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May 30[edit]

2 cos^2 x = 1[edit]

i know how 2 solve if its 2 cos x = 1 (cos x = 1/2, then cos^-1) but how to solve when its cos^2? cos^2 x = cos x^2 or = (cos x)^2 or what? —The preceding unsigned comment was added by 202.172.220.10 (talk • contribs) 09:04, May 30, 2007 (UTC) – Please sign your posts!

First I'll answer your question about notation. Whenever you have a superscript number on a trigonometric function it means the result of the function is being raised to some power. Thus,

\cos ^{2}x=(\cos x)^{2}

. So to solve the equation

2\cos ^{2}x=1

,

you have to follow steps similar to the ones you described, but with one extra step in the middle to get rid of the exponent. nadav (talk) 09:23, 30 May 2007 (UTC)[reply]

thanks. so 2 cos^2 x = 1 2 (cos x)^2 = 1 (cos x)^2 = 1/2 cos x = 1/root 2 & i know cos 45 = 1/root 2 so x = 45. correct? —The preceding unsigned comment was added by 202.172.220.10 (talk • contribs) 09:37, May 30, 2007 (UTC) – Please sign your posts!

You should add the degree symbol, as in 45°, to make clear that you are not using radians. x = 45° is one solution, but there are several more! --Lambiam ^Talk 09:52, 30 May 2007 (UTC)[reply]

you beat me in time, so i won't add my answer. Cthulhu.mythos 09:55, 30 May 2007 (UTC)[reply]

what are the others? —The preceding unsigned comment was added by 202.172.220.10 (talk • contribs) 09:56, May 30, 2007 (UTC) – Please sign your posts!

The cos^a x notation is disgusting, by the way.

—The preceding unsigned comment was added by 149.135.29.71 (talk • contribs) 10:32, May 30, 2007 (UTC) – Please sign your posts!

First, let's put c = cos x. You solve c² = 1/2 by c = 1 / √2. Think: is this the only root of this quadratic equation in c? Then, in solving cos x = y, you use x = cos⁻¹ y. (I prefer the notation x = arccos y.) Again, is this the only solution? Take a look at the graph of the cosine function (note that π = 180°), and search for which values of x the value of cos x is equal to 1 / √2 ≈ 0.7071. For the general case, see the section General solutions in our article Inverse trigonometric function. --Lambiam ^Talk 10:43, 30 May 2007 (UTC)[reply]

[Edit conflict] Cosine is both periodic (cos(θ+2πn) = cos(θ)) and even (cos(−θ) = cos(θ)), which gives many options. Another approach might also be of interest: Use the fact that cos(2θ) = 2cos²(θ)−1 The answer(s) should be the same with either approach, so this gives a way to check your work. --KSmrq^T 10:54, 30 May 2007 (UTC)[reply]

First, let's put c = cos x. You solve c² = 1/2 by c = 1 / √2.

Eh? There are two solutions to the above equation.

Solution 1 : c = +1 / √2

solution 2 : c = -1 / √2

For each solution, there are infinite number of valid values for x (as in c = cos x).

202.168.50.40 22:34, 30 May 2007 (UTC)[reply]

202.168, as has been pointed out before, giving away answers doesn't help anyone. The OP will have a much richer learning experience if he thinks about the solution himself, and you can rest assured that the respondents are aware of the answer... -- Meni Rosenfeld (talk) 22:56, 30 May 2007 (UTC)[reply]

Summation of a vanishing series[edit]

My own response to a question above has got me thinking. Is anyone aware of a meaningful way to assign values to infinite sums $\sum _{n=0}^{\infty }a_{n}$ , where $\lim _{n\to \infty }a_{n}=0$ , but the sum does not exist in the classical sense? The values will, of course, not be real numbers, but I would hope them to form an ordered field, and hold some natural properties such as coincidence with the classical sum when it exists, linearity, additivity and so on. Thanks. -- Meni Rosenfeld (talk) 14:38, 30 May 2007 (UTC)[reply]

I have a distant memory in Gouvea's p-adic Numbers of divergent series that converge p-adically. I claim no understanding though. iames 14:53, 30 May 2007 (UTC)[reply]

Yes, unfortunately, this is a distinct problem from what I am considering here (the p-adics are an alternative way of proceeding from rational numbers, with a different notion of distance; Here I'm aiming at proceeding further from the reals). -- Meni Rosenfeld (talk) 14:58, 30 May 2007 (UTC)[reply]

Also distinct but interesting are Banach limits. Under the topic of summability theory, see Divergent series -- I think Cesaro and Abel sums are the closest to what you seek. Or try the good people at sci.math. iames 15:32, 30 May 2007 (UTC)[reply]

I will definitely study the linked articles, thanks. I will be grateful for any additional input in the meantime (since this could take a while). -- Meni Rosenfeld (talk) 22:40, 30 May 2007 (UTC)[reply]

I remember attending a course on something called "resurgent series"; while I do not remember what it was about in detail, it had something to do with dealing with assigning a "reasonable" limit to function series which do not converge. But this is all I remember, and there is no Wikipedia page about this. You could try Google. Cthulhu.mythos 09:59, 31 May 2007 (UTC)[reply]

Added to the list. Thanks. -- Meni Rosenfeld (talk) 16:50, 1 June 2007 (UTC)[reply]

stuf[edit]

how many sq.acres are in a sq.mile —The preceding unsigned comment was added by Sivad4991 (talk • contribs).

1 square mile is equivalent to 640 acres. x42bn6 Talk Mess 18:49, 30 May 2007 (UTC)[reply]

An acre is a measure of area, so a "sq. acre" is redundant. Or maybe it's a cube. Also, Google's calculator is always good for such questions, just type "acres in 1 square mile" into the search box and it will give the result. --LarryMac | Talk 19:04, 30 May 2007 (UTC)[reply]

1 mile^4 = 409600 square acres. Tesseran 07:05, 31 May 2007 (UTC)[reply]

The magic of hypercubes. Black Carrot 05:46, 1 June 2007 (UTC)[reply]

Hash collisions[edit]

Birthday attack gives a formula for the number of inputs that must be tried before a hash collision is generated, with a certain probability and hash length. (For example, to have a 50% chance of finding an input that produces the identical MD5 - which is 128 bits - as input X you must try 2.2 × 10^19 random inputs.) I was actually interested in adding a formula to hash collision describing the probability, given a certain number of random inputs and a hash length, that any two outputs will collide. For example, if I have 200,000 random inputs, what are the chances that any two of them will have the same 128-bit MD5? -- Beland 20:15, 30 May 2007 (UTC)[reply]

I've modified a bit of the content to Birthday attack. Is the math clearer? It would seem you have the answer, assuming the article's veracity, you just didn't understand it ( which is understandable considering its previous odd wording). You want p(n), not n(p). If MD5 can come up with any and all 128 bit combinations, then 2^128 is your H. Root⁴(one) 22:00, 30 May 2007 (UTC)[reply]

I suppose I should also point out here I am also assuming MD5 output is randomly distributed as well, which might be a big assumption. Root⁴(one) 22:15, 30 May 2007 (UTC)[reply]

I haven't done a randomness analysis, though a colleague at my former job (cryptographic library implementations) rejected the full outputs of MD5 and SHA1 for evenly distributed RNGs after checking the distribution with a couple of thousand outputs. We settled on cutting out a piece "in the middle" for the random generator we wanted at that point. So, I'm pretty certain that it's a very big assumption that MD5 distributes according to some nice random distribution. Michiexile 09:57, 3 June 2007 (UTC)[reply]

Math/physics help needed -- # of Professional Lights off of a Circuit Breaker?[edit]

I'm not sure if this is one for the Science desk or for here. I guess we'll try here first. -- To make a long story short, I need to figure out how many professional lights I can run off of a circuit breaker in a house without overloading the circuit breaker. The circuit breaker seems to be rated for 200 amps. The technical specifications for the lights I'm working with - available here - are 650 watts off of a 230 Volt power supply.

Keeping in mind that I have been out of high school physics for all too many years -- according to the volt article, volt = watt / amp. This led me to come up with the following two equations:

$volt=(watt*n)/amp$

$n=(amp*volt)/watt$

Where n is the number of lights. Plugging in the numbers I mentioned, I ended up with an answer of 70.769 lights - provided that there are no other demands on the circuit breaker. Can anyone tell me if I managed this correctly? Thanks! --Brasswatchman 20:45, 30 May 2007 (UTC)[reply]

Your math looks right. Though usually in household wiring, there is a main circuit breaker which feeds into multiple 15-20 amp circuits, so you may need to spread the lights across several smaller circuits. These sounds like hot lamps, so be careful about not catching nearby objects on fire. Presumably the manufacturer has more specific recommendations. -- Beland 20:52, 30 May 2007 (UTC)[reply]

I would work it like this:

Each light consumes 650W / 230V = 2.826 Amperes

The fuse can handle a maximum of 200 Amperes.

So the maximum number of lights is 200A / 2.826A = 70.77 lights

WARNING!!! I have never heard of a 200 Amperes fuse @ 240 Volts for domestic household usage. I sincerely think that you have made a mistake somewhere. If you are not a qualified electrician then you may probably burn the whole house down as a result of an electrical fault.

PS If you are a qualified electrician and you ask such question on Wikipedia reference desk then I would recommend that your practicing licence be stripped from you immediately.

202.168.50.40 22:22, 30 May 2007 (UTC)[reply]

I'd be very careful before depending on advice given here. I can't speak for all of us, but I myself have no idea about this stuff. Can we assume, for example, that you can just add up the voltage of the individual lights to get the total voltage? If I recall from high school, this is not generally true. I would ask this question at the Science desk. But even then, I think it would be much wiser ask a qualified expert on these matters. nadav (talk) 22:36, 30 May 2007 (UTC)[reply]

Do not request medical, legal, or electrical advice. Ask a doctor, a lawyer, or a qualified electrician instead :-).

Seriously, though, as others have pointed out, you shouldn't do anything risky based solely on the advice you get here. While the arithmetic seems to hold up, any number of things can go wrong...

Oh, and since the lights are assumed to be connected in parallel, it is their current that adds up, not their voltage. -- Meni Rosenfeld (talk) 22:49, 30 May 2007 (UTC)[reply]

This is not advice. I just want to discuss this as a theoretical problem. First, the main breaker to the house may be 200 amps, but then there are many individual breaker circuits attached to that. The big one is just the main switch and 200A is a common main breaker for households. However, you should not try and rig up your lights so that they all run 200A through wires coming off of the main switch. That would be suicide. If you try and hook up a whole bunch of lights in one area of the house (from outlets), you'll throw one of the individual 20A or so breakers. Theoretically, the only way you are gonna hook up 70 lights in one area is if you run extension cords from all over the house from each of the individually broken circuits. Don't do that. Also, are you sure the outlets are 240V? Is there an air conditioner or water heater that is active? These are things to consider. About the calculations: They don't take into account all of the resistance that will come from varying lengths of wire throughout the house and the wire running from the outlet to the fixture. More resistance = more current required to meet the demands of the bulb. This means one would have to estimate the effect of this resistance and would have to greatly underestimate the number of lights that could actually be hooked up. So, don't they make generators for this sort of thing?Mrdeath5493 18:01, 31 May 2007 (UTC)[reply]

Yes, but that would require more money than I have. :) Anyway, just to clarify here: I'm *NOT* planning to hook up 70 or so lights within this one house. That would be... not smart of me. I'm just trying to figure out a rule of thumb to keep me from blowing (literally) this particular shoot. Thanks for the warning about the 20A breakers - so that means that I can safely hook up, what, a maximum of three or so 650 W lights to each breaker? Probably keeping it down to one or two 650W lights per breaker to be safe (and even less if the breaker has some other appliance attached to it)? - Yes, I think the outlets in this particular house are standard for American homes - which would be 240V, right? -- Of course I'm not a certified electrician. I'm just a filmmaker, trying to figure out a way to handle a low-budget shoot. Thanks for your help, everyone. Appreciate it. --Brasswatchman 18:17, 31 May 2007 (UTC)[reply]

Standard American outlets deliver 110-120V. Most outlets that provide more are specially-shaped so as to not toast normal appliances. --Tardis 18:49, 31 May 2007 (UTC)[reply]

Standard European outlets are nominally 230V. In reality, UK electricity tends to be 240V and continental europe thens to be 220V for historical reasons - it just gets called 230V. Looking above, the sum 650W/240V = 2.7A is the easiest way to do the calculations - thus a 20A outlet will supply 7.4 lamps. -- SGBailey 22:32, 31 May 2007 (UTC)[reply]

I don't understand the technical specifications: "Lamp type: ... 120V ... / Supply Voltage: 230V". Are there two lamps in series in each lamphead? If the required voltage of the power supply is 230V, and you hook them up to house outlets delivering the standard American household voltage of 120V, then it's just not going to work anyway, even if you don't electrocute anybody and don't blow up the house or any breakers or fuses. Don't you have a gaffer for the shoot who is familiar with the local conditions? --Lambiam ^Talk 15:33, 1 June 2007 (UTC)[reply]