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July 23

Estimating population standard deviation in the limit of small numbers

I have a problem where I need to measure the standard deviation of a population that is difficult to sample, and so I would like to use as few samples as I can practically get away with.

In the limit of large numbers, and with the assumption that the underlying distribution is normally distributed, I know that the appropriate estimates are:

${\bar {y}}={\sum {y_{i}} \over N}$

$s(y)={\sqrt {\sum {(y_{i}-{\bar {y}})^{2}} \over N-1}}$

Where the standard error in the standard deviation is

$SE(s(y))={s(y) \over {\sqrt {2N}}}$

Which implies that if I had 200 samples, I would expect to know the standard deviation to about 5%.

But this is devised in the limit of large N. I would like to know how the uncertainty might change in the limit of small N (e.g. N = 5 or 10). Applied as is, the formulas suggest at N = 5, the estimate of the population standard deviation will have an error of about 30% in the typical case. But is that really true, or in considering such small numbers would my error be significantly worse than that (and how much worse)?

Also, in the limit of small numbers, are there any procedures that can improve the estimate the population standard deviation. For example, would the interquartile range be less subject to fluctuations. I doubt it, but it is probably worth asking. Dragons flight (talk) 18:03, 23 July 2011 (UTC)

A lot of this depends on your assumptions about the distribution of the data, and whether you take a frequentist or Bayesian approach. If Bayesian there's no easy way out, just choose your prior and do what Bayes says, the result will probably not have a nice closed form.

The formula you give is the square root of an unbiased estimator of the variance. It is not, however, an unbiased estimator of the standard deviation. Those also exist if we assume normality.

If our estimator for the standard deviation takes the form $s={\sqrt {c(N)\sum (y_{i}-{\bar {y}})^{2}}}$, then it is always the case that the variance of s is exactly (if my calculations are correct) $c(N)(N-1)\sigma ^{2}-\mathbb {E} [s]^{2}$. -- Meni Rosenfeld (talk) 08:51, 24 July 2011 (UTC)

July 24

Continuity Correction

Hi everyone, I write here regarding to Normal Distribution theory issue. It states that Normal Distribution can be used as an alternative to Binomial Distribution in case number of trials n is large (n>50), and mean np>5 then a discrete random variable of the Binomial Distribution can be approximated by a continuous random variable of the Normal Distribution through Continuity Correction:

If P(X=A)(discrete) then P(A – 0.5 < X < A + 0.5) (continuous)

If P(X>A) (discrete) then P(X > A + 0.5) (continuous)

If P(X≤A) (discrete) then P(X < A + 0.5) (continuous)

If P (X<A) (discrete) then P(X < A – 0.5) (continuous)

If P(X ≥A) (discrete) then P(X > A – 0.5) (continuous)

And my question is that, in the 5 statements above, what kinds of principle have been utilised to build up such statements? Is it possible to prove them? Or those are solely consequences from practical problems? Thanks in advance.Torment273 (talk) 11:18, 24 July 2011 (UTC)

The first thing to realize is that all these statements are really the same. First let's improve the notation and let X be the binomial variable and Y be a normal variable approximating it. Then the first statement becomes $\mathrm {Pr} (X=A)\approx \mathrm {Pr} (A-0.5<Y<A+0.5)$. From this all the others follow - for example,

$P(X>A)=\mathrm {Pr} (X=A+1)+\mathrm {Pr} (X=A+2)+\mathrm {Pr} (X=A+3)+\cdots \approx$

$\approx \mathrm {Pr} (A+0.5<Y<A+1.5)+\mathrm {Pr} (A+1.5<Y<A+2.5)+\mathrm {Pr} (A+2.5<Y<A+3.5)+\cdots =\mathrm {Pr} (Y>A+0.5)$.

Now, the rule $\mathrm {Pr} (X=A)\approx \mathrm {Pr} (A-0.5<Y<A+0.5)$ is a result of how we choose Y, and we choose it this way because we want Y to be as close to X as possible. Since A is an integer, for Y in the range $A-0.5<Y<A+0.5$ the closest integer is A. -- Meni Rosenfeld (talk) 11:41, 24 July 2011 (UTC)

Amazing. Thanks for a quick reply. Have you worked it out on your own?Torment273 (talk) 13:24, 24 July 2011 (UTC)

Yes. Actually to me it's clear why these rules are valid and how to show they're all equivalent. Hopefully with experience it will become as obvious to you as well. It's good that you're asking these questions instead of taking things for granted. -- Meni Rosenfeld (talk) 14:25, 24 July 2011 (UTC)

To make this into five separate statements is to make a simple thing extremely complicated. Just remember that for integer-valued variables,

X < 6

and

X ≤ 5

are the same thing and the continuity correction uses the value half-way between them. And similarly

X > 8

and

X ≥ 9

are the same thing. Michael Hardy (talk) 04:24, 25 July 2011 (UTC)

July 25

An explanation of chirality.

I've read both articles that explain the concept of chirality here on Wikipedia (Chirality (chemistry) and Chirality (mathematics)), but I am still having trouble understanding the concept. Could somebody please explain it to me?190.25.14.117 (talk) 22:56, 25 July 2011 (UTC)

Probably not in any way that is better than the lead of Chirality (mathematics). Where in that article do you first encounter difficulties? Looie496 (talk) 23:08, 25 July 2011 (UTC)

Suppose you paint your left hand purple, and I had a picture of you and another picture of your reflection in a mirror. Then I could tell which was which by looking for the purple. So you're chiral. If you wash your hand off, then I can't tell which is which, so you're no longer chiral.--Antendren (talk) 23:12, 25 July 2011 (UTC)

A left shoe looks different from a right shoe. Clockwise is different from counterclockwise. That's chirality. Michael Hardy (talk) 00:13, 26 July 2011 (UTC)

A screw is another example. You could create a screw with the spiral thread in the opposite direction (and they do), but no matter how you turn it, one screw never changes into the opposite spiral. StuRat (talk) 22:31, 26 July 2011 (UTC)

July 26

Triple Trouble

Greetings, and great to be back after all this time. I am having trouble with a mathematics question where we are asked to use a triple integral to find the volume of the ellipsoid, where x²/a²+ y²/b²+ z²/c² = 1 .

I have looked at all the ways to do it, and am still confused, having been working on this for days. If I integrate, where does the pi term come from to form the volume which I do know to be pi times abc  ? Chris the Russian Christopher Lilly 07:45, 26 July 2011 (UTC)

Generally it would come as a result of the arcsin function which comes up when you integrate factors of the form ${\sqrt {1-x^{2}}}$. -- Meni Rosenfeld (talk) 09:02, 26 July 2011 (UTC)

It may be easier to start with determining the area of the ellipse x²/a²+ y²/b² = 1 using a double integral, similar in principle but with fewer steps.→86.155.185.195 (talk) 10:24, 26 July 2011 (UTC)

The difficult bit is working out the limits on the integrals. Once you've done that, it's all fairly standard integrals. Say you integrate with respect to x first. You need to find the limits of x within which x²/a²+ y²/b²+ z²/c² < 1. --Tango (talk) 19:04, 26 July 2011 (UTC)

I would recommend changing from Cartesian coordinates to spherical coordinates. — Fly by Night (talk) 19:17, 26 July 2011 (UTC)

Use the change of variables u=x/a, v=y/b, w=z/c. The Jacobian is abc, so the volume is abc times the volume of the unit sphere: $4\pi abc/3$.Sławomir Biały (talk) 21:41, 26 July 2011 (UTC)

I thought about something similar, i.e. using a special linear transformation to change the ellipsoid to a sphere and then applying the formula for the volume of a sphere. But the question wants derivation of the formula of an ellipsoid using integration. There is a general formula for the volume of an ellipsoid. If the OP can't use that, then we can assume that s/he can't use the formula for the volume of a sphere. — Fly by Night (talk) 22:13, 26 July 2011 (UTC)

Ok, in principle one could evaluate the triple integral

$\int _{-a}^{a}\int _{-{\sqrt {b^{2}-x^{2}/a^{2}}}}^{\sqrt {b^{2}-x^{2}/a^{2}}}\int _{-{\sqrt {c^{2}-x^{2}/a^{2}-y^{2}/b^{2}}}}^{\sqrt {c^{2}-x^{2}/a^{2}-y^{2}/b^{2}}}\,dz\,dy\,dx.$

--Sławomir Biały (talk) 00:16, 27 July 2011 (UTC)

Ouch! There's no need for language like that ;-) Wouldn't spherical coordinates be the natural choice? Trying to integrate as you suggest presents all kinds of problems. The signs of the arguments of the radicals is one. Besides that, the integrals themselves are less than elegant, e.g.:

$\int {\sqrt {1-t^{2}}}\,\operatorname {d} \!t={\frac {1}{2}}t{\sqrt {1-t^{2}}}+{\frac {1}{2}}\arcsin t\,.$

I won't even try to write the answer to

$\int \int {\sqrt {1-u^{2}-v^{2}}}\,\operatorname {d} \!u\,\operatorname {d} \!v\,,$

but it involves logs, inverse tangents and absolute values. P.S. Did you miss the upper limit from the middle integral? — Fly by Night (talk) 00:18, 27 July 2011 (UTC)

Well, I think the natural coordinate system to carry out the integration is the one I suggested. Compute the integral of the Jacobian over the unit ball in the uvw space. (That integral can be done using spherical coordinates, or just with the volume formula.). But I thought you were saying we weren't allowed to change coordinates, though. Hence I suggested the nasty brute-force approach. Sławomir Biały (talk) 01:50, 27 July 2011 (UTC)

I guess it's just a matter of taste. I prefer to see the volume of a unit sphere, centre the origin, is expressed in spherical coordinates as

$V=\int _{0}^{2\pi }\int _{0}^{\pi }\int _{0}^{1}r^{2}\sin \theta \,\operatorname {d} \!r\,\operatorname {d} \!\theta \,\operatorname {d} \!\varphi ={\frac {4\pi }{3}}\pi \,.$

instead of the Cartesian coordinate system that you suggested

$V=\int _{-1}^{1}\int _{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}\int _{-{\sqrt {1-x^{2}-y^{2}}}}^{\sqrt {1-x^{2}-y^{2}}}\!\!\!\operatorname {d} \!z\,\operatorname {d} \!y\,\operatorname {d} \!x\,.$

I suppose that one might argue that your expression is easier to calculate than mine. — Fly by Night (talk) 03:41, 27 July 2011 (UTC)

You misunderstand me. I'm not opposed to using spherical coordinates to calculate $\iiint abc\,dV$ after one applies the Jacobian change of variables (although it isn't necessary to do so for hopefully obvious reasons). The reason I wrote the above is that you seemed to suggest that changes of coordinates were off limits. That just leaves us with the original Cartesian integral, which can (in principle) be calculated using elementary methods by brute force. Sławomir Biały (talk) 11:41, 27 July 2011 (UTC)

Added: Although I see I probably misunderstood your reply as well. Reading it again, I see you meant that we can't use the standard formula for the volume of the sphere. (Instead we should calculate this volume by some standard method: spherical coordinates being the easiest, and cylindrical the next easiest.) Ok. I guess we agree, then. Sławomir Biały (talk) 11:46, 27 July 2011 (UTC)

Thanks all so much, and he, that is I, am very impressed by all of this, which should be of much use indeed. Chris the Russian Christopher Lilly 01:20, 27 July 2011 (UTC)

(edit conflict) My solution would be to make a volume presrving, linear change or coordinates. The equation (x/a)² + (y/b)² + (z/c)² = 1 gives an ellipsoid that cuts the x–axis at x = ±a, it cuts the y–axis at y = ±b and it cuts the z–axis at z = ±c. We can make a linear transformation of xyz–space that preserves the x–, y–, and z–axes; which also preserves volume. Such a transformation is given by (x,y,z) → (αx,βy,γz) where α, β and γ are real numbers such that the product αβγ = 1. We want the linear transforation to carry the ellipsoid to a sphere. A sphere has an equation of the form x² + y² + z² = r² for some real number r > 0. We must choose α, β and γ so that the equation (x/a)² + (y/b)² + (z/c) = 1 is transformed into the equation x² + y² + z² = r². If we make the linear transformation (x,y,z) → (rx/a,ry/b,rz/c) then the image of the set of points given by the equation (x/a)² + (y/b)² + (z/c) = 1 will be transformed into the set of points given by the equation x² + y² + z² = r² or, if you prefer (x/r)² + (y/r)² + (z/r)² = 1. We still don't know what r is. We can determine r from the fact that we want (x,y,z) → (rx/a,ry/b,rz/c) to be volume preserving. The determinant of (x,y,z) → (rx/a,ry/b,rz/c) is given by r³/abc and r³/abc = 1 if and only if r = (abc)^1/3. Thus, your problem reduces to evaluating the volume of the sphere x² + y² + z² = (abc)^2/3. To solve this you can use the formula for the volume of a solid of revolution. The formula (which can easily be proven if necessary) is as follows:

$\pi \int _{\ell _{1}}^{\ell _{2}}\!\!f(x)^{2}\,\operatorname {d} \!x$

gives the volume of the solid given by rotating the graph y = ƒ(x), for $\ell _{1}\leq x\leq \ell _{2}$, given in the xy–plane by a full 2π radians about out the x–axis in xyz–space. As a result, the volume of the sphere (and so the volume of the ellipsoid) is given by calculating

$V=\pi \int _{-k}^{k}\left({\sqrt {k^{2}-x^{2}}}\right)^{2}\,\operatorname {d} \!x={\frac {4}{3}}\pi k^{3}\,,$

where k = (abc)^1/3. It follows that V = 4⁄3πabc. I know that it might seem long winded; but I wanted to explain everything properly. The main details are regarding the linear transformation, which is the easiest bit once you understand it. But it's not always so clear for the the first time. Anyway, I hope this helps. — Fly by Night (talk) 01:46, 27 July 2011 (UTC)

Yes, thank You. I see the key is in trying to simplify the ellipsoid equation into that of a sphere, but being in mind of the differences between the two. To be sure, as seems obvious, an ellipsoid is just a skewed kind of sphere, where in the end it makes sense that their volumes should be of a similar nature. I have spent most of this week on this problem, which it appears is one given to unsuspecting maths students all over the world, designed to get us to think. From this I can begin to understand how the triple integral works, and this shall assist me a lot in giving a proper answer, thank You.Chris the Russian Christopher Lilly 23:36, 27 July 2011 (UTC)

Eigenvalues of real skew-symmetric matrices are imaginary. Does this work in the other direction?

All eigenvalues of real skew-symmetric matrices are imaginary or zero. Does this work in the other direction, i.e. are all real matrices whose eigenvalues are imaginary or 0 similar to a real skew-symmetric matrix? I.e. if all eigenvalues of some real matrix A are imaginary, does this imply the existence of a transformation P such that P^(-1)AP is skew-symmetric? 83.134.166.168 (talk) 16:56, 26 July 2011 (UTC)

No. Any nilpotent matrix is a counterexample. Sławomir Biały (talk) 21:39, 26 July 2011 (UTC)

Thanks! Apparently nilpotent matrix have only zero complex eigenvalues. What if all eigenvalues are imaginary and there's at least one nonzero complex conjugate eigenvalue pair? Is that a sufficient condition for the matrix to be similar to a real skew-symmetric matrix? (clearly if the matrix is real and all its eigenvalues are imaginary, then the polynomial whose zeros are eigenvalues factors like lambda^(number of zero eigenvalues)*(lambda^2+c1)*(lambda^2+c2)*... with all c real and positive, so the eigenvalues do come in complex conjugate pairs ) 83.134.166.168 (talk) 06:18, 27 July 2011 (UTC)

No. You need to assume diagonalizability as well, since any skew symmetric matrix is diagonalizable. If you assume the eigenvalues of a diagonalizable matrix are imaginary and come in conjugate pairs, then the matrix is similar to a skew symmetric matrix. (Hopefully the reason for this is clear.) Sławomir Biały (talk) 10:46, 27 July 2011 (UTC)

July 28

Value of the world's hoarded coins

Greetings all. I don't visit this desk often. Math isn't something I probably grep past simple, square arithmetic, but I understand (I think) that $BigMath$ is numbers and the things they represent ((whe{rn}+/-^e) applicable) working out simple, square arithmetic over and over again, with different applications reducing anomalies into the equation until they fit with >>Reality<<, so I ask this question here, instead of Humanities or Computing (which desk I also tend not to frequent) where it might also go.

This, also, is my first question to any of the WPRef desks, since I like to think I'm smart, and toss in my ₡10.0553042 to show it every now and then & consider myself xth above equals. So I'm humbled.

My question is this: what is the value of all the coins that each person currently living on Earth hoards. By hoard I mean keep in physical form accessible without too much effort -- a jar, a book, a safe or safe deposit box, a drawer, a rafter, &c are all places one might hoard -- for some reason. The reasonings I'm thinking would be along the lines of 'valuable now, & always will be,' "not valuable here, but valuable elsewhere," "not valuable now, but possibly valuable later", 'not valuable here or elsewhere, but interesting. Nice to have around.', &c. Valuable would mean it could be traded for some good, service, or physical currency of equal or higher value. In other words, what are the:

• 1990 loonie
• 1989 colón
• 1965 quarter that must have fallen into a gearbox recently, or someone tried to make a slug out of it, or something; whatever it was, it's never going to go kindly into a coin machine ever again.
• 1984 20 centime piece
• 2005 €2
• 1954 half crown

sitting in a teapot in Someplace USA worth, in $US, against the value of all coins in similar circumstances across the world?

It seems like there should have been some attempts to figure this out made, money counting being as popular as it is, but I've no idea how to search for them. Is there a way to sift all the variables down into something that functions, even within a certain sluggish margin of error, in >>the real world<<?

--some jerk on the Internet (talk) 04:36, 28 July 2011 (UTC)

Perhaps you can get some indication of the incidence of hoarding by studying public statistics on how much pre-Euro currency remains in circulation. According to the Bundesbank there are Deutschmark coins worth 6.91 billion Dm in circulation as of mid-2011. Assuming that the coins are mainly kept by German citizens, this means a hoarding rate on the order of 100 Dm per person. Of course the hoarding rate may vary in different cultures, and may be entirely different in the case of coins that are currently legal tender, but you've got to start somewhere. 188.117.30.209 (talk) 20:39, 31 July 2011 (UTC)

what is remainder??:

what is remainder after this division: [(47)^43]/47.. and how..?? — Preceding unsigned comment added by 203.88.131.62 (talk) 06:40, 28 July 2011 (UTC)

Assuming you have copied the question correctly, this is very straightforward. Notice that 47^43 = (47^42) x 47. That should give you a big clue. Gandalf61 (talk) 07:49, 28 July 2011 (UTC)

[(47)^43]/47 is exactly 47^42 which is an integer, so the remainder is 0. Widener (talk) 10:46, 29 July 2011 (UTC)

Diagrams for Young Tableau

How do we import or create diagrams for Young Tableau in a Wikepedia article?--Profvk (talk) 07:02, 28 July 2011 (UTC)

Go to Special:Upload and follow the instructions there. Then go to the article and insert the picture as [[File:Name.jpg|thumb|Caption]]. Rcsprinter (talk) 10:48, 28 July 2011 (UTC)

I think I might just use tables and forget about special images. Dmcq (talk) 17:19, 28 July 2011 (UTC)

There's a LaTeX package, youngtab, which does some nice Young tableaux. Sadly it doesn't do polytabloids (row or column), but you can make them by editing an outputted PDF file. Then you can convert to an image, say SVG. It would be nice if they could add the Youngtab package as part of the LaTeX system, but I'm sure it could get out of hand with everyone recommending their own package, and having to debug compatibility problems. --Sam^Talk 19:56, 28 July 2011 (UTC)

name of series

What is the name of this series which defines y? a=1 to infinity y=x^(a/a+1) --DeeperQA (talk) 15:32, 28 July 2011 (UTC)

The question doesn't make any sense. If "series" means an infinite sum, it will not converge. If "series" means a limit, the limit is just y = x. Looie496 (talk) 17:21, 28 July 2011 (UTC)

It may be y=y+(a/a+1), with either a=1 to infinity or a=0 to 1. I saw "series" but the power was out and all I had were matches. I left my magnifier at home. --DeeperQA (talk) 19:49, 28 July 2011 (UTC)

You're still not making any sense... --Tango (talk) 21:33, 30 July 2011 (UTC)

Let's remember to not bite people posting questions. A person wouldn't need to post a question if s/he understood all of the concepts and vocabulary. We're here to help people; not to judge them. To DeeperQA, could you take another look, and try to give us the question another time? At the moment it's hard to understand what you're asking. Take a look at our sequence and series articles too. They might help you to phrase the question. Also, maybe include a reference to the place that you found the problem. — Fly by Night (talk) 03:45, 31 July 2011 (UTC)

Converting dollars to Indian money

i want to know how to convert 1 ounce gold from us dollar to indan money in full fromul — Preceding unsigned comment added by 117.207.113.247 (talk) 17:14, 28 July 2011 (UTC)

At today's conversion rate, multiply by 44 to get the value in rupees→86.155.185.195 (talk) 19:05, 28 July 2011 (UTC)

July 29

shooting arrows and finding angles

I have been given this problem about angles:

Archer A is positioned at the top of a cliff face, 40 meters directly above Archer B. Both archers release arrows at the same same time.

Archer A fires horizontally from the cliff top at 28 m/s. Archer B fires at an unknown angle.

Both arrows strike an identical spot on the ground, at the same time.

Show that the angle between the embedded arrows will be approximately 18.5 degrees.

It has been a long time since I learned how to do this type of question back in high school, I need help refreshing my memory.

Is it something to do with gravity being 9.8 m/s squared, and creating some sort of quadratic parabola...?

I don't know where to start, not even sure if it can be solved with the information given. Don't we need to know either Archer B's speed or angle? I'm clueless, please help.

118.208.128.137 (talk) 00:13, 29 July 2011 (UTC)

Have you tried to search our articles, e.g. Trajectory and Projectile motion? Take a look at these articles and the let us know how you get on. — Fly by Night (talk) 00:56, 29 July 2011 (UTC)

Does archer B also fire at 28m/s? Otherwise I think there's insufficient data. -- Meni Rosenfeld (talk) 06:50, 29 July 2011 (UTC)

Let's see:

1. We know that archer A fires horizontally, so the time taken for their arrow to fall 40 meters is sqrt(80/g) - let's call this time T.
2. The vertical velocity of A's arrow at time T is gT (which is sqrt(80g)), and its horizontal velocty is a constant 28 m/s, so the angle it makes with the vertical when it strikes the ground (assuming it always points along its velocity vector) is tan^-1(28/gT).
3. During time T, B's arrow goes up and comes down again. Its change in vertical velocity is gT (just the same as for A's arrow) so, by symmetry, it's initial vertical velocity is gT/2 upwards, and its final vertical velocity is gT/2 downwards.
4. Since A's arrow and B's arrow are fired at the same time and hit the ground at the same point and the same time, the horizontal velocity of B's arrow must be 28 m/s, just the same as A's arrow (so the two arrows always have equal horizontal displacements throughout their motion).
5. So when B's arrow hits the ground, it makes an angle tan^-1(56/gT) with the vertical.

Now just calculate the two angles and find their difference. If you use the value of 9.8 for g then gT turns out to be a convenient whole number. Gandalf61 (talk) 09:24, 29 July 2011 (UTC)

July 30

mathematics

my hundreds digit is 3 more than my ones digit and my tens digit is 1 more than my hundreds digit what numbers can i be? — Preceding unsigned comment added by 112.206.73.68 (talk) 01:41, 30 July 2011 (UTC)

Sooner or later I will learn to stop looking at Wikipedia on Friday evening. Looie496 (talk) 01:59, 30 July 2011 (UTC)

That's less than a thousand cases to check by hand; no biggie. Get on it, anon! --COVIZAPIBETEFOKY (talk) 02:20, 30 July 2011 (UTC)

For future reference, please avoid giving questions such a nonspecific title as mathematics. Every question asked here is supposed to be about mathematics; your title should say what is different about your particular question. --Trovatore (talk) 03:00, 30 July 2011 (UTC)

The complete list of three digit numbers satisfying those conditions is 340, 451, 562, 673, 784, 895. Any number with more than three digits will be valid if the last three digits are any of those combinations too.Widener (talk) 03:51, 30 July 2011 (UTC)

I feel for you, I am not a number, I am a free man Dmcq (talk) 07:43, 30 July 2011 (UTC)

Quaternion/matrix differential equation

I am trying to solve ${\frac {d}{dt}}\log(A(t)B)=0$ for $B$, where $A(t)$ and $B$ are quaternions. Does anyone know how I should approach this problem? I don't believe that I can assume commutativity of $A(t)$ and $B$, but in case I'm wrong, does anyone know a simple way of finding a solution if they do commute? I have code to test such a solution numerically, so any ideas are welcome.--Leon (talk) 09:31, 30 July 2011 (UTC)

It might be better if you could tell us why you need to solve this problem. What is the motivation? Why do you need to solve this problem? What progress have you made towards solving this problem? Remember that the reference desk is not a homework solution resource. If you have encountered this problem in your own work then you will be able to give us ample motivation. If it is a homework problem then you will be able to give us your current ideas and attempt. — Fly by Night (talk) 01:23, 31 July 2011 (UTC)

Okay, I'm trying to extend the notion of curvature for curves in $\mathbb {R} ^{3}$ to the (double-cover of the) space of rotations, ${\text{Spin}}^{3}$, and was hoping to define a radius of curvature by the amount of rotation (angle) associated with a spherical-linear motion from some orientation $B$ to an orientation on the (arbitrary) rotational motion $A(t)$ ($A(t)$ is an orientation that varies smoothly with time and defines a rotational motion). To meaningfully define a radius of curvature, the angle of rotation associated with a motion from $A(t)$ (which is the rotational motion I'm ultimately interested in) to $B$ should be approximately a constant for a small variation in $t$. Now if THAT sounds like homework then education must have changed a lot since I went to school!--Leon (talk) 11:32, 31 July 2011 (UTC)

This question looks nothing like homework, and even if it is, its phrasing is a far cry from the "solve my HW for me" crowd. Explaining the motivation for a question was never a requirement here. -- Meni Rosenfeld (talk) 08:22, 31 July 2011 (UTC)

I don't have much experience with quaternions, but wouldn't you be able to say that since ${\frac {d}{dt}}\log(A(t)B)=0$, $\log(A(t)B)$ is constant and so $A(t)B$ is constant? So $B=A(t)^{-1}c$, and if B is supposed to be a constant then $A(t)$ must be constant? -- Meni Rosenfeld (talk) 08:22, 31 July 2011 (UTC)

That's what I figured shortly after writing the question. However, given the above, have you any ideas where I may be going wrong?--Leon (talk) 11:32, 31 July 2011 (UTC)

July 31

Triple Double Trouble

Greetings once again, and thank you all very much for those of You who helped with my previous question on triple integration. I now have a second such question which also has me stumped, and I am perplexed, since I thought the technique we were given was meant to simplify things. We have been asked to express and evaluate the following triple integral in terms of Spherical Polar Coordinates, and I have had a number of goes at it, but cannot seem to figure it out.

$\int _{-1}^{1}\int _{0}^{\sqrt {1-x^{2}}}\int _{0}^{\sqrt {1-x^{2}-y^{2}}}\,dz\,dy\,dx.$

of the function e^ -(x^2 + y^2 + z^2 ) ^ (3/2)

Now there seem to be a number of formulae about and I am not sure about the order of integration, but one I had been given was as follows, since  :

as x = ρcosθsinφ y = ρsinθsinφ and z = ρcosφ while dzdydx = ρ2sinφdρdφdθ then we would have

$\int _{-1}^{1}\int _{0}^{\sqrt {1-\rho ^{2}\cos ^{2}\theta \sin ^{2}\phi }}\int _{0}^{\sqrt {1-\rho ^{2}\cos ^{2}\theta \sin ^{2}\phi -1-\rho ^{2}\sin ^{2}\theta \sin ^{2}\phi }}\,\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta$

of the polar version of the exponential, which is e^ -(rho squared cos squared theta sin squared phi + rho squared sin squared phi sin squared theta + rho squared sin squared phi ) to the three over two.

But then if this is right, how do I proceed ? I have worked at it, and been able to simplify some of it, but then when I integrate I end up with what seem to be horrible integrals and I am not sure I have done it right. Thank You, Chris the Russian Christopher Lilly 08:20, 31 July 2011 (UTC)

The most obvious problem is you didn't change the limits of integration with the substitution. For example the outermost integral is still using the range for x, not the range for theta.--RDBury (talk) 19:33, 31 July 2011 (UTC)

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The expression

$x^{2}+y^{2}+z^{2}\,$

should simply become

$\rho ^{2}.\,$

The bounds of integration need to be those for the appropriate corresponding variables. Thus $\rho$ goes from 0 to 1, and $\theta$ goes from 0 only to $\pi /2$ since you have only the upper hemisphere. And $\varphi$, the "longitude", goes all the way around from 0 to $2\pi$. Hence you need

$\int _{0}^{\pi /2}\int _{0}^{2\pi }\int _{0}^{1}e^{-\rho ^{2}}\rho ^{2}\sin \phi \,d\rho \,d\phi \,d\theta .$

Since there's no $\theta$ in the function being integrated, this becomes

${\frac {\pi }{2}}\int _{0}^{2\pi }\int _{0}^{1}e^{-\rho ^{2}}\rho ^{2}\sin \phi \,d\rho \,d\phi$

Then in the inner integral, $\sin \varphi$ is a constant that pulls out:

${\frac {\pi }{2}}\int _{0}^{2\pi }\left(\sin \varphi \int _{0}^{1}e^{-\rho ^{2}}\rho ^{2}\,d\rho \right)\,d\phi$

Finally, the now-inner integral is itself a constant factor that pulls out:

${\frac {\pi }{2}}\left(\int _{0}^{1}e^{-\rho ^{2}}\rho ^{2}\,d\rho \right)\left(\int _{0}^{2\pi }\sin \varphi \,d\phi \right).$

You largely missed the point of the transformation to spherical coordinates.

(Clearly something is wrong in the details above; I suspect I'm confusing the longitude and latitude. The answer should be a positive number.) Michael Hardy (talk) 19:40, 31 July 2011 (UTC)

Rather, the integral is over the interior of a quarter of a sphere of unit radius. In the spherical coordinate system, the limits of integration would be $\rho \in (0,1),\theta \in (0,\pi ),\varphi \in (0,\pi /2)$. Note also the typo above: $x^{2}+y^{2}+z^{2}=\rho ^{2}$, so the function being integrated is $\exp\{-(x^{2}+y^{2}+z^{2})^{3/2}\}=\exp\{-(\rho ^{2})^{3/2}\}=\exp\{-\rho ^{3}\}$. Nm420 (talk) 19:55, 31 July 2011 (UTC)