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July 25

in MATLAB, the absolute value of the Pearson correlation coefficient is much higher than sqrt(R^2) of least squares fit

According to our article Pearson product-moment correlation coefficient, R^2 in linear regression should be the square of Pearson's r. However in MATLAB, I have data sets where my R^2 value (for my least squares fit) is something like 0.28, but r is something like -0.74 or -0.85. Is polyfit() not finding the best fit? The original data I believe follows a power law, so I have taken the log of both x and y to make a log-log data set, and used linear least squares regression on it. 76.104.28.221 (talk) 05:40, 25 July 2012 (UTC)

In general, engineers, statistician, and mathematicians may use slightly different definitions of correlation, covariance, autocorrelation, etc. They are all related via scaling or centering, but this can be a pain to work out. See for instance Autocorrelation#Signal_processing for an alternate definition. As a wild guess, try passing the 'coeff' flag into the corr function, and that might force the definition that you are expecting. Here's MathWorks' info on xcorr [1]. SemanticMantis (talk) 15:16, 25 July 2012 (UTC)

It's simply impossible for r to be -0.74 for the best linear fit. The worst that r could ever be is zero. Something is not right here. Looie496 (talk) 16:15, 25 July 2012 (UTC)

r implies a negative correlation, i.e. it is quite a good fit.

I've been using corr2() -- I guess this is not the Pearson correlation coefficient, but a "2D correlation coefficient". (However I am comparing linear vectors). Does anybody know why the 2D correlation coefficient is so much stronger for correlating one-dimensional vectors versus Pearson correlation? What method is used in 2D correlation? Does it allow cross-correlation? My two variables are time-series variables -- velocity (predictor) and curvature (predicted) of fruit flies. I do suspect sometimes that velocity predicts curvature a little later (e.g. 1/15 seconds later) rather than the curvature of that very moment. (Predictor and predicted could be reversed of course, I am just using one hypothesis.)

I should mention that the trajectories of fruit flies have several interesting properties-- for example literature mentions that paths are often self-similar (fractal-like) regardless of scale (at least up to the temporal resolution limit of conventional digital imaging, i.e. 1/30 seconds). However, I am a biochemistry student, not an engineering / signals and systems student, so I am just figuring this out. (My engineering friend explained autocorrelation to me a few months back.) 137.54.30.45 (talk) 16:47, 25 July 2012 (UTC)

The closest equivalent of corr2() I can find on Wikipedia is Digital image correlation, which allows for stretching and various other transformations of the vectors. Is this what corr2() does? If I apply digital image correlation between two one-dimensional images (so to speak -- allowing for stretching and translation?) how do I find an equation that allows one to determine the other? What I am trying to determine is a power law, after all.... 137.54.30.45 (talk) 16:57, 25 July 2012 (UTC)

I would strongly advise that you find somebody local to help you. You're using words without really understanding what they mean, and so the questions you are asking don't make sense. A local person could look at your data and help you to find the right words to describe it. Looie496 (talk) 17:18, 25 July 2012 (UTC)

Okay, but I am absolutely certain that a Pearson r of approximately -0.74 should yield an R^2 of approximately ~0.55. I mean, that's what our article seems to imply. I'm using definitions described in documentation and am simply trying to find the Wikipedia articles describing similar concepts. The MATLAB documentation for corr2 says that it finds the 2D correlation coefficient -- that is indeed digital image correlation is it not? Digital image correlation is mentioned as an application of corr2(). It seems that corr2() returning a negative-r should be possible...137.54.30.45 (talk) 17:24, 25 July 2012 (UTC)

Are you sure you included in your data a constant column? For example if I want to regress x on y the command is b=regress(y,[ones(size(x)) x]). Yaniv256 (talk) 23:40, 28 July 2012 (UTC)

polyfit() not finding the best fit?

So explicitly called the Pearson's correlation coefficient function in MATLAB using corr() rather than corr2(), but the results are completely the same. However, in many cases the square root of the R^2 of the least squares fit found via polyfit() is significantly below Pearson's r, which is always around -0.7 to -0.85. (In one case, r is -0.77, but the polyfit() R^2 is 0.28?) Is polyfit() not finding the best fit? 137.54.30.45 (talk) 18:07, 25 July 2012 (UTC)

Term for visual appearance of slope

Taken from Misleading graph

Misleading line graph with no scale

Volatility	Steady, fast growth	Slow growth
The lack of a scale allows the graph to be easily manipulated.

What is the term for the visual appearance of a slope? I'm writing the article misleading graph and want to express that visually, it appears that the slope is increasing/decreasing. I don't want to use slope or gradient as the underlying data is not changing. I'm currently using the terms volatility and growth, but these are imprecise economic terms. Anyone know how to better express Misleading_graph#Axis_changes and Misleading_graph#No scale? Smallman12q (talk) 12:07, 25 July 2012 (UTC)

Articles should be based on sources. What do the sources say? If they don't then we shouldn't either. Dmcq (talk) 12:10, 25 July 2012 (UTC)

The sources say that graphs without scales are misleading...and then give several examples. The economics texts say that one graph shows growth while the other one doesn't. I'm hoping there's some better way to express this... Smallman12q (talk) 12:22, 25 July 2012 (UTC)

I think it is valuable to make progress, even if citations are currently lacking (they can always be added later). I don't think there's a widely-used formal term for this. I would just describe the phenomenon as accurately as possible. Something like "Even though the actual slope of the (x,y) data is the same in both graphs, the way that the data is ploted can change the visual appearance of the angle made by the line on the graph. This is because each plot has different scale on its vertical axis. Because the scale is not shown, these graphs can be misleading." Now that I think of it, slope is independent of any plotting scheme, but angle can (and does in this example) depend on plotting scheme. So that might be the best distinction to make. SemanticMantis (talk) 15:03, 25 July 2012 (UTC)

Steepness. 207.224.43.139 (talk) 02:18, 30 July 2012 (UTC)

How does curvature correlate to velocity in a random walk with randomly-distributed speeds (i.e. Brownian motion)?

While investigating fly path curvature in a 2D arena as a statistical parameter, I have found that the curvature-speed relationship follows a power law where curvature in (1/mm) is predicted by the relation C = 10^a * x^b, where a is ~0.8 and b is ~ -1.6 and x is absolute speed in mm/s. Firstly, if behavior is random, doesn't the physical equation for curvature already "totally account" for velocity, i.e. curvature should be statistically independent of velocity in a random walk of random speeds? Curvature is |y"x' - x"y'| / speed^3, but both the numerator and denominator are dependent on velocity and should cancel each other out, right? Thus any correlation between curvature and velocity should be due to non-random behavior-- is this correct? 137.54.30.45 (talk) 17:24, 25 July 2012 (UTC)

The word "random" just means unpredictable in some way -- it doesn't mean anything specific. When you have a physical object, if it is moving very fast, momentum makes it impossible for it to curve very sharply, so there is always going to be a correlation between velocity (in the sense of speed) and curvature. But more basically, can I ask again that you look for somebody local to advise you? We can try to answer specific and well-formulated questions, but we really can't perform the functions of a mentor, because we can't look at your data. Looie496 (talk) 17:34, 25 July 2012 (UTC)

I'm talking about a random walk, where time progresses in fixed intervals so speed is distance/t, but since distance travelled is random, so speed should have the same random distribution. Consider a random walk model that factors in momentum and dampening parameters. A fruit fly I think has low momentum and high dampening (there is probably some dimensionless number describing this), so a fruit fly behaving randomly would be well-approximated by Brownian motion, no? In that case, is the curvature of a Brownian-motion particle statistically independent of its velocity?

Also, my PI is very busy, I have to do all the work and literature research myself; however I am finding it difficult to search the literature on this question. I think my questions are very well-defined -- if they are too general, I can narrow them down further. I think my questions are reasonable? 137.54.30.45 (talk) 18:29, 25 July 2012 (UTC)

The relation between speed and curvature that you would expect just depends on the model you use to explain the fly's movement. If your model says that the two should be independent, but your data shows some relation, then it's not a good model. This doesn't have anything to do with whether the fly's behavior is random or not. First, whether the fly moves in a way that's truly random is question of metaphysics that you probably are not trying to answer. But more importantly, any model is going to have randomness built into it since a deterministic model is out of reach, but there are many different random models. If one fails you just need to find a better one. Rckrone (talk) 20:52, 25 July 2012 (UTC)

Oh no, the random model is the null hypothesis. I am creating a model which implies non-random behavior but I need to compare it against a "null" model, in this case the random walk / Brownian hypothesis. I do not think a random model would predict a power law with an exponential parameter of -1.6? 137.54.30.45 (talk) 00:32, 26 July 2012 (UTC)

So, you have (or at least "approximately have") something like the "ideal" Brownian motion, when the particle (in this case - fly) moves straight all the time with exception of the points when it instantly changes direction? In such case the curvature of the trajectory is zero in the straight segments and infinite for turning points. Thus, if it is so, it makes little sense to look for any "interesting" relationship between curvature and anything else. In other words, if you want to get curvature that makes sense, first of all make sure you have enough points to see how the trajectory is curved.

A second point. "Curvature is |y"x' - x"y'| / speed^3, but both the numerator and denominator are dependent on velocity and should cancel each other out, right?" - it is going to be easier if you will forget the velocities for the moment. The curvature describes a point in a curve. In this case the curve is trajectory. Trajectory does not tell you anything about the speed. Neither will curvature (unless there is something more interesting in the underlying process).

A third point. Look how you find the velocities and curvatures. If you use numerical differentiation, remember that it is not resistant to noise. --Martynas Patasius (talk) 23:16, 25 July 2012 (UTC)

The resolution I'm looking at is 1/15 s-- I think this is high enough to examine curvature. The fly actually "wobbles" and we can measure things like yaw or absolute angular difference from some orientation but this results in a more noisy measure of curvature. One measure of curvature I used initially was angular velocity over linear velocity, but this was noisy compared to using the |y"x' - x"y'| / (x'^2 + y'^2)^3/2 definition. 137.54.30.45 (talk) 00:30, 26 July 2012 (UTC)

In Brownian motion the speed and curvature are everywhere infinite. -- Meni Rosenfeld (talk) 16:38, 29 July 2012 (UTC)

July 27

Lagrangian mechanics

This is probably a dumb question, and I'm probably over thinking this, but it's been nagging me for a while and for some reason I can't resolve it.

This question is part physics and part math, but I think the math desk is the most appropriate place for it.

Consider a (classical mechanical) system, describable by the generalized coordinates $(q_{1},q_{2},....,q_{N})={\vec {q}}$ and the Lagrangian $L({\vec {q}},{\dot {\vec {q}}},t)$. Suppose we have a trajectory ${\vec {q}}(t)$ which satisfies the equations of motion, which is to say that ${\frac {\partial L}{\partial {\vec {q}}}}({\vec {q}}(t),{\dot {\vec {q}}}(t),t)={\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\vec {q}}}}}({\vec {q}}(t),{\dot {\vec {q}}}(t),t)\right)$. I'm using a funny but common notation where ${\frac {\partial L}{\partial {\vec {q}}}}=\left({\frac {\partial L}{\partial q_{1}}},{\frac {\partial L}{\partial q_{2}}},...\right)$ and ${\frac {\partial L}{\partial {\dot {\vec {q}}}}}=\left({\frac {\partial L}{\partial {\dot {q}}_{1}}},{\frac {\partial L}{\partial {\dot {q}}_{2}}},...\right)$.

Now we transform the trajectory ${\vec {q}}(t)$ to a new trajectory ${\vec {Q}}(t)$. I'd like to show that if the Lagrangian remains invariant under this transformation ie $L({\vec {q}}(t),{\dot {\vec {q}}}(t),t)=L({\vec {Q}}(t),{\dot {\vec {Q}}}(t),t)$, then ${\vec {Q}}(t)$ also obeys the equations of motion ie ${\frac {\partial L}{\partial {\vec {q}}}}({\vec {Q}}(t),{\dot {\vec {Q}}}(t),t)={\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {\vec {q}}}}}({\vec {Q}}(t),{\dot {\vec {Q}}}(t),t)\right)$. 65.92.7.148 (talk) 05:53, 27 July 2012 (UTC)

It depends on the transformation. Check out the articles canonical transformation and symplectomorphism. — Fly by Night (talk) 14:52, 27 July 2012 (UTC)

The symplectomorphism article was above my head, but I think I understand the canonical transformation article, and I don't think it's what I'm looking for. I don't dispute that only some special types of transformations (like canonical transformations) preserve the equations of motion despite changing the Lagrangian and Hamiltonian. But what I'd like to show is that all transformations which do preserve the Lagrangian also preserve the equations of motion. 65.92.7.148 (talk) 18:56, 27 July 2012 (UTC)

So you're trying to prove the statement: "If a transformation preserves the Lagrangian then it preserves the equations of motion." To be honest, my knowledge of this subject is more mathematical, e.g. symplectic geometry. Have you got access to the book V. I. Arnold (1989). Mathematical Methods of Classical Mechanics. Springer.? That text's a gold mine. It starts from the basics and ends up on some very deep stuff, including Hamiltonian mechanics. — Fly by Night (talk) 20:01, 27 July 2012 (UTC)

Yes, I have a copy of that book. Most of it is Arabic to me, but fortunately I was able to find something relevant. On page 88, in the section on Noether's theorem, he writes: "First, let $M=\mathbb {R} ^{n}$ be coordinate space. Let $\phi :\mathbb {R} \rightarrow M,\mathbf {q} =\phi (t)$ be a solution to Lagrange's equation. [If] $h_{*}^{s}$ preserves L, the translation of a solution, $h^{s}\circ \phi :\mathbb {R} \rightarrow M$ also satisfies Lagrange's equations." $h^{s}:M\rightarrow M,s\in \mathbb {R}$ is defined as a "one-parameter group of diffeomorphisms", whatever that means. I'm not sure how $h_{*}^{s}$ is defined. At any rate, he seems to be asserting what I'm trying to prove, but unfortunately he himself does not provide a proof, unless it's buried somewhere else in the text. That said, perhaps with the problem translated into the language of higher mathematics you can solve it, and then translate the solution into something that I, with my feeble knowledge of math, can understand. 65.92.7.148 (talk) 21:43, 27 July 2012 (UTC)

We have a diffeomorphism article. A one parameter group of diffeomorphisms is just the integral curves of a vector field, i.e. the paths followed by a particle under the incluence of smoothly varying force field. Take a look at Integral curve, diffeomorphism and Flow (mathematics). — Fly by Night (talk) 23:23, 27 July 2012 (UTC)

I appreciate your help, but unfortunately those articles don't help me much. Really, this question should be solvable by simple calculus. I think learning about diffeomorphisms is a bit overkill. 65.92.7.148 (talk) 04:15, 28 July 2012 (UTC)

I had a quick look in Landau and Lifshitz's Mechanics (3rd Ed) - s45 discusses canonical transformations but as a prelude it mentions point transformations, they talk about the choice of generalised coordinates being unrestricted and how we can change from $q_{1},q_{2},$ to $Q_{1},Q_{2},$ by $Q=Q(q,t)$. There's no detail but I web-searched for point transformations and got this PDF [2] titled "Coordinate Invariance of Lagrange's Equations" which looks helpful, maybe give that a look? Although I don't think the Langrangian itself is invariant but the Euler-Lagrange equations should be...83.100.173.200 (talk) 15:01, 28 July 2012 (UTC)

I had a look through the PDF, and I don't believe it solves the problem (though I confess I may be mistaken). From what I understand, the author expresses the same system in terms of a different set of generalized coordinates (ie makes a passive transformation), and shows that the trajectory of the system as expressed in terms of these new coordinates still obeys Lagrange's equation. What I'm trying to do is make an active transformation of the coordinates ie change the system, and show that Lagrange's equations are still applicable if the Lagrangian is the same. That said, I appreciate the help. 65.92.7.148 (talk) 03:48, 29 July 2012 (UTC)

Sorry it wasn't helpful. Diffeomorphisms are a very common topic and I'm suprised you're doing Hamiltonian mechanics but haven't been taught about diffeomorphisms. They're just transformations that, along with their inverses, are differentiable, i.e. the smudge things in a smooth way. The beauty of the general setting, i.e. using vector fields and differential forms, means that everything is coordinate free: you don't even need to use coordinates! But I understand that it might be bewildering trying to learn it on your own. — Fly by Night (talk) 17:19, 28 July 2012 (UTC)

Don't be sorry. I suspect you mistook me for a math major, but alas I'm just a lowly physics undergrad. And when we learn analytical mechanics, we don't use any fancy-shmancy mathematics, just you're everyday multivariable cal. So that's why I was a bit put-off by this diffeomorphism business. 65.92.7.148 (talk) 03:48, 29 July 2012 (UTC)

What do you mean by "transform the trajectory ${\vec {q}}(t)$ to ${\vec {Q}}(t)$"? Taking your proposition at face value, it isn't true. Let $L={\tfrac {1}{2}}m{\dot {x}}^{2}+{\tfrac {1}{2}}m{\dot {y}}^{2}$. Then $x(t)=vt,y(t)=0$ (for some constant v) satisfies the equations of motion, but $X(t)=r\sin(v/r)t,Y(t)=r\cos(v/r)t$ doesn't, even though $L({\dot {x}}(t),{\dot {y}}(t))\equiv L({\dot {X}}(t),{\dot {Y}}(t))$. -- BenRG (talk) 17:42, 28 July 2012 (UTC)

You're right, I didn't state the question well. Here's a second try: suppose we make the transformation ${\vec {q}}\rightarrow {\vec {Q}}({\vec {q}},t),\ {\dot {\vec {q}}}\rightarrow {\dot {\vec {Q}}}({\vec {q}},{\dot {\vec {q}}},t)$ which preserves the Lagrangian. Then, if ${\vec {q}}(t)$ is a solution to Lagrange's equations, then so is ${\vec {Q}}({\vec {q}}(t),t)$. In this case, your counter-example no longer works, because the transformation $x\rightarrow X=r\sin {\frac {x}{r}},\ y\rightarrow Y=r\cos {\frac {x}{r}},\ {\dot {x}}\rightarrow {\dot {X}}={\dot {x}}\cos {\frac {x}{r}},\ {\dot {y}}\rightarrow {\dot {Y}}=-{\dot {y}}\sin {\frac {x}{r}}$ doesn't preserve the Lagrangian. 65.92.7.148 (talk) 03:36, 29 July 2012 (UTC)

I think you need to be clear about what you mean by "preserves the Lagrangian". Your transformation seems to be a change of trajectory, but not a change of generalised co-ordinates. The Lagrangian is a function of the generalised co-ordinates (and, sometimes of time too). So the form of the Lagrangian is preserved since you are not changing your system of co-ordinates. In BenRG's example, he has two trajectories, both described in Cartesian (x,y) co-ordinates, and along both trajectories the value of the Lagrangian is constant, and has the same value. One trajectory is $(vt,0)\,$ i.e. particle moves along the x axis at constant speed v. The other is $\left(r\sin(vt/r),r\cos(vt/r)\right)\,$ i.e. the particle travels around a circle of radius r at constant speed v. The Lagrangian in this example is just the kinetic energy of the particle (there are no external forces or constraints on the particles motion), which has a constant value of $mv^{2}/2$ along both trajectories. However, the first trajectory satisfies the equations of motion (which are $m{\ddot {x}}=0,\ m{\ddot {y}}=0\,$), and the second one doesn't. In other words, a particle moving in the absence of external forces or constraints will have constant kinetic energy, but the converse is not true. Gandalf61 (talk) 10:29, 29 July 2012 (UTC)

By "preserves the Lagrangian" I mean this: if you define a transformation that changes the values of the generalized coordinates (and hence the generalized velocities as well, usually), the new value of the Lagrangian will always be numerically equal to the original value of the Lagrangian. By always I mean for any values of the original generalized coordinates and velocities.

Example: suppose you have a particle moving in a 2D plane, with a potential proportional to the distance from the origin. So, in Cartesian coordinates, $L={\frac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2})-U\left({\sqrt {x^{2}+y^{2}}}\right)$. If I make the transformation $x\rightarrow X=x+a,\ y\rightarrow y$ ie move the particle in the x direction by an amount a, then the Lagrangian will in general by different, because the value of the potential energy will change. In the particular case that $x=-{\frac {a}{2}}$, the Lagrangian will be the same under this transformation, but that's just a special case.

On the other hand, if I make the transformation $x\rightarrow X=x\cos \theta +y\sin \theta ,\ y\rightarrow Y=y\cos \theta -x\sin \theta$, then the Lagrangian will be the same no matter what x, y, dx/dt, and dy/dt are: ${\frac {1}{2}}m({\dot {X}}^{2}+{\dot {Y}}^{2})-U\left({\sqrt {X^{2}+Y^{2}}}\right)={\frac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2})-U\left({\sqrt {x^{2}+y^{2}}}\right)$ for all x, y, dx/dt, dy/dt.

In BenRG's example, $L(X,Y,{\dot {X}},{\dot {Y}})={\frac {1}{2}}m({\dot {X}}^{2}+{\dot {Y}}^{2})=m{\dot {x}}^{2}$, which is only equal to the original Lagrangian $L(x,y,{\dot {x}},{\dot {y}})$ if dy/dt = 0. 65.92.7.148 (talk) 16:00, 29 July 2012 (UTC)

You need to think carefully about the difference between changing the trajectory of a particle and changing the generalised co-ordinates that are used to describe that trajectory. In BenRG's example, he described two different trajectories:

$(x_{1}(t),y_{1}(t)=(vt,0)$

and

$(x_{2}(t),y_{2}(t)=\left(r\sin(vt/r),r\cos(vt/r)\right)$

because you appeared originally to be asking about changes to trajectories that preserve the value of the Lagrangian along the trajectory. The form of the Lagrangian is the same for both trajectories:

$L={\tfrac {1}{2}}m({\dot {x}}^{2}+{\dot {y}}^{2})$

because the form of the Lagrangian depends only on the system of generalised co-ordinates, not on the trajectory. BenRG does not change the system of generalised co-ordinates, he only changes the trajectory through phase space. So along the first trajectory we have

$L_{1}(t)={\tfrac {1}{2}}m({\dot {x_{1}}}^{2}(t)+{\dot {y_{1}}}^{2}(t))={\tfrac {1}{2}}mv^{2}$

and along the second trajectory we have

$L_{2}(t)={\tfrac {1}{2}}m({\dot {x_{2}}}^{2}(t)+{\dot {y_{2}}}^{2}(t))={\tfrac {1}{2}}m\left(v^{2}\sin ^{2}(vt/r)+v^{2}\cos ^{2}(vt/r)\right)={\tfrac {1}{2}}mv^{2}$

and so the value of the Lagrangian is the same (and is constant) along both trajectories. But the first trajectory satisfies the equations of motion, whereas the second one doesn't.

But now you seem to have switched to asking about changing the system of generalsied co-ordinates - your last post does not mention trajectories at all. You have simply observed that given a potential that is radially symmetric about a point P, a translation of co-ordinates changes the form of the Lagrangian, but a rotation of co-ordinates about the point P does not change the form of the Lagrangian. And, of course, if the form of the Lagrangian is unchanged then the equations of motion are not only satisfied by the same trajectories through phase space (which we would expect on physical grounds), but they also take the same form in the new generalised co-ordinates system i.e. you can just substitute X for x and Y for y. Which also tells us that if we find a trajectory (x(t), y(t)) that satisfies the equations of motion, then any rotation of that trajectory through an angle θ about P will also satisfy the equations of motion - just as we would expect from the symmetry of the system. Gandalf61 (talk) 17:51, 29 July 2012 (UTC)

Yes, you're right, after reading BenRG's counterexample I realized that I had made a mistake and didn't ask the question I was trying to ask. So, here's what I'm really asking: suppose the configuration of a system is described by the vector ${\vec {q}}$ in configuration space. Next, we define some transformation ${\vec {q}}\rightarrow {\vec {Q}}({\vec {q}},t)$ which changes the configuration of the system from q to a new configuration Q eg by rotating the entire system, or by shifting the entire system, etc. Now we assume that the Lagrangian of the system in the new configuration has the same value as the system in the old configuration. My question is, if q(t) obeys Lagrange's equations, will Q(t)=Q(q(t),t) too? This is a little different than the way you phrased my question, because you're talking about a change of coordinate system (eg rotating the x-y axes) rather than changing the system (eg rotating the system).

That said, despite what you wrote I still think the claim that Q(t)=Q(q(t),t) satisfies the equations of motion is a non-trivial one. Now if I were talking about a change in the coordinate system, then yes it would be trivial because Lagrange's equations hold for any coordinate system: that's why they're so powerful! But it doesn't seem so obvious if we consider an active transformation of the coordinates.

Let's look at the previous example. Lagranges' equations are $m{\ddot {x}}=-{\frac {\partial U}{\partial x}}$ and $m{\ddot {y}}=-{\frac {\partial U}{\partial y}}$. The question is, if x(t) and y(t) satisfy the EOMs, do X(t) = x(t)cosθ + y(t)sinθ and Y(t) = y(t)cosθ - x(t)sinθ satisfy them as well? I can verify that this is the case by plugging them in, but it isn't obvious to me that this should always be true. 65.92.7.148 (talk) 22:59, 29 July 2012 (UTC)

If you rotate or translate the whole system configuration, including external forces or constraints, then in general the form of the Lagrangian and the equations of motion will change, but a new rotated or translated trajectory will produce the same values of the Lagrangian along its path as the original trajectory. If the original trajectory satisfied the original equations of motion then the new trajectory will satisfy the new equations of motion. This is clear because a rotation or translation of the whole system is equivalent to a rotation or translation of the co-ordinate system of the same magnitude but in the opposite direction.

If you just rotate or translate the positions and velocity vectors of the particle(s), but do not rotate or translate forces and constraints, then the form of the Lagrangian and equations of motion do not change (because they do not depend on the particle configuration), but a new rotated or translated trajectory will, in the general case, produce different values of the Lagrangian along its path from the original trajectory. The new trajectory will not necessarily satisfy the equations of motion, even if the original trajectory did.

In your example, the potential (representing external forces) is radially symmetric about some point P, so U has the same form whether we rotate it or not. So a rotation of the system configuration about P leaves the form of the Lagrangian unchanged (because we can regard U as not being rotated), and a rotated trajectory also produces the same values of the Lagrangian along its path as the original trajectory (because now we can switch our view point and regard U as having rotated along with the trajectory).

On the other hand, a translation of particle positions and velocity vectors without translating U leaves the form of the Lagrangian unchanged, but the values of the Lagrangian along a translated trajectory are different from the values along the original trajectory, and in general the translated trajectory will not satisfy the equations of motion.

To sum up:

1. If two trajectories x₁(t) and x₂(t) produce the same values of the Lagrangian at the same time t along the whole of their paths, must one trajectory satisfy the equations of motion if the other does ? Answer: No - BenRG has shown a counterexample.
2. If two trajectories x₁(t) and x₂(t) produce the same values of the Lagrangian at the same time t along the whole of their paths because of a symmetry in the external forces or constraints, must one trajectory satisfy the equations of motion if the other does ? Answer: Yes, because we can exploit the symmetry of the external forces (or, if you like, the symmetry of the Lagrangian) to regard these forces as having moved along with the trajectory, but nevertheless still leaving the form of the Lagrangian unchanged. Gandalf61 (talk) 07:41, 30 July 2012 (UTC)

July 28

Very important question

If $x=7+{\sqrt {40}}$, then, find ${\sqrt {x}}+1/{\sqrt {x}}$. I found this question in a competition book. I solved almost every question of that book but I could not solve this one. As I don't take any tuition class, will you, please, solve the given question. Sunny Singh (DAV) (talk) 15:40, 28 July 2012 (UTC)

What have you tried so far? As a hint, ${\sqrt {7+2{\sqrt {10}}}}$ is of a form that denests quite nicely. --Kinu ^t/_c 16:09, 28 July 2012 (UTC)

To denest, look for a and b such that ${\sqrt {7+{\sqrt {40}}}}={\sqrt {a}}+{\sqrt {b}}$. The first step towards doing this is to square both sides giving you $7+{\sqrt {40}}=a+b+{\sqrt {4ab}}$. Can you find a and b such that a + b = 7 and 4ab = 40? — Fly by Night (talk) 22:15, 28 July 2012 (UTC)

The obvious solution is just to use a calculator to find that x = 13.32455532, and plug that in to find your answer, but I assume they want you to solve this in a more convoluted manner. BTW, please try to use more descriptive titles. In this case, "Square root problem" might work. StuRat (talk) 22:48, 28 July 2012 (UTC)

I think the question wants you to simplify the radicals rather than evaluate them as decimals, using the suggestions above if ${\sqrt {x}}={\sqrt {a}}+{\sqrt {b}}$

then

${\sqrt {x}}+{\frac {1}{\sqrt {x}}}={\sqrt {a}}+{\sqrt {b}}+{\frac {1}{{\sqrt {a}}+{\sqrt {b}}}}$

$={\sqrt {a}}+{\sqrt {b}}+{\frac {{\sqrt {a}}-{\sqrt {b}}}{a-b}}$

$={\frac {(a-b+1){\sqrt {a}}+(a-b-1){\sqrt {b}}}{a-b}}$ . 83.100.173.200 (talk) 09:44, 29 July 2012 (UTC)

I'm skeptical of your last step there. —Tamfang (talk) 10:05, 29 July 2012 (UTC)

Thank you, I've hopefully corrected it! 83.100.173.200 (talk) 10:23, 29 July 2012 (UTC)

Better! —Tamfang (talk) 05:06, 1 August 2012 (UTC)

Tests

There is a name for tests that require an absolute level of ability and you are not judged against others, such as a driving test, and tests where your result if indicative of your ability relative to others taking the test, such as A-levels, but I cannot remember what the names are. Can someone help? Thanks. 92.14.213.234 (talk) 17:04, 28 July 2012 (UTC)

Tests where you are measured relative to others are "graded on a curve". However, the grade before the curve is also your absolute competence level.

One problem with just measuring absolute competence level is that it's highly dependent on the test. Thus, a poorly worded test or one which tests skills not needed in real life ("What was the name of Benjamin Harrison's childhood goldfish ?") does not produce useful results.

A problem with grading on a curve results from too small of a sample size. Thus, a student measured against one class may rank better than if the same student is ranked against another class. The solution, of course, is to compare with a larger pool. StuRat (talk) 22:35, 28 July 2012 (UTC)

I assume that the distinction is between norm-referenced assessment and criterion-referenced assessment.→86.139.64.77 (talk) 17:01, 29 July 2012 (UTC)

July 29

under a coordinate transformation

$a(x,y){\frac {\partial ^{2}u}{\partial x^{2}}}+2b(x,y){\frac {\partial ^{2}u}{\partial x\partial y}}+c(x,y){\frac {\partial ^{2}u}{\partial y^{2}}}+d(x,y,u,{\frac {\partial u}{\partial x}},{\frac {\partial u}{\partial y}})=0$.

Under a coordinate transform $\xi =\xi (x,y),\eta =\eta (x,y)$ we magically get the following:

$\alpha (\xi ,\eta ){\frac {\partial ^{2}u}{\partial \xi ^{2}}}+2\beta (\xi ,\eta ){\frac {\partial ^{2}u}{\partial \xi \partial \eta }}+\gamma (\xi ,\eta ){\frac {\partial ^{2}u}{\partial \eta ^{2}}}+\delta (\xi ,\eta ,u,{\frac {\partial u}{\partial \eta }},{\frac {\partial u}{\partial \eta }})=0$ where $\gamma (\xi ,\eta )=a\eta _{x}^{2}+2b\eta _{x}\eta _{y}+c\eta _{y}^{2}$.

Could somebody please explain. --150.203.114.14 (talk) 07:20, 29 July 2012 (UTC)

I don't think you should be able to prove it.

So, someone here refused to suppose that a prime, j, was the sum of a certain number of perfect numbers. This makes sense if the perfect numbers are all even, as, with the first one being 6, you could never get to an odd prime by adding even numbers. (I say "odd prime" because above the first perfect number they're all odd.)

So, I wonder if he tacitly was refusing to suppose that there was an odd perfect number. I looked at the article and there is no proof - only strong heuristics that there wouldn't be.

I personally think that it can't ever be proven. 84.3.160.86 (talk) 08:38, 29 July 2012 (UTC)

What is your question? --Tango (talk) 14:55, 29 July 2012 (UTC)

as stated. I've given you my reasoning. 84.3.160.86 (talk) 23:51, 29 July 2012 (UTC)

well, I guess I haven't given my reasoning. The question is I personally think that it can't ever be proven? 84.3.160.86 (talk) 23:54, 29 July 2012 (UTC)

That isn't a question. Can you rephrase? Are you asking if we think it can be proven? I believe consensus is that it will someday be proven that there is no odd perfect number, but this is just a (educated) guess. Staecker (talk) 00:47, 30 July 2012 (UTC)

Why? Don't mathematicians expect loads of statements to turn out to be true-but-only-statistically-provable, being able to produce a proof that it's not actually provable one way or another, while having an actual reality and not being independent? 84.3.160.86 (talk) 01:35, 30 July 2012 (UTC)

Yes, but perfect numbers aren't the kind of thing we would expect to run into logical difficulties with. They aren't weird enough for that! --Tango (talk) 20:04, 31 July 2012 (UTC)

I've also asked another question below which you may be interested in. 84.3.160.86 (talk) 00:01, 30 July 2012 (UTC)

is "can't be proof" synonymous with independent?

Suppose I produce this fine glowing proof on one sheet of crisp high-quality slightly rosy-colored paper that it is impossible to prove that there are no odd perfect numbers. Mathematicians look over the proof but find no flaw and accept it. (And not because you can't argue with high-quality paper.)

So, we now have an accepted proof that we can't prove that there are no odd perfect numbers. Does the existence of this correct proof ==> that suddenly this question has become independent and ==> now I can make a mathematics that is completely the same as the standard mathematics accept that it assumes there is an odd perfect number?

Or, even though there is a proof that we can't prove that there are no odd perfect numbers, is it still possible for it to be impossible for it to be independent? (for example due to the incredibly strong heuristics.)

I guess what I am saying is, is a proof that something can't be proved at once a proof that under the same system it cannot be wrong (i.e. must be independent) to assume that it is true? 84.3.160.86 (talk) 23:59, 29 July 2012 (UTC)

Well, you need to specify what axioms are incapable of proving it. I can construct a very short proof that there are no odd perfect numbers, by taking it as an axiom. Once you've specified your axiom system, then yes, if you can prove a statement unprovable, you can add its negation to your axiom system and have a fine new system. This is either a consequence of Godel's completeness theorem, or a consequence of the fact that classical logic supports proof by contradiction, depending on what you require of an axiom system.--119.225.45.242 (talk) 00:09, 30 July 2012 (UTC)

Can you explain more about the last part? Let's take our normal axioms, nothing extra. Now we examine the "no odd perfect numbers" (nopn) statement, which there is incredibly compelling heuristics for believing. It is vanishingly unlikely to be false. Suppose that we have a proof, P, that the nopn is unprovable. Now might it not still be wrong to assume either nopn or to assume aopn, since, in fact, there could be a reality? (the reality for which we have incredibly statistical certainty or its opposite reality).

(I guess what I'm asking is that if you prove that nopn is unprovable, but the reality of the system - if you could do an infinite check real quick - is that the numbers that match opn are {} - then you, thinking nopn is unprovable, might feel free to add an axiom that is it is false -- as you've just said is allowed -- but in the physical sense under that axiom you've just taken a false statement as an axiom!!! which is obviously horrifically bad.

(it's like this: if under some weak system it is impossible to prove something trivial - but you can prove this non-provability - that doesn't make it independent - it could still have a truth-value, couldn't it? Therefore, if it has a truth value, then if you pick the wrong side you've just made your system include an axiom that is false under the other axioms. So it's not really independent at all, even though you can't prove it. Only by not including it do you have a consistent system.

(So, assuming our current system is consistent, couldn't it be made inconsistent by trying to add the axiom as you suggest, after proving it's unprovable?) 84.3.160.86 (talk) 01:32, 30 July 2012 (UTC)

You're confusing several different notions here. When you talk about "a reality", or about statements "having a truth-value", you're talking about whether or not statements are true in the standard model of arithmetic. This is the setting we all know and love--the natural numbers with the usual arithmetic operations on them.

Suppose you've proven that nopn can neither be proven nor disproven in PA. Then you've shown that nopn is independent of PA, and so PA + aopn is consistent. This new system doesn't describe the standard model of arithmetic, but that doesn't make horrifically bad or not really independent. The axioms of group theory don't describe the standard model of arithmetic, but that's no reason to condemn them.--119.225.45.242 (talk) 02:09, 30 July 2012 (UTC)

How do you go from "Suppose you've proven that nopn can neither be proven nor disproven in PA. Then you've shown that nopn is independent of PA" is what I want to know.

Definition. Independence (mathematical logic).--119.225.45.242 (talk) 02:35, 30 July 2012 (UTC)

This is interesting. So, as long as you can't prove a false statement or it's opposite, you can assume the false statement as an axiom and nothing bad will happen? 84.3.160.86 (talk) 08:58, 30 July 2012 (UTC)

No, that's not correct. You won't be able to prove a contradiction. But that doesn't mean "nothing bad will happen". You will be able to prove false things (the false axiom, as one example, will have a trivial proof), and it is bad to be able to prove false things, because that means your proof system is not reliable. --Trovatore (talk) 10:26, 30 July 2012 (UTC)

So you're saying independence is independent of truth, so to speak. So, is the axiom of choice actually probably true under the rest of zfc or actually probably false under the rest of zfc. (not that I'm not asking about our world,but the world defined only by zfc and nother else). 84.3.160.86 (talk) 14:04, 30 July 2012 (UTC)

It might very much be dependent on PA - after all, PA is what produces all the statistical evidence we have for it!!!! - and have one monstruously definite truth value. However PA might be just weak enough that that truth value cannot be proven. How do you then go and say it could be the opposite of what it would be if PA were slightly stronger and could prove it?

Basically, you are saying that the moment I make a system just dumb enough to not be able to prove that Pi is irrational - but we do have pi in that system, which has the same value there as anywhere else - then you might as well assume the axioms that lead to the normal value of pi, and also that it terminates. But this is now inconsistent.

What I'm saying is just because you're a shade too weak to prove something doesn't mean it can't still be true. So there could be a special case of you being just strong enough to prove that you're just weak enough not to be able to prove something, but also that you're strong enough for these axioms to definitely imply a truth-value under the same axioms.

It's like saying that if I prove to myself that I'm not smart enough to prove whether there is an integer that is both divisible by 2 and divisible by 3, but not divisible by 6, then suddenly this is independent. Well, no. There is no such number, regardless of whether I've just proven to myself that I'm not smart enough to prove it. 84.3.160.86 (talk) 02:22, 30 July 2012 (UTC)

This looks suspiciously like it's veering off into a discussion of the nature of truth, but I'll try to explain the point in the third paragraph. The point is you have "you're strong enough for these axioms to definitely imply a truth-value under the same axioms". The question is what exactly is meant by "imply". Assuming we're (sufficiently strong - note that strength here has got nothing to do with how much you know about arithmetic, and is dealing with things like A->B and B->C implying A->C) in first-order logic of some kind (such as weak PA), Godel's completeness theorem says that imply in the sense of "can be proved from" (syntactic) and in the sense of "must be true if _ are" (semantic) are exactly the same. Now I've written that, I've realised you might not be in weak PA but in PA with a full induction axiom (all subsets containing 0 and successors of everything they contain are the whole model): our article on the Peano axioms#Models says that this any model of this is ordinary arithmetic. For the record I also ought to point out that strong heuristic evidence is not proof, especially in number theory. Does that help at all? Straightontillmorning (talk) 18:35, 30 July 2012 (UTC)

I'm having trouble understanding what you're saying. Let me tell you more about my thinking. Suppose I the following axiom: a sibling is a boy or a girl but not both. Now suppose I take as axiomatic that "I have two siblings". Can you prove (from this and the other axiom, plus the rest of zfc+) they are both girls? No. Can you prove they are both boys? No. Can you prove that one is a girl, one is a boy? No. But can you prove that at least one is a boy or a girl? Yes. So returning to the first two statements "Both are girls" is neither false nor true under the axiomatic system - it really is totally independent and has NO truth value under just those axioms.

So this is one kind of "independence" - lack of enough "implicative" power in the axioms to pigeonhole the thing to a truth value. But there is the other kind -> where the axioms DO pigeonhole the thing into having an actual truth-value. Thus if some axiomatic system comes up with Pi but can't determine whether it's rational or irrational - well, despite the fact that they can't determine it, if it's the same pi we are talking about then the answer is "it's irrational". So there is a truth-value there, JUST from the rest of the axioms. It's impossible to come up with a true system (i.e. one where you can't prove statements that are actually false in that system) where Pi has the same value it has in our system but is proved to be rational. Sure, with the right weak system of axioms maybe they can come up with a non-constructive proof of hte false statement that there is an exact numerator and denominator that is the reduced value of Pi - but it will be false despite being consistent there. So whereas I can have two sisters or I can have two brothers, and you can't prove it from what I've told you and EITHER could be correct, if pi could be rational and pi could be irrational and "EITHER" could be correct, in fact one "can't". One is really false. Do you see my thinking? That there are two kinds of independent axioms: ones that actually have a pigeon-holed value (such as the truth that pi is irrational in any system that produces its true - our - value) and others that actually are free to be one way or another (such as having two brothers or two sisters in the axiomatic system that says siblings are just male or female. likewise in that system it is simply "false" to say someone could have 1 sibling, and at least one sister and one brother - even if they can't prove it because their axiomatic system is too weak for the pigeonhole principle.) DO you get where I'm coming from? 84.3.160.86 (talk) 20:56, 30 July 2012 (UTC)

I understand what you're saying - you're talking about the difference between proving and entailment (our article on this doesn't make a lot of sense to me but that's the right concept) - but Godel's completeness theorem says that in the cases we normally care about these are the same - if you know what pi is, you know it is irrational. If you can't establish it's irrational, you certainly can't establish that it's the same pi. Straightontillmorning (talk) 12:37, 31 July 2012 (UTC)

A first order theory is consistent iff it has a model, a statement is independent if it and its negation is consistent with the theory; so if p is independent of T, then there should be a model for T+p and T+~p. This doesn't seem to jive with the above at all; then again, I'm at work and having a dreadfully unpleasant day, maybe I missed something or am having a moment:-)Phoenixia1177 (talk) 11:38, 1 August 2012 (UTC)

July 30

suppose a man produces a proof P that statement Q cannot be proven under Some Axioms but definitely has a certain Truth Value (unknown) under those axioms.

A man produces a proof, P, that under a certain set of axioms, TCA (the cool axioms), there is a statement, Q, that cannot be proven under TCA, but definitely has a concrete truth value (unknown) under those axioms and so assuming it is independent would certainly be an error as it would lead anyone to DEFINITELY produce an inconsistent system wherenever the wrong value was chosen for the "independent" axiom - it's not independent at all. 84.3.160.86 (talk) 01:38, 30 July 2012 (UTC)

the question here is, some people are saying this isn't possible. why not? why couldn't I prove that it is definitely unprovable under standard axioms whether there is an odd perfect number, but that this also DEFINITELY has a truth-value that no one can prove under these axioms? Why would provability be conflated with truth? I find it quite easy to imagine that an axiomatic system has a truth that cannot be proven; so assuming its opposite would be false and inconsistent. 84.3.160.86 (talk) 01:39, 30 July 2012 (UTC)

You're right that provability is not the same as truth. What's confusing, then, is talking about truth "under a set of axioms". Axioms don't determine truth, only provability (at least in the ordinary first-order logic paradigm).

It is quite possible for a statement to have a definite truth value in spite of being independent of some given set of axioms. Which statements you think those are is likely to depend on your philosophy of mathematics.

For example, ZFC neither proves nor refutes Con(ZFC), the statement that ZFC is consistent (unless ZFC is actually inconsistent, in which case ZFC both proves and refutes Con(ZFC)). The large majority of mathematicians would accept Con(ZFC) as a statement with a definite truth value. (Whether it's an "unknown" truth value is another matter and depends on what you mean by "known".)

I personally would go much further. I would assert, for example, that the continuum hypothesis is a statement with a definite, but still unknown, truth value. --Trovatore (talk) 01:52, 30 July 2012 (UTC)

You say that axioms don't determine truth, but how can they be? Any axioms that determine Pi do determine a billionth, trillionth, and so on digit. If it can't be proven whether Pi switches at a certain point to being the digits of e and then continues thereon like that, doesn't mean that this isn't actually true or actually false for actual pi: what does provability have to do with truth? Of course there are truths beyond provabilities - why would you say axioms don't have truth when they clearly do, under that system of assumptions some things are true and others false. 84.3.160.86 (talk) 02:13, 30 July 2012 (UTC)

I didn't say axioms don't have truth. I said they don't determine truth. From the mathematical realist point of view, mathematical objects are real things, existing without reference to our reasoning about them. We choose axioms that we believe to be true about those objects, and then use them to prove other statements that must also be true, assuming we didn't make an error in the choice of axioms in the first place. So mathematical truth comes before the axioms, not after. --Trovatore (talk) 02:29, 30 July 2012 (UTC)

It's certainly possible -- that's just what Gödel's theorem does. Gödel proved that in any axiom system that meets certain basic conditions, it is possible to construct a statement that is (a) true, and (b) impossible to prove within that axiom system. Looie496 (talk) 02:05, 30 July 2012 (UTC)

Yes, but this "true" is in an absolute sense; it is not relative to the axioms. So that would be my quibble with the way the OP put it. --Trovatore (talk) 02:07, 30 July 2012 (UTC)

You miss that I require that it be provable that it's unprovable, from within the axiomatic system. You don't add that - we are talking about the proof of not some general "must-be-some-true-statement-can't-be-proven" we are talking about "I just proved that THIS statement is unprovable and that (THIS statement xor OPPOSITE THIS STATEMENT) is true." Okay, I just felt like a complete idiot writing that out. Obviously (THIS statement xor OPPOSITE THIS STATEMENT) is always true. What I mean is that one of them is definitely false and it's not independent, you can't just assume one of them. 84.3.160.86 (talk) 02:13, 30 July 2012 (UTC)

In other words: I just made a proof that THIS statement is unprovable but that (either assuming THIS statement produces an inconsistent new set xor assuming OPPOSITE THIS statement produces an inconsistent new set): in other words, IT'S NOT INDEPENDENT AT ALL, but dependent in some unprovable way on the other axioms. 84.3.160.86 (talk) 02:16, 30 July 2012 (UTC)

You keep equating "inconsistent" with "doesn't describe intuitive reality", but they're very different concepts. If TCA + X is inconsistent, then TCA proves "not X". This is the basic proof by contradiction of classical logic. So in the above you are supposing a contradiction; it's no wonder strange consequences are following.--119.225.45.242 (talk) 02:21, 30 July 2012 (UTC)

Okay, you've gotten at the crux here. You're saying "If [you can prove] TCA + X is inconsistent, then TCA proves "not X"". But my whole point is you CAN'T prove TCA + X is inconsistent, and you CAN'T prove TCA + Not X is inconsistent...you can PROVE that you CAN'T prove these things. But, at the same time, you can also prove that ONE of them must be false just under your previous axioms. (They can't both be false.) 84.3.160.86 (talk) 08:46, 30 July 2012 (UTC)

A trivial example: Some of the numbers 2,3,4,5,6,7,8,9,... are composite (such as 2×2=4 and 2×3=6). The others are prime. Now let's make this true statement an axiom: "Even numbers beyond 2 are composite". This axiom is insufficient. It cannot prove that 9 is composite even if it is true that 9=3×3 is composite. Nor can it prove the negation, that 9 is prime. It is the rule rather than the exception that axioms are insufficient. Bo Jacoby (talk) 07:30, 30 July 2012 (UTC).

I'm trying to follow your argument. You give an example of a true statement being taken as an axiom; but my question is about a false statement that you can't prove. Can you take it or its opposite as an axiom, even though it is false under the rest of the axioms in an unprovable way? Indeed is that possible? FOr example, in a system that is too weak to prove that 7 is prime, but strong enough to prove under itself that it SURELY CAN'T prove that, under such a system it would seem that "7 is prime" is independent. But in fact we know that it is NOT prime. So it has a definite truth-value. Here, let me give you a case that is apparently as simple as that; http://en.wikipedia.org/wiki/Transcendental_number#Numbers_which_may_or_may_not_be_transcendental

Catalan's_constant is a number we can 'count to'. Suppose that we prove that we can't prove whether it's irrational. That doesn't mean it's ACTUALLY not rational!! So, people here are saying hte moment you prove you can't prove whether Catalan's constant is irrational, you can take "Catalan's constant is irrational" as an axiom and go about your way as it's completely independent. But in fact that doesn't change that, there it is, perhaps it is rational and has a very nice integer numerator and very nice integer denominator that are simply outside of our system's ability to prove. So, we have an unbeknonst-truethness, that it's rational, coupled with a proof that we can't prove it way way or another. It's not really independent; it's just unprovable. If you pick the wrong side as an axiom you've created an unbeknownst-false-system. (Since some of your axioms imply something that you take the opposite of as a new axiom.). Do you see what I'm getting at? Proving that Catalan's constant can't be proved irrational or rational doesn't mean that it definitely ISN'T one of these. Isn't there a problem if you assume something that is false under the rest of your arithmetic, merely because you've proven you can't prove it false or true? 84.3.160.86 (talk) 09:08, 30 July 2012 (UTC)

specific example

For a long time a proof was sought for Euclidean parallel postulate, until it was proved independent. In other words, if Euclidean parallel postulate was "x", then it was proven that neither TCA + X nor TCA + Not X was inconsistent.

Now imagine that if instead of this proof, for a similar postulate this was proved: "1. You cannot prove TCA + X is consistent or inconsistent and you cannot prove TCA + Not X is consistent or inconsistent, under TCA. 2. Secondly, through some other means you prove that one of them is in fact (Unprovably) inconsistent. [but obviously non-constructively; you can't prove which, just that one of the two are inconsistent. obviously if you could prove which it would contradict (1).]"

Obviously if #2 specified WHICH one was inconsistent, it would be tantamount to a proof. BUt without specifying WHICH one is inconsistent, just that ONE of them is (unprovably), in an unconstructive way that doesn't tell you which one is false. In other words: it's dependent but unprovable how. 84.3.160.86 (talk) 08:53, 30 July 2012 (UTC)

So that is the mistake in the above poster: ". If TCA + X is inconsistent, then TCA proves "not X". This is the basic proof by contradiction of classical logic." -- I didn't say you prove that TCA + X is incosistent; only that you prove that either TCA + X is inconsistent or TCA + Not X is inconsistent, and, therefore, it's not indendependent. However the proof is not a constructive one: you are only able to prove that there is a truth-value inside the system. You can't prove that truth-value; and in fact you proceed to prove that you can't prove it. So, please explain to me why this state of affairs is not possible. 84.3.160.86 (talk) 08:52, 30 July 2012 (UTC)

Standard proof for identity:

Is there a standard proof for this identity?: ${\sqrt {a^{2}+b^{2}}}\leq |a|+|b|\leq {\sqrt {2}}{\sqrt {a^{2}+b^{2}}}$. Intuitively it makes sense and you can prove it using polar coordinates if you assume some trigonometric identities, but there is probably a better way. --130.56.93.90 (talk) 04:29, 30 July 2012 (UTC)

1. Show all three expressions are non-negative.
2. As they are nonnegative, you can square them and inequalities will hold, so you'll get $a^{2}+b^{2}\leq \left(|a|+|b|\right)^{2}\leq 2(a^{2}+b^{2})$.
3. HTH --CiaPan (talk) 05:26, 30 July 2012 (UTC)

To write out a formal proof, you need to start from an identity and derive the given expression (although informal exploration often proceeds in the opposite direction). There are two parts to your given expression, so your proof comes in two parts:

1. Start from $(|a|+|b|)^{2}=a^{2}+2|a||b|+b^{2}\,$ and derive ${\sqrt {a^{2}+b^{2}}}\leq |a|+|b|\,$.
2. Start from $(|a|-|b|)^{2}\geq 0\,$ and derive $|a|+|b|\leq {\sqrt {2}}{\sqrt {a^{2}+b^{2}}}\,$. Gandalf61 (talk) 06:49, 30 July 2012 (UTC)

If you have a series of statements connected by "iff"'s, then it doesn't matter which direction you write them in. They are logically equivalent. --Tango (talk) 20:33, 31 July 2012 (UTC)

we just proved we can't prove it one way or another or know which one it is, but this statement is definitely false or true - not independent.

what's wrong with this statement? It looks fine to me. — Preceding unsigned comment added by 84.3.160.86 (talk) 09:35, 30 July 2012 (UTC)

What's wrong with it is that "independent" does not mean "not false or true". Rather, "independent" means "not provable nor refutable". You seem to have picked up a distorted impression of what "independence" means — not hard to do; the popularizers make this sort of mistake all the time. --Trovatore (talk) 10:29, 30 July 2012 (UTC)

Could you be more explicit. So, it can be independent while being false not in a physical sense or in our world, but in the world of a few axioms which are enough to "make" it true? So, for example, in some very simplistic system that is barely able to count 1, 2, 3, 4, 5, it can be "independent" whether there is another integer between 1 and 2 - even though, since it successfully enumerated the integers, it has produced a world in which that statement is very much physically false. So, we could add an axiom to that system that assumes an integer between 1 and 2, even though it is physically false EVEN within that simplistic system? 84.3.160.86 (talk) 11:03, 30 July 2012 (UTC)

I also find it very interesting that we can do a heuristic/statistical analysis on, e.g. the GOldbach conjecture, "Every even integer greater than two 2 can be expressed as the sum of two primes". Suppose someone proves it independent. You're saying it could still be true, and that the system where it's opposite is taken as an independent axiom is simply false mathematics - it assumes something that is false according to actual reality as established by the rest of the axioms? 84.3.160.86 (talk) 11:03, 30 July 2012 (UTC)

There's some interesting discussion of what it would mean for the Goldbach conjecture to be undecidable here. AndrewWTaylor (talk) 16:59, 30 July 2012 (UTC)

Trovatore: okay, what if I emend the statement to: "{we just proved we can't prove it one way or another or know which one it is, but this statement is definitely false or true - but our proof of this fact is not constructive and we don't know which." Is there anything wrong with it then? What if I were to add... "so INSOFAR as it is independent (not saying it is), there is a right/true and a wrong/false value for it as a new axiom. We just don't know which is which, but we have the non-constructive proof that it definitely has a specific truth-value we can't prove." — Preceding unsigned comment added by 84.3.160.86 (talk) 11:07, 30 July 2012 (UTC)

example of an axiom that is statistically probably false

have there been examples of an axiom for which the previously accepted axioms were eventually sufficient to give a high statistical/heuristic likelihood that it had a particular truth value without defining it as such by taking it axiomatically? — Preceding unsigned comment added by 84.3.160.86 (talk) 11:17, 30 July 2012 (UTC)

popular description of result of independence

in the popularization of mathematics, it sounds like when a mathemtician proves - usually surprisingly - that a statement is 'independent', it is taken to mean that it could be true or false without leading you to prove anything false by assuming either one. But that isn't right then is it - you could very much prove wrong results, using independent axioms, if in fact you picked an axiom that happened to actually be false, right? In other words: INDEPENDENCE guarantees your proofs won't be inconsistent: not that you won't prove something that is false. Do I have it right? 84.3.160.86 (talk) 14:09, 30 July 2012 (UTC)

Indeed, is it possible that out of the 5 axioms we hold most dear, some of them are actually false in every logical world where the others are true? But it simply can't be proven as it's independent, and in fact even though false (more specifically, you can prove false things), the story remains "consistent". 84.3.160.86 (talk) 14:09, 30 July 2012 (UTC)

The Axiom of Choice is false. Count Iblis (talk) 17:21, 30 July 2012 (UTC)

Can you also specify the relationship between this fact and its independence, and how you surmise that it is false? Also is it false in a physical sense, i.e. "our" geometry or the "best" world-approximating axioms we might pick, or false on some deeper level and probably notwithstanding a little fidging with what particular axioms we're working with? How do you know, or think you know, about it's falseness? This gets to the center of my questions above and perhaps I should have just begun with this example. 84.3.160.86 (talk) 17:42, 30 July 2012 (UTC)

See this article. Count Iblis (talk) 18:05, 30 July 2012 (UTC)

I think the question I have to ask in all of these is "independent of what?" - independence is not a term that makes sense of a statement on its own. Straightontillmorning (talk) 18:44, 30 July 2012 (UTC)

zfc or other normal axiomatic systems that are strong enough to prove many questions about the same things that are of interest to us. 84.3.160.86 (talk) 20:27, 30 July 2012 (UTC)

The thing is, a lot of things that are independent of ZFC are not independent of natural strengthenings of ZFC. The large-cardinal axioms provide a natural hierarchy of such strengthenings. You could potentially identify a category of statements that are independent of ZFC plus all known large-cardinal axioms (for example, the continuum hypothesis), but it's still a little artificial; there is no guarantee that all future natural strengthenings will behave the same way. --Trovatore (talk) 23:51, 31 July 2012 (UTC)

sigh. "a lot of things that are independent of ZFC are not independent of natural strengthenings of ZFC". Is that because they're already true under ZFC as well, so it's natural to strengthen it so they can be proven instead of just sitting there all true? 84.3.160.86 (talk) 12:47, 1 August 2012 (UTC)

The axiom of choice, of course, is true. It is self-evident; anyone who understands the intended interpretation (and agrees that the objects of that interpretation actually exist) is almost forced to agree that it is true. I take it that Iblis's point is, he doesn't agree that those things actually exist (I know that's what Zeilberg thinks). They're wrong, of course, but this is discovered by a more a posteriori process. --Trovatore (talk) 23:39, 31 July 2012 (UTC)

This is exactly what attracts me in mathematics, and I think this is the true heart of math. People arguing by bold statement over what we should definitely believe a priori. The rest just follows. 84.3.160.86 (talk) 12:50, 1 August 2012 (UTC)

Integral proof

$\int _{0}^{\infty }{\frac {\ln(x)}{1+e^{x}}}dx={\frac {-1}{2}}\ln ^{2}(2)$

This is according to Wolfram Alpha, but no closed form of the indefinite integral is given. How can I derive this? 149.169.218.127 (talk) 18:01, 30 July 2012 (UTC)

Looks pretty nasty but it is obviously a contour integration problem. I believe you'll need a special 'trick' to generate logs by substituting x^t for the log, solving that general problem and then differentiate by t to generate the log x. On a quick try there seems a couple of other problems but I believe that should get the answer somehow. Dmcq (talk) 18:55, 30 July 2012 (UTC)

This is number 4.354 in Gradshteyn-Ryzhik, a table of integrals that always gives references to places where proofs can be found. They give the reference GW (324) (86a), but I can't tell you what that means from the Google Books snippet. See here for the book. —Kusma (t·c) 19:14, 30 July 2012 (UTC)

Some more searching gives this book as a source. It probably contains a proof. —Kusma (t·c) 19:26, 30 July 2012 (UTC)

Kusma, looking for the answer in books is cheating :(

Dmcq's suggestion seems to work, but you need to do additional work here as you need to consider the limit of the derivative of (1-1/2^s)Zeta(s+1) Gamma(s+1) for s to 0. Count Iblis (talk) 19:34, 30 July 2012 (UTC)

$\int _{0}^{\infty }{\frac {\log(x)}{1+e^{x}}}dx=-{\frac {(\log 2)^{2}}{2}}$. (clarifying $\ln ^{2}(2)$). Bo Jacoby (talk) 07:39, 31 July 2012 (UTC).

Made all the "ln"s in this thread operators for consistency. OCD and all... --Kinu ^t/_c 09:24, 31 July 2012 (UTC)

Um yeah that's right Count Iblis but I'm afraid I funked out after that and did a google search of 'contour integral logarithm exponential' and the second entry after the wikipedia one of contour integrals was [3] which is this exact problem with a complete solution! The main delay was typing the query. I'm getting seriously scared that google is both able to read my mind and do maths far better than me ;-) The thing is contour integrals weren't used directly but I wouldn't have found that discussion if I hadn't stuck it in the query. Dmcq (talk) 09:57, 31 July 2012 (UTC)

July 31

Matrices are unique?

Why is the mattrix corresponding to a linear transformation $L:\mathbb {R} ^{n}\rightarrow \mathbb {R} ^{m}$ unique ?--130.56.71.52 (talk) 11:09, 31 July 2012 (UTC)

It is unique with respect to the choice of basis. The columns of the matrix correspond to where the basis vectors map to, so once you know the linear transformation those columns are uniquely specified. Staecker (talk) 12:26, 31 July 2012 (UTC)

The key fact is that every vector can be written as a unique linear combination of the basis vectors. (Which is true by definition.) This guarantees that the columns that Staecker mentions are themselves unique. Which tells you that your n-by-m matrix is itself unique, with respect to the fixed basis. — Fly by Night (talk) 18:53, 31 July 2012 (UTC)

Quality of polling

The Gallup Organization, Pew Research Center, Angus Reid Public Opinion, and ComRes. These are four of many listed in Opinion_poll#Polling_organizations. Is there a site anywhere that compares their Survey methodology? Like a consumer reports listing them by rank of quality? Some may just phone 10 people a week and sell those numbers. Two I found two that claim 1000 a day.--Canoe1967 (talk) 12:25, 31 July 2012 (UTC)

I think I figured it out. Depending on the value of the data they pay more for a better sample. For public opinion in a news story they don't need much. For building a mall or where to election campaign harder they need better numbers. I will resolve this unless someone else wants to kick it.

Resolved

--Canoe1967 (talk) 13:13, 31 July 2012 (UTC)

Axioms for Euclidean geometry

I'd like to explore what axioms of perpendicularity need to be added to the affine structure of R² in order to obtain Euclidean geometry.

More precisely, I'll assume we have a real affine plane P, that is, a set on which a two-dimensional real vector space V acts simply and transitively. Now I would like to add axioms concerning the relation of perpendicularity between lines in order to reach Euclidean geometry (with a scalar product that is well-defined up to a constant multiple).

I've discovered that the following axioms work to get us to the final stage in Euclidean geometry (unless I'm mistaken), but I'm not entirely satisfied with them for reasons I'll explain.

1. The set of lines of V is partitioned into classes of cardinality 2. In each class we say that one line is perpendicular to the other. The relation of perpendicularity is then extended in the obvious way to lines of P. In particular, this allows us to define the "perpendicular bisector" of any segment.

2. The perpendicular bisectors of the three sides of a nondegenerate triangle meet at a single point.

The reason I'm not satisfied with 2 is that the proof I know relies on the notion of "congruence" of segments. That is, there is no way I know to make Axiom 2 intuitively clear without an elaborate argument appealing to the concept of distance.

I'd like to find a set of axioms for perpendicularity that are visually obvious, that don't rely for their intuitive justification on the additional concept of congruence (beyond what is present already in the affine structure), and that give us all of Euclidean geometry. Can anyone help? 96.46.200.71 (talk) 03:38, 1 August 2012 (UTC)

I think perhaps your terminology is confused. You don't have to add any axioms to the affine structure of R² to get Euclidean geometry -- you can define perpendicularity in terms of coordinates. Also your axiom 1 seems to be equivalent to a statement that for each line there is precisely one other line that is perpendicular to it, which is obviously false. Looie496 (talk) 04:51, 1 August 2012 (UTC)

August 1

Factorization

Find the factors of (factorize):
x² + 4y² + 4y - 4xy - 2x - 8 106.199.10.161 (talk) 09:44, 1 August 2012 (UTC)

Try doing little bits, the bits with two variables then the bits with one variable and you should see it get simpler. After that personally I'd add and subtract a 1 rather than invoking the formula for solving quadratics, can you see why? Dmcq (talk) 10:00, 1 August 2012 (UTC)

This is as far as I get:

x² + 4y² + 4y - 4xy - 2x - 8

x² + 4y(y+1-x) - 2x - 8

x² - 2x - 8 + 4y(y+1-x)

(x-4)(x+2) + 4y(y+1-x)

StuRat (talk) 10:23, 1 August 2012 (UTC)

(x-2y+2)(x-2y-4). Bo Jacoby (talk) 10:39, 1 August 2012 (UTC).

You can keep y as a parameter and solve the quadratic equation for x. The solutions are 2y+4 and 2y-2, which gives the factorization Bo Jacoby posted a minute before I finished calculating. —Kusma (t·c) 10:43, 1 August 2012 (UTC)

[4] :-). Bo Jacoby (talk) 10:48, 1 August 2012 (UTC).

simple toy axiomatic system that something is true in?

So, I am still back at square 0, trying to understand "entailment". Could you give me a simple axiomatic system that something is true in, but it is so simple (e.g. a toy system) that it can't prove anything at all? For example, maybe there are enough axioms to start counting 1, 2, 3, 4, 5, but the system is so simple that it can't prove that by continuing it won't ever get back to 1 again. (Which is nevertheless "true" in the system). Or any other example you want. Thank you. 84.3.160.86 (talk) 12:41, 1 August 2012 (UTC)

In other words, my example is such that from "inside" it would seem independent. Well, if our only guarantee is that if we take the next number we always get a number, we never run out, that doesn't proven the whole thing isn't cyclical. But in fact that IS the truth. (This is just my foolish laical attempt to come up with something where something is trivially true to an outsider. 84.3.160.86 (talk) 12:56, 1 August 2012 (UTC)