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July 29[edit]

August 9[edit]

August 11[edit]

August 20[edit]

August 21[edit]

Deriving the Taylor series of the secant function[edit]

Basically, how does one show that the explicit formula(s) of the Euler numbers (given in the article) define the coefficients of the Taylor series of the secant function (up to a change in sign)?

One guess I have is to take advantage of the integral definition of the inverse secant, \arcsec(x) = \int_1^x \frac{dt}{t\sqrt{t^2-1}}, expand the square root with a binomial series, antidifferentiate, and then apply the Lagrange inversion theorem, but I do not have experience applying that.

Once this series is found, the Cauchy product of it with the series for sine produces the tangent function's series.--Jasper Deng (talk) 01:40, 21 August 2014 (UTC)

Is secant(t) = \frac{1}{\cosh (it)} = \sum_{n=0}^\infty  \frac{i^nE_n}{n!} \cdot t^n\! helpful?--Wikimedes (talk) 18:45, 27 August 2014 (UTC)

August 22[edit]

Taylor /Maclaurin Series number of roots[edit]

When applying Taylor Series or any other expansions to a [[|Function (mathematics)|function]] such as \frac 1 x, then we will have a polynomial with infinite number of terms. Can we then consider this infinite terms polynomial to have a degree of n the approaches infinity and so, has infinite number of roots? Or is considered as a mathematical fallacy since we applied the series to a point that is not necessarily near the roots?--Almuhammedi (talk) 06:11, 22 August 2014 (UTC)

The problem is that you're playing with infinity here. The fundamental theorem of algebra strictly applies only to finite degree polynomials. It also says nothing about whether the set of roots of any of the partial sums must have a limit as n→infinity. Naturally there are cases such as the trigonometric functions that do have infinitely many roots. But this is certainly not true in general, such as with the exponential function, which is nowhere zero.--Jasper Deng (talk) 06:59, 22 August 2014 (UTC)
Here's an interesting paper on the location of the n roots of the nth partial sum of the Taylor expansion of the exponential function as n goes to infinity: Ian Zemke Zeroes of the Partial Sums of the Exponential Function. --catslash (talk) 12:40, 22 August 2014 (UTC)

August 23[edit]

Determining whether 4 points are a square viewed from somewhere...[edit]

Watching the logo of windows made me think of the following: If I have four points on a plane, how can I tell whether those four points could be the projection of a square from some point onto the plane. I guess this is two different questions because a rectangle could only be a projection of a square if the point is at infinity, so I guess, one question without points at infinity and one with.Naraht (talk) 14:51, 23 August 2014 (UTC)

Isn't the rectangle formed by the ordered triplets (1,1,1), (1,2,3), (2,1,1), (2,2,3) the projection of the square formed by the ordered pairs (1,1), (1,2), (2,1), (2,2) onto the plane containing that rectangle? To be clear, the rectangle has the square as its regular projection onto the x-y plane (i.e. a surface integral over the rectangular surface is (most easily) evaluated by a double integral over the square); through a rotation of coordinates, one could make a similar argument for the rectangle being a projection of a square.
Clearly, though, your definition of a projection is obviously different from mine. If you mean projection in the optical sense, then the above example is basically a projection viewed from infinitely far away (at least in Euclidean geometry). In the case of a finite distance, it is sufficient to show that the four points do form a square if you already know them to be coplanar (but not necessary - if you view from an angle relative to the plane, I'm not entirely sure how to handle it).--Jasper Deng (talk) 17:41, 23 August 2014 (UTC)
This question seems to fit into projective geometry. Since the cross-ratio is essentially the only invariant under projective transformations, I expect that you'd find a similar cross-ratio between four of the six lengths of the segments defined by the four points that would determine whether the original figure could have been a square. —Quondum 19:06, 23 August 2014 (UTC)
Any four random points on a plane could be the projection of a square from some point onto the plane. The image of every other point is determine by the four points though. Dmcq (talk) 20:22, 23 August 2014 (UTC)
There appear to be four conditions to meet and at least five degrees of freedom, so it is likely that there is always a solution (irrespective of whether the given points are coplanar). You can choose the size of the square too. --catslash (talk) 21:18, 23 August 2014 (UTC)
Wouldn't any projection of a square from a point to a plane at least be convex? That would imply that not every four points can be such a projection. AndrewWTaylor (talk) 21:58, 23 August 2014 (UTC)
The projection plane could intersect the square and produce a projection that is not convex. Dmcq (talk) 22:28, 23 August 2014 (UTC)
For any projection plane that intersects the square, you can draw parallel planes that do not intersect it, on which the projected image ought to be the same except for scale. —Tamfang (talk) 00:32, 24 August 2014 (UTC)
The four points of the square need not all be projected to the same side of the projection point. Consider a square with the projection point above it, then we'll project onto another plane going through one of the corners and almost perpendicular to the first plane but leaning away a little facing a little above the projection point. Then the opposite corner will project to the opposite side of the projection point high up and the two sides ones down below. The original corner point will be within the three other points. Dmcq (talk) 12:01, 24 August 2014 (UTC)
I don't quite understand the construction here, but it cannot produce a non-convex quadrilateral. The projection map will send line segments to line segments and, vice versa, the inverse projection will map line segments to line segments. So if a line segment in the projected quadrilateral is drawn between two of its interior points, the inverse image of that line segment must be interior to the square (because the square is convex), and so its projection back down into the quadrilateral must also be interior to that quadrilateral. Since the line segment between any two points of the quadrilateral lies inside the quadrilateral, it is convex. Sławomir Biały (talk) 13:42, 24 August 2014 (UTC)
concave projection of a square
Okay I've set up and uploaded a slightly better construction. ABCD is the original square and O is the point to project througgh and is above the centre of the square. the new plane is given by the points B and D and a point A' which is on the opposite side of O. B and D project to themselves. Then the point C is projected to C' which is on intersection of the line OC, and the line between A' and the mid-point M of BD - this is because the line is given by A'BD and the plane AA'C. As you can see this new point C' is inside the triangle formed by A'BD. Dmcq (talk) 16:09, 24 August 2014 (UTC)
Yes, it looks like that exchanges the interior of the square with the exterior. So although straight lines are sent to straight lines, the lines may pass through a point at infinity (so the quadrilateral in the image has a "convex" exterior rather than interior). Clever. Sławomir Biały (talk) 23:31, 24 August 2014 (UTC)
A convex outside, that's a good one ;-) Anyway I think a simple modification of that construction can give a way of projecting a square ABCD to a square ABD'C' which is also a bit surprising. Dmcq (talk) 00:30, 25 August 2014 (UTC)
Convex exterior
Anyway I've drawn a picture of a convex exterior with a circle showing the line at infinity. Thanks for that idea. As a puzzle a person could try doing one for the square ABD'C' I described just above. Dmcq (talk) 10:06, 25 August 2014 (UTC)

August 24[edit]

August 25[edit]

Comparison of two integrals.[edit]

Hi there,

I want to know how two integrals given below are related. Let's assume m is a natural number, perhaps m = 40; then n = 360/m and n = 9; Function f(φ) is smooth, differentiable in its domain. The domain is a small fraction of circumference: {0,360/m} or {0,n}. Then the integral (1) will have the integrand as product of the function f(φ) and one whole period of cosine function cos(mφ) where function cos(mφ) is in fact defined on the whole circumference:

\int\limits_{0}^{n}f(φ)cos(mφ) dφ        (1)

Then I "stretch" the function and its domain m times to the extent that it will coincide with the whole circumference. The cosine function or rather the portion thereof defined initially on {0,n} interval will be stretched as well. The amplitude of the function f(φ) will be preserved, the same can be said of the cosine function. Then I will take the integral:

\int\limits_{0}^{360}f(ω)cos(ω) d(ω)      (2)

I want to know how the integrals (1) and (2) are related. My hunch is that I will have to multiply the integral (1) by m to get integral (2) but not sure about it. Not a homework :-) It is a practical problem I am facing in some computations I need to perform. Will appreciate any insight. Thank you. --AboutFace 22 (talk) 20:56, 25 August 2014 (UTC)

This is a simple change of variables (except for the "rescaling of f"). If one takes u = mx, du = m dx, \int_0^n f(x) \cos(mx) dx = \frac{1}{m} \int_0^{360} f(\frac{u}{m}) \cos(u) du. I'm not sure about what you mean by "stretch" here, but your hunch is rather close, except that I think your second integral is not exactly what you're talking about (you shouldn't use the same letter f to denote two different functions - you should write it as \int_0^{360} g(u) \cos(u) du with g(u)=f(u/m)).--Jasper Deng (talk) 00:18, 27 August 2014 (UTC)

Thank you much for your note. You are correct. I should have changed the letter for the function. Actually I have been thinking about doing this for two days but somehow other things intevened and I missed the opportunity. Sure it is a different function after the "dilation." So, could you say with certainty that the two integrals are related in such a way that I can divide the value of the second integral by m to get the value of the first integral? It seems the formula you wrote says this clearly, right? Your "rather close" makes me a little bit nervous, though :-) --AboutFace 22 (talk) 13:28, 27 August 2014 (UTC)

@AboutFace 22: I know the expression in my previous comment to be correct. The reason why I say "rather close" is that you have not made "dilation" well defined; if you mean it as I did above, then that is correct. I think you know this, but m need not be a natural number. The equality is valid for all real nonzero m.--Jasper Deng (talk) 19:02, 27 August 2014 (UTC)

Mr. Deng, thank you. I know that my definitions are lame, I also know that any real non-zero number could do the same job as the whole number m. In my application though, m is a natural number, that is why I stated that. I am completely satisfied now. I appreciate your help very much. --AboutFace 22 (talk) 01:06, 28 August 2014 (UTC)

August 26[edit]

Commutative magmas?[edit]

Have commutative magmas been studied? Is there something like partially associative (and everywhere commutative) magma like there is a trace monoid (for partially commutative but everywhere associative magma)? JMP EAX (talk) 11:53, 26 August 2014 (UTC)

Derive variable values from formulae[edit]

Given the following 9 formulae with known solutions, is it possible to derive the values of the other variables?

  • A = B + C + D
  • E = 2(F + G) + C
  • H = I + J + K
  • L = B + I + J
  • M = C + D + J
  • N = O + C + D + K
  • P = C + I + J
  • Q = B + C + I + J
  • R = S + B + D + J

To be clear, the values are known for A, E, H, L, M, N, P, Q, and R. Is it possible to discover the values of any other variable for certain? ΣΑΠΦ (Sapph)Talk 13:03, 26 August 2014 (UTC)

Sure. Note that F and G only occur together, so you can replace F + G with some new variable T. Now you've got 9 equations and 9 unknowns, so in general you'd expect to be able to solve for the unknowns using Gaussian Elimination or a similar technique. And indeed, you can in this case. So the only values you can't get are F and G, but you can at least get F + G.-- (talk) 13:19, 26 August 2014 (UTC)
If you only want to know whether you can discover at least one value then consider Q-P. PrimeHunter (talk) 13:46, 26 August 2014 (UTC)
As a point of clarification, it is not necessarily the case that any assignment of values to the variables listed will lead to solutions for the whole system. The system may be inconsistent. In the vein of PrimeHunter's suggestion, you can also see by inspection that Q-L=C. SemanticMantis (talk) 19:46, 26 August 2014 (UTC)
This particular set of equations however is consistent and has a solution for any values of A, E, H, L, M, N, P, Q, and R (and you can choose any value for either F or G too). --catslash (talk) 21:28, 26 August 2014 (UTC)
You have 1 free variable (G). The solutions for your other variables are: B = Q - P, C = Q - L, D = A + L + P - 2*Q, F = E/2 + L/2 - G - Q/2, I = A + L - M - 2*Q + 2*P, J = M + Q - A - P, K = H + Q - L- P, O = L + N - A - H, S = P + R - L - M (talk) 22:28, 26 August 2014 (UTC)

-- (talk) 08:48, 27 August 2014 (UTC)

Let's see how far we can get by inspection on this...
Turns out it is consistent, and we can get all the way (hatted to keep it short, and in case anyone else wants a try) MChesterMC (talk) 08:37, 28 August 2014 (UTC)

Why is Regression analysis called regression analysis?[edit]

Regression means to go back but I don't see what that has to do with regression analysis. (talk) 16:42, 28 August 2014 (UTC)

This is explained in the history section of the article. This came from using the word "regression" for the phenomenon (also known as "regression towards the mean") that children go back to the average relative to their parents. The statistical procedures that came out of this observation have therefore been called "regression". -- Meni Rosenfeld (talk) 16:50, 28 August 2014 (UTC)