I saw a guy on a train wearing a fairly plain black jacket but there was a mathematical formula embroided on the back.

${\displaystyle {\delta V \over \delta t}+{1 \over 2}\sigma ^{2}S^{2}{\delta ^{2}V \over \delta S^{2}}+rS{\delta V \over \delta S}-rV=0}$

Does anyone recognize what this is and why would someone have it on their jacket? I'm guessing it's some math "in joke". Vespine (talk) 03:36, 20 November 2015 (UTC)[reply]

It's the Black-Scholes equation. As for what the intended message of putting it on a jacket was, your guess is as good as mine.--Antendren (talk) 08:29, 20 November 2015 (UTC)[reply]

This was the equation that all the financial experts were using until it was realised recently that there was a flaw in the assumptions. Sorry I can't remember the subtle details. Dbfirs 23:10, 21 November 2015 (UTC)[reply]

That's as much of an answer as I was expecting, thank you!

Resolved

The standard Black-Scholes equation computes the theoretical price of a European call option, which is a simple type of financial derivative. The jacket formula you saw is the generalized form of the equation, which is a partial differential equation that estimates the theoretical price of any "derived" asset in terms of the change in its "strike" asset. It's not flawed per se, but it assumes the asset return is normally distributed, which is not always the case in practice. My guess is the person you saw is a quant. OldTimeNESter (talk) 21:31, 23 November 2015 (UTC)[reply]

Hello! I'm a first year university student, and while practicing integration I made up a question ${\displaystyle \int x^{2}e^{x^{2}}dx}$ and tried to solve it using integration by parts. I got nowhere after a short while and, after looking the integral up, it turns out there is no elementary integral for the problem. However, integration by parts gives me this infinite series:

${\displaystyle {\frac {1}{3}}x^{3}e^{x^{2}}-{\frac {2}{3}}\left({\frac {1}{5}}\right)x^{5}e^{x^{2}}+{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {1}{7}}\right)x^{7}e^{x^{2}}-{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {2}{7}}\right)\left({\frac {1}{9}}\right)x^{9}e^{x^{2}}+{\frac {2}{3}}\left({\frac {2}{5}}\right)\left({\frac {2}{7}}\right)\left({\frac {2}{9}}\right)\left({\frac {1}{11}}\right)x^{11}e^{x^{2}}\pm \cdots }$

I was just wondering, is there a way to write this expression as an infinite sum? I can kind of see a pattern but I don't have enough experience with infinite sums to generate one (if it even exists!). Thanks for your help. 70.54.112.243 (talk) 04:15, 21 November 2015 (UTC)[reply]

You just wrote down an infinite sum, so it is not clear to me what you want. You may put the first term outside parentheses to get the expression

${\displaystyle {\frac {1}{3}}x^{3}e^{x^{2}}\sum _{i=0}^{\infty }\prod _{k=1}^{i}{\frac {-2x^{2}}{2k+3}}}$

Try also [[1]]

Bo Jacoby (talk) 05:59, 21 November 2015 (UTC).[reply]

This can also written without the product notation as ${\displaystyle e^{x^{2}}\sum _{i=0}^{\infty }{\frac {(-1)^{i}4^{i+1}(i+2)!}{(2i+4)!}}x^{2i+3}}$. -- Meni Rosenfeld (talk) 22:49, 21 November 2015 (UTC)[reply]

I'm sure you are familiar with Capital-sigma notation, but if you haven't seen Capital Pi notation before, you may wish to check out that article. (And while reading on the subject, you should familiarize yourself with Empty product.) Finaly, the erfi in the indefinite integral on the Wolfram Alpha page Bo Jacoby linked above is the imaginary error function. -- ToE 14:18, 21 November 2015 (UTC)[reply]

In J the function looks like this:

f=.13 : '3%~(y^3)*(^y^2)*+/*/\1,(-2*y^2)%3+2*1+i.10'

and a test run looks like this:

f&>0 0.5 1 1.5 2 2.5

0 0.0485128 0.627815 5.0843 46.7366 1398.4

Bo Jacoby (talk) 18:46, 22 November 2015 (UTC).[reply]

The Maclaurin series of this is straightforward to derive: ${\displaystyle \int x^{2}e^{x^{2}}dx=\int x^{2}\sum _{i=0}^{\infty }{\frac {(x^{2})^{i}}{i!}}dx=\int \sum _{i=0}^{\infty }{\frac {x^{2i+2}}{i!}}dx=\sum _{i=0}^{\infty }{\frac {x^{2i+3}}{i!(2i+3)}}+C}$. Using the doubling formulas for the gamma function, this can be easily shown to be equivalent to the series you derived. --Jasper Deng (talk) 23:33, 23 November 2015 (UTC)[reply]

I started looking at how many ways a season could go with every game being part of a 2+ game winning streak or a 2+ game losing streak. After calculating the examples for up to 6 games, I found OEIS: A006355, which has as one if its examples "a(6)=10 because we have: 000000, 000011, 000111, 001100, 001111, 110000, 110011, 111000, 111100, 111111. - Geoffrey Critzer, Jan 26 2014". So this is what I'm looking for. This sequence is a Fibonacci sequence starting with positive numbers: 0,2,2,4,6,10,16, etc. (So if you only count those where the team started out by winning, you get the standard Fibonacci series, offset)

So my question is, is there any obvious way to see in calculating that (for example) the number of 8 game seasons is equal to the sum of the number of 7 game seasons and the 6 game seasons?Naraht (talk) 08:49, 22 November 2015 (UTC)[reply]

Take all possible 7-digit sequences, repeat the last digit, and you get all possible 8-digit sequences ending in 000 or 111. Take all possible 6-digit sequences, append the opposite of the last digit twice, and you get all possible 8-digit sequences ending in 011 or 100. Combining these two cases, you get all possible 8-digit sequences. Egnau (talk) 13:17, 22 November 2015 (UTC)[reply]

A direct proof is this: Remove the initial and final digit from each sequence. Now for a given sequence, replace each 01 or 10 with a 2, and each digit between two of the same digit with a 1. For example 001110000110011111→0111000011001111→21211222111. This gives a 2 to 1 correspondence between the sequences of length n+2 and ways of writing n as a sum of 1's and 2's. If you count the latter you get the Fibonacci sequence. There are many similar correspondences between strings and sums satisfying certain criteria; some come from simple rewrite rules and some are more difficult. A surprising one is between strings of 0's and 1' that do not contain "101" (OEIS: A005251) and sums of 2's and 3's (OEIS: A000931) . --RDBury (talk) 15:23, 22 November 2015 (UTC)[reply]

@Egnau: That makes sense, thanx!Naraht (talk) 04:21, 24 November 2015 (UTC)[reply]

I'm wondering how, given a differentiable function of two spatial variables, how to compute the steepest paths of descent on its graph. All I know is that such a path must have the gradient as a tangent vector at every point, and thus should be itself a smooth curve. I also know that that means that the x and y components of the derivative of the function (of one parameter) describing the path should be parallel (as a vector) to the gradient. I'd much rather not have to use the calculus of variations.--Jasper Deng (talk) 00:16, 24 November 2015 (UTC)[reply]

I think what you're asking for are essentially orthogonal trajectories. --RDBury (talk) 00:51, 24 November 2015 (UTC)[reply]

Discrete mathematics is opposed to continuous mathematics, which is basically numerical analysis, which is opposed to symbolic manipulations. How is the relation between the discrete math, and symbolic manipulations? Is there something like an internal opposition in discrete math - symbolic discrete methods vs. numeric discrete methods? Or, is discrete math just symbolic manipulation (or a subset of this latter)?--3dcaddy (talk) 15:12, 24 November 2015 (UTC)[reply]

Er... why should continuous mathematics be identified with numerical analysis? Numerical analysis usually means approximating continuous problems by discrete ones. —Kusma (t·c) 15:23, 24 November 2015 (UTC)[reply]

I have changed the redirect to go to mathematical analysis instead. —Kusma (t·c) 15:30, 24 November 2015 (UTC)[reply]

Yes, that's right. Anyway, the first part of the question is still more confusing than explanatory. So, I am leaving the bottom part and striking it.--3dcaddy (talk) 16:13, 24 November 2015 (UTC)[reply]

I think you may be referring to something like the discrete element method which, despite the name, would fall under numerical analysis and therefore continuous mathematics and not discrete mathematics. Typically, the term discrete mathematics refers to finite (countable sometimes) structures while analysis or continuous mathematics refers to structures involving real numbers in some way. But there are areas that don't fall into either camp and our Areas of mathematics divisions cut across this line. Mathematics Subject Classification has main divisions for discrete math/algebra and analysis, but it also has divisions for foundations, geometry, and applied mathematics. But Lie groups, classified in with other areas of group theory under discrete math/algebra, might more properly be put under analysis. And while most of geometry would fall under continuous mathematics, there is a whole area of discrete geometry that wouldn't. The boundaries between different areas of mathematics are usually fuzzy and often based more on tradition than anything else, so it's difficult to give a precise meaning to the distinction. --RDBury (talk) 18:07, 24 November 2015 (UTC)[reply]

Previous answer is about mathematics aspect of the question. From the computing point of view, there are three main classes of mathematical computation: numerical computation (sometimes improperly called numerical analysis or even scientific computation), which essentially deals with real numbers represented as floating point numbers; symbolic computation, also called computer algebra for which the objects of computation are essentially mathematical expressions; and discrete mathematics computations that deals essentially with finite sets and finite structures. These three classes of methods, although using very different algorithmic methods, are not completely independent. For example, computational geometry makes heavily use of the three kinds of methods.

In your question, you use "symbolic discrete method" and "numeric discrete methods". Personally I find these terms confusing and useless: All computer science may be qualified as "discrete", as, inside computers, everything is expressed in terms of the finite set with two elements. Also, you are wrong when searching oppositions between methods. Many of the best recents achievements were obtained by importing methods of one of the above classes of computation for improving solutions given by other classes. D.Lazard (talk) 13:08, 25 November 2015 (UTC)[reply]

Teaching complex number to school kids should not be hard. Complex number addition is just equivalent to vector addition. Complex number multiplication is just treating one complex number as a vector and using the other complex number as a procedure to perform scaling and follow by an angle rotation. This can be done in the argand diagram. When intepreted in the geometrical format there is nothing special about I (imaginary number) which is nothing but a rotation of 90 degrees counter-clockwise in the argand diagram.

It is not just teachers in high school. Look at the wikipedia article on complex numbers, it does not mention that addition of two complex numbers has similar properties to vector addition nor does it mention that multiple of two complex number to be consider a form of scaling and rotation of a complex number (about the origin) in the complex plane. The smart student may figure this out all by themselves but there is no reason not to mention this to all readers of the wikipedia article.

This is nothing but a global elitist conspiracy to prevent the understanding of the true nature of complex numbers from the dumb masses. Instead of allowing them to understand complex numbers, the students are forced to memorize complex procedures (pun not intended) and endure deliberate obfuscation. Where as what the students need to know is what the complex numbers are and then the procedures for dealing with them which can be programmed into a smart phone similar to calculators which can perform addition , subtraction , multiplication and division. I can't even remember when I last did complex multiplication and complex division by hand. My point is that the conspiracy to prevent the understanding of complex numbers is prevents literally billions of people from using complex numbers in their daily lives. — Preceding unsigned comment added by 175.45.116.59 (talk) 23:33, 24 November 2015 (UTC) 175.45.116.59 (talk) 23:02, 24 November 2015 (UTC)[reply]

Note: The redirect Argand diagram -> Complex plane is tagged {{R with possibilities}}, and while "Argand diagrams" is bolded in the lede, they are not defined there and are not mentioned in the article's body. -- ToE 23:28, 24 November 2015 (UTC)[reply]

As part of the global elitist conspiracy why would people pay us anything but peanuts if they understood how easy it all is ;-) Actually if you do look at the complex number article you'll see vector addition in the diagram beside the description of addition. The corresponding diagram showing multiplication is equivalent to adding the angles and scaling is in the section about polar representation. That was a bad oversight leaving that in, ah well too late I'd have people complaining if I removed it. Dmcq (talk) 00:02, 25 November 2015 (UTC)[reply]

But aren't complex numbers already part of the standard school curriculum? Sławomir
Biały 00:31, 25 November 2015 (UTC)[reply]

Please do elaborate on what use billions of people would find for complex numbers in their daily lives. That would have given me a more convincing answer when I used to teach College Algebra and the students wanted to know why they needed to learn about complex numbers. The actual answer I gave, "you should learn about complex numbers because they're really cool", is one I stand by, but I'm not sure they totally bought it. And maybe they shouldn't have, because my mental reservation was, "... they're really cool, but if you take College Algebra and then stop, you're never actually going to find out why they're cool". --Trovatore (talk) 00:44, 25 November 2015 (UTC) [reply]

I think the IP's point, the incivilities aside, is that most people are not taught that complex numbers can also be thought of as vectors in R². However, that is quite an advanced treatment for high school and below... I'm not even sure if the connection really can be done without going further than vectors. In particular, division of complex numbers does not have a straightforward vector counterpart. (I'm also about 1/3-sure that the IP is also just trolling.)--Jasper Deng (talk) 03:23, 25 November 2015 (UTC)[reply]

Division in complex number is easy. If A times B equal C. And this means, a vector A is scaled and rotated (via B) to obtain the result C. So division just means, given the resultant vector C and the "scaling and rotation" B, find the original vector A. 175.45.116.59 (talk) 03:35, 25 November 2015 (UTC)[reply]

This isn't a result that is readily apparent to most high school students, for it requires the notion of the argument of a complex number. The closest high school usually gets to that is de Moivre's formula (even then, only calculus classes delve into its derivation from Maclaurin series), hence why I don't consider it straightforward from the standpoint of a high school student. I also think that the concept of a bivector and geometric algebra is useful when talking this, since the imaginary unit has a very straightforward representation as a bivector.

Whether this should be taught, I'm not so sure. You're assuming that high school students intrinsically want to use advanced mathematics in their everyday life. You must remember that complex numbers are introduced as roots of otherwise irreducible polynomials, and that only a small subset go on to the point where complex numbers become useful in this geometric sense. I also would highly caution you against making highly general statements such as a "global conspiracy". I know plenty of highly rigorous high schools which have honors classes dealing with this (for students who are truly interested).--Jasper Deng (talk) 04:31, 25 November 2015 (UTC)[reply]

Choose a zero point, 0, and a unit point, 1. Similar triangles have proportional sides. If triangle (0,1,a) is equiangular to triangle (0,b,x), then x/b=a/1, or x=a·b. That's all. Bo Jacoby (talk) 12:22, 25 November 2015 (UTC).[reply]

• I do not think teaching complex numbers as vectors will make them simpler to understand. On contrary it over-complicate everything. In addition, the vector interpretation of them is not practically useful. Ruslik_Zero 13:45, 26 November 2015 (UTC)[reply]

What's the name for the technique in statistics that was used to generate this map of the "most disproportionately popular cuisine" in each US state?

http://www.huffingtonpost.com/2015/01/14/most-popular-cuisine-state_n_6457252.html

Thanks! --50.35.66.30 (talk) 07:02, 26 November 2015 (UTC)[reply]

Unless they state which methods they used, we can't tell for sure, but to me it looks like they divided restaurants into a few dozen categories, determined the national average for percentage of each type, then compared each states's averages with the national average. For example, if 1% of all restaurants nationally are Taiwanese, but 2% of those in California are, then that's 200% the national average. If that was the highest percentage they found, then it wins for California. (I assume they had more significant digits than my example, BTW.) StuRat (talk) 07:48, 26 November 2015 (UTC)[reply]

The article explains what method it used! The categorizations were taken from Yelp, but otherwise it is as Stu says. But that isn't what we were asked. --70.49.170.168 (talk) 08:26, 26 November 2015 (UTC)[reply]

Thanks for your help. I do understand the method that was used to derive the results, I was just wondering if there's a concept or term in statistics for the subset of data points that are "most disproportionately x." Does that make sense? I looked at articles like statistical distance, for instance, but that didn't seem directly relevant. `--50.35.66.30 (talk) 09:32, 26 November 2015 (UTC)[reply]

I saw something about a college curriculum that I don't know whether it was a spoof or not. Within multivariate calculus, do you have the concept of Calculus over Fractional (like 3.5) dimensions?Naraht (talk) 14:50, 26 November 2015 (UTC)[reply]

Analysis on fractals is certainly a thing. Perhaps most fundamentally, fractals have a self-similar hausdorff measure, which allows functions on fractals to be integrated. Fractals are metric measure spaces, and so differential calculus is also possible in some sense. Probably the most important results concern the study of diffusion on fractals (e.g., percolation theory). But this all well outside of any standard "college curriculum" that one might teach to undergraduates in a multivariable calculus course. Sławomir
Biały 15:02, 26 November 2015 (UTC)[reply]

Fractal derivative is possibly relevant. Additionally, Lebesgue integration can be applied to the Hausdorff measure of a fractional-dimensional space. -- Meni Rosenfeld (talk) 15:12, 26 November 2015 (UTC)[reply]

Also, while this is a completely different thing than what you asked, I think an answer to a question that includes the words "fractional" and "calculus" should at least mention fractional calculus. -- Meni Rosenfeld (talk) 15:12, 26 November 2015 (UTC)[reply]

Is it worth sometimes to play a lottery? I know that they normally have negative expected value (in the same way as roulette, sport betting or horse races, among others).

However, what if a lottery has a jackpot that accumulates at each draw without a winner?

If the lottery has 1,000,000 numbers, $700,000 in prizes and each ticket costs $1, it is not worth playing without a jackpot. But how about buying 100,000 numbers in the same lottery with an accumulated jackpot of $2,000,000.--YX-1000A (talk) 16:17, 26 November 2015 (UTC)[reply]

Generally, as the jackpot goes up, more people play, so it never gets to the point where there's a positive ROI. This is especially true because the ROI is so low to begin with (lower than just about any other form of gambling). Also, once you figure in the tax implications, that makes the average ROI even worse. Then the advertised jackpot value is sometimes spread over many years, meaning you have to figure in inflation between now and then.

Also, buying more tickets does increase your odds, but at a decreasing rate. Consider if there were just 2 tickets sold, in a contest where tickets are drawn from a barrel, and you had one and somebody else had the other. Then you would have a 50% chance of winning. But if you spent twice as much and bought 2 tickets, then you wouldn't double your chances of winning, but only increase the chance to 2/3. StuRat (talk) 16:30, 26 November 2015 (UTC)[reply]

A gamble can be worthwhile even if its expected value is negative, if your utility function has a nonconvex region. Usually the utility function is convex, so the variance of a gamble adds insult to the injury of its negative expectation, but in some case it can be beneficial. This can happen if there is something you need badly, but can't afford.

Example: Your life-long dream is to travel to space. Your utility will increase by 1 unit if you manage to do it at least once in your life. Other uses of money will give you utility equal to the logarithm of your net worth (in $).

There are companies that will allow you to do this if you pony up the $$$ - $250K to be exact. But you can't afford $250K - paying this amount is either impossible or decrease your utility by more than 1.

If your net worth prior to paying for space travel is $X, then your utility without space travel is ${\displaystyle \log X}$, and after buying space travel, ${\displaystyle 1+\log(X-250000)}$. So your optimal strategy will give you the maximum between these two values, and bends at ${\displaystyle X=395494}$. If you are close to this bend (to its left), you can increase your utility by gambling. For example, if your net worth is $395,000, then buying a ticket which costs $200 and has a 1% chance to give you $10K, will increase your utility by 0.000131. -- Meni Rosenfeld (talk) 17:36, 26 November 2015 (UTC)[reply]

• At least one lottery operator put this argument succinctly: "You can't win if you don't buy a ticket." --70.49.170.168 (talk) 22:18, 26 November 2015 (UTC)[reply]

The example I like to give is thus: "You owe X dollars to a loan shark, who is waiting outside the casino you are inside and is going to kill you unless you pay it back, in full, when you leave. The casino is about to close, you have less than X now, and nobody is willing to lend you money or buy anything you have. Thus, it's in your interest to gamble, with the best odds you can find, even if those odds are against you. (But for God's sake, if you manage to get X dollars, don't decide you are on a winning streak and let it ride !)" StuRat (talk) 19:35, 26 November 2015 (UTC)[reply]

See Run, Lola, Run. Or for that matter, Casablanca. --70.49.170.168 (talk) 22:18, 26 November 2015 (UTC)[reply]