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July 19

Number that isn't a scalar

As I understand it, a scalar is a number which doesn't depend on the coordinate system. For example, no matter what coordinate system you use, the temperature of an object remains the same, so temperature is a scalar.

Are there examples of numbers which aren't scalars? 65.92.7.148 (talk) 03:12, 19 July 2012 (UTC)

The x-component of a vector is not a scalar. Bo Jacoby (talk) 03:31, 19 July 2012 (UTC).

I think you're talking about the definition in physics. Scalar has another meaning in mathematics, see Scalar (mathematics) and Scalar multiplication — Fly by Night (talk) 03:44, 19 July 2012 (UTC)

Indeed. I think what the OP is really asking about is cases where the mathematical definition and the physics definition disagree. Bo's example is a good one. Speed is another example. --Tango (talk) 03:55, 19 July 2012 (UTC)

Also agree. When physicists say "scalar" they (often) mean a scalar field. So:

1. A temperature distribution is a scalar field that associates a scalar (in the mathematical sense) to every point in space.
2. As Bo says, the x-component of a vector field associates a scalar (in the mathematical sense) to every point in space but is not a scalar field (unless it means "x component relative to a given fixed co-ordinate system" in which case it is a scalar field, but not a very natural one).
3. On the other hand, x-component² + y-component² + z-component² of a vector field is a scalar field because it represents the square of the magnitude of the vector field at each point, and so is invariant under co-ordinate transformations.
4. To be really pedantic, you have to keep unit length the same in the previous example. If you allow transformations that change the unit length then the magnitude of a vector field is a scalar density or relative scalar. Gandalf61 (talk) 08:37, 19 July 2012 (UTC)

Scalars are quantities which do not depend on certain group of symmetry transforms (see Symmetry for mathematical context and Symmetry (physics) for physical). But the term "scalar" is confusing because there is not always clear from the context, which symmetry group is assumed. For example, a function corresponding to what physicists call a scalar field is a scalar corresponding to the space transforms (such as different coordinate systems), and hence is a scalar function from the PoV of differential geometry. But if that physical theory provides gauge symmetry, the function is not a scalar value (with respect to gauge symmetry). For example, complex-valued ψ from Ginzburg–Landau theory is a geometrical scalar, but it is not a gauge scalar – only its absolute value, a real number, is a scalar in both senses. Oppositely, the magnetic field F_1 2 = $\partial _{1}A_{2}-\partial _{2}A_{1}$ from the same theory (2-dimentional + 1 time) is a gauge scalar (F is a 2-form of curvature), but it is not a geometrical scalar (relatively to Lorentz transforms). So, the notion of "scalar" is not absolute. Incnis Mrsi (talk) 11:53, 19 July 2012 (UTC)

Extendible cardinals vs Reinhardt cardinals

How are they related? Or are they the same thing? Rich (talk) 08:08, 19 July 2012 (UTC)

rotating a sphere in higher dimensions

how many dimensions is this tesseract rotating in?

If you rotate a circular object such as a bicycle tire about its axis (the axis at its center and normal to it) a force will be exerted that will tend to make it want to expand outward equally in all directions in the plane normal to the axis of rotation. If you spin an elastic spherical object about an axis passing through its center it will expand outward at its equator in the plane normal to its center of rotation. Would it be mathematically possible to spin an expandable sphere in some higher dimension so that all points on its surface would move outward equally in three dimensions (like a balloon expanding) away from the point at its center, rather than just at its equator? If so, what would this rotation be about, (obviously not about a two-dimensional linear axis) and in what dimension would the rotation have to be? Thanks. μηδείς (talk) 19:57, 19 July 2012 (UTC)

The first thing I will say is that rotation in even and odd dimensional spaces behaves very differently. I think you'll need to consider the cases $\mathbb {R} ^{2n}$ and $\mathbb {R} ^{2n+1}$ separately. — Fly by Night (talk) 21:00, 19 July 2012 (UTC)

You can do this in four dimension. It's enough to cook up a rotation that moves each point on the sphere by the same amount. You can find such a rotation by representing rotations in four dimensions by left and right multiplication by unit quaternions. Sławomir Biały (talk) 21:18, 19 July 2012 (UTC)

Thanks. I have to say that I am not familiar with anything more than one year of high school euclidean geometry and can understand how a tesseract works from Sagan and Flatland and comprehend its rotation from this wonderful animation. But the meaning of the terms $\mathbb {R} ^{2n}$ and quaternion are entirely unfamiliar to me. I will read the link to quaternion. My assumption is that if you can rotate a sphere in two dimensions about a one-dimensional axis in a three dimensional space and have its equator expand in a plane, by analogy you can rotate a sphere in three dimensions about a two dimensional plane in a 4d hyperspace and have its surface expand in three dimensions. Is that right? If so, can someone help me visualize what it is to rotate about a plane? And would it actually be a plane, or just a circle that bisects the sphere? What articles should I read?μηδείς (talk) 21:39, 19 July 2012 (UTC)

You might look at Plane of rotation and Rotations in 4-dimensional Euclidean space. To answer your question in four dimensions an isoclinic rotation has the property you desire: every point on the 3D surface of a sphere in four dimensions rotates at the same speed with such a rotation. The tesseract in the animation is actually rotating isoclinally in 4D, though I don't know if that helps visualise it: both the shape of the tesseract and the projection make hard to see what's going on.

In higher even dimensions you get analogous rotations, I guess also called isoclinic, where every point on a hypersphere is rotating with the same speed. In odd dimensions you always have a non-rotating axis so at least two fixed points.--JohnBlackburne^words_deeds 21:53, 19 July 2012 (UTC)

You can visualize the "isoclinic" rotations in 4 dim easily with the aid of the Hopf fibration, but this sort of exceeds my ability to explain. Sławomir Biały (talk) 22:06, 19 July 2012 (UTC)

Ok, so here's how you can visualize a sphere in four dimensions. At each point of a spherical globe, place a circle (a "clock") flat against the globe (so the face is pointing outward from the center). The four dimensional sphere is then parametrized by picking a point on the sphere and a clock value at that point. Then simultaneous rotation of every clock by the same amount is an isoclinic rotation ("passage of time", if you like). Sławomir Biały (talk) 00:22, 20 July 2012 (UTC)

Isn't that $S^{2}\times S^{1}$? That's not the same as $S^{3}$ is it? (In the same way that $S^{1}\times S^{1}$ is a torus, not a sphere.) --Tango (talk) 17:04, 20 July 2012 (UTC)

No, what I have described is the unit tangent bundle of S^2, which is the 3-sphere. Note that you cannot smoothly orient all of the clocks, so there is no preferred global time. This is something of a small technical point that is likely to be of little interest to the OP though. Sławomir Biały (talk) 19:42, 20 July 2012 (UTC)

And I thought the OP was talking about rotating a regular 3D sphere in a 4 dimensional space $(S^{2}$ embedded in $\mathbb {R} ^{4})$. Even after re-reading, the wording is ambiguous... SemanticMantis (talk) 19:21, 20 July 2012 (UTC)

That's how I read it too. 86.179.1.131 (talk) 20:18, 20 July 2012 (UTC)

July 20

If you know the average of a group bigger or equalt to 2 ...

...you do not know anything about its elements (right?). Is there a name for this 'rule'? OsmanRF34 (talk) 17:54, 20 July 2012 (UTC)

Well, you might know some things. For example, if you have 2 elements and the average is 2.75, you know that at least one of the numbers isn't an integer. You also know at least one of the numbers is larger than or equal to the average and one is smaller or equal. So, for example, if the average is negative, at least one of the elements is negative. StuRat (talk) 18:53, 20 July 2012 (UTC)

"at least one of the numbers is larger than the average and one is smaller" -- being very pedantic, unless the numbers are equal... 86.179.1.131 (talk) 20:20, 20 July 2012 (UTC)

Fixed. StuRat (talk) 23:44, 20 July 2012 (UTC)

Such a rule wouldn't be too useful. First, you'll have to restrict it maybe to positive numbers like age, size, and so on. Second, if you do this, you end up with other additional information - you know that ages and sizes have a certain range. If I give you a group of 2 humans whose average is 2 meter, you can deduct that they are not very far away from 2 m each one. Third, in the same way that you obtained the average (from a series of measures), you can obtain other statistics - mean, mode, and all sorts of distribution patterns. 88.9.110.244 (talk) 23:35, 20 July 2012 (UTC)

If you just know that the average is at least 2 (but not what it is), there are some things you can say. Some element is greater than or equal to 2. If any element is less than 2, then some element is greater than 2. Maybe in a stretch you could say that these are the pigeonhole principle.

If you know the average precisely and that all the values are not negative, then you get Markov's inequality, which says that for any value x, at most a fraction of average/x elements have value greater than or equal to x. Rckrone (talk) 16:58, 21 July 2012 (UTC)

July 21

System of bilinear equations

We know that system of linear equations can be solved in polynomial time (in terms of input bits). I want to know how we solve bilinear equations and if we can solve, can we solve in polynomial time (in terms of input bits)? — Preceding unsigned comment added by Karun3kumar (talk • contribs) 15:32, 21 July 2012 (UTC)

See System of polynomial equations. Bo Jacoby (talk) 18:45, 21 July 2012 (UTC).

Here is a PDF of a 1997 paper called Systems of bilinear equations that discusses the general problem and how it can be solved. Looie496 (talk) 19:19, 21 July 2012 (UTC)

(2x)^y=x

what is y?

example: 9^y=4.5

thank you — Preceding unsigned comment added by 79.180.141.120 (talk) 19:13, 21 July 2012 (UTC)

$y=\log _{2x}x={\frac {\ln x}{\ln 2x}}$ (for x > 0)--Wrongfilter (talk) 20:24, 21 July 2012 (UTC)

$={\frac {\ln x}{\ln 2+\ln x}}={\frac {1}{1+{\frac {\ln 2}{\ln x}}}}$ --CiaPan (talk) 20:40, 21 July 2012 (UTC)

Function inversion via control theory

I have an unknown function $f:\mathbb {R} ^{2}\to \mathbb {R} ^{2}$; writing $(\xi ,\eta )=f(x,y)$, we have ${\frac {\partial \xi }{\partial x}}>0$ and ${\frac {\partial \eta }{\partial y}}>0$ everywhere (a sort of monotonicity). I suspect that f is convex in the same sense, but doing without that assumption would of course be more powerful.

I would like to evaluate (with a precision dependent on the computational effort expended) $f^{-1}(\xi ,\eta )$ and the simpler $f_{x}^{-1}(\eta )$ where $f_{x}(y):=f(x,y)$. The numerical tool I have available takes the form of a dynamical system on $\mu =(x,y,\ldots )$ (where the dots indicate many additional dimensions). In this dynamical system ${\frac {dx}{dt}}={\frac {dy}{dt}}=0$ (the reason for calling them dimensions will be given in a moment), and there exist known functions $\Xi (\mu )$ and $H(\mu )$ such that $\lim _{t\to \infty }{\frac {1}{t}}\int _{0}^{\infty }{\bigl (}\Xi (\mu (\tau )),H(\mu (\tau )){\bigr )}\,d\tau =f(x,y)$. However, $\lim _{t\to \infty }\Xi (\mu (\tau ))$ and $\lim _{t\to \infty }H(\mu (\tau ))$ do not exist (they oscillate chaotically and thus serve as something like pink noise), and they may differ significantly from $f(x,y)$ for some time after whatever initial state.

So far, the obvious approach is to choose a $\mu$ for each of a number of pairs $(x,y)$, evaluate $f(x,y)$ by integrating $\mu$ for some period of time (dependent on desired accuracy), and then obtain some sort of fit ${\tilde {f}}(x,y)$ and thence ${\tilde {f}}^{-1}(\xi ,\eta )\approx f^{-1}(\xi ,\eta )$.

However, the reason x and y were included in $\mu$ is that there may exist a better algorithm that approaches the solution continuously (a sort of optimal control and/or stochastic filter) by varying them during one (long) integration. The system will take time to "recover" towards $f(x,y)$ after such a change, and it's easy to overcontrol by reacting to fluctuations in $\Xi (\mu )$ and $H(\mu )$, so what's the best approach here? --Tardis (talk) 22:53, 21 July 2012 (UTC)

${\frac {dx}{dt}}={\frac {dy}{dt}}=0$ or $1$? 75.166.200.250 (talk) 06:41, 22 July 2012 (UTC)

0. The only way they change is by external intervention in the controlling algorithm. --Tardis (talk) 07:12, 22 July 2012 (UTC)

I feel terrible that I can't wrap my mind around the textual description of this. I always do this on the Math Desk, but if you could explain a little more about the application I might be able to help more, but for now, all I can offer is to suggest going through PID controller and seeing if anything jumps out at you. 75.166.200.250 (talk) 03:57, 23 July 2012 (UTC)

All that really jumps out at me from the PID scheme is that the P term considered to be the most fundamental seems questionably useful; the solution of $u=K_{p}(S(u)-S^{*})$ will give $S(u)\approx S^{*}$ only if it happens that $S(0)\approx S^{*}$. It's the integral term that's useful, as can be trivially seen by differentiating: ${\dot {u}}(t)=K_{p}{\dot {e}}(t)+K_{i}e(t)+K_{d}{\ddot {e}}(t)$, so that u decays exponentially to whatever value causes $e=0$. Lots of linked articles seem relevant, though, like lead-lag compensator and integral windup, which can occur here not through saturation but via the lag in the approach of $(\Xi (\mu ),H(\mu ))$ to $(\xi ,\eta )$.

As for the application, it's a molecular dynamics simulation: x is density, y is specific energy, $\xi$ is pressure, and $\eta$ is temperature. You can change density and specific energy (within reason) at any time by scaling coordinates (for density) or velocities (for specific energy). Then the system will adopt a new structure (producing a new pressure) and rebalance its mix of potential and kinetic energy (producing a new temperature). However, the instantaneous measure of temperature is simply kinetic energy (up to a scalar constant), so it oscillates forever around the new value; pressure behaves similarly. The goal is to adjust the density and specific energy in response to produce a goal average pressure and temperature without being mislead by noise or oscillations and without overcorrecting because of the delay before a new oscillation is adopted. Standard barostats and thermostats exist, but they tend to hold the pressure and temperature constant (and thus drive fluctuations in the density and energy) rather than allowing them to oscillate naturally. --Tardis (talk) 06:38, 24 July 2012 (UTC)

July 22

Texniccenter

When I type something in Texniccenter, e.g. \frac{}{} or \begin{equation}, before I finish typing it gives a suggestion. What key do I press to make it automatically use the suggestion? Enter doesn't work. Money is tight (talk) 08:19, 22 July 2012 (UTC)

Tab? 75.166.200.250 (talk) 21:41, 22 July 2012 (UTC)

No doesn't work. Money is tight (talk) 23:11, 22 July 2012 (UTC)

I don't think it did that last time I used TeXnicCenter, but if Tab doesn't work, maybe you can try Ctrl+Enter. -- Meni Rosenfeld (talk) 03:45, 23 July 2012 (UTC)

It's done that every time since I downloaded it around 2008. — Fly by Night (talk) 00:22, 24 July 2012 (UTC)

Right arrow? 75.166.200.250 (talk) 03:47, 23 July 2012 (UTC)

Ctrl+Space should do it. I had this same problem and had to google it. Rckrone (talk) 18:13, 23 July 2012 (UTC)

This worked when I tried it. I'd also always wondered how to do this. Good work Rckrone! — Fly by Night (talk) 00:22, 24 July 2012 (UTC)

Thanks Rckrone!!! Money is tight (talk) 07:41, 24 July 2012 (UTC)

Commission

A used car salesperson can be paid using two methods of commission. Method X uses straight commission of 3.4% of selling price of all vehicles sold. Method Y uses a ixed amount of ₤250 per week plus commission of 1.5% of the selling price of all vehicles sold. If the total selling price of the cars sold in each week is on average ₤20000,calculate which of the two methods of commission the salesperson would prefer. — Preceding unsigned comment added by Ommykiz (talk • contribs) 14:18, 22 July 2012 (UTC)

Which would you prefer, £680 or £550? But I don't know how to calculate a preference.←86.139.64.77 (talk) 16:05, 22 July 2012 (UTC)

Sorry, we don't do people's homework for them. Looie496 (talk) 16:27, 22 July 2012 (UTC)

Calculate the amount of money you would make with each method when the total sale is $20000. --130.56.93.112 (talk) 23:17, 22 July 2012 (UTC)

Show that the hemisphere is connected.

Show that the set $\{{\vec {x}}\in \mathbb {R} ^{n}:||{\vec {x}}||=1,{\vec {x}}_{n}\geq 0\}$ is connected. --130.56.93.112 (talk) 23:14, 22 July 2012 (UTC)

Sorry, we don't do people's homework for them. Looie496 (talk) 01:51, 23 July 2012 (UTC)

Also, that's not a hemisphere. In 2D it's a quarter circle, and in 3D it's an eighth of a sphere. 75.166.200.250 (talk) 03:51, 23 July 2012 (UTC)

The notation is a little odd, but I think ${\vec {x}}_{n}$ is supposed to be the nth component of ${\vec {x}}$ (I would call it $x_{n}$). If that's correct, then it is a hemisphere. --Tango (talk) 06:56, 23 July 2012 (UTC)

Start with the definition of connectedness and see what you can find. --Tango (talk) 06:56, 23 July 2012 (UTC)

July 23

Prime Pascal

Do we know any interesting facts about the distribution on the primes amongst Pascal's triangle, and more specifically when it comes to the ratio of primes to composites in each subsequent row. — Fly by Night (talk) 01:06, 23 July 2012 (UTC)

I'm pretty sure the only binomial coefficients that are prime are of the form ${\binom {p}{1}}$ and ${\binom {p}{p-1}}$ for prime p, although I'm having trouble coming up with a proof. Rckrone (talk) 04:44, 23 July 2012 (UTC)

${\binom {n}{r}}$ is greater than n for 1 < r < n-1, so it can't be prime as no prime factors >n are involved in the calculation of ${\binom {n}{r}}$ as n(n-1)...(n-r+1)/r! For prime p, ${\binom {p}{r}}$ is divisible by p for 0 < r < p, is prime iff r = 1 or p-1, so Rckrone's conjecture is correct. AndrewWTaylor (talk) 12:48, 23 July 2012 (UTC)

Thanks Rckrone and AndrewWTaylor. I don't think I thought about it properly before I posted it. But thanks for taking the time to reply. — Fly by Night (talk) 00:19, 24 July 2012 (UTC)

It's trivial as the answers show. For a variation, if the decimal digits in a row are concatenated then [1] says the result is prime for row 2, 9, 30 and no others in the first 696 rows. I wrote "The search stopped before row 697 which has 104568 digits if anyone feels lucky." PrimeHunter (talk) 13:42, 23 July 2012 (UTC)

Different non-touching edge choices on Regular polyhedra question...

For the Pyramid, there is only one way to choose the maximum number of edges which don't share a vertex, for the Cube there are two ways (all four parallel, and then with two "twisted"), and for the Octahedron there are two that are mirror images. Does anyone have any advice for trying to count the unique ways for the 6 edges on the Icosahedron or the 10 edges on the Dodecahedron?Naraht (talk) 13:49, 23 July 2012 (UTC)

One way to view this question is as counting the number of maximal matchings of the graph, although you seem to be counting only up to orientation preserving symmetries of the polyhedron. The article mentions the FKT algorithm for planar graphs (which yours are). The icosahedron and dodecahedron are small enough that the FKT algorithm can be done by hand except for computing a large determinant. However once you get the total number of maximal matchings, modding out by the symmetries might be complicated. Rckrone (talk) 17:37, 23 July 2012 (UTC)

After messing around with it a bit, I think there are only 3 matchings up to rotation for the dodecahedron. Around a given pentagon, there are only 3 arrangements possible: 0 edges from the pentagon in the matching, 1 edge, or 2 edges. If any pentagon has 0 edges, there is only one possible configuration. Otherwise exactly 4 pentagons will have 1 edge in the matching, and there is only one arrangement but with two orientations. I may have missed something though. Rckrone (talk) 04:58, 24 July 2012 (UTC)

By "pyramid" do you mean square pyramid? I see two fundamentally different ways (excluding mirror images): two parallel edges on the base, and one edge on the base plus one between vertex and base. I'm not sure if I am misunderstanding the question... 86.129.16.51 (talk) 23:35, 24 July 2012 (UTC)

I'm pretty sure "pyramid" was referring to the tetrahedron. He/she is trying to answer this question for each Platonic solid. Rckrone (talk) 01:42, 25 July 2012 (UTC)

Ah, yes, of course. Unless otherwise specified, I tend to think of "pyramid" as meaning square pyramid... 86.129.16.51 (talk) 01:50, 25 July 2012 (UTC)

I just did the icosahedron since I was curious. I broke it up into cases similar to the dodecahedron example. There are always going to be 12 triangles that contain an edge in the matching, and then 8 that don't. You can enumerate the ways that the edges can be arranged around a given triangle that doesn't contain any edges, and then for each arrangement list out the possible ways that the other 3 edges can be chosen. I came up with 8 different matchings up to rotation: 2 that are invariant under flips, and 3 pairs of mirror images. Rckrone (talk) 02:40, 25 July 2012 (UTC)

Error in article but cannot edit it

http://en.wikipedia.org/wiki/Elementary_algebra

In the article above, there is an example of the commutativity of polynomials (2) and while it states that it may be obvious that (3+5) = (7+5) for the commutative property, obviously that is an incorrect example. The typed example is correct when I went to edit it but the pictures of the numbers, which I cannot edit, are the incorrect numbers. The picture numbers should also show that (3 + 5) = (5 + 3).

Thank you! — Preceding unsigned comment added by Jessicamoreau (talk • contribs) 16:58, 23 July 2012 (UTC)

The latex code was recently fixed (as you discovered when you tried to edit) but your browser must have the old image cached, so try refreshing that. Rckrone (talk) 17:43, 23 July 2012 (UTC)

The error was fixed after the message was posted here. The article is being developed pretty actively at the moment -- the error was introduced today and fixed a short time later. Looie496 (talk) 17:47, 23 July 2012 (UTC)

Primorials

What's the sum of the inverse primorials (1/2+1/6+1/30+1/210+...)? Since it's less than the inverse factorials, it clearly converges, but I can't find the sum. --146.7.96.200 (talk) 21:32, 23 July 2012 (UTC)

0.70523... Looie496 (talk) 23:04, 23 July 2012 (UTC)

Or if you want more precision, 0.7052301717918009651474316828882485137435776391091543281922679138139197811480028635861190519840274766...

isc (the new Plouffe's inverter) doesn't say anything about this number other than that it is the sum of inverse primorials. -- Meni Rosenfeld (talk) 09:49, 24 July 2012 (UTC)

I doubt there is a closed form expression. You'll only be able to approximate it. — Fly by Night (talk) 00:16, 24 July 2012 (UTC)

This raises a sort of interesting philosophical issue. Even if there was a closed form expression, it's unlikely that it would be easier to evaluate numerically than the series itself, which converges very rapidly. Looie496 (talk) 16:29, 24 July 2012 (UTC)

The interest in a closed-form expression is presumably driven by mathematical curiosity rather than a practical need to calculate the number to great accuracy... 86.179.3.247 (talk) 17:50, 24 July 2012 (UTC)

For mathematical curiosity and correctness. These numerical methods never give the correct answer; they give a sequence of incorrect answers that get closer to the correct answer. Numerical approximations are more than adequate for any real-life application. However, this is quite a pure question, with very little real-world application. I would argue that a numerical solution goes against the spirit of such a pure maths type question. — Fly by Night (talk) 17:54, 24 July 2012 (UTC)

In a way, though, the expressions that are allowed as "closed-form" are somewhat arbitrary. For example, we would, I assume, be happy with 2/e as a closed-form answer, yet e is only calculable as the unreachable limit of a series of approximations. In a different universe, the sum of inverse primorials might be an important number with a designation f, and then we would be happy with, say, 2/f as a "closed form" answer to some problem. 86.129.16.51 (talk) 20:49, 24 July 2012 (UTC)

I agree totally with the first part. There is something arbitrary about it all. There's nothing stopping you to define a number that is the answer to the OP's sum, and then claim it's a closed form expression. — Fly by Night (talk) 20:54, 24 July 2012 (UTC)

In reply to talk:Looie496. Yes it is an interesting question, it seems to be the subject of Transcendence theory. Its basically an extension of the idea of transcendental numbers. I guess most numbers will be of this form.--Salix (talk): 20:36, 24 July 2012 (UTC)

Rate of convergence is a relative notion. No matter how many terms you sum, you'll still have infinitely many incorrect digits. — Fly by Night (talk) 16:38, 24 July 2012 (UTC)

I doubt it, I'd only have a finite number of digits so only have a finite number wrong. Just to continue the philosophical debate with a bit of reality. ;-) Dmcq (talk) 17:36, 24 July 2012 (UTC)

Under the convention that 0.5 = 0.5000…, you would have infinitely many incorrect digits. Alternatively, consider the difference between any finite partial sum and the actual limit. All of these differences, i.e. errors, will be infinitely-long, non-trivial chains of digits. — Fly by Night (talk) 17:42, 24 July 2012 (UTC)

Well a potential infinite then rather than an actual infinite. Dmcq (talk) 18:03, 24 July 2012 (UTC)

təˈmeɪtoʊ, təˈmɑːtəʊ. — Fly by Night (talk) 20:34, 24 July 2012 (UTC)

July 24

Solve the problem

Solve this problem I was unable to solve it.
"If x³ + 1/x³ = 110, find the value of x + 1/x." --Sunny Singh 14:15, 24 July 2012 (UTC) — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs)

Try (x + 1/x)³ and see where that gets you. --Wrongfilter (talk) 14:30, 24 July 2012 (UTC)

But after using (x+1/x)³ I don,t get the answer. Please, you solve it.--Sunny Singh 14:40, 24 July 2012 (UTC) — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs)

Er, no. --Wrongfilter (talk) 14:41, 24 July 2012 (UTC)

Hint: If you are only interested in solutions in the real numbers, then x + 1/x is an integer. Gandalf61 (talk) 15:48, 24 July 2012 (UTC)

I also want to see steps, please, explain this problem with steps. I tried hard but I couldn't solve it. — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs) 15:58, 24 July 2012 (UTC)

So what did you get when you expanded (x + 1/x)³ ? Gandalf61 (talk) 16:01, 24 July 2012 (UTC)

I think it is easier if you multiply through by x³ and consider the result as a quadratic. Dmcq (talk) 17:31, 24 July 2012 (UTC)

I tried that, but it didn't give a very nice solution. You get the cube-root of a radical expression:

$x={\sqrt[{3}]{55\pm 12{\sqrt {21}}}}\,.$

However, these solutions can be expressed as

$x={\frac {1}{2}}(5\pm {\sqrt {21}})\,.$

I think that the methods above lend themselves to finding these latter solutions. Besides, the OP wants $x+1/x$. — Fly by Night (talk) 18:00, 24 July 2012 (UTC)

$x^{3}+1/x^{3}=\left(x+{\frac {1}{x}}\right)\left(\left(x+{\frac {1}{x}}\right)^{2}-3\right)$. Now use the prime factorization of 110, along with the assumption that x+1/x is an integer. Sławomir Biały (talk) 18:15, 24 July 2012 (UTC)

Not assuming it is an integer gives the other solutions ${\frac {-5\pm 3i{\sqrt {7}}}{2}}$. Perhaps the OP can try working that out. Dmcq (talk) 20:58, 24 July 2012 (UTC)

I just tried out Wolfram Alpha at http://www.wolframalpha.com with "solve x^3=110+3x" and it immediately gives the simplified solution. That's another easy way of getting an answer but it is better in the first instance whilst learning to try out for yourself. Dmcq (talk) 21:08, 24 July 2012 (UTC)

I think Wrongfilter is arguing that

$(x+{\frac {1}{x}})^{3}=x^{3}+3x+3{\frac {1}{x}}+{\frac {1}{x^{3}}}$, we know that x³ + 1/x³ = 110 so we have:

$(x+{\frac {1}{x}})^{3}=110+3(x+{\frac {1}{x}})$, so we have a cubic in (x+1/x) compare with $y^{3}-3y-110=0$ . This has a solution y=5 as 5^3 = 125, 3*5 = 15 and 125-15 = 110. You need to prove/satisfy yourself this is the only real root - is this any help? 83.100.173.200 (talk) 19:01, 24 July 2012 (UTC)

Word numerical

Can you solve this problem for me? I was unable to solve it. "A machine gun fires 25g bullets at the rate of 600 bullets per minute with a speed of 200m/s. Calculate the force required to keep the gun in position." — Preceding unsigned comment added by Sunnysinghthebaba (talk • contribs) 14:58, 24 July 2012 (UTC)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. —Kusma (t·c) 15:15, 24 July 2012 (UTC)

That's a very puny machine gun, with a muzzle velocity of only 200m/s - 750m/s to 850m/s seem to be more typical. Anyways, as you should know from Isaac Newton, every force has an equal and opposite force. Some force accelerates 600*25g from zero to 200m/s every minute. The reaction to that action would accelerate the machine gun backwards if it weren't held in place. You should be able to take it from there. --Stephan Schulz (talk) 16:38, 24 July 2012 (UTC)

Wikigraphs

Could someone check this math?

• http://www.brianckeegan.com/2012/07/2012-aurora-shootings/ (just kidding)--Canoe1967 (talk) 20:01, 24 July 2012 (UTC)

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)