Wikipedia:Reference desk/Archives/Mathematics/2007 May 18

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May 18

e ^ i*π

Why in the world would this combination of iconic numbers wind up being equal to -1? Is this proof of God? Gbgg89 01:42, 18 May 2007 (UTC)

It is rather elegant! </understatement> It really just falls out as a special case of Euler's formula. You might find the Euler's identity article illuminating. (I fixed the above sig.) Icthyos 01:42, 18 May 2007 (UTC)

Thanks for fixing my signature and for the info! I am 11 years old and find this sort of thing very interesting. Gbgg89 01:48, 18 May 2007 (UTC)

This formula is one of the reasons I fell in love with mathematics. It is incredibly beautiful, and it unites so many fields of mathematics into one single statement. There are other elegant identities involving e and π, but none is quite like this. nadav 02:17, 18 May 2007 (UTC)

It can also be rendered as e ^ i*pi + 1 = 0, thus uniting 5 fundamental concepts, not just 4. JackofOz 02:55, 18 May 2007 (UTC)

e^(i*Pi) + 1^inf = 0/C , thus uniting 7 fundamental concepts.

C being the speed of light, I presume? JackofOz 05:08, 18 May 2007 (UTC)

That's weak. Here's another equation uniting 6 fundamental concepts (what does 1^inf mean? inf is not a number): 0/e + 0/i + 0/pi + 0/1 + 0/c = 0. fixed

Indeed; especially since 1^∞ is an indeterminate form. There's nothing interesting about adding in additional constants - they can be stripped away or added in arbitrarily. The initial equation is as reduced as possible...depending on your preference for the right hand side. Icthyos 09:57, 18 May 2007 (UTC)

The expression above is not an equation — no equals sign in it. --CiaPan 10:10, 18 May 2007 (UTC)

True, but I'm guessing the intention was to equate it to zero, since that is what the expression is equal to. I'm rather dubious about calling the speed of light a mathematical constant, too. Icthyos 10:23, 18 May 2007 (UTC)

It was equated to -1 ("Why in the world would this combination of iconic numbers wind up being equal to -1?"), which is what it does equal. Zero only comes into the picture if you transfer the -1 to the left hand side, as per my alternative. -- JackofOz 12:09, 18 May 2007 (UTC)

I think CiaPan was referring to "Here's another equation uniting 6 fundamental concepts (what does 1^inf mean? inf is not a number): 0/e + 0/i + 0/pi + 0/1 + 0/c, which initially was not equated to anything. Icthyos 12:30, 18 May 2007 (UTC)

If you set up your definitions right, then $e^{i\pi }=-1$ is part of the definition of $\pi$. The miracle then is that this arbitrary number has something to do with circles. Algebraist 12:39, 18 May 2007 (UTC)

That is really intense. I wonder if anyone can find a connection between either $\pi$ or $e$ and the golden ratio $\varphi$? —Preceding unsigned comment added by 69.241.236.12 (talk • contribs) 21:10, 2007 May 18

The golden ratio, Φ = (1+√5)/2, appears in a regular pentagon, so we have an easy connection:

$\Phi =e^{\mathbf {i} \pi /5}+e^{-\mathbf {i} \pi /5}.\,\!$

This again exploits Euler's formula. --KSmrq^T 22:06, 18 May 2007 (UTC)

That isn't really remarkable, though. Any complex number is expressible as r exp(i t) -- but what is remarkable is the underlying fact that exponential function should have anything at all to do with circles, of all things.

Yes, the relationship between the exponential function and circles is very surprising. But the fact that the golden ratio appears in the regular pentagon, which is what this formula embodies, is also interesting to many people. nadav 07:36, 19 May 2007 (UTC)

Unfortunately, the only really interesting thing about $\varphi$ (to me at least) is that it's the least-well-approximated-by-rationals real. Challenge: can anyone find a simple expression relating $e$, $\pi$, and my own favourite number, $\omega _{1}$? Algebraist 13:01, 19 May 2007 (UTC)

I have to admit that I have never really understood why Euler's identity is considered so great. It seems to just be a consequence of a definition. Since we defined exponentiation with complex exponents in such a way, we happened to get this identity. It doesn't seem to be surprising, in the sense that the fundamental theorem of calculus is (it's quite astonishing). It doesn't seem to be two well-understood but at first glance largely unrelated areas that turn out to be strongly connected in surprising ways. It just seems that we defined e^z in such a way that we got this result rather trivially, rather than it being an amazing connection of concepts that was waiting for us to discover it. That said, I'm happy to be enlightened on why so many think it's special. Is there more to Euler's identity than meets my eye? Maelin (Talk | Contribs) 10:25, 20 May 2007 (UTC)

Well, if the definitions were just pulled out of thin air and this identity fell out, it wouldn't be as impressive. From your point of view, the fact that the definition of complex exponentiation happens to give you repeated multiplication (for appropriate values) is pretty miraculous. Or I'm wrong. 81.183.151.25 13:20, 20 May 2007 (UTC)

Yes. One common method is to define e^z by its power series, then defining $\pi$ as the least positive real such that $e^{i\pi }=-1$. With these definitions, Euler's identity is trivial, but the fact that $\pi$ is connected with circles and $e^{z}$ corresponds somehow to regular exponentiation by integers are both reasonably miraculous. Algebraist 20:45, 20 May 2007 (UTC)

I must incorporate the fantastic oxymoron "reasonably miraculous" into my writing - seriously. Something that has been observed as a fact in the natural world either can be explained by science/reason, or it can't. If not, it's called a miracle. But mathematicians have a way of using words in specialised (often unintuitive) ways, so maybe "miraculous" has a different meaning here. Are there degrees of miraculosity, mathematically speaking? JackofOz 00:46, 21 May 2007 (UTC)

The fact that pi has anything to do with circles is pretty obvious. But that exp(z) has anything to do with circles is "reasonably miraculous" indeed.

I'm still none the wiser. Can you give me an example of something that would be unreasonably miraculous? -- JackofOz 00:57, 21 May 2007 (UTC)

The same concepts, minus the ironic understatement. As I understand it, the arc of magical discovery goes Real Numbers -> i -> Complex Numbers -> Hey, these are algebraically closed! -> Hey, multiplication and addition correspond to translation, rotation, and scaling of vectors! -> Hey, the most natural extension of exponentiation also corresponds to rotation! -> Cool, a sine wave! -> pi -> WTF? 23:47, 22 May 2007 (UTC)

To answer the original question, which got lost somehow, this is not quite proof that God exists. According to Descartes, it's proof that you exist, otherwise you couldn't think about it, and according to certain postmodernist viewpoints, life is an illusion and everything you see is a figment of your imagination. So, it does prove the existence of the person who created your universe. Whether, based on that, you consider yourself a god will take some deep personal reflection. Black Carrot 23:55, 22 May 2007 (UTC)

Swimming pool?

My friend says that his swimming pool is bigger but i'd like to prove him wrong.My swimming pool is 10 meters long,6 meters wide and 2 meters deep.My friends swimming pool is 12 meters long , 5 meters wide and 1.5 meters deep.Is'nt my swimming pool bigger?

Define bigger. Also, define homework question. Hint to the first: if you mean volume, then do you have a formula for the volume of a rectangular prism? Confusing Manifestation 04:32, 18 May 2007 (UTC)

These are not pools; they are barely puddles. The length of an Olympic size swimming pool is 50 m and its width is 25 m, so your two pools could be placed end-to-end sideways (22 m) in an Olympic pool with room to spare. As for volume, it is unlikely that these are simple rectangular boxes, so multiplying the three dimensions is of questionable accuracy. On the chance that this is disguised homework, I leave the calculations (or measurements) to you and your friend. --KSmrq^T 04:35, 18 May 2007 (UTC)

For the requisite formulas for the homework, see Cuboid.  --Lambiam^Talk 08:55, 18 May 2007 (UTC)

At least you tried to disguise your homework question, you get bonus pts for that. :-) StuRat 16:37, 18 May 2007 (UTC)

I always award bonus points for honesty, rather than the opposite. Tesseran 01:29, 19 May 2007 (UTC)

To answer your question, I made a drawing. An user called Rockpocket decided to make my drawing very small, as you can see. The good thing about this is that the page takes less time to load. The bad thing is that you will need to click on the image to see the picture with full resolution on a different page.

I am not going to do your homework for you, but, hopefully, you will be able to do it yourself and post your answer here for us later! Here is the drawing. The first image is a meter. The second one is a square that is one meter "high" and one meter "wide". It's called a square meter. Square meters are good to measure areas, such as the floor of your house, or a big sheet of paper.

To measure volume, there is the cubic meter, as you can see. It is one meter deep, one meter wide and one meter long.

The fourth image is just 4 cubic meters. Imagine that you draw a square meter on the floor. Then you draw another square meter right next to it. Then you dig it to make a hole. When your hole is two meters deep, it will be just like this image. You can fill it with water to make a pool. This pool will be as deep as your swimming pool, because it is two meters deep, and so is yours. But this pool is not as wide as your pool, and also not as long as your pool. We need more cubic meters to draw a pool just like yours and figure out how many cubic meters your pool has.

The fifth image is just 8 cubic meters. It is a pool that is two meters deep, two meters wide and two meters long. Still much smaller than your pool.

The sixth image has 12 cubic meters. It is two meters deep, three meters long and two meters wide. Much smaller than your pool, still.

The seventh image has 36 cubic meters! It is already as deep and as wide as your pool. It is two meters deep and six meters wide, but only three meters long. We need to make it much longer.

The eight image is, finally, your pool! As you can see, it has 120 cubic meters!! Now, you just have to do the same with your friends pool, and see how many cubic meters it has. If your friend's pool has more than 120 cubic meters, then it is bigger than yours. If it has less than 120 cubic meters, then it is smaller than yours. If it has exactly 120 meters, then both pools have the exact same volume! Try to find that out yourself, and then come back here, if you can, to tell us if you could do it. Good luck!

And... That last image was an attempt to make it look like an actual pool, but I wanted to be still able to see the cubic meters, so just imagine that two of the walls are made of glass! A.Z. 03:03, 19 May 2007 (UTC)

Thank you very much. To tell the truth the reason i posted this question is because on a test I got this question wrong and when I checked it I saw that I actuelly got it right but I was still afraid I might be wrong so I posted it here.When I did aproach my teacher she just ignored me and did'nt listen.So now i have a new problem , what should I do to convince the teacher to look at my test and see that she marked it wrong? Oh yeah and the answer is "My friends house has a swimming pool with a volume of 90 meters squared."

You're welcome!

I don't really know how you can convince your teacher to look at your test. Maybe other people here are better suited to help you with that, since I too have a hard time interacting with other people and trying to convince them of things.

But I have a guess about your test. Maybe what you got wrong is that you said that the volume of your friend's pool is 90 square meters, but it is 90 cubic meters instead. Could that be it? What answer did your teacher claim to be the right one? A.Z. 23:02, 19 May 2007 (UTC)

(Please sign your posts.) Thanks for the extra information. I suspect important details are still missing from the story as told here; however, I will suggest an alternative argument for this specific question. The dimensions of "your" pool are 10×6×2, and the dimensions of the other are 12×5×1.5, both in meters. Take a one meter strip from the side of your pool, 10×6×2 = 10×5×2 + 10×1×2. Split it in half, and rearrange it to fit on the end: 10×5×2 + 10×1×2 = 10×5×2 + 2×5×2 = 12×5×2. Now it should be obvious that your pool has the same surface area as the other pool, but since it is a half meter deeper, it has greater volume. This is a watertight argument! (Sorry.) Now, for some politics. The way to present this to your teacher is to avoid the accusation "You graded my test wrong!", but rather to say, "Can you help me see the flaw in my reasoning?" And if a teacher will not do that, then speak to their boss. --KSmrq^T 08:11, 20 May 2007 (UTC)

I agree with "Can you help me see the flaw in my reasoning ?". However, if that doesn't work, I'd take it to your parents. Assuming they are able to do this basic math problem and see that you are right, they will likely be upset that your teacher is marking correct answers wrong and will make sure she listens to them. Also, if you have a friend who gave the same answer and got it marked right, that's good evidence to bring. StuRat 21:07, 20 May 2007 (UTC)

Motion from action

δʃΦds=0

can someone show me how the equation of motion was drivable from the action principle? —Preceding unsigned comment added by 69.241.236.12 (talk • contribs) 05:32, 2007 May 18

Something about averages

If you have an average number, and then an average number of all the values that are greater than the original average and the same with less than, what does that tell you? I don't know how I can make this question more specific, but I'm basically wondering what that says about values that go further from the original, like if you had average +/- N and N is a random number how could you use the greater and lower averages to come up with random number that isn't somehow incorrect? Thanks, Jeffrey.Kleykamp 16:11, 18 May 2007 (UTC)

You could look at percentile, this can be used to show what the value is at 25%, 50%, 75% stages of a data-set, also try looking at mean, mode and median. I'm not sure what it would tell you other than the average of your original data set, then the average of your 2 reduced data-sets. I guess the difference between the upper/lower and original mean will show you a little about the shape of the dataset, but you could figure that out better with a percentile function I think. ny156uk 17:03, 18 May 2007 (UTC)

Assuming for simplicity that no value is equal to the mean, then we can start off by saying that knowing those two sub-means tells you what proportion of the values are above and below the overall mean, since $m_{b}n_{b}+m_{a}n_{a}=m(n_{a}+n_{b})\Rightarrow m_{b}{\frac {n_{b}}{n_{a}}}+m_{a}=m\left(1+{\frac {n_{b}}{n_{a}}}\right)\Rightarrow {\frac {n_{b}}{n_{a}}}={\frac {m-m_{a}}{m_{b}-m}}$. Let's suppose that the ratio is near unity (which discards one of the submeans), and further suppose that the data follow a normal distribution. Then the submeans should be the mean $\pm {\sqrt {2 \over \pi }}\sigma$, so you can get from your data that $\sigma ={\sqrt {\pi \over 2}}(m_{a}-m)$. You could of course suppose any distribution, not just a normal one, and solve for its parameters. To use both means, we need to use a trial probability distribution with skewness; I'll leave the selection of such as an exercise. What you're doing is in effect a moment analysis of the data, if that helps. --Tardis 17:48, 18 May 2007 (UTC)

I just don't get it, I don't know what I don't get but it, in general, doesn't make sense to me, could I maybe have an example? Jeffrey.Kleykamp 19:19, 18 May 2007 (UTC)

Given {-3, -1, 0, 1, 2, 3, 5, 7, 10, 30} as data, we calculate a mean of $m=5.4$, so the sublists are {-3, ..., 5} with mean $m_{b}=1$ and {7, 10, 30} with mean $m_{a}\approx 15.667$. From this we can see that there are ${\frac {n_{b}}{n_{a}}}={\frac {7}{3}}$ as many values lower than the overall mean as there are higher; in this case those numbers are in fact 7 and 3, but we don't know that from the means, just that they have that ratio: 14 and 6 would be just as valid (indeed, consider simply duplicating every element in the list). We can also attempt to guess how wide the distribution of values is: from my earlier equation, I get $\sigma _{b}=5.51,\sigma _{a}=12.9$. It's somewhat broken to calculate a separate standard deviation on each side, not least because we're attempting to "ignore" one side when we do so and yet changing values on that side would change the overall mean and would change where the cut between the two sublists went, etc. However, we do see here evidence that the distribution is wider on the right than the left — that is, that its average is where it is because of a bulk of values close by on the left and a few values far away on the right. This is true; moreover, just looking at the arbitrary edge values I picked, we see that $a-m=30-5.4=2.93(5.4-(-3))=2.93(m-b)$, and $\sigma _{a}=2.333\sigma _{b}$, so the "widths" do correspond to the data relatively well.

So we can recover a good deal of information about the distribution of the values just from our three means; however, this is largely because I chose the data to have a "bunch" (-1 ... 3) and then gradually widen only on one side to give skewness. While our conclusions about another set of data would be just as valid for the model, the model might not happen to work so well as it does here. In particular, if you take four values, two on each side of the mean, and move the inner two in and the outer two out, all by the same amount, you will not affect any of our statistics (so long as no point crosses the overall mean). As such, I surmise that our three-means sampling is relatively insensitive to kurtosis. This makes sense; with three numbers we can only hope to determine three parameters of the data, and mean, variance (or standard deviation), and skewness are three such parameters that seem relevant. We have the first, so we have two more with which to determine the other two; however, we are not likely (with arbitrarily-chosen information like the two submeans) to be able to determine those exactly, but rather to get partial information about them and partial information about other parameters (like the median, perhaps). Does that make it any clearer? If the last bit about degrees of freedom is interesting, I suggest reading up on information theory, which can elucidate and quantify the notion of the partial knowledge one can get from other partial knowledge. --Tardis 21:14, 18 May 2007 (UTC)

I'll try asking the question again, I have an average size of a list, and I wish to come up with two values representing a range, so one number is less than the average (but greater than or equal to zero) and the other number is greater than the average, I decided to use the greater and lower averages as the two sizes however the numbers will always be in-between those two values yet the fact that the averages are there means that the list size was, at least once assuming the sample is made up of at least two values, outside that range. Can you see my problem? Jeffrey.Kleykamp 23:03, 19 May 2007 (UTC)

Adding your comment here to your original post (with the phrase "random number" in it) leads me to suspect that you have a list of values and want to produce more random values that are "characteristic" of the list but are not necessarily members of it. Moreover, you are suggesting (as an approximation to this goal) drawing numbers uniformly from $[m_{b},m_{a}]$, but realize that since (for $n>2$) at least one of the two submeans is not an extreme value of the overall list, you are truncating the distribution of results. Am I right?

Supposing that I am, you really have two distinct problems: producing a probability distribution that "matches" some given data in some sense, and then producing values that have that distribution. The first part is largely what I have been addressing in my previous responses, albeit in a somewhat indirect fashion: I was discussing how to get a best guess from only m, m_b, and m_a. Given access to the entirety of the data, we can proceed more directly: calculate the mean, variance, skewness, kurtosis, etc., of the data, and look for a probability distribution that has (in a very general sense) the same shape as your data. For instance, if you calculate a mean of 48, a variance of 16, a skewness of -0.004, and a kurtosis of $1.2\times 10^{-5}$, your data would look very much like the normal distribution; if you instead have a mean of 0.5, a variance of 0.23, a skewnewss of 0.4, and a kurtosis of -1.5, it sounds like you're looking at the output from a biased coin (as a sequence of 0s and 1s, with more 1s). So you pick the class of distribution (e.g., "normal" or "gamma"), then calculate as many moments (or other functions) of your data (whose values are known for the distribution) as the distribution has parameters. Solve for the parameters and you have the distribution.

The second part is producing values given a probability distribution; this is relatively straightforward, although it can require some numerical computation. A general method is inverse transform sampling; for some distributions, there exist easier methods (see, for example, Laplace distribution#Generating Laplace variates). Does that help? --Tardis 22:21, 21 May 2007 (UTC)

No, it doesn't help because it's too technical and possibly unspecific to my problem because I just have the average, higher and lower averages and a list of things that I wish to search and cut between a range that is based on those averages. But thank you anyway because your help is voluntary, Jeffrey.Kleykamp 14:43, 22 May 2007 (UTC)

OK, so I suspected wrong. You'll then have to define "search and cut between a range". While you're at it, am I right in understanding that you have access to the entire list of values? If so, the means and submeans are completely redundant, of course, although it seems that you want to use them, which is fine. --Tardis 16:12, 22 May 2007 (UTC)

The list is a list of objects containing a string (also called a word) and a number (the list is sorted by the number and moving down to lower values) for a computer program that I am writing, I wish to search the list by looking for the highest difference in the current number minus the next number in the list and cutting or disposing of all items after that drop, however, I can't start from the beginning because the output is supposed to be the same, on average, to the input (however the highest difference is always moving), I'm starting at the lower average and moving up to the higher average (which is furthest away from the beginning of the list), is that clear now? Jeffrey.Kleykamp 16:31, 22 May 2007 (UTC)