Wikipedia:Reference desk/Archives/Mathematics/2007 September 26

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[edit] September 26

[edit] Square mean root?

{\color{white}\Bigg|}RMS=\sqrt{\frac{A^2+B^2}{2}};\,\!
SMR=\left(\frac{\sqrt{A}+\sqrt{B}}{2}\right)^2=\frac{\left(\sqrt{A}+\sqrt{B}\right)^2}{4}=\frac{A+2\sqrt{AB}+B}{4};\,\!

The Root Mean Square is also known as the quadratic mean.
Is the inverse "Square Mean Root" recognized as a unique function (since it equals the mean of the "A + B" median and the geometric mean) and, if so, does it have an alternative name, equivalent to RMS's "quadratic mean"?  ~Kaimbridge~00:08, 26 September 2007 (UTC)

It's covered by Generalized mean but not given a name there, you could call it a power mean with power = 0.5,
The Heronian mean is similar..87.102.23.3 01:52, 26 September 2007 (UTC)
Ah, okay, it looks like that is just what it is (well, actually more like the three point "trapezoidal Heronian mean"! P=)  ~Kaimbridge~20:57, 26 September 2007 (UTC)

[edit] probability of a set containing a particular subset

I as wathcing a TV show a saw a lottery game that works as follows: 20 numbers are chosen randomly from a set 1 ≤ n ≤ 80. Players select 10 numbers from the same set of 80. If the set of 10 numbers is a subset of the 20, the player wins. What would be the probability of winning? I know (assuming both sets are chosen randomly) there are

\binom{80}{20} \approx 3.53531614 \times 10^{18}

ways of choosing the set of 20 and

\binom{80}{10} = 1646492110120

ways of choosing the set of 10, as well as

\binom{20}{10} = 184756

possible subsets of size ten within the set of twenty. I'm still not sure how to determine the probability that a particular set of 10 (chosen from the set of 80) is in the set of 20. How do I determine this? - SigmaEpsilon → ΣΕ 03:55, 26 September 2007 (UTC)

How many 10-sets will win? \binom{20}{10}.
How many 10-sets are possible? \binom{80}{10}
What's the probability of winning? Left as an exercise for the reader
PaulTanenbaum 04:27, 26 September 2007 (UTC)
Thanks. It's late at night here and I just needed a nudge in the right direction. - SigmaEpsilon → ΣΕ 04:30, 26 September 2007 (UTC)
An alternative approach is as follows. The probability that the first number chosen by the player is one of the 20 is 20/80. If this is successful, then the player is now choosing their second number from a set of 79. The probability that this matches one of the remaining 19 from the set of 20 is 19/79. So the probability that the first two numbers are both in the set of 20 is (20.19)/(80.79). The pattern becomes clear. You can continue this to find the exact probability that 10 numbers chosen are all in the set of 20 - you get the same result outlined by Paul Tanenbaum above. Or to get a rough estimate of the probaility, you can note that each of the 10 probabilities that we are multiplying together is less than or equal to 1/4, so the final probability is less than (1/4)^10 i.e. less than the probability of flipping a coin and getting 20 heads in a row. Gandalf61 08:21, 26 September 2007 (UTC)

[edit] MS Reference Specialty font has weird and pointless characters?

MS Reference Specialty is a font bundled with WindowsOffice. It has some weird characters, such as a "39 divided by 6" character, and a "14 divided by 07" character. Why? --24.17.13.197 20:50, 26 September 2007 (UTC)

I have no idea ... but ... isn't that just a generic character that allows you to change it to any number you want divided by any other number you want? That is, the 39 and the 6 are just place holders that can be changed into any specific numbers that one might need to be represented in a long division format. Just a guess on my part. (Joseph A. Spadaro 01:17, 27 September 2007 (UTC))
No, they're not those things. At least, I can't figure out how to replace the characters if they are. Weird, eh? Besides, the "1/27" and the "1/28" characters look otherwise identical. --24.17.13.197 01:14, 28 September 2007 (UTC)
I don't think this is bundled with Windows. My guess is that it's bundled with something like Encarta, and contains precisely those special symbols which happen to be needed in some Encarta article. -- BenRG 11:38, 27 September 2007 (UTC)
Ah, you're partly right; it's not included with Windows. My bad. There are some fonts with similar names included with Encarta and Bookshelf, but it looks like this one comes from Office 2007. Why would Office need this? --24.17.13.197 01:14, 28 September 2007 (UTC)

[edit] Tangent-line problem

I'm being asked to find the equation of the line that is tangent to the graph of f(x) = x3 and parallel to 3x − y + 1 = 0. To find the equation of the tangent line, I first need to find the slope, which is the derivative of the function, and I keep getting 3x2—I don't know if that's right or not. y − y1 = m(x − x1), so I plugged in 3 for the slope (the tangent line has to have the same slope as the other line, which it's parallel to), and if 3x2 is indeed the derivative, then I should be able to substitute the point (2,12) for x and y in the final equation. But that's not right! Help! Thanks, anon. —Preceding unsigned comment added by 70.23.76.70 (talk) 23:57, 26 September 2007 (UTC)

Looks good up until you mentioned (2,12). I don't know where you got that from. The point (2,12) is on the graph of the derivative, but it's not on the graph of the original function so there's no reason to expect the tangent line to go through that point. You just need to find the point on that curve where its derivative is 3, then find the line going through that point whose slope is 3. --tcsetattr (talk / contribs) 00:31, 27 September 2007 (UTC)

There are two line equations

(1) y = 3 x + 1
(2) y = x^3

A line that is tangent to (2) and have the same slope as (1) have the generic equation

(answer) y = m x + c

We know that (1) have a slope of 3 so we have

(answer) y = 3 x + c

Next we find the slope of (2) is equal to the slope of (1) which is 3 so we have

dy/dx of (2) is 3 x^2
dy/dx of (2) is 3
3 x^2 = 3
x = 1 or x = -1

So the answer must be

(answer) y = 3 (x==1) + c

But accord to (2) y = x^3 so y = 1^3 = 1

(answer) (y==1) = 3 (x==1) + c
c = -2
(answer) y = 3 x - 2

The alternate answer is for x = -1

(answer alternate) y = 3 x + 2

You can double check by plotting y = x^3 , y = 3 x - 2 , y = 3 x + 2

202.168.50.40 03:06, 27 September 2007 (UTC)

Since the problem is as much as solved already, let's have a little fun and perhaps learn some ancient elementary algebraic geometry.
We will work in the real projective plane, using homogenous coordinates. For this, we must introduce w into the equations enough times in each term to achieve a consistent total degree, making each polynomial a homogeneous polynomial.
\begin{align} y &= x^3 & \quad &\to & x^3 - y w^2 &= 0 \\ 3x - y + 1 &= 0 & \quad &\to & 3x - y + w &= 0 \end{align}
When w is 1 we recover the original equations, but we can now consider "points at infinity". Parallel lines share a common point at infinity. We can find this point for our given line by intersecting it with the line at infinity, whose equation is
w = 0 . \,\!
The solution is clearly the point with homogeneous coordinates (1:3:0).
Our next step is to find the "polar curve" for the cubic with respect to an arbitrary point p∗ = (x∗:y∗:w∗). (The "polar" here has nothing to do with polar coordinates.) We take the derivative with respect to x and multiply by x∗, the derivative with respect to y and multiply by y∗, the derivative with respect to w and multiply by w∗, and sum the three. That is, we apply the differential operator
\begin{align} P &= (x_{\ast} \tfrac{\part}{\part x} + y_{\ast} \tfrac{\part}{\part y} + w_{\ast} \tfrac{\part}{\part w}) \\ &= \bold{p}_{\ast} \cdot \nabla \end{align}
to get
P [x^3 - y w^2] = 3 x_{\ast} x^2 - y_{\ast} w^2 - 2 w_{\ast} y w . \,\!
We want the polar curve because it intersects the original curve precisely at those points where a line from p∗ will be tangent. This works even with a point at infinity.
Substituting p∗ = (1:3:0) in the polar equation gives the parabola
3 x^2 - 3 w^2 = 0 . \,\!
In this example we will not have tangency "at infinity", so we may safely substitute w = 1 and look for (x,y) values satisfying
\begin{align} 3 x^2 - 3 &= 0 , \\ x^3 - y &= 0 . \end{align}
But this is trivial; the only solutions are (x=1,y=1) and (x=−1,y=−1), homogeneous p1 = (1:1:1) and p2 = (−1:−1:1).
Now we are left with a basic task: find the equation of a line through two given points. Homogeneous coordinates form a vector space, so we need merely observe that a point p is on the line through p∗ and p1 precisely when the vectors p, p∗, and p1 are linearly dependent. We can detect this using a matrix determinant. Thus the desired equations are
\begin{align} 0 &= \det (\bold{p},\bold{p}_{\ast},\bold{p}_1) \\ &= \det \begin{bmatrix} x & x_{\ast} & x_1 \\ y & y_{\ast} & y_1 \\ w & w_{\ast} & w_1 \end{bmatrix} \\ &= \det \begin{bmatrix} x & 1 & 1 \\ y & 3 & 1 \\ w & 0 & 1 \end{bmatrix} \\ &= 3x - y - 2w , \\ 0 &= \det (\bold{p},\bold{p}_{\ast},\bold{p}_2) \\ &= \det \begin{bmatrix} x & x_{\ast} & x_2 \\ y & y_{\ast} & y_2 \\ w & w_{\ast} & w_2 \end{bmatrix} \\ &= \det \begin{bmatrix} x & 1 & -1 \\ y & 3 & -1 \\ w & 0 & 1 \end{bmatrix} \\ &= 3x - y + 2w . \end{align}
Setting w = 1 gives the two equations without homogeneity:
\begin{align} y &= 3x - 2 , \\ y &= 3x + 2 . \end{align}
And we're done. Really, the only subtle part is the use of a polar curve.
This is algebraic geometry as it was done in the late 1800s. Today the field has become something entirely different, rewriting its foundations in terms of commutative algebra and a powerful abstraction called schemes. Another significant development is the use of computer algebra systems employing a tool for polynomials called the Gröbner basis. --KSmrqT 08:17, 27 September 2007 (UTC)
I'm totally mystified - is it true that in the 1800's they would do all the above, but couldn't solve the problem in a few lines using the equality of the two differentials and simple algebra???83.100.254.236 11:16, 27 September 2007 (UTC)
Computation has always been an essential skill for mathematicians, even more so before the advent of electronic digital computers; they could probably have solved the problem six ways from Sunday. If you look closely at this "fancy" method, you'll see that it is much the same as the simple one, but with a different language and more generality. That's a Good Thing, because it helps us better understand and appreciate the more sophisticated tools. If we strip out the explanatory verbiage, the computation itself is brief.
For those who know very little, this is a (hopefully enticing) hint that there is a wider universe to explore. For those who know more, this is a reminder that there can be interesting mathematics in elementary questions. For those who don't know and don't care, this is supplementary material that can be ignored; I won't be offended (only disappointed). --KSmrqT 19:20, 27 September 2007 (UTC)
Doh! I was so excited when I pressed that link - I thought it was going to be an introduction to the methods you used, but no just a definition of 'hopefully' Eh!. Oh well at least it was relevent.87.102.83.163 16:40, 28 September 2007 (UTC)
I included the link for the usage note, not the definition. Everyone one knows what "hopefully" means here, yet some people insist that we should never use the word like this, and that I'm an ignorant boob if I do so. The link mocks them. ;-)
Online sources to pursue this would be great; I wish I could offer more. Homogeneous coordinates are heavily used in computer graphics, but rarely with a discussion of points at infinity and the projective plane. The projective plane is classic geometry, but use of coordinate methods is not so popular. As for pole/polar ideas used this way, discussions do exist but are hard to find. Printed sources are Semple & Roth, Introduction to Algebraic Geometry (ISBN 978-0-19-853363-4), though the 1949 style is uninviting; and Salmon, A Treatise on the Higher Plane Curves (online at the Internet Archive), with an even more archaic style. I know computer vision and computer graphics have both had a few publications, such as "An ellipse detection method from the polar and pole definition of conics" (doi:10.1016/0031-3203(93)90039-Y) and "Scan line display of algebraic surfaces" (doi:10.1145/74334.74348). Hope this helps! --KSmrqT 22:13, 28 September 2007 (UTC)
Maybe for your next trick you'd like to derive the boltzmann distribution?83.100.254.236 12:06, 27 September 2007 (UTC)