October 24[edit]

A flyer from Scientific American invites me to buy books about 0, φ, π, e and i. It says that π was mentioned in testimony in the murder trial of O. J. Simpson. Anyone know how? —Tamfang 02:31, 24 October 2007 (UTC)[reply]

I found it with a little Googling. Search "FBI" in [1]. It quotes The Joy of Pi by David Blatner. An FBI Special Agent is quizzed on pi and the area of a circle. His suggestions for pi were 2.12 and 2.17. Judge Ito got closer with 3.1214. PrimeHunter 22:17, 24 October 2007 (UTC)[reply]

At [2], in the upper left hand corner, there's a picture of a crumpled up surface. What is it? Black Carrot 05:44, 24 October 2007 (UTC)[reply]

I believe I've seen that same picture labeled as a minimal surface, but I can't find a reference. 75.3.82.67 06:30, 24 October 2007 (UTC)[reply]

It is indeed: namely a gyroid, made by Bathsheba Grossman. I have one beside my monitor. —Tamfang 08:42, 24 October 2007 (UTC)[reply]

Is it really metal? does it weigh a bit.. Wow - has anyone got any links to how these are made - it's not immediately apparent how I would print with metal - are chemicals involved - or some sort of ion beam?87.102.94.157 12:58, 24 October 2007 (UTC)[reply]

These things can be made with Rapid prototyping systems. Basically the shape is built up lay by layer using special chemicals which are hardened by lasers (or similar). The end result is a plastic model which can then be used to cast a metal object. An alternative could be CNC (computer numerical control). --Salix alba (talk) 15:38, 24 October 2007 (UTC)[reply]

Actually it might be done with Direct metal printing, Direct metal deposition is the closest WP article. --Salix alba (talk) 15:42, 24 October 2007 (UTC)[reply]

the artist's description of the technology. I have two of her pieces: the mini gyroid weighs perhaps an ounce, and the mini Nexus a bit more. They're quite sturdy. —Tamfang 16:22, 24 October 2007 (UTC)[reply]

Thanks those links answered it for me.. ... from the picture it looked like the thing was six inches across and weighed about a kilo... (not 28g)87.102.94.157 19:16, 24 October 2007 (UTC)[reply]

Please help me in knowing the process of writing a number as sum of four squares ( if there is any ) according the Lagrange's theorem. Will such splitting is unique? Also let me know the way of writing the following numbers as the sum of four squares.

33102; 33215; 103993; 104348 Kasiraoj 13:29, 24 October 2007 (UTC)[reply]

I think the proof of Lagrange's four-square theorem given here is constructive, and can be turned into a method of finding a four squares representation of any positive integer. The representation is not usually unique, even if we disregard differences of order, because identities such as

${\displaystyle 7^{2}+4^{2}=8^{2}+1^{2}}$

often allow you to replace one pair of squares with another pair. If you want a simple pragmatic method of finding a representation, I would try a greedy algorithm approach - start with the largest square less than n, then take the largest square less than the remainder, and see if that leaves you with something that is obviously a sum of two squares. If not, backtrack and try a smaller square on either of the first two steps. For example:

${\displaystyle 33102-181^{2}=341}$

${\displaystyle 341-18^{2}=17}$

${\displaystyle 17=4^{2}+1^{2}}$ Gandalf61 16:34, 24 October 2007 (UTC)[reply]

The applet at [3] does it, explained at [4]. PrimeHunter 22:01, 24 October 2007 (UTC)[reply]

What are MCA's? Are they worth while for most home owners or are they only good in a few situations?

Thanks

Don Dutton —Preceding unsigned comment added by 130.76.32.182 (talk) 14:27, 24 October 2007 (UTC)[reply]

A quick google search brings up lots of relevant results. The short version is that your paycheck deposits directly against your mortgage (instead of into a traditional checking account) and you're allowed to make withdrawals against your mortgage. Money that would otherwise be unspent counts toward your mortgage, paying it off faster (and reducing the interest you're paying). Additionally, there are suggestions that you can fiddle with credit card balances and such to gain an edge in interest accumulation. A significant caveat is that the system demands careful money management, particularly with regard to long-term savings. As for their overall merit, you would be well-advised to consult a professional. — Lomn 15:02, 24 October 2007 (UTC)[reply]

I wouldn't do something like that. Unless you have a very high mortgage rate, the benefits of the liquidity of cash-equivalent investments such as CDs outweigh the savings of paying down a mortgage in most circumstances. In a declining real estate market, it especially makes little sense to focus on paying down a mortgage ahead of schedule as those who are in danger of being underwater on their mortgage are also unlikely to be able to pay off the mortgage at a sufficient rate to fix their situation. I would imagine that in an MCA situation the bank is going to look very unfavorably at someone's mortgage balance exceeding the market value of the home so you could see the house next door being sold in a foreclosure auction quickly eliminating any savings that you might have managed to store in an MCA. Having a stash of ready cash will be helpful for handling such situations as getting into an apartment (between security deposit, first/last month's rent and moving costs, it can easily be $5-6K to move into an apartment in Los Angeles). Basic money management would put paying down the mortgage at the bottom of the priority list. Donald Hosek 16:33, 24 October 2007 (UTC)[reply]

I performed the same Google search as Lomn, and after seeing a lot of words but not a lot of information, I added the word "scam" to my search. It seems that a lot of money is being charged for these accounts, and there are also a lot of "paid referrers". I don't have any statistics at hand, but I'd guess that most mortgage loans these days allow additional payments of principal at any time, without resorting to some "new" scheme that has a bit of on odor to it. --LarryMac | Talk 17:52, 24 October 2007 (UTC)[reply]

First of all, I'm not sure if this is the right place to ask this. But here it is anyway.

Wikipedia can be represented as a graph (where pages are vertices, and links are edges). Has anyone ever done any research about wikipedia in terms of graph properties? For example, it would be interesting to know whether the English wikipedia has more than one nontrivial connected component. Since it's more closely represented by a digraph, maybe "strongly connected" would be a better term. I find graph theory quite interesting, and I wonder if this type of analysis has ever been done with wikipedia. This sparked my interest when I heard about a game in which you pick any two topics, and try to find the shortest path from one to the other by only clicking links in the article. --BennyD 15:15, 24 October 2007 (UTC)[reply]

• Hello BennyD, did you try this one : wikimindmap ? Just take a look and tell me if it helps. -- DLL ^{.. T} 17:11, 24 October 2007 (UTC)[reply]

Well there is Special:Lonelypages which are pages not linked from anywhere else. Whether there are isoleted components with more than one node would take a fair bit of work on a database dump to find out. Wikipedia:Six degrees of Wikipedia may be of interest showing that the large mass is very well connected. --Salix alba (talk) 23:23, 25 October 2007 (UTC)[reply]

Someone did a similar analysis using the pages of the original wiki as nodes in a graph -- see WikiWikiWeb:WikiLinkStructureAnalysis. Hope that helps. --75.19.73.101 09:24, 27 October 2007 (UTC)[reply]

I have to work out Taylor series of various functions, and there's one that seems a little tricky. It's not that I can't do Taylor series, but the only way I can think of to differentiate ${\displaystyle {\frac {1}{2+3x}}}$ is to use the quotient rule, which works fine for the first differential, but is totally unwieldy for the 3rd (having to square a massively long polynomial is a complete pain). To make things harder, I'm not making the series around 0, so I can't cancel out all the x terms later. Is there a simpler way to differentiate this function (it seems really simple, so I'm sure I'm missing something obvious...) Laïka 16:49, 24 October 2007 (UTC)[reply]

Firstly, it's easier to use the chain rule rather than the quotient rule for this: the derivative of 1/f(x) is -f'(x)/(f(x)^2) (of course, the quotient rule gives the same answer). Secondly, don't expand the polynomial in the denominator. So the first derivative is ${\displaystyle {\frac {-3}{(2+3x)^{2}}}}$, and the 2nd derivative is ${\displaystyle {\frac {-3*-3*2}{(2+3x)^{3}}}}$ - can you take it from there? AndrewWTaylor 17:14, 24 October 2007 (UTC)[reply]

Differentiation is the hard way to find the Taylor series of this function. I'll give a hint for the easy way: The Taylor series of ${\displaystyle {\frac {1}{1-x}}}$ around 0 is ${\displaystyle \sum _{n=0}^{\infty }x^{n}}$. -- Meni Rosenfeld (talk) 17:21, 24 October 2007 (UTC)[reply]

Thanks; I'd tried the chain rule but a stupid mistake in my working screwed up the answer; I knew there was something obvious I'd missed. Laïka 19:46, 24 October 2007 (UTC)[reply]

This is probably a bit of a dunce-ish question but never mind. If I know the cross section of a river, assumed constant, and the speed of the water flowing in the river, again assumed constant, how can I estimate the volume of water flowing in a given time frame. (This is a homework question but I tried to be vague. If it doesn't make sense as it is, I can give more specific info.) Thanks asyndeton 20:08, 24 October 2007 (UTC)[reply]

What do you get if you multiply the area of the section by the speed? Note the units. —Tamfang 20:24, 24 October 2007 (UTC)[reply]

The speed is m/s, the area is m^2, so I would get m^3/s? And then I just mulitply by, in my case, 60 to get the volume in a minute. Thanks asyndeton 20:44, 24 October 2007 (UTC)[reply]

Yep. It's somewhat impressive how many problems reduce to simple multiiplications and divisions once you look at the units. Donald Hosek 23:12, 24 October 2007 (UTC)[reply]

A tad embarassing really. But then again, I suppose we all learn from our mistakes. asyndeton 23:14, 24 October 2007 (UTC)[reply]

Blindly trusting the power of dimensional analysis is dangerous, as there might be other physical quantities or constants effecting the problem, the units of which can change the result. I will always remember a book I've read which uses dimensional analysis to "prove" that drag in a liquid is proportional to the square of the speed, where in practice the relation is much more similar to linear. The discrepancy is due to ignoring a unit-bound property of the liquid.

This particular problem can be solved easily without resorting to such unsound methods. -- Meni Rosenfeld (talk) 12:24, 25 October 2007 (UTC)[reply]

Well I'm looking to learn. How would you go about it? asyndeton 12:26, 25 October 2007 (UTC)[reply]

But knowing about that constant (and its units) again makes it clear what the formula is. Dimensional analysis isn't a universal solution, by any means, but it solves a very large number of problems. It's been a couple decades since I've taken freshman physics, but as I recall, knowing the units on the constants revealed many of the formulae that were necessary in solving the problems. Maybe one of these days I'll find an updated copy of H&R and re-learn all that physics I've since forgotten. (And yes, I know that dimensional analysis didn't solve all the problems, just a surprisingly large number of them... and for the high school students out there, it's an indispensable tool for the ACT/SAT). Donald Hosek 16:34, 25 October 2007 (UTC)[reply]

But the truth is that dimensional analysis never works, in the sense that it never guarantees that some formula is correct. Textbooks might give a biased picture of its utility by only using it on those occasions when they know a priori that it will happen to lead to the correct result. Even in the best possible scenario, when you are absolutely certain that you have isolated all the relevant quantities, and they are few enough to force a formula by dimensional analysis, it is only guaranteed to be correct up to a constant factor (which is useless if we wish to make quantitative calculations).

Of course, it is still valuable as a sanity check for a result obtained by other means. -- Meni Rosenfeld (talk) 20:39, 25 October 2007 (UTC)[reply]

Sanity check, right. The truth is that checking units discovers many errors. The pendulum formula problem: find a combination of mass (M), lenght (L), and acceleration (g=L·T⁻²) having the dimension of time (T). The solution is M⁰·L^1/2·g^−1/2, showing that the period (T) is independent of the mass (M) and proportionate to the square root of the length (L^1/2). This result did not require solving a differential equation. Bo Jacoby 21:20, 25 October 2007 (UTC).[reply]

And what makes you so confident that the period is independent of the amplitude (which it isn't, by the way)? I could just as easily say the solution is ${\displaystyle A^{1/2}g^{-1/2}}$ or any other crazy combination.

This demonstrates my point perfectly. If you already know the answer, then you can selectively pick only those quantities which you know end up effecting the result, and then demonstrate how dimensions miraculously produce the correct formula (up to a constant). But if you genuinely don't know the result, there are too many potential factors to consider to make this possible. -- Meni Rosenfeld (talk) 21:36, 25 October 2007 (UTC)[reply]

There are several equivalent ways to think about it - one is to consider the "end" of the river, where the water enters the ocean. The volume of water flowing in a given time frame (${\displaystyle \Delta t}$) is the volume of water exiting the river in that period. The water that will exit the river from now until ${\displaystyle \Delta t}$ from now is exactly the water that can reach the end in time less than ${\displaystyle \Delta t}$, and since it flows at speed v, this is exactly the water at distance up to ${\displaystyle v\Delta t}$ from the end - and it forms a prism of cross section S and height ${\displaystyle v\Delta t}$. The volume of this is ${\displaystyle Sv\Delta t}$. -- Meni Rosenfeld (talk) 12:38, 25 October 2007 (UTC)[reply]