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April 23

Iterative method for e?

Is there any iterative method for finding e? How do you actually find iterative methods for irrational numbers? Thanks. --wj32 ^t/c 00:30, 23 April 2008 (UTC)[reply]

Well, you can turn the infinite series representation of e into an iterative method as follows:

a_{1}=1\,;\,b_{1}=1

a_{n+1}=a_{n}+b_{n}

b_{n+1}={\frac {b_{n}}{n+1}}

and then a_n converges to e. You might also be able to construct an iterative algorithm from the continued fraction representation of e, which has quite a simple pattern, even though it is not periodic. Another method that can work for some irrational numbers is to find an equation that is solved by the required number, and then apply an iterative method such as Newton's method. Gandalf61 (talk) 06:43, 23 April 2008 (UTC)[reply]

...or, also from the infinites series representation:

a_{0}=1\,;\,b_{0}=1

a_{n+1}=(n+1)a_{n}+1

b_{n+1}=(n+1)b_{n}

and now the ratio a_n/b_n converges to e. With this algorithm you only need to use integer arithmetic until the very last step. Gandalf61 (talk) 07:22, 23 April 2008 (UTC)[reply]

You can use Newton's method for

\ln x-1

to get:

x_{n+1}=x_{n}(2-\ln x_{n})\;\!

Of course, calculating ln is not really easier than calculating e with the usual method. -- Meni Rosenfeld (talk) 09:12, 23 April 2008 (UTC)[reply]

Thanks. Is there any really fast way (like the Gauss-Legendre method for pi)? --wj32 ^t/c 00:35, 24 April 2008 (UTC)[reply]

Okay, so what you really want is an efficient method. This paper describes various algorithms for calculating the value of e - it may have what you are looking for. Gandalf61 (talk) 08:23, 24 April 2008 (UTC)[reply]

Three houses and three different sources

Please help me for finding a solution for the following problem:

Three different houses needs separate connections to their supplies of water, electricity, and gas in such a way that no wires nor pipes cross together.

Is it possible?

Also, is this problem in the scope of topology? Alexius08 is welcome to talk about his contributions. 02:50, 23 April 2008 (UTC)[reply]

It is not possible. Here is a link cut-the-knot.org. HYENASTE 05:11, 23 April 2008 (UTC)[reply]

This is an example of a problem in graph theory or, more specifically, in topological graph theory. Connecting the three utilities to each of the three houses creates the complete bipartite graph denoted by K_3,3. One of the properties of K_3,3 is that it is not a planar graph i.e. it cannot be drawn on a 2-dimensional plane without at least two edges crossing each other. Gandalf61 (talk) 06:22, 23 April 2008 (UTC)[reply]

...then it's a good thing we live in a 3-dimensional world (or 4-, counting time). StuRat (talk) 01:40, 25 April 2008 (UTC)[reply]

Including time is an interesting point - that's what allows us to have level crossings. You could solve this problem in the plane if you allowed the graph to be dynamic (add some kind of switch between different pipes so the utilities can share one route where needed - of course, then you would have a power cut every time you turned the tap on, but that's just the real world getting in the way of the maths!). --Tango (talk) 15:59, 25 April 2008 (UTC)[reply]

We have an article on this very problem: water, gas, and electricity. — Kieff | Talk 06:35, 23 April 2008 (UTC)[reply]

Non-perturbative effects in Euler-Maclaurin expansion

I need to calculate

\sum _{l=1}^{\infty }e^{-al^{2}}\ ,

which is of course a Jacobi theta function. But theta functions are not fit for what I want to do with them afterwards, so I have to work around them. The Euler-Maclaurin expansion seemed to me to be the way of doing this. It gives:

\sum _{l=1}^{\infty }f(l)\sim \int _{0}^{\infty }f(l)dl-{\frac {f(0)}{2}}-\sum _{k=1}^{\infty }{\frac {B_{k+1}}{(k+1)!}}f^{(k)}(0)

if f and all its derivatives go to zero at infinity (which is the case). The first two terms give

{\begin{aligned}\int _{0}^{\infty }e^{-al^{2}}dl&={\frac {1}{2}}{\sqrt {\frac {\pi }{a}}}\\-{\frac {f(0)}{2}}&=-{\frac {1}{2}}\ .\end{aligned}}

This looks like a good start to a Taylor expansion in a. But then I run into problems with the subsequent terms. All odd derivatives of my function are zero in {{{1}}}, and all odd Bernoulli numbers are also zero. There are no more nonzero terms. But when plotting the "series" I already have, I do not find the correct result (it goes to -1/2 for big a, which should be zero, as the theta function is). The difference between what I have and what I should have seems to behave as e^–1/a. What's happening? David 08:55, 23 April 2008 (UTC)[reply]

Shouldn't it be

\sum _{l=1}^{\infty }f(l)\sim \int _{1}^{\infty }f(l)dl-{\frac {f(1)}{2}}-\sum _{k=1}^{\infty }{\frac {B_{k+1}}{(k+1)!}}f^{(k)}(1)

(Where the lower limit of the integral is 1 instead of 0) ? -- Xedi (talk) 14:28, 23 April 2008 (UTC)[reply]

No, definitely not. When computing the sum from 1 to n-1, an integral from 0 to n is needed. (See Euler-Maclaurin formula, where the term in (f(0)+f(n))/2 was brought to the other side.) It's certainly not that which could explain the discrepancy. David

07:36, 25 April 2008 (UTC)[reply]

Compactness in Functional Analysis

There are plenty of examples of compactness in well-known spaces such as $\ell _{p},L_{p}[0,1],L_{p}(R),L_{p}(R^{n}),L_{p}(X)$ where X is a locally compact space. The space of (continuous linear) operators are also well-known. Of course when p=2, it is more interesting. The operators on $\ell _{p}$ are naturally identified as some infinite matrices. As an encyclopedia, does Wiki collect the criteria and/or characterizations of compact subsets of these spaces somewhere (URL) ? Thank you in advance. twma 09:38, 23 April 2008 (UTC)[reply]

Go figure the math game

PLZ CAN U HELP ME OUT BY GIVING ME KNOWLEDGE ABOUT HOW TO DOWNLOAD GO FIGURE MATH GAME. THE VERSION WHICH IS SHOWN IN YOUR SITE.

THNX IF U CAN HELP ME OUT.

SOHIT BAKSHI. —Preceding unsigned comment added by Sohitbakshi (talk • contribs) 10:59, 23 April 2008 (UTC)[reply]

I'd imagine it would be illegal to do so, but you can buy the Microsoft Entertainment Pack (if it still exists) which contains it. x42bn6 Talk Mess 12:06, 23 April 2008 (UTC)[reply]

Ebay or Craigslist (or local equivalent) is where I would start looking. -- 128.104.112.85 (talk) 21:41, 23 April 2008 (UTC)[reply]

hi

If a 10-pound sack of potatoes costs $6.55, how much would 1 pound cost? Round your answer to the nearest cent. Would it be 0.65 or 0.66?? —Preceding unsigned comment added by 67.55.5.83 (talk) 19:57, 23 April 2008 (UTC)[reply]

You can take a look at rounding, but really it's going to depend on what you have been taught in class. --LarryMac | Talk 20:05, 23 April 2008 (UTC)[reply]

I was taught that 5 rounds up, so it would be 0.66, but I've never been convinced of the merit of that convention, myself. A 5 follow by anything other than 0's should certainly round up, but just a 5 is bang in the middle. --Tango (talk) 20:17, 23 April 2008 (UTC)[reply]

I rather like the convention that you round in the direction to make the final digit even, i.e. ...55 -> ...6 and ...45 -> ...4, so that any subsequent division by 2 doesn't re-introduce the problem. —86.132.239.122 (talk) 20:31, 23 April 2008 (UTC)[reply]

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn how to solve such problems. Please attempt to solve the problem yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Thank you. Adam (Manors) 22:16, 23 April 2008 (UTC)[reply]

Also, most greengrocers charge less per pound/unit in "bulk" quantities (10 pounds of potatoes or a package of 12 oranges for example). I wouldn't be surprised to to find the cost of un-sacked potatoes to be 70¢ to 75¢ per pound while the 10 pound sack costs $6.55. Additionally, I doubt that any vendor would round down. --hydnjo talk 22:34, 23 April 2008 (UTC)[reply]

We're trying to do abstract reasoning here in our ivory tower. You can't just waltz in and confuse us with facts from this so-called "real life" of yours. :) -- Meni Rosenfeld (talk) 23:01, 23 April 2008 (UTC)[reply]

My tower is made of renewable resources, not ivory. I choose to make my tower out of the souls of my enemies, since there will always be more of those. -mattbuck (Talk) 23:14, 23 April 2008 (UTC)[reply]

Ivory is a renewable resource, if you maintain your own herd of elephants, and harvest ivory at no faster than its replacement rate. Unregulated hunting of wild elephants, of course, is unlikely to be sustainable due to the Tragedy of the Commons. See also: Overfishing. I would imagine that advances in biotechnology will enable ivory to grow in nutrient solutions someday, allowing for much higher efficiency at converting feed into ivory, compared to using elephants, which are very large compared to their tusks. An elephant is an inefficient device for manufacturing ivory, not to mention slow, and it generates a lot of manure. Although you might appreciate the manure if you practice organic farming. Elephants can be handy for moving logs. --Teratornis (talk) 04:47, 26 April 2008 (UTC)[reply]

Manors, this template message isn't really appropriate here. Yes, this is probably a homerwork problem, but (unless the "Would it be 0.65 or 0.66?" is a part of the question statement) it's not like the OP asks us to solve anything. He was just asking about the convention regarding ambiguous rounding. -- Meni Rosenfeld (talk) 22:58, 23 April 2008 (UTC)[reply]

Agreed, the OP has narrowed it down to two possible answers and is, understandably, unsure which is correct. It's a perfectly legitimate question for this page. --Tango (talk) 23:21, 23 April 2008 (UTC)[reply]

So with the above settled, the convention I learned in high school (especially in Chemistry, but in math as well) is that you would round a 5 up. I think the reason for this is because the 5 numbers 0,1,2,3,4 all round down, and the four numbers 6,7,8,9 clearly round up, so 5 would seem to belong to the smaller group, though the choice is completely arbitrary. Since 0 really doesn't round. Significant figures will answer your question in a much more general case. (I think) A math-wiki (talk) 06:09, 24 April 2008 (UTC)[reply]

The reason we round .5 to 1 is that while .5 is equally-distant from 0 and 1, .51, .501, .500000001 etc are closer to 1 than to 0. So using the rule that .5 rounds to 1, you only have to look at one digit, as in the case of any other rounding problem. I do like the rule of rounding .5 up half the time, and down half the time, but since I'm an engineer and therefore believe that it is going to be extremely rare to have anything ever be exactly .5, rounding .5 up to 1 is a good rule. moink (talk) 19:29, 24 April 2008 (UTC)[reply]

The problem with that argument is, OK, so you have an approximation ending in a 5, that you have to round to one less digit. Why do you think the true value is more likely to end in 50123... than, say, 49876...? --Trovatore (talk) 07:16, 25 April 2008 (UTC)[reply]

Well, that assumes there's some sort of sequential rounding going on. That the number you're looking at has been rounded first to .5 and then you round it to 1. That would be a bad idea, unless you're trying to do quick math in your head (in which case 2 is 1, 3 is 1, pi is 1, and pi squared is 10). moink (talk) 17:27, 25 April 2008 (UTC)[reply]

There's pretty much guaranteed to be sequential rounding involved, if the number you start with is a measured value -- you rounded it the first time when you read it off the gauge. --Trovatore (talk) 17:49, 25 April 2008 (UTC)[reply]

"Round your answer to the nearest cent." Sounds like homework to me, but I digress. Going by what I have been taught the answer would be 0.66. Adam (Manors) 14:45, 25 April 2008 (UTC)[reply]

We're not denying that it is homework, but read the guidelines more carefully:

Do your own homework. The reference desk will not give you answers for your homework, although we will try to help you out if there is a specific part of your homework you do not understand. Make an effort to show that you have tried solving it first.

The questioner has a problem with a very specific part of his homework, and I reckon it's ok to give an answer to that. He's got to one of two answers, so has made some effort, and the answer is rather ambiguous and slightly confusing to someone new to it. -mattbuck (Talk) 14:51, 25 April 2008 (UTC)[reply]

It's not so much that the answer is confusing - the answer is completely arbitrary. There's no way for the OP to work out the answer, someone just has to tell them what the convention is (their teacher would be the obvious person, but we can do it just as easily). --Tango (talk) 15:56, 25 April 2008 (UTC)[reply]

Didn't I pretty much say that in the first answer? --LarryMac | Talk 17:35, 25 April 2008 (UTC)[reply]

Yeah, give or take. --Tango (talk) 18:31, 25 April 2008 (UTC)[reply]

Yes LarryMac you did. However, if this is a question from the son of a greengrocer who is trying to "learn" the business and is also trying to one-up his dad then these abstracts are merely a diversion. Without context any of the above may meet the OP's needs. If this is homework then those who assumed so (LM in particular) are quite correct but if it's "real world" then the spectrum of responses wouls be richer indeed ;-) hydnjo talk 19:02, 25 April 2008 (UTC)[reply]

Another reason to round .5 to 1 is if you round it down, numbers ending in 01234 or 5 will go down and 6789 will go up, that means in a random set of numbers 3/5 will round down and 2/5 up. Rounding 5 up splits it evenly so that in a long enogh sequence of random numbers the errors should cancel out. Mad031683 (talk) 20:43, 25 April 2008 (UTC)[reply]

Only in the case where the numbers all have exactly one more digit than you want - in most real world cases, the numbers are far longer than required so you virtually never have the case of a 5 followed by only 0's. Even if you only have 2 digits more than required the odds go to 51/100 and 49/100, which isn't a large discrepancy. The fact that rounding 5's up results in an upwards bias on the mean is more significant, and is why in cases where it matters (accountancy, for example, where numbers often are only 1 digit longer than required) people use the round-to-even method. --Tango (talk) 21:24, 25 April 2008 (UTC)[reply]