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October 7

Name

What's the name given to numeric 6's and 9's that have their thingos swinging out, like the ones in this font? February 15, 2009 (talk) 00:23, 7 October 2008 (UTC)

What font? Wikipedia text is shown in your browser's default sans-serif font, which is probably not the same as my browser's default sans-serif font. Anyway, perhaps you're referring to text figures? —Ilmari Karonen (talk) 04:23, 7 October 2008 (UTC)

I believe that "seriff" is the "feet" on letters such as i, l, t etc... But I believe the rule might be similar to whether or not a lowercase Q or a lowercase J gets curved at bottom, or just a straight line. Guessing from memory (and yes I had to learn this for a computer class) it might be Times, but there's also a rule of whether or not a W gets the same amount of width, as a lowercase L. Its interesting stuff, but your question about 6's for example, is probably the same answer as to the stipulation about lowercase Q and lowercase J. That might help you find it easier, since you can expand your question, i.e. a google search. Sentriclecub (talk) 10:45, 7 October 2008 (UTC)

Try the article on font. Sentriclecub (talk) 10:46, 7 October 2008 (UTC)

Swash (typography) that should be it. Each typeface has its own rules of whether or not to give swashes onto a 6 or 9. Sentriclecub (talk) 10:56, 7 October 2008 (UTC)

Also see Regional handwriting variation#Arabic numerals --71.106.183.17 (talk) 04:36, 10 October 2008 (UTC)

Compass and Midpoint

I recently encountered a problem related to geometry drawing, but did not find an article on it.

So basically, you have a compass. No ruler or anything else. This means you cannot draw a straight line. So given a line segment, use the compass to find the midpoint of it.

Anyways if there is any relevance in this, I thought it might be important to write an article related to geometry drawing? Any tips on the problem let me know ;).

--electricRush (T C) 03:23, 7 October 2008 (UTC)

Are you absolutely sure that you do not have access to a straightedge? Because if you did, you could draw two congruent circles centered on the endpoints of the line segment. Then you would draw a line between the points of intersection between these circles, and then that line would intersect with the original segment at its midpoint. 206.117.40.10 (talk) 03:55, 7 October 2008 (UTC)

Yes, for this problem you may not use a straightedge. All you can do is use the compass to draw circles. This problem is supposed to be very hard though... --electricRush (T C) 04:08, 7 October 2008 (UTC)

Can you access this file (PS)? It describes how to perform the operation. It also speaks of the (for me unknown until tonight) Mohr-Mascheroni theorem. Pallida  Mors 06:52, 7 October 2008 (UTC)

In case you can't handle the postcript, here there's a web site. Pallida  Mors 07:04, 7 October 2008 (UTC)

Even more amazingly see Weisstein, Eric W. "Matchstick Construction". MathWorld., you only need matchsticks to do anything a normal compass and straightedge can do. Dmcq (talk) 07:55, 7 October 2008 (UTC)

Infinite increase

If I have an integer, let's take 1, and I increase that integer at an infinite speed (or rate), does my number become infinite in a zero time period? —Anonymous Dissident^Talk 07:40, 7 October 2008 (UTC)

It's hard to increase an integer; they're pretty stubborn about staying just what they are. (An integer variable is another matter.)

Extrapolating from your question, it's quite possible for a function to have infinite first derivative at a point and yet be pretty well behaved generally. (OK, technically, the derivative won't be defined at that point, but it's pretty obvious what "infinite first derivative" means.) An example is the cube-root function, whose derivative becomes infinite at zero, but (the original function) takes only finite values on the real line. --Trovatore (talk) 07:54, 7 October 2008 (UTC)

From the concept of a massless string, the answer is yes. If you apply a net force to a massless string, it will accelerate infinitely and reach an infinite velocity, instantly. But you can't apply a force, before the force is applied, which is why your Q is a bit tricky. As soon as the instant you apply your force, say at t=2.7 seconds, then yes it instantly reaches infinite velocity. Oops, upon second thought, you would need an infinite acceleration, not an infinite speed (or rate). But since you're asking about the number reaching an infinite "position" (which i'm butchering your question now), then yes if you apply an infinite velocity to an object, it will instantly reach an infinite position. Just the same that a tiny net-force applied to a massless string will cause it to reach an instant infinite velocity, then the reasoning holds true about your original question. So the answer is yes but you could better clearify the last 5 words, since the word "instantly" is what you are asking, right? Sentriclecub (talk) 10:39, 7 October 2008 (UTC)

Actually the answer is "no" (or more exactly, "not necessarily"). That's what my first answer said, but apparently I didn't say it clearly enough. --Trovatore (talk) 12:50, 7 October 2008 (UTC)

Correlation between shapes

I am looking for a formula which determines a correlation coefficient which measures the correlation between two shapes in a two-dimensional space. Each shape is given by connecting the heads of six vectors all pointing in independent directions with angles of 2*pi/6 radians between them. There are two shapes. If the two shapes are identical then the correlation should be one. If the shapes are pointing in opposite directions we would expect the correlation to be negative. The correlation coefficient should between minus one and plus one. Does anybody know how to construct such a formula? HowiAuckland (talk) 10:34, 7 October 2008 (UTC)

I'm afraid you can't. Think again what 'the shapes are pointing in opposite directions' means.

Imagine a regular hexagon, centered at the coordinate system's origin. Let P and Q be two opposite vertices of the figure. Make two new shapes by moving P and Q, respectively, a bit outside from their original positions. I suspect you would call it 'shapes pointing in opposite directions', thus assigning a correlation of −1 to that pair. However if the vertices displacement is small enough, the shapes stay 'almost' identical to each other (and to the original hexagon) so their 'correlation' ought to be close to +1.

So we have a pair of shapes which deserve a correlation of −1 and +1 at the same time. How are you going to solve it?

CiaPan (talk) 13:28, 7 October 2008 (UTC)

Hi CiaPan. Thanks for your reply and you may very well be correct in your assertion. However, in your example just noted the second polygon isn't actually pointing in an opposite direction as the first polygon. In fact, the second polygon is equally pointing, the same amount, in both opposite directions, which is very similar to the first polygon, so I would expect the correlation to be close to +1. Your example has no assymetry, and therefore I expect the correlation to be close to +1. The negative correlation should come from an asymmetry between the two polygons. An example of two polygons which I think should give a negative asymmetry is if you took your example, and instead of stretching out both vertices P and Q equally, you simply stretch one side, say P. Now I would think this would have a negative correlation between these two shapes.

A quandary that I have come across which has made me think that the problem might not be well-defined is if you take as your first polygon, a polygon with P stretch outward significantly in one direction, and the second polygon stretched outward at Q in the opposite direction, then you should get a negative correlation. However if you then look at the limit as the P and Q vertices are shrunk back to the same length as the other vertices from the center, then the correlation should get progressively negative but closer to zero. The quandary is that in that limit one might also expect the correlation to approach +1 since in that limit the two shapes are identical. Does anybody think that this example also indicates the possible non-well-posedness of the problem? HowiAuckland (talk) 20:53, 7 October 2008 (UTC)

You could calculate the mean of the six points and measure the angle between the means for each shape (considered in polar co-ordinates). That would give you a number between 0 and pi, call it x. Then consider 1-2x/pi, does that satisfy your requirements? It would consider two shapes that are just simple scalings of eachother to be identical, is that a problem? --Tango (talk) 23:09, 7 October 2008 (UTC)

This quantity would be undefined when the means for either shape add up to zero. HowiAuckland (talk) 23:36, 7 October 2008 (UTC)

That's a good point, but I think that's due to your question. You talk about whether or not they are pointing in the same direction, but a regular hexagon doesn't point in any direction. --Tango (talk) 23:49, 7 October 2008 (UTC)

Very true. Regular hexagon wouldn't point in any direction. However, the correlation between two identical shapes should always be +1, and if they are very similar it should be close to +1. (I think, subject to the quandary I listed above, which I can't resolve.) HowiAuckland (talk) 23:58, 7 October 2008 (UTC)

If you're willing to skip regular hexagons, you could just literally use the correlation of the two sextuples of vector lengths. Then any other shape would have correlation 1 with itself and would typically have negative correlation with itself rotated or reflected. But you're not going to get perfect -1 in general. --Tardis (talk) 00:37, 8 October 2008 (UTC)

Yes, I think this might be a good idea, except it seems to give strange results. As an example, in the case where you have all the lengths zero except one vector equal to 1 in the first shape, and then in the second shape you have all lengths to be zero except for the opposite direction with length of 1. Then this gives a correlation of −0.2. Similarly, if you take all lengths to be zero except for a single entry, regardless of position, to be one, you will always get a correlation of −0.2, even when the vectors in the two shapes are pointing nearly in the same direction. I would expect this case to give a positive correlation. Even though the sign is curious, certainly they shouldn't all give the same value of the correlation. They are entirely different shapes. HowiAuckland (talk) 01:35, 8 October 2008 (UTC)

OK, then take Tango's idea but use the cosine of the angle (that is, the dot product of the normalization of the sum of the vectors for each shape): ${\vec {X}}:=\sum \nolimits _{i}{\vec {x}}_{i}$, $c(\{{\vec {x}}\},\{{\vec {y}}\}):={\hat {X}}\cdot {\hat {Y}}$. Then rotating a shape by any angle will give a correlation with the original equal to the angle's cosine. It's still undefined if either of the shapes is regular, but you can perturb the angles slightly and get a function (of the 12 scalars) that's always defined but is very sensitive near the line $s_{1}=s_{2}=\dots =s_{12}$. You could also consider the (cosine of the) angle between the centroids of the two shapes; usually these are nearly the same thing, of course, so I don't know how to tell which one is better for your purposes. --Tardis (talk) 16:54, 8 October 2008 (UTC)

It's a good idea, but we are looking for a formula which works in all cases. As you have pointed out, in certain limits , i.e. regular, or small sums of vectors which are perturbed, the procedure is undefined, or not well defined or doesn't give answers that you should expect, such as +1. We had actually tried this idea previously and saw the defects of the procedure. Thanks so much for your opinions in this matter.HowiAuckland (talk) 22:21, 8 October 2008 (UTC)

I've been coming to the opinion, as CiaPan indicated above, that such a finalized formula for a single real number between −1 and +1, which works for all cases is most likely impossible to generate. I have been thinking that in order to study the structure of such a system, at least two numbers of correlation should be defined. A measure of asymmetry $(-1\leq c_{1}\leq +1)$ and a measure of similarity $(0\leq c_{2}\leq +1)$. I know how to determine a measure of similarity. I extracted this from similar concepts in hyper-spherical coordinates: Consider an angle defined by the formula $\tan \alpha _{i}={\frac {y_{i}}{x_{i}}}$ (with $\alpha _{i}$ set to be 0 in both cases of either $x_{i}=0$ or $y_{i}=0$, and $\tan \alpha _{i}={\frac {x_{i}}{y_{i}}}$ is equally good). Now consider $c_{2}={\frac {1}{6}}\sum _{i=1}^{6}\sin ^{2}(2\alpha _{i}).$ This works very well for a coefficient of similarity, but this measure does not capture the concept of asymmetry. Now I just need a formula, which is well-defined in all limits, to compute $c_{1}$ a coefficient of asymmetry. However, there is still the possibility that I might have missed something in my analysis and that a single formula for a single number might be well-definable. I still don't know for sure. HowiAuckland (talk) 22:21, 8 October 2008 (UTC)

side, edge of a circle

what is the difference between or the meaning of side of a circle and edge of a circle... How many sides does a circle have? How many edges does a circle have?

116.75.181.164 (talk) 10:43, 7 October 2008 (UTC)

The boundary of a circle is neither called side nor edge, but circumference. So the short answer to your question is: zero sides and zero edges. However it does make some sense to say that the circle has an infinite number of sides, as it is approximated sufficiently well by a polygon having sufficiently many sides. Bo Jacoby (talk) 11:21, 7 October 2008 (UTC).

But then again, infinity is not a number. —Preceding unsigned comment added by 81.187.252.174 (talk) 11:54, 9 October 2008 (UTC)

The 1-D Heat Equation

Help! I don't understand this Heat Equation stuff.

I have a problem with a semi-infinite cylinder of metal on the positive x-axis with sides and face at x=0 insulated. The heat will therefore diffuse according to the 1-D heat equation. The initial temperature distribution is $u(x,0)=u_{0}\exp(-x/L)$, the end is insulated which indicates that $u_{x}=(0,t)=0$.

I'm trying to solve this using Green's function, extended to the infinite axis as an even function, and then changing variables to get the error function. But I have been given the solution and no matter what I try I can't get to it. I don't think I understand exactly how to incorporate the initial value function as part of the integrand.

The solution given is $u(x,t)=(1/2)u_{0}exp(c^{2}t/L^{2})[exp(-x/L{1+erf((x-2c^{2}t/L)/{\sqrt {(}}4c^{2}t)}+exp(x/L){1-erf((x+2c^{2}t/L)/{\sqrt {4c^{2}t}}}]$

If anyone can explain where this solution comes from I would really appreciate it. —Preceding unsigned comment added by 122.108.235.248 (talk) 13:49, 7 October 2008 (UTC)

I suppose you mean $u_{x}(0,t)=0$. The c is unexplained. There are some missing parentheses in the formula which might have been

$u(x,t)=(1/2)u_{0}\exp(c^{2}t/L^{2})[\exp(-x/L)(1+\operatorname {erf} ((x-2c^{2}t/L)/{\sqrt {4c^{2}t}}))+\exp(x/L)(1-\operatorname {erf} ((x+2c^{2}t/L)/{\sqrt {4c^{2}t}}))]$

where

$\operatorname {erf} (x)={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}dt$

is the error function.

Simplify the computations by choosing the unit of temperature to be $u_{0}$, and the unit of length to be $L$, and the unit of time such that $c=1$. Then the formula looks like this:

$u(x,t)=e^{t}[e^{-x}(1+\operatorname {erf} ((x-2t)/{\sqrt {4t}}))+e^{x}(1-\operatorname {erf} ((x+2t)/{\sqrt {4t}}))]/2$.

The initial condition at t→0⁺ is

${\begin{aligned}u(x,0^{+})&=[e^{-x}(1+\operatorname {erf} (x/0^{+}))+e^{x}(1-\operatorname {erf} (x/0^{+}))]/2\\&=[e^{-x}(1+\operatorname {erf} (+\infty ))+e^{x}(1-\operatorname {erf} (+\infty ))]/2\\&=[e^{-x}(1+1)+e^{x}(1-1)]/2\\&=e^{-x}.\end{aligned}}$.

OK! Then check the other initial condition

$u_{x}(0,t)=0\,$

and the heat equation.

$u_{t}-u_{xx}=0\,$

Bo Jacoby (talk) 20:02, 7 October 2008 (UTC).

Continuing. The u-function can be written compactly

$u={\frac {u_{+}+u_{-}}{2}}\,$

where $u_{+}$ and $u_{-}$ are defined by

$u_{\pm }=e^{t\mp x}\cdot (1\pm \operatorname {erf} (z_{\pm }))$

and $z_{+}$ and $z_{-}$are defined by

$z_{\pm }=t^{-{\frac {1}{2}}}({\frac {x}{2}}\mp t).$

The differential is

$du={\frac {du_{+}+du_{-}}{2}}$

where

${\begin{aligned}du_{\pm }&=d(e^{t\mp x}\cdot (1\pm \operatorname {erf} (z_{\pm })))\\&=d(e^{t\mp x})\cdot (1\pm \operatorname {erf} (z_{\pm }))+e^{t\mp x}\cdot d(1\pm \operatorname {erf} (z_{\pm }))\\&=e^{t\mp x}\cdot (1\pm \operatorname {erf} (z_{\pm }))\cdot d(t\mp x)\pm e^{t\mp x}\cdot 2\cdot \pi ^{-{\frac {1}{2}}}\cdot e^{-z_{\pm }^{2}}\cdot dz_{\pm }\\&=u_{\pm }\cdot (dt\mp dx)\pm e^{t\mp x}\cdot 2\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}-t\pm x}\cdot ({\frac {1}{2}}t^{-{\frac {1}{2}}}dx-{\frac {1}{4}}xt^{-{\frac {3}{2}}}dt\mp {\frac {1}{2}}t^{-{\frac {1}{2}}}dt)\\&=u_{\pm }dt\mp u_{\pm }dx\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}\cdot (t^{-{\frac {1}{2}}}dx-{\frac {1}{2}}xt^{-{\frac {3}{2}}}dt\mp t^{-{\frac {1}{2}}}dt)\\&=(\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})dx+(u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (-{\frac {x}{2}}\mp t))dt\\&=\mp (u_{\pm }-\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})dx+(u_{\pm }+\pi ^{-{\frac {1}{2}}}e^{-{\tfrac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (\mp {\frac {x}{2}}-t))dt\end{aligned}}$

This means that

${\frac {\partial u_{\pm }}{\partial x}}=\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}$

${\frac {\partial u_{\pm }}{\partial t}}=u_{\pm }+\pi ^{-{\frac {1}{2}}}e^{-{\tfrac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (\mp {\frac {x}{2}}-t)$

The initial condition

$u_{x}(0,t)=0\,$

is checked:

${\begin{aligned}u_{x}(x,t)&={\frac {1}{2}}\left({\frac {\partial u_{+}}{\partial x}}+{\frac {\partial u_{-}}{\partial x}}\right)\\&={\frac {1}{2}}\left(-u_{+}+\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}+u_{-}-\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}\right)\\&={\frac {1}{2}}\left(-u_{+}+u_{-}\right)\\&={\frac {1}{2}}\left(-e^{t-x}\cdot (1+\operatorname {erf} (z_{+}))+e^{t+x}\cdot (1-\operatorname {erf} (z_{-}))\right)\\&={\frac {e^{t}}{2}}\left(-e^{-x}\cdot (1+\operatorname {erf} (t^{-{\frac {1}{2}}}({\frac {x}{2}}-t)))+e^{+x}\cdot (1-\operatorname {erf} (t^{-{\frac {1}{2}}}({\frac {x}{2}}+t)))\right)\end{aligned}}$

The limit for $x\rightarrow 0$ is

${\begin{aligned}u_{x}(0,t)&={\frac {e^{t}}{2}}\left(-(1+\operatorname {erf} (t^{-{\frac {1}{2}}}(-t)))+(1-\operatorname {erf} (t^{-{\frac {1}{2}}}(+t)))\right)\\&={\frac {e^{t}}{2}}\left(-\operatorname {erf} (-t^{\frac {1}{2}})-\operatorname {erf} (t^{\frac {1}{2}})\right)\\&=0\end{aligned}}$

because

${\begin{aligned}\operatorname {erf} (-z)=-\operatorname {erf} (z)\end{aligned}}$

So far I have shown you that the u-function satisfies the two boundary conditions. It remains to be shown that it also satisfies the heat equation. Some other day, perhaps. Is this helpful at all? Bo Jacoby (talk) 20:08, 9 October 2008 (UTC).

Here we go again. A second order partial derivative has to be computed.

${\begin{aligned}{\frac {\partial ^{2}u_{\pm }}{\partial x^{2}}}&={\frac {\partial }{\partial x}}{\frac {\partial u_{\pm }}{\partial x}}\\&={\frac {\partial }{\partial x}}(\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})\\&=\mp {\frac {\partial {u_{\pm }}}{\partial x}}\pm \pi ^{-{\frac {1}{2}}}t^{-{\frac {1}{2}}}{\frac {\partial {e^{-{\frac {x^{2}}{4t}}}}}{\partial x}}\\&=\mp (\mp u_{\pm }\pm \pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}})\pm \pi ^{-{\frac {1}{2}}}t^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}{\frac {\partial {-{\frac {x^{2}}{4t}}}}{\partial x}}\\&=u_{\pm }-\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {1}{2}}}\mp {\frac {1}{2}}\pi ^{-{\frac {1}{2}}}e^{-{\frac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}x\\&=u_{\pm }+\pi ^{-{\frac {1}{2}}}e^{-{\tfrac {x^{2}}{4t}}}t^{-{\frac {3}{2}}}\cdot (\mp {\frac {x}{2}}-t)\\&={\frac {\partial u_{\pm }}{\partial t}}\end{aligned}}$

So $u_{+}$ and $u_{-}$ each satisfies the heat equation, which is linear, and therefore the linear combination

$u={\frac {u_{+}+u_{-}}{2}}\,$

also satisfies the heat equation

${\frac {\partial ^{2}u}{\partial x^{2}}}={\frac {\partial u}{\partial t}}$

I hope this answered your question. I got routine in writing wikipedia formulas. Bo Jacoby (talk) 15:40, 10 October 2008 (UTC).

Simple statistics question

(related to Talk:Mongolia): Assuming that 800 children are born in a hospital in one year, and assuming that the probability that a newborn is a boy is 0.5 (feel free to calculate with a more correct value, but then tell me so), what is the probability of more than 430 of the newborn being girls?

My impression is that this should be roughly 0.025 (normal distribution, E=400, sigma square = 200). Anything wrong about this? Yaan (talk) 18:50, 7 October 2008 (UTC)

Using a normal approximation should get a pretty good answer for n=800 (I haven't checked to see if you calculated it right, but the method is good). You could also do it directly as a binomial distribution (that article should have the formulae you need). The normal approximation is easier, though, if you don't need an absolutely precise answer. --Tango (talk) 18:59, 7 October 2008 (UTC)

From human sex ratio, "The natural sex ratio at birth is estimated to be close to 1.1 males/female", so you need to take that into account. But your stat is the reverse!? Saintrain (talk) 19:48, 7 October 2008 (UTC)

Yes, but the numbers are quite low, so it's actually not that unlikely of you repeat the "experiment" over a number of years, or a number of hospitals. The actual numbers from a statistical yearbook were 395 boys vs. 424 girls, i.e. a deviation from the mean of 14.5 if you use a sex ratio of 1. Standard deviation would then be 14.31. Though if you assume a ratio of 1.05 (p("boy")=0,512), the deviation would be 24.5, with the standard deviation slightly lower than before, and if you assume a ratio of 1.1 (p("boy")=0,524), the deviation from the mean would be 34, and the standard deviation only around 14.2. Yaan (talk) 14:43, 8 October 2008 (UTC)

The standard deviation is 200^1/2=14.1421. The deviation of the observation from the mean is 430−400=30. And 30/14.1421=2.12133. The probability that a normally distributed observation is greater than the mean + 2.12133 standard deviations is 0.016947, or roughly 1/59, (as compared to 0.025=1/40). Bo Jacoby (talk) 21:17, 7 October 2008 (UTC).

If you're going to claim so much accuracy, you should do a continuity correction using the fact that in this context, "> 430" is the same as "≥ 430.5". So

${\frac {430.5-400}{\sqrt {200}}}=2.1567,$

and the probability that the standard normal exceeds that is about 0.015515, or about 1/64. Michael Hardy (talk) 01:08, 8 October 2008 (UTC)

You are absolutely right! Do you know the binomial probability without the error due to the normal distribution approximation? Bo Jacoby (talk) 05:51, 8 October 2008 (UTC).

The exact answer is given by the binomial distribution. The R idiom would be sum(dbinom(431:800 , 800 ,prob=0.5)), giving 0.01548346. Robinh 11:13, 8 October 2008 (UTC) [what is the correct markup for verbatim?]

Try pre or code tags, one of them should do what you want. You may also require nowiki tags. I've never quite got my head round the different combinations of those tags. --Tango (talk) 12:50, 8 October 2008 (UTC)

Thanks a lot everybody. Yaan (talk) 14:43, 8 October 2008 (UTC)

October 8

Minimizing first-time comparisons

What's the best sorting algorithm that minimizes the number of first-time comparisons? This generally involves data that is expensive to initially compare (or otherwise can't be algorithmically compared), but once the initial comparison is made, the result can be cached.

After a new comparison is made, my system looks for things similar to A < B < C and ensures that A < C. It then searches the cache to find which pair eliminates the most combinations. Even though this works, I feel this may be the bottleneck, since a full search (probably O(n^4+)) could make the sorting much longer than it could, and a limited search (currently O(n^3)) may be less accurate.

I'm aware of a monte-carlo algorithm, but I'm not sure whether to go for that or the full-search in the comparison matrix. --Sigma 7 (talk) 02:29, 8 October 2008 (UTC)

If the elements are in random order initially then you can't really beat a good standard sort by anything significant for smallest average number of comparisons. The Comparison sort article also mentions that 42 is the least maximum number of comparisons needed for 15 items, I don't know if it is best on average though - some hand crafted algorithm like this could be used in the middle of another sort to improve it slightly - about five is as much as people deal with specially normally. Dmcq (talk) 08:41, 8 October 2008 (UTC)

Merge sort comes very close to the lower bound on the number of comparisons. It uses n ceiling(log₂n) - n + 1 comparisons, while the comparison tree lower bound is approximately n log₂n - 1.4427 n comparisons. It's not optimal, but it's straightforward and good. —David Eppstein (talk) 17:07, 9 October 2008 (UTC)

The Ford-Johnson algorithm mentioned in the article on comparison sorts is a kind of merge sort, not very efficient internally but very close to optimal in the number of comparisons. Dmcq (talk) 17:36, 9 October 2008 (UTC)

Possible to solve this calculus problem?

I saw [1] on Uncyclopedia and I was wondering if it really could be solved.

Can it? (If you take out the Tuesday thing) What if you substituted values?

Would anybody be willing to solve it? (You can use your own units or whatever) Or is it lacking vital information for that?

~Cheers!~ User: ECH3LON

P.S. -If this would be considered "spam", I apologize, just ignore it if that's the case :) —Preceding unsigned comment added by ECH3LON (talk • contribs) 23:46, 8 October 2008 (UTC)

I think the 'problem' is pretty obviously a joke, as might be expected given the source. AndrewWTaylor (talk) 09:36, 9 October 2008 (UTC)

The problem is complete nonsense, it's just a random sequence of mathematical words, they have no meaning in that combination. --Tango (talk) 10:30, 9 October 2008 (UTC)

I don't know. I found a really complicated quantum physics equation on Uncyclopedia one day. It's in my sandbox if you would like to see. :) Ζρς ι'β' ^¡hábleme! 21:32, 9 October 2008 (UTC)

A lengthy discussion on that very equation is in the ref desk archives (maths, I think, possibly science)! Just because that's a genuine equation doesn't mean "Matrix the antiderivative to the sixth power where needed" has any actual meaning. :) --Tango (talk) 22:04, 9 October 2008 (UTC)

Yep, science archives. Oh, I know that thing doesn't have any real meaning, I was just saying that there are some legitimate things on Uncyclopedia. Ζρς ι'β' ^¡hábleme! 22:19, 9 October 2008 (UTC)

October 9

Order of subgroups

I am studying abstract algebra and group theory, and came across a problem I'm not sure how to solve. Given subgroups H and K of a finite group G, we want to show that |HK| = (|H||K|)/|H intersect K|, where |H| denotes the order (number of elements) of H. I haven't studied cosets yet, and have been told by several math professors that we need them to solve this. Does anyone have advice on where to start, without using cosets? Many thanks. 134.129.57.188 (talk) 02:16, 9 October 2008 (UTC)

The question uses the notion of left cosets anyway! Left cosets are very simple to understand once you know what a subgroup is. Basically, left cosets are defined as follows: Let H be a subgroup of G. Then a left coset of H in G is:

a*H = {a*h | h is in H}

for some a in G. A right coset is similarly defined where we right multiply a with every element of H. Note that the collection of all left (or right) cosets form a partition of G. Basically this means:

• The union of all the left cosets of H equals G

• If two left cosets intersect, they must be equal

The first property is very easy to see (try to see it yourself). To see the second property, suppose that two left cosets a*H and b*H intersect. Then there exists h₁ in H and h₂ in H such that a*h₁ = b*h₂ (don't make the mistake of assuming that h₁ = h₂; this need not be the case). It follows that a*h₁*h₂^-1 = b. From this one can deduce that a*H equals b*H (try to see this yourself by showing that b*H is a subset of a*H and conversly).

Now to solve your problem notice that HK = U {h*K | h is in H} so that the cardinality of HK is just the cardinality of K multiplied by the no. of h's that are there (not counting twice h₁ and h₂ if h₁*K = h₂*K). This is just your expression. To see this formally, we define an equivalence relation on H to set h₁ ~ h₂ iff h₁*K = h₂*K. Then the cardinality of K times the number of equivalence classes that are there is simply the cardinality of HK. Furthermore, this expression equals yours (I will leave you to work out why since I am not really allowed to help you with your homework. Just show that the cardinality of H divided by the cardinality of H intersection K is the number of equivalence classes that are there).

Topology Expert (talk) 05:54, 9 October 2008 (UTC)

Just one note. If H intersection K is empty, the problem does not make any sense. Are there any other additional assumptions involved in the problem (i.e assuming that K and H are not disjoint)?

Topology Expert (talk) 06:04, 9 October 2008 (UTC)

Their intersection contains the identity.76.126.116.54 (talk) 06:28, 9 October 2008 (UTC)

How could I have said that! The problem is OK. Just go ahead and use my method (or any other method) and you should come up with the answer (note: the only possible methods to solve this problem would be to use cosets since that is how the problem is defined).

Topology Expert (talk) 09:37, 9 October 2008 (UTC)

Polarities

Can someone point me to a proof that every polarity of a finite projective plane has absolute points? In other words, that any bijection between the points and lines of the plane which preserves incidence takes some point to a line containing it. Black Carrot (talk) 06:43, 9 October 2008 (UTC)

Have you tried looking in Coxeter's books? I don't really know projective geometry so I can't help. – b_jonas 11:57, 15 October 2008 (UTC)

Combinations

Hello. Is there a way to find r choose n but the set (n) has i identical elements? Sort of like choosing combination of three letters from the word CANADA. I know how to do it for a given problem, but I wonder if there is something general. Thx Brusegadi (talk) 06:30, 10 October 2008 (UTC)

If you want non-identical elements, you just have to remove the duplicates from the set and then choose from that, so instead of rCn it's (r-a)Cn where a is the # of elements removed. -mattbuck (Talk) 12:09, 9 October 2008 (UTC)

So in the example above, if I want to choose three, I will do 4 choose three, which is 4. But, I can come up with more than that: AAA, AAC, AAN, AAD, CND... The real problem are cases like AAN, which I would count twice if I took out one A and inserted another A. Did I miss something obvious? Brusegadi (talk) 06:30, 10 October 2008 (UTC)

Since you are concerned about identical elements in combinations, don't you mean that you would not count AAN twice? Since if you do count AAN multiple times, once for each pair of distinct A's that could be chosen, you end up with the standard combination count ${\binom {6}{3}}=20$. 84.239.160.166 (talk) 09:59, 12 October 2008 (UTC)

There is a formula for exactly what you want, but I can't remember it. I think it's the usual choose formula but with an extra factor in the denominator. I'll try and find it/work it out. --Tango (talk) 17:09, 9 October 2008 (UTC)

Check out the multinomial theorem. 84.239.160.166 (talk) 20:10, 9 October 2008 (UTC)

Note, that is talking about permutations rather than combinations, so doesn't exactly answer the OP's question. --Tango (talk) 21:02, 9 October 2008 (UTC)

Thanks for your help. I also remember seeing a formula somewhere, and it is killing me! I will try to remember during the weekend, since I am preoccupied with work :) Brusegadi (talk) 06:30, 10 October 2008 (UTC)

Ok, sorry about the confusion. Given that there are specific numbers of identical elements, the number of combinations can be counted as follows. Suppose there are elements of $k$ types, with the number of elements of each type given by $n_{1},\dots ,n_{k}$, and you want to count combinations of $m$ elements. In the CANADA example we have $k=4$ types of elements, $n_{1}=3$ for the letter A and $n_{2}=n_{3}=n_{4}=1$ for the letters C, N and D.

Each combination of $m$ elements can be identified with a vector of non-negative integers $i_{1},\dots ,i_{k}$ such that $\sum _{j=1}^{k}i_{j}=m$ and $i_{j}\leq n_{j}$ for all $j=1,\dots ,k$. Denoting the number of such combinations by $C(m;n_{1},\dots ,n_{k})$, and by considering the effect of removing all elements of type $k$ we get the recursion

$C(m;n_{1},\dots ,n_{k})=\sum _{i=0}^{\min(n_{k},m)}C(m-i;n_{1},\dots ,n_{k-1})$

with the base cases $C(m;n_{1})=1$ for $m\leq n_{1}$ and $C(m;n_{1})=0$ for $m>n_{1}$.

The recursion formula isn't as simple as one might hope for, but it is relatively easy to implement on a computer. In the CANADA example even computation by hand is feasible:

$C(3;3,1,1,1)=C(3;3,1,1)+C(2;3,1,1)$

$=C(3;3,1)+2C(2;3,1)+C(1;3,1)$

$=C(3;3)+3C(2;3)+3C(1;3)+C(0;3)$

$=1+3*1+3*1+1=8$

The 8 different combinations are AAA, AAC, AAN, AAD, ACN, ACD AND CND. In this particular case, the fact that $n_{2}=n_{3}=n_{4}=1$ makes the first three steps resemble a binomial expansion, and the fourth step follows directly from the base case of the recursion. In fact, this gives a hint about a less general formula that would directly address the original question... but I'll leave that as an exercise for the reader. 84.239.160.166 (talk) 09:59, 12 October 2008 (UTC)

Thanks. This problem was bothering me one night because I was convince there was a simple plug and chug out there. But it seems like the summation is as general as it gets. Thanks for generalizing it, by the way. You were very helpful. Good day, Brusegadi (talk) 07:24, 14 October 2008 (UTC)

simultaneous equations

Is substitution the best way to find the solution for this set of simultaneous equations or is there a better method? (all results are zero) -- Taxa (talk) 20:08, 9 October 2008 (UTC)

You have far more variables than you do equations, so there won't be a unique solution. Adding and subtracting equations or multiples of equations from eachother is the standard way of solving simultaneous equations, although substitution can also help. You'll only be able to get so far though since you'll have used up all the equations and will still have lots of variables left. --Tango (talk) 21:00, 9 October 2008 (UTC)

Probably a little more efficient would be to rearrange the equations so that the right-hand sides all equal zero, then put that into a matrix and apply row reduction to help you determine the solution space (which, as Tango says, will involve a whole continuum of possible solutions, not just one unique one). Confusing Manifestation(Say hi!) 22:18, 9 October 2008 (UTC)

To be clear, it's more efficient notationally, but it's the same actual method. --Tango (talk) 22:43, 9 October 2008 (UTC)

Actually, in this particular case, substitution probably is fastest. Unless I've made a mistake, there is only one variable, u, that appears on both the left and right -- it appears on the left in the first equation, and on the right in the last and second-to-last equation. If you just substitute for u, eliminating the first equation, you're done. A solution is characterized by choosing any values you like for the variables remaining on the right, and then plugging those values in to get the remaining variables on the left. --Trovatore (talk) 00:42, 10 October 2008 (UTC)

No, the mistake is mine and corrected. The u on the left in the first equation should have been xx since u rather than t was the last variable used on the right side of the equations. All variables on the left should have been zeros instead as now shown in the corrected image. -- Taxa (talk) 02:44, 10 October 2008 (UTC)

Limit of an α-sequence

In my Introductory Set Theory course, we just started talking about α-sequences. We defined a limit to be $\lim _{\gamma \rightarrow \alpha }\beta _{\gamma }=\sup\{\beta _{\gamma }|\gamma <\alpha \}$. This seems counterintuitive to me, since if we talk about the identity sequence, $\lim _{\gamma \rightarrow 3}\beta _{\gamma }=2$, while $\beta _{3}=3$. I think. I know that, for example, $\lim _{\gamma \rightarrow \omega }\beta _{\gamma }=\omega$ and $\beta _{\omega }=\omega$. Am I confused, or is this what's expected? Thanks!Djk3 (talk) 22:31, 9 October 2008 (UTC)

I've never heard of an α-sequence, but my guess would be that that less than should be a less than or equal to, then the problem goes away. Double check the definition in your textbook. --Tango (talk) 22:46, 9 October 2008 (UTC)

An an α-sequence is generally only defined for γ<α. The notation is counterintuitive, but only for successor ordinals. For limit ordinals, you clearly need strict inequality to get anything remotely interesting. siℓℓy rabbit (talk) 23:04, 9 October 2008 (UTC)

What you've been given isn't really the definition; it's something equivalent to the definition when β is nondecreasing and α is a limit ordinal. Topologically, successor ordinals are isolated points; the notion of the limit of a function as its argument approaches an isolated point is not well-defined. And if the sequence β fails to be non-decreasing, then it could do something silly like take its maximum value at β₀ and then never get close to it again, and then the supremum is clearly different from the limit. --Trovatore (talk) 23:15, 9 October 2008 (UTC)

Okay, sure. That makes sense. I see now that I have "If β is non-decreasing," and I probably just missed "and α is a limit ordinal." Thanks a lot. Djk3 (talk) 00:10, 10 October 2008 (UTC)

October 10

The real deal with fitting lines to observed data points

I have a set of (x, y) data points with corresponding experimental uncertanties (dx, dy) which are different for the different points. I want to estimate the slope of the line that best approximates them. I want an estimation method with the following properties:

1. It should take into account the errors in both x and y (unlike linear regression).
2. It should be symmetric under the exchange of x and y (unlike linear regression).
3. The method should be scale-invariant; that is, even if x and y have different units, the best-fit line should not change if they are measured in different scales (unlike total least squares).
4. More accurate points (those with smaller uncertainties) should contribute more to the fit.

This doesn't seem like much to ask, but I can't find any method that actually satisfies these.

Something called "diagonal regression", "geometric mean functional relationship", "least areas line", "least products regression", "line of organic correlation", "neutral data fitting", or "reduced major axis" (do we have an article on it?) seems close, but I need a weighted version of it to satisfy the last criterion. —Keenan Pepper 06:54, 10 October 2008 (UTC)

I numbered your points. Point 3 is satified even by linear regression. To allow for point 2, try minimize Σz_i² where z_i=f(x_i,y_i)=ax_i+by_i+c and a²+b²=1. This is symmetrical in x and y. It provides the same result as linear regression, however, unless the line is vertical. Take account of point 1 = point 4 by minimizing a weighted sum Σa_iz_i² where a_i is the weigth of the (x_i,y_i). Bo Jacoby (talk) 08:02, 10 October 2008 (UTC).

No no no. This clearly does not satisfy 3, because if x has units of length and y has units of time, then a must have units of 1/length and b must have units of 1/time to make the first equation dimensionally consistent. In the second equation, however, a² and b² are added together, so they should have the same units, which is a contradiction. Therefore your method is not dimensionally consistent, and if I change the scale of x or y, the best-fit line will change. Also, there must be some mistake when you say it's symmetrical, yet provides the same result as linear regression. The result given by linear regression is NOT symmetrical (i.e. the slope of y as a function of x and the slope of x as a function of y are NOT inverses), except in the special case when all the points lie exactly on a line already. —Keenan Pepper 15:11, 10 October 2008 (UTC)

Have you looked at Errors-in-variables model? 155.91.45.231 (talk) 17:03, 10 October 2008 (UTC)

Keenan Pepper may be right but I am not quite convinced. Do you have a simple example where 'the slope of y as a function of x and the slope of x as a function of y are NOT inverses' ? Bo Jacoby (talk) 23:20, 10 October 2008 (UTC).

{(0,0), (1,1), (1,0)}. If you apply linear regression, i.e. minimize:

$\chi ^{2}=\sum (y_{i}-a*x_{i}-b)^{2}$ or its inverse $\chi ^{2}=\sum (x_{i}-a*y_{i}-b)^{2}$

Then the normal solution is y = 0.5*x, and the inverse solution is x = 0.5*y + 0.5. This is because this way of formulating the problem intrinsically treats uncertainty in x and y asymmetrically. Dragons flight (talk) 00:00, 11 October 2008 (UTC)

You can get a correct solution by minimzing the function:

$\chi ^{2}=\sum _{i}{(y_{i}-ax_{i}-b)^{2} \over (\sigma _{y_{i}}^{2}+a^{2}\sigma _{x_{i}}^{2})}$

You'll note that the goodness-of-fit function is dimensionless and changing the dimensions of x or y can always be exactly compensated by the corresponding dimensional changes in a and b. You can also show it works under x, y interchange by writing the inverse form and showing that identifying a' = 1/a and b' = -b/a restores the original fit function, exactly as using y = a x + b => x = a' y + b'. This is the form most often seen in research applications, and satisfies your conditions. You can also minimize the derivatives of $\chi ^{2}$ to get algebraic solutions for a and b. Dragons flight (talk) 00:40, 11 October 2008 (UTC)

This is exactly what I'm looking for. Thanks, Dragons flight. I think this method is called element-wise weighted total least squares or EW-TLS. Does that sound right? —Keenan Pepper 01:19, 12 October 2008 (UTC)

Thank you. I stand corrected. Bo Jacoby (talk) 09:15, 11 October 2008 (UTC).

Use of atan2 in a maths article

Bit of a dispute I'd like some advice on I'd like to use atan2 as used in complex number#Conversion from the Cartesian form to the polar form in the form there in another article

$\varphi =\arg(z)=\operatorname {atan2} (y,x)$

In fact I wanted to have the multivaluedness explicit so I could refer to k by saying

$\varphi =\arg(z)=\operatorname {atan2} (y,x)+2\pi k$

And another editor undoes the edit saying atan2 is not mathematical saying

I understand quite well why the arctangent is not the correct function to use. However "atan2" is a not a commonly used mathematical function. I also know there are several computer programming languages that have this function in their library. Using atan2 in a math article is fully unnecessary and makes a strange impression. The intention of the angle is much clearer explained by defining it as one that together with the radius yields back the coordinates by the cosine and sine functions

and

I am not against mentioning the atan2 function as a way to calculate the angle, but it should not appear in the definition.

My main point is

The article is not solely for pure maths people, it is for a general audience including people who want to actually use the information. Defining a value by saying the conditions it must satisfy rather than giving a well defined function that just produces it is not very useful in applied mathematics.. And arg is not as well defined as atan2 as it could be multivalued or map to [0, 2π) even if the principle value is conventionally (-π,π].

What do you feel about using atan2 the way it is in the complex number article? Could you frame it differently to make it more mathematical and less programmerish? Dmcq (talk) 22:44, 10 October 2008 (UTC)

What the hell is atan2? Algebraist 22:47, 10 October 2008 (UTC)

atan2. My question is, why is this being asked here rather than at Talk:Complex number? siℓℓy rabbit (talk) 22:50, 10 October 2008 (UTC)

It has been used in the complex number article for a while, I was going to use it in another article logarithm. It would be rather prejudicing the issue to raise it in a place where people seemed quite happy with it and anyway it seems a general issue - how do you represent how to get the principal value of the angle when converting to polar form? And should a computerish function be used if there isn't something mathematial which expresses it easily - tan^-1(y/x) goes wrong for negative x if converting to polar form Dmcq (talk) 23:00, 10 October 2008 (UTC)

Another problem with the tan^-1(y/x) is that it blows up if x is zero. Arg is formally undefined for z=0 but the conventional value is 0. atan2 just gives all the conventional values without the messing around. Dmcq (talk) 23:06, 10 October 2008 (UTC)

This should really be on Wikipedia:Wikiproject Mathematics, but oh well... I would be against using atan2 in an article, since only programmers are going to have the faintest idea what you're talking about. Don't be afraid to use words when describing mathematics - just explain how to choose the appropriate quadrant in English. --Tango (talk) 23:08, 10 October 2008 (UTC)

Sorry, will do that if I ever come across a problem like this again. I've been trying to think of a good way to frame it and I also worry about giving a rigmarole to people when most good calculators have the function on them. Dmcq (talk) 23:25, 10 October 2008 (UTC)

I just had a look via google and atan2 had more hits than atan, 12 mllion versus 7 million. Dmcq (talk) 23:36, 10 October 2008 (UTC)

So? Neither atan nor atan2 are common terms in mathematics. The common term is arctan, which has 25 million ghits. Algebraist 23:38, 10 October 2008 (UTC)

Re Dmcq: I'm not surprised about that; there's a reason atan2 is included in so many programming languages. But mathematicians by and large will never have heard of it. They will be familiar with the ordinary arctangent, which is a function of a single variable, not two.

The main concern for me isn't mathematicians, but high school students. Most of them will also not know about atan2, even if they have learned about arctangent in trigonometry class. But I had left the atan2 function in the complex number article when I edited it; I think I may be biased by already knowing what atan2 is. — Carl (CBM · talk) 23:51, 10 October 2008 (UTC)

My concern is people using tan^-1(y/x). See for example this one I found on the first page with a google of 'polar form' at [2] where a science and engineering encyclopaedia encourages people to get it all wrong. Dmcq (talk) 00:07, 11 October 2008 (UTC)

Yes, that page is all bad. — Carl (CBM · talk) 00:16, 11 October 2008 (UTC)

Well, that is the correct formula, you just need to include details of how to choose between the multiple values it returns (and what to do if x=0). --Tango (talk) 00:25, 11 October 2008 (UTC)

I think of arctan as a normal, single-valued, function. There are other ways of defining the argument of a complex number than using arctan, and I expect that's how a lot of math texts do it. — Carl (CBM · talk) 01:30, 11 October 2008 (UTC)

The reasons programmers love atan2 is that it's so well behaved. According to the C manual page:

"The atan2() function calculates the arc tangent of the two variables x and y. It is similar to calculating the arc tangent of y / x, except that the signs of both arguments are used to determine the quadrant of the result."

Using a regular atan function produces all sorts of annoying practical nightmares. I can understand that mathematicians don't use it - but it's really their loss. It's a perfectly reasonable function - it's just a matter of history that it's not a part of the standard canon. SteveBaker (talk) 02:23, 11 October 2008 (UTC)

As a mathematician, I'd say we do use it, we just call it 'the principal value of the argument' and have it act on complex numbers rather than pairs of reals. Algebraist 07:32, 11 October 2008 (UTC)

True, but I'd been looking at arg (mathematics) too and really I think that article at least should refer more to the multivalued form and use Arg for the principal value. Also it refers to the complex numbers, if doing that we might as well remove the square root equivalent of the modulus at the same time, it seems silly to have one and not the other. Dmcq (talk) 08:01, 11 October 2008 (UTC)

Where I come from, arg and log are single-valued while Arg and Log are multi-valued. Are other practices in use? Algebraist 08:28, 11 October 2008 (UTC)

The Principal value article uses the convention that Arg gives the principal value. MathWorld and any books I've seen do the same. Dmcq (talk) 13:23, 11 October 2008 (UTC)

My university uses Arg and Log for the principal values, arg and log for multi-valued. Eric. 131.215.159.187 (talk) 03:48, 12 October 2008 (UTC)

October 11

4th grade physics question (fluids, force, pressure)

A partially evacuated airtight container has a tight fitting lid (of negligible mass) over a airtight container.

• Surface area of the lid = 77 meters-squared
• Pressure above the lid = 100,000 Pa
• Pressure below the lid = unknown
• Force required to move the lid up = 480 N

My book gives the wrong answer (38,000 Pa), I just need a verification that its 99,993 Pa, or am I just using a wrong measurement such as meters instead of millimeters. Thanks, i'm embarassed to even ask this question, its easy algebra. The only reason I ask, is that I missed the first 4 questions in a row before this question (all for careless mistakes) but this one is playing tricks on my mind. 38kPa is the correct answer if the conversion factor between a netwon per square meter and a Pascal was some other amount. This was from yesterday's 11th hour of a study session and it caused me a great deal of headache, and I look forward to starting today's marathon session with closure. Sentriclecub (talk) 10:25, 11 October 2008 (UTC)

Are you sure you have the question right? The given answer is right if the surface area is 77 square centimetres (which is also a much more reasonable size for an airtight container!). Algebraist 10:57, 11 October 2008 (UTC)

Thanks. I'm looking at the book right now Fundamentals of Physics 8E Jearl Walker and I bought the detailed solutions manual. Now I can have my sanity back. Sentriclecub (talk) 11:09, 11 October 2008 (UTC)

October 12

Multiplying matrices of functions/arrays (in MATLAB)

I am looking for a good way to multiply matrices of functions, or matrices of arrays in Matlab. The description of my problem is as follows:

(In the following, uppercase letters denote matrices and lowercase letters denote scalars.)

I want to find Z = A(x)*B(x) where,

A(x) = [a11(x), a12(x), ..... , a1n(x);
        a21(x), a22(x), ....., a2n(x);
        ..............................;
        an1(x), an2(x), ....., ann(x)];

B(x) = [b11(x), b12(x), ..... , b1n(x);
        b21(x), b22(x), ....., b2n(x);
        ..............................;
        bn1(x), bn2(x), ....., bnn(x)];

Here, (.)(x) means (.) is a function of x.

For example, the first element of Z would be

z11(x) = a11(x)*b11(x) + a12(x)*b21(x) + ... + a1n(x)*bn1(x)

Furthermore, I need to integrate Z(x) from x = a to b to get ZI, which would be a simple nxn matrix.

The functions aij(x) and bij(x) are known numerically, so each one of them is a 1D array (say, of size m).

As an example, consider 2x2 matrices

A = [a11(x), a12(x);
     a21(x), a22(x);]

B = [b11(x), b12(x);
     b21(x), b22(x);]

Where,

x = [1,2,3,4,5];

a11(x) = x + 1 = [2 3 4 5 6]

a12(x) = x.^2 = [ 1 4 9 16 25]

a21(x) = x - 1 = [ 0 1 2 3 4]

a22(x) = 1.5*x = [ 1.5000 3.0000 4.5000 6.0000 7.5000]

b11(x) = sin(x) = [ 0.8415 0.9093 0.1411 -0.7568 -0.9589]

b12(x) = cos(x) = [ 0.5403 -0.4161 -0.9900 -0.6536 0.2837]

b21(x) = exp(-x) = [0.3679 0.1353 0.0498 0.0183 0.0067]

b22(x) = [1 1 1 1 1]

Then

Z = A*B = [a11(x)*b11(x) + a12(x)*b21(x), a11(x)*b12(x) + a12(x)*b22(x);
         a21(x)*b11(x) + a22(x)*b21(x), a21(x)*b12(x) + a22(x)*b22(x)]

That is, the terms of Z are the functions of x, such that for x = [1,2,3,4,5]

z11 = a11.*b11 + a12.*b21 = [2.0508 3.2692 1.0126 -3.4910 -5.5851]

z12 = a11.*b12 + a12.*b22 = [ 2.0806 2.7516 5.0400 12.7318 26.7020]

z21 = a21.*b11 + a22.*b21 = [ 0.5518 1.3153 0.5063 -2.1605 -3.7852]

z22 = a21.*b12 + a22.*b22 = [1.5000 2.5839 2.5200 4.0391 8.6346]

and integral of Z over x = 1 to 5 will be approximately (using trapezoidal rule)

ZI = [-0.9763, 34.9147;
     -1.9556,  14.2103]

I am looking for a simple way to do this in the general case. It must be possible to do it somehow using a matrix with three indices, but so far I have not been able to come up with a way.

Any help will be *greatly* appreciated. deeptrivia (talk) 02:53, 12 October 2008 (UTC)

Whether this is simple or not will depend on how your functions are defined, but it certainly possible to store values in three index arrays. The drawback is that they don't have natural arithmetic the way that two index matrices do.

For example on could implment it as:

A(1,1,:) = a11;
B(1,1,:) = b11;
A(2,1,:) = a21;
B(2,1,:) = b21;
etc... 

for k = 1:n
   Z(:,:,k) = A(:,:,k)*B(:,:,k);
end
Z = sum(Z*dx,3);

You'd still have to manually define the lists a_kj and b_kj, but you are going to have to do that anyway in telling Matlab what the functions are for each index. The "3" in the final sum tells Matlab to sum over the third index after multiplying by the increment dx and results in a n x n matrix of the integral you are looking for. You can of course do more complicated approximations to the integral, but generally that will require defining the integration process inside a for loop. Dragons flight (talk) 04:04, 12 October 2008 (UTC)

Thanks Dragon's flight. I have to define complicated expressions, and I would like to use operators like "*" to do multiplication. Is there any operator overloading in Matlab? I just figured out a cumbersome way to do transposes for these kind of "3d" matrices, but I'd like to overload that method onto the " ' " operator. And, I want to overload the method you suggested to the "*" operator. Is that possible? Thanks a ton! deeptrivia (talk) 05:11, 12 October 2008 (UTC)

Recent versions of Matlab do allow you define data classes with overloaded operators, though this is not an area I have much experience with. Each operation in Matlab has a builtin function name associated with it. For example "A + B" is translated by the interpreter to "plus(A,B)", and within a class you can define a new "plus" function that will then become the "+" operation for that class. This provides an introduction to class programming on Matlab, though it is somewhat cumbersome. Unless you will be working with these things alot I'd think it would be simpler to write functions to deal with your specific cases rather than writing a data class and overloaded operators to deal with general cases. Dragons flight (talk) 05:53, 12 October 2008 (UTC)

Thanks, Dragons flight. What if I just modified the default mtimes.m to check if the operands are 3D arrays. If so, use my routine, and if not, do the normal thing. mtimes.m is the following

function [varargout] = mtimes(varargin)

if nargout == 0
  builtin('mtimes', varargin{:});
else
  [varargout{1:nargout}] = builtin('mtimes', varargin{:});
end

How can I suitably modify this? Thanks, deeptrivia (talk) 16:49, 12 October 2008 (UTC)

Proving inequality for derivates (defined in Royden's Real Analysis)

Hello, I am working on a homework problem and I am not necessarily even looking for a hint but it seems to be wrong and I just want to see what others think. It's number 3 from Chapter 5 of Royden's book, 3rd edition. It says

 If $f\,$ is continuous on [a, b] and assumes a local maximum at $c\in (a,b)$, then
 $D_{-}f(c)\leq D^{-}f(c)\leq 0\leq D_{+}f(c)\leq D^{+}f(c)$.

Any way, the outer two inequalities are immediate, so the inner two are the problem. In case you do not know what a "derivate" is, because it does not seem to be a very common term, the definitions for the two needed ones are

$D_{+}f(x)=\liminf _{h\rightarrow 0+}{\frac {f(x+h)-f(x)}{h}}$

and

$D^{-}f(x)=\limsup _{h\rightarrow 0+}{\frac {f(x)-f(x-h)}{h}}$

So, my problem is this. If there is a local maximum at $c\,$, then $f(x)\leq f(c)$ for all $x\in [a,b]\cap N_{\delta }(c)$, for some $\delta >0\,$. Then, $f(c+h)-f(c)\,$ should be negative for $h\,$ close enough to $c\,$ so that $D_{+}f(c)\,$ should be nonpositive, whereas we are asked to prove it is nonnegative. Similarly, $f(c)-f(c-h)\,$ should be positive so that $D^{-}f(c)\,$ should be nonnegative, whereas we are asked to prove it is nonpositive.

Am I completely missing something here? I'm not looking for a solution. I just want to understand this part. Thanks StatisticsMan (talk) 03:04, 12 October 2008 (UTC)

You look right; it seems that either the definition provided or the homework problem have a sign flipped somewhere. Eric. 131.215.159.187 (talk) 03:43, 12 October 2008 (UTC)

Differentiation problem

Find the coordinates of the stationary points on the graph $y=(x-2)^{3}-12(x-2)$.

So I differentiated to get ${\frac {dy}{dx}}=3(x-2)^{2}-12$ And then found the x coordinates to be 0 and 4 and therefore the coordinates are (4, -16) and (0, 16)

However then I have to give the points of intersection with the axes which I dont know how to do. I know the intersection with the y axis from above is (0,16) but how do i get the x intersections.

This is what i tried: $0=(x-2)^{3}-12(x-2)$

Multiplied out the using the binomial theorem to get: $0=x^{3}-6x^{2}+16$

Is this correct? If so what now? --RMFan1 (talk) 12:55, 12 October 2008 (UTC)

Where a graph intersects the x-axis is where the function equals zero, so your method is right. It's equivalent to saying "find the roots of the polynomial". --Tango (talk) 13:26, 12 October 2008 (UTC)

Yes I know how you find intersections i just dont know how i can simplify the above. I could also have $0=(x-2)^{3}-12(x-2)$ so that $0=(x-2)^{2}-12$ but when I try to solve with the quadratic formula, the discriminant is negative so I cant simplify it. --RMFan1 (talk) 13:34, 12 October 2008 (UTC)

If the discriminant is negative then there are no real solutions, so it doesn't intersect the x-axis. That means x=2 is your only intersection point. --Tango (talk) 14:07, 12 October 2008 (UTC)

First think visually. Sketch the graph. You already know two points on it, (0,16) and (4,-16). When x is large and negative then y is also negative, so the graph comes up from the bottom left. It must cross the x axis at least once in the range x<0 to get up to (0,16). Then it must cross the x axis at least once more in the range 0<x<4 to get down to (4,-16). And when x is large and positive then y is also positive, so there is at least one more intersection with the x axis in the range x>4. But a cubic equation can have at most three real roots - so we have one root less than 0, a second root between 0 and 4, and a third root greater than 4.

Now suppose we recast the equation in terms of t=x−2. As a function of t, we have

$y=t^{3}-12t$

The (t,y) graph looks just like the (x,y) graph except it is shifted to the left by 2 units - so in the (t,y) graph the maximum and minimum are at (-2,16) and (2,16). Finding the three real roots of

$y=t^{3}-12t$

is quite easy - one root is obviously t=0. To find the correspoding values of x, just use x=t+2. Gandalf61 (talk) 14:12, 12 October 2008 (UTC)

Linear algebra puzzler

Let A be a 3×2 matrix and B a 2×3 matrix. Suppose that

$AB={\begin{bmatrix}5&-1&2\\-1&5&2\\2&2&2\end{bmatrix}}.$

Find BA. Your solution should also show that there is a unique right answer. siℓℓy rabbit (talk) 15:15, 12 October 2008 (UTC)

My first thought would be to consider transposes. --Tango (talk) 15:24, 12 October 2008 (UTC)

Not that that seems to work... I can't get anywhere with it (but I may be missing something, the question still screams "take transposes" to me). You could always try it the hard way - write A=(a_ij), B=(b_ij), multiply out both ways round and try and solve the horrible mess of simultaneous quadratic equations - I wouldn't recommend it, though! --Tango (talk) 15:38, 12 October 2008 (UTC)

I'm pretty sure the uniqueness of $BA=6I$ can be proved by considering all possible singular value decompositions of $AB$. $A$ and $B$. 84.239.160.166 (talk) 16:29, 12 October 2008 (UTC)

On a quick look I'd say it was probably 6 times the 2x2 identity matrix since multiplying the matrix by itself gives itself multiplied by 6. Just did that because I got BA in the middle of AB.AB. Dmcq (talk) 16:52, 12 October 2008 (UTC)

I suppose I should give the answer, although I was expecting this to generate somewhat more interest.

Method 1. Note that AB²=6AB. That is ABAB=6AB. Since AB is of rank 2, both A and B are also rank 2, and so A:R²→R³ is injective and B:R³→R² is surjective. Thus A is cancellable on the left and B is cancellable on the right of ABAB=6AB, and so BA=6I.

Method 2. From ABAB=6AB, we have

$A^{T}ABABB^{T}=6A^{T}ABB^{T}.$

Since rank(A^TA) = rank(A) = 2 and rank(BB^T) = rank(B) = 2, we have

$BA=6(A^{T}A)^{-1}A^{T}ABB^{T}(BB^{T})^{-1}=6I.$

Remark. I seem to recall that there was a way to solve this problem using orthogonality somehow (e.g., the SVD or QR decomposition). I was unable to recall the details for the purposes of this solution, but feel free to add a "Method 3" if you can get this to work. Cheers, siℓℓy rabbit (talk) 14:36, 13 October 2008 (UTC)

cost of transporting goods

Con you tell be the cost to transport 100 lbs of material in a car that gets 25 mpg / gas @ $3.50 gal. My husband feels that the 'junk' I was to take is worth less than the gas. We will travel 1000 miles. Many ThanksTwinkle2toes (talk) 21:48, 12 October 2008 (UTC)

Can't be done with mathematics only. If I were to assume that fuel consumtion is proportional to the loaded mass of the vehicle, and the vehicle weighs around 2,000 lbs, then you'd end up with around $7 additional cost. But I'm pretty sure it's not proportional, and I wouldn't be surprised if the physics behind it is complicated. Fuel economy in automobiles only says that engine efficiency varies with the mass of the automobile and its load, but not how. We need input from someone who knows something about cars... -- Jao (talk) 22:51, 12 October 2008 (UTC)

Not just cars, but that specific car. It's going to vary widely from car to car, for example a lightweight car is going to affected by an extra 100lbs more than a very heavy car (I would imagine, anyway). --Tango (talk) 23:03, 12 October 2008 (UTC)

The cost to transport "100 pounds of material" is like talking about the cost to bring along a small teenager or a box or two of treasured family photos and such . The answer is more subjective than practical remembering that "one person's junk is another's treasure". Neither the actual dollar cost of transporting nor the emotional cost (even though you didn't ask) of leaving the "material" behind can't be calculated with the information given. -hydnjo talk 23:51, 12 October 2008 (UTC)

Fuel efficiency would be more or less proportional to the mass assuming things are ideal (like the engine is a carnot engine). This makes some sense because both kinetic and potential energy are proportional to mass, and so is any frictional loss during braking, etc. So, Jao's estimate is the best we can do unless we decide to go really fancy. 128.118.150.91 (talk) 00:20, 13 October 2008 (UTC)

Yeah, $5 to $10 seems about right. -hydnjo talk 00:31, 13 October 2008 (UTC)

The fuel economy depends on whether the extra 100 hundred pounds is being driven in town or on the highway. Once you're at a steady highway speed (the 1000-mile trip), air drag dominates and an 100 extra pounds isn't going to matter much. Take the stuff out for around-town driving, though. Saintrain (talk) 12:55, 13 October 2008 (UTC)

$5 sounds about right. Let's say you get a 2% drop in gas mileage, down from 25 mpg to 24.5 and that means instead of using 40 gallons of fuel, you will use instead 41 gallons of fuel. Sentriclecub (talk) 15:20, 13 October 2008 (UTC)

October 13

Volume ratio

The ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube is? —Preceding unsigned comment added by NAKABBOSS (talk • contribs) 00:20, 13 October 2008 (UTC)

Well, the formulae for the volume of a cube and a sphere are fairly simple, you just need to know what you use for the radius and the side length. -mattbuck (Talk) 00:34, 13 October 2008 (UTC)

The ratio is: click here. -hydnjo talk 00:39, 13 October 2008 (UTC)

1/x integral

Is the integral of 1/x from 0 to 1 infinity or undefined? —Preceding unsigned comment added by 165.124.138.143 (talk) 02:06, 13 October 2008 (UTC)

Infinity (you might find proving this formally using the epsilon delta definition interesting).

Topology Expert (talk) 02:20, 13 October 2008 (UTC)

If you want to get pedantic, then the Riemann integral does not exist, but the improper integral (and the Lebesgue integral, IIRC) evaluates to +infinity. Confusing Manifestation(Say hi!) 04:44, 13 October 2008 (UTC)

help in editing gelfand pairs

I hope it's the rite plase to ask this requst. I'm curently editing the artical gelfand pair and I wont to put ther a section abut its aplications. so I rote it and put it in the discation page. However my english and riting stayl are to poor to put it in the artical, so I'll b grateful if somebody will fix this section and put it in the artical.

Thenks alot Aizenr (talk) 10:47, 13 October 2008 (UTC)

Calculating a growth (or interest) rate

If I have a starting population of X, a number of periods P, and a final population of Y, how do I calculate the population growth per period assuming it's a smooth growth rate? — PhilHibbs | talk 15:41, 13 October 2008 (UTC)

Assuming you are talking true exponential growth, then its easiest to manipulate your compounding period to satisfy f(1)=e. Setting up the problem with no aim, is difficult and unintuitive...

The population P at any time is given by the formula... $P(t)=Ae^{kt}\,$

A represents the starting value (but you chose X which is no problem) however k is very tricky. I'll give you several examples of k, until you see a pattern.

For a bacteria population that triples every seven years, $k={\frac {ln3}{7}}\,$ and your period is 7 years.

For a sample of Uranium that has a half life of 4.5 billion years,$k={\frac {ln{\frac {1}{2}}}{4.5*10^{8}}}\,$ and your period is 4.5 billion years.

For a human population that goes up by a factor of 1.001400763 each generation, $k={\frac {ln1.001400763}{1}}\,$ and your period is 1 generation.

For a rabbit population that goes up 10 fold, year over year, $k={\frac {ln10}{1}}\,$ and your period is 1 year.

Now lets answer your question...

• starting population, we'll assign the symbol A = 5,000
• final population, we'll assign the symbol Z = 6,000,000,000
• Final population divided by Starting population = 1,200,000 (because ${\frac {Z}{A}}$ = 1.2 million)
• number of periods, we'll assign the symbol N = 10,000
• length of period, we'll assign the symbol L = 1 generation
• The growth rate per period is $e^{k}\,$ =
• e is called euler's number
• The value of $e^{k}\,$ that when multiplied by itself ten thousand times = 1.2 million, super easy
• $e^{k}\,$ = 1.001400763 and you're done!

This can be verified by checking the original equation. $P(t)=Ae^{kt}\,$ I needed this refresher--use it or lose it! Sentriclecub (talk) 16:38, 13 October 2008 (UTC)

Yes, I meant a smooth exponential, e.g. starting with 2000, over 1000 generations, a final population of 42,000,000 requires a growth rate of 1%. I calculated this example the easy way around! 1% growth actually gives 41,918,311. Oh, and this isn't homework, I left school 22 years ago. I could probably have done this in my head back then. — PhilHibbs | talk 15:51, 13 October 2008 (UTC)

A little more info - I'm trying to demonstrate how many "Mitochondrial Eve"s there should be given a given population bottleneck at any point in the past. So for a population bottleneck of 5,000 people 10,000 generations ago, I need a population growth rate that would result in 6,000,000,000 people after 10,000 generations of growth. I can trial-and-error it and get 0.149267437184% but it takes ages of fiddling around to get each answer. — PhilHibbs | talk 16:04, 13 October 2008 (UTC)

$Y=Xr^{P}$

You know Y, X and P, so solve to find r. Gandalf61 (talk) 15:58, 13 October 2008 (UTC)

Yeah, I don't know how to reverse that $r^{P}$ bit. — PhilHibbs | talk 16:05, 13 October 2008 (UTC)

• click* - it's $r^{(}-P)$ isn't it? So that gives $r=(Y/X)^{(}-P)$? (if you forgive the broken math tags) — PhilHibbs | talk 16:08, 13 October 2008 (UTC)

D'oh, no, it's r^(1/P) — PhilHibbs | talk 16:13, 13 October 2008 (UTC)

(ec) Suppose it grows by a factor k each period. Then

Y = X k^P

so

$k=\left({\frac {Y}{X}}\right)^{\frac {1}{P}}$

That gives you the growth factor over a particular period.

Alternatively, if you want the current instantaneous rate of growth, consider

$Y(t)=Xk^{\frac {t}{T_{P}}}$

$\qquad \qquad =Xe^{\frac {t\ln k}{T_{P}}}$

where T_P is the duration of each time period.

Therefore, at any time t the rate of growth is

${\frac {dY(t)}{dt}}=X{\frac {\ln k}{T_{P}}}e^{\frac {t\ln k}{T_{P}}}$

And so the proportional rate of growth is

${\frac {1}{Y}}{\frac {dY(t)}{dt}}={\frac {\ln k}{T_{P}}}$

Substituting in k from above gives

$\qquad \qquad ={\frac {\ln Y-\ln X}{PT_{P}}}$

You can get to the second-to-last step more quickly, once you've realised that that is what you want, by noting that

${\frac {1}{Y}}{\frac {dY(t)}{dt}}={\frac {d\ln Y(t)}{dt}}$

Jheald (talk) 16:17, 13 October 2008 (UTC)

Dumb question about infinity

Are there an infinite number of infinities? Sorry if this is a dumb question.

There are no dumb questions about infinity, it's a very confusing topic! The answer is "it depends". There are different things "infinity" can mean. It can be used to mean "going on forever" (when doing limits and things), so if you're talking about a sequence indexed by natural numbers, there's just one infinity. If you're talking about a function on the real numbers, there are two, positive and negative, if you're talking about a function on the plane or some high dimensional space, then there are infinitely many, one in each direction (the infinity being the same as the number of real numbers or the "cardinality of the continuum"). On the other hand, there are infinite ordinals (numbers used for ordering things) and cardinals (numbers used for counting things), there are infinitely many of each of those too (I'm not sure which infinity - I think there are more ordinals than any infinity you can think of, so I guess that's a kind of super-infinity!). --Tango (talk) 18:07, 13 October 2008 (UTC)

Solving equations

Hi, I'm a hobbyist programmer writing games, and I need a way to solve equations of the form (with complex constants A, B, C, D, E):

$AB^{x}+Cx+D=E^{x}$

Is there a good (ie, fast) way of telling if there is a real solution quickly (because there might not be), and if so finding it? 79.78.73.72 (talk) 18:24, 13 October 2008 (UTC)