

It would be Bernoulli if you were just waiting for the first win, but the fact that when you lose you have to make it up with a win to get back to where you started complicates the problem. I would be surprised though if the problem isn't already well studied. The coefficients of the probabilities from Catalan's triangle whose generating function is known. To respond to the comment above, I don't there's any reason to assume the result is linear in x. --RDBury (talk) 01:58, 22 September 2013 (UTC)[reply]

Actually it would be linear, good point. --RDBury (talk) 02:01, 22 September 2013 (UTC)[reply]

A standard approach for this kind of problems is conditioning on the first step. Denoting by

f(x)

the expected number of flips when the goal is x, then after a flip our new relative goal will be either

x-1

or

x+1

, so we have the recurrence relation

f(x)=pf(x-1)+(1-p)f(x+1)

From which it is possible to deduce that

f(x)

is linear. For figuring out the actual slope, I can't currently think of a better way than RDBury's. -- Meni Rosenfeld (talk) 20:20, 23 September 2013 (UTC)[reply]

September 22

Tonelli-Shanks extensions

How can the algorithm be extended to, for example, x^2 - x == n (mod p)? Or to any polynomial function of x? 68.0.144.214 (talk) 00:40, 22 September 2013 (UTC)[reply]

You can solve the quadratic equation in the usual way, division mod p is well defined, and when you have to take the square root, you have available the Tonelli-Shanks algorithm. Count Iblis (talk) 01:33, 22 September 2013 (UTC)[reply]

Rank and crank of a partition

I'm just starting to learn about this, and I see that the rank of a conjugate partition is the negative of the original rank, so in particlar, the rank of a selfconjugate partition is zero, and the selfconjugates are a subset of rank zero partitions, so i figured this skew-symmetry or odd-functionness with respect to conjugates would be important. But the crank doesn't have this property at all. Is there a way of tweaking the crank function to make it skew-symmetric? Thanks, Rich Peterson64.134.220.186 (talk) 20:27, 22 September 2013 (UTC)[reply]

September 23

Iterated exponentiation

I wanted to ask a general question about mathematics. For example: x^x^x^x^x^... = 2

The way forward would be. x^y = 2 where y=x^x^x^x^...

But the value of y would be 2 as it is the premise.

So x^2=2

Here is why I got confused. There are two solutions to the above equation. x=root2 and x= - root2

However only one of the solution works for x^x^x^x^x^... = 2

Why is it that even though it satisfy x^2=2, only one of the solution works? Is satisfying x^2=2 only a requirement and not a guaranty of a solution? 220.239.51.150 (talk) 12:57, 23 September 2013 (UTC)[reply]

Let me clarify my confusion. Look at these equations and statements.

(Equation 1) x^x^x^x^x^… = 2
(Statement A) Every numeric value of x that satisfy Equation 1 is a valid solution to the problem

(Equation 2) x^y = 2 where y=x^x^x^x^x^…
(Statement B) Every numeric value of x that satisfy Equation 2 is a valid solution to the problem

OK, I'm gonna stop you right there: this statement is false. Why ? Because the implication in question is only one of an infinite number of implications which result from the initial equation. So the logical conclusion would be that the values which satisfy the original equation are to be found among those that satisfy the latter... In other words, inclusion rather than equality. — 79.113.229.58 (talk) 13:23, 24 September 2013 (UTC)[reply]

(Equation 3) x^y = 2 where y=2
(Statement C) Every numeric value of x that satisfy Equation 3 is a valid solution to the problem

(Equation 4) x^2 = 2
(Statement D) Every numeric value of x that satisfy Equation 4 is a valid solution to the problem

Statement A is CORRECT but Statement D is INCORRECT and yet every step of the way is perfect and flawless.

That is why I am confused. 220.239.51.150 (talk) 13:34, 23 September 2013 (UTC)[reply]

I don't really understand what exactly you are trying to do. Note that for Equation 1 there should be a unique solution for each number of xs (at least I think so). For example, there is a unique real solution to x^x = 2. -- Toshio Yamaguchi 14:01, 23 September 2013 (UTC)[reply]

"... there is a unique real solution to x^x = 2"

I have to think about that claim again. This seems to be equivalent to the claim that the real numbers are closed under exponentiation, but I am actually not sure whether that claim is true. Shouldn't be the case. On the other hand, if we restrict x to be positive, it might be true. -- Toshio Yamaguchi 14:39, 23 September 2013 (UTC)[reply]

See our article on tetration. Iterated exponentiation converges for values of x between e^-e and e^1/e, but since the positive square root of 2 is within this interval, this solution is okay. Note that the reasoning above shows that if x satisfies x^x^x^... = 2 then x^2 = 2; you cannot conclude from this that if x^2 = 2 then x^x^x^... = 2. Gandalf61 (talk) 15:43, 23 September 2013 (UTC)[reply]

I think the problem you're having is assuming that the "series" x^x^x^x… taken to it's infinite limit converges under any given input, and that 2 is in the (output) range where it converges. That is, you're neglecting the possibility that there doesn't actually exist any x for which x^x^x^x^x^… = 2 exists. It's a well known property of certain normal series (the conditionally convergent ones) that if you alter the way you compute what they converge to, you change what they converge to. [2] If you reorder the series, you change its value. I don't know anything about the properties of your exponentiation "series", but it could suffer from much the same problems. You can't just blithely chop off an x and assume the value stays the same. Especially not when negative numbers are involved. -- 67.40.209.200 (talk) 17:05, 23 September 2013 (UTC)[reply]

Statement B should be Every pair of numerical values of x, y that satisfy Equation 2 is a valid solution to the problem. But y = 2, x = minus root 2 does not satisfy the last part of Eq. 2 because the iterated expectation does not converge. So I think the logical mistake is that you've lost restricting information in writing the last part of Eq. 3. Duoduoduo (talk) 18:11, 23 September 2013 (UTC)[reply]

Statement B says "if x^y = 2 and y = x^x^..., then x^x^... = 2". This is true. However, statement C says "if x^y = 2 and y=2, then x^x^... = 2". This does not follow, because from the facts that x^y = 2 and y=2 you can't deduce that y = x^x^... and execute statement B. -- Meni Rosenfeld (talk) 20:05, 23 September 2013 (UTC)[reply]

Integrate (x^2+1)^10

Integrating a function can be like unscrambling eggs! Sometimes, there are 2 functions that look almost alike, but their integration technics are not the same. An example is to integrate each of:

$(x^{2}+1)^{10}$ and $((x^{2}+1)^{10})x$ .

Let's try both functions. To integrate $((x^{2}+1)^{10})x$ , you just relate this function to the derivative of $(x^{2}+1)^{11}$ by a constant multiplier, because the derivative is $22x(x^{2}+1)^{10}$ . Thus, the integral of $((x^{2}+1)^{10})x$ is $((x^{2}+1)^{11})/22+C$ .

But to integrate $(x^{2}+1)^{10}$ , we would need a variable multiplier. The integral of a product does not equal the product of the first function and the integral of the second unless the first function is a constant, so we cannot integrate it this way. Any way it can be done other than expanding the binomial power and integrating each term?? Georgia guy (talk) 14:38, 23 September 2013 (UTC)[reply]

Use integration by parts. Sławomir Biały (talk) 17:13, 23 September 2013 (UTC)[reply]

Will this help? For a general exponent it seems the expression involves a hypergeometric function, so I doubt there is anything simpler than expansion with a specific exponent. -- Meni Rosenfeld (talk) 19:53, 23 September 2013 (UTC)[reply]

The question was whether it can be done another way, not whether it should be done another way. Since binomial coefficients are "precomputed" you save effort over repeated integration by parts. Strangely, the standard way of handling the equivalent problem of integrating sec²²t would be integration by parts rather than doing a reverse trig substitution and applying the binomial theorem. --RDBury (talk) 21:25, 23 September 2013 (UTC)[reply]

It's a fair point that it doesn't really "help" per se. The fact is that the integral has many terms, and any integration method will have to compute those terms one way or another. Sławomir Biały (talk) 12:39, 24 September 2013 (UTC)[reply]

I am not great at all with complex analysis, but it sounds like this might also be the same as $\int (x+i)^{10}(x-i)^{10}dx$ which can be integrated by tabular integration after letting $u=x-i,du=dx$ . It's still a lot of integration by parts, but tabular integration is relatively effortless, at least for me.--Jasper Deng (talk) 22:28, 23 September 2013 (UTC)[reply]

You can define a generating function as follows. If we put

$I_{n}(x)=\int \left(x^{2}+1\right)^{n}dx$

then we can calculate the generating function:

$f(x,t)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {I_{n}(x)t^{n}}{n!}}={\sqrt {\frac {\pi }{2t}}}\exp(-t)\operatorname {*} {Erf}\left({\sqrt {t}}x\right)$

Count Iblis (talk) 00:00, 24 September 2013 (UTC)[reply]

Use Newton's binomial, and the properties of the integral, namely that the integral of a sum is the same as the sum of integrals, ultimately arriving at the following identity:

\int (x^{2}+1)^{10}\ dx\ =\ \ldots \ =\ \sum _{k=0}^{10}C_{10}^{k}\cdot {\frac {x^{2k+1}}{2k+1}}

— 79.113.233.194 (talk) 01:57, 24 September 2013 (UTC)[reply]

September 24

Diagram

What you can or cannot conclude from a diagram. — Preceding unsigned comment added by Mike Boissonnault (talk • contribs) 00:50, 24 September 2013 (UTC)[reply]

That would rather depend on the diagram, wouldn't it ? Care to give us an example ? StuRat (talk) 13:09, 24 September 2013 (UTC)[reply]

True or False ?

Would anyone be so kind as to verify the following equality, either confirming it or infirming it ?

\int _{0}^{\infty }{\frac {f(x)}{F(x)}}\ dx\ =\ \int _{0}^{\infty }{\frac {x!}{x^{x}}}\ dx

where

f(x)\ =\ \sum _{n=1}^{\infty }{\frac {n^{x}}{n^{n}}}\qquad ,\qquad F(x)\ =\ \sum _{n=1}^{\infty }{\frac {n^{x}}{n!}}

Thank you ! — 79.113.233.194 (talk) 01:46, 24 September 2013 (UTC)[reply]

The equation is meaningless as written as the factorial function only applies to non-negative integers, so can't be used on a variable of integration. You could perhaps replace

x!

by

\Gamma (x+1)\

. AndrewWTaylor (talk) 13:22, 24 September 2013 (UTC)[reply]