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September 29

For which $\alpha \in \mathbb {R} ^{+}$ does $\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{n}}{n!^{\alpha }}}$ tend to zero for $x\to \infty$ ?

Count Iblis (talk) 00:21, 29 September 2016 (UTC)[reply]

Presumably you mean for x ∈ ℝ (as opposed to ℂ), and that it tends specifically to positive infinity. You could also include α = 0 if you permitted analytic continuation. It is easy to find some values (0, 1, 2) of solutions of α, but I suppose you're not interested in isolated solutions. —Quondum 03:37, 29 September 2016 (UTC)[reply]

Yes, 0, 1, 2, yield 1/(1+x), exp(-x) and J0(2x) respectively, so I'm interested in the general case. Count Iblis (talk) 21:47, 29 September 2016 (UTC)[reply]

For

α = 2

you seem to have mistyped the expression

J 0 (2 \sqrt x)

. For larger integer values (3, 4, etc.), the limit seems to be

\infty

. So a strawman hypothesis (considering the numeric observation below) might be that there is a range

0 \leq α \leq ω

for which the limit is zero, and above the boundary

ω

it is infinite, where

2 \leq ω \leq 3

. The series is clearly convergent for all finite

x

with

α > 0

. I can't help feeling that it should not be difficult to find

ω

, or at least a moderately good upper bound on it, after which one would show that it definitely goes to infinity due to the fast convergence, though I don't know how to formulate that. —Quondum 06:29, 30 September 2016 (UTC)[reply]

Numerically it seems to go to zero for all

0\leq \alpha \approx \leq 1.75

. 24.255.17.182 (talk) 05:01, 29 September 2016 (UTC)[reply]

Crocodile problem: are all these solutions right and acceptable in an exam?

This problem was widely reported by several media, some even claiming it was unsolvable. A description of the problem can be found here: [1].

I wonder whether there are several ways of solving it. All three solutions below lead to the same value, 8m, one is simple trial-and-error, one is trigonometric, and the other derivative. In the exam, at no point there was any indication about what methods to use.

Trial-and-error: plug integer values in the equation, once you found the lowest (that is, 8m) try values around it, that is 7.999999 or 8.000001. That will assure you a certain level of precision, but it's obvious that maybe it could still be a number with many decimal places like 8.00000000001.

Trigonometric: calculate speed in land and water. That's a pace of 4 and 5 tenth of second per meter respectively. The river is 6m wide. Going over land/going through water are in a relationship of 4/5. That is, 4 on land is equal to 5 through water. Divide 4/5 to obtain 0.8, which is the sine of the angle 53.13. Get the cosine of the same angle, that's 0.6, 6/h = . 0.600. The hypotenuse is 10m. x is, per Pythagoras theorem, 8m.

Derivative:

$T(x)=5*(36+x^{2})^{1/2}+(80-4x)$

$T'(x)=5/2*(36+x^{2})^{-1/2}*2x-4$

$T'(x)={\frac {5x}{\sqrt {36+x^{2}}}}-4$

${\frac {5x}{\sqrt {36+x^{2}}}}-4=0$

$5x=4*{\sqrt {36+x^{2}}}$

$25x^{2}=16*(36+x^{2})$

$25x^{2}=576+16x^{2}$

$x^{2}=64$

$x=8$

Llaanngg (talk) 19:46, 29 September 2016 (UTC)[reply]

I've edited the derivative calculation to remove about six typos that presumably occurred when translating into wikitext. --69.159.61.230 (talk) 22:19, 29 September 2016 (UTC)[reply]

The problem looks like a test of calculus skills. Ruslik_Zero 20:13, 29 September 2016 (UTC)[reply]

Yes, it seems clear that the derivative method is the intended solution and is valid. It identifies a local extremum and from consideration of other points on the (x,T) curve that must be a minimum. As to trial-and-error, Llaanngg himself or herself has explained why it is not a mathematically valid solution.

The trigonometric method is interesting, but is based on an unstated and unproven premise. This solution involves a right triangle where one side is along the river bank (on land) and the hypotenuse is the diagonal that the crocodile swims (in water). The unstated premise is that the overall time is minimized if these two sides are in proportion such that the side on land is to the side in water as the speed in water is to the speed on land. Generalizing the problem by substituting arbitrary constants for the numerical ones, I see that this does indeed always produce the right result, but I have to say it is not obvious to me why it should, so I think some proof of that premise is required.

--69.159.61.230 (talk) 22:19, 29 September 2016 (UTC)[reply]

This is a very old problem. A similar problem appears in chapter 26 of The Feynman Lectures on Physics, as the problem of a person on shore trying to reach a drowning girl in the minimum amount of time. (He uses it as an analogy for the refraction of light passing between two media with different indices of refraction). The problem as presented in the link above is almost trivial, since they hand you the equation relating time to the shoreline position, instead of the usual presentation which just gives you the speed on land and the speed on water and lets you work out the equation yourself. It's pretty sad comment on the level of mathematical literacy in society if some media are claiming it is unsolvable, or even particularly difficult. CodeTalker (talk) 23:26, 29 September 2016 (UTC)[reply]

I would think that the expected answer would probably have been the one involving calculus, though I suppose you might get credit for the trigonometric one if you proved the unstated assumption that it makes. Double sharp (talk) 01:37, 30 September 2016 (UTC)[reply]

The simplest method, by far, is to use a graphing calculator and take the minimum (8,98). While an advanced class may not allow this, it's a good way to solve the problem in a basic math class. That is, it shows the student knows how to use the tools available to solve problems. StuRat (talk) 15:52, 1 October 2016 (UTC)[reply]

Actually I daresay it would be the advanced class that would allow it, and the basic class that would not: the idea being that you're only allowed to use the tools once you are very clear about what you're actually doing with them, and you know how you could do it yourself if the situation arose. Double sharp (talk) 15:59, 1 October 2016 (UTC)[reply]

You might be right, but we do need to change the way education is done. For people who aren't math or science majors, it's a waste of time and resources to try to teach them techniques they won't use in their jobs, and will soon forget. What's more important, for those students, is to teach them how to access the available info and use the tools available to them, to solve real world problems.

To reverse the situation, it may be important for an arts major to memorize many artists and their works, but is it critical for math and science majors to do so ? If they ever need to know who wrote a particular sonata, they can just Google it. StuRat (talk) 16:06, 1 October 2016 (UTC)[reply]

Education is not only (perhaps not even primarily) about job preparation, nor should it be. (Not that this tangent has anything to do with the question asked ....) --JBL (talk) 04:43, 2 October 2016 (UTC)[reply]

October 1

Question about volumes of a cube section.

Let a cube be divided into three pieces by two parallel planes, each of which go through exactly three vertices. (Think of a cube with a vertex at the top, a vertex at the bottom and two horizontal planes) What fraction of the cube is above the top plane, between the planes and below the bottom plane (obviously the first and last answer are the same). I think this can be done by calculating the volume of the pyramid above the top plane with edge of 1 unit, the height of the pyramid is 1/3 the solid diagonal or sqrt(3)/3 and the base of the pyramid is a triangle made up of face diagonals, etc. Is this the way to do it, or is there a clean fast way (preferably one that could be extended to do the equivalent slices of a hypercube...Naraht (talk) 02:34, 1 October 2016 (UTC)[reply]

Are you seeking a higher dimensional version of Parallelepiped#Corresponding tetrahedron? If so, see Parallelotope. -- ToE 14:52, 1 October 2016 (UTC)[reply]

Simpler if you think of the pyramid as having a base formed by two edges of the cube and a face diagonal. The area of the pyramid's base is then a²/2 and its height is a where a is the length of one edge of the cube. Gandalf61 (talk) 15:35, 1 October 2016 (UTC)[reply]

For the n-cube with side 1, the volumes between consecutive slices are given by the Eulerian numbers divided by n!, a fact which is noted at (sequence A008292 in the OEIS). (You can also think of this as the probability that n independent variables distributed uniformly on [0, 1] have sum between k and k+1.) The correspondence is not obvious since the Euler numbers are given in terms of the number of decreases in a permutation. But I believe if you take x₀=0, then the map y_i=x_i-x_i-1 (if x_i≥x_i-1, y_i=1+x_i-x_i-1 (if x_i<x_i-1) is volume preserving and if you slice up the cube into simplices according to the order of the coordinates then it maps the slices with k decreases to the region where the sum is between k and k+1. --RDBury (talk) 00:22, 2 October 2016 (UTC)[reply]

Calculating Sine and Cosine using nothing but a single complex number

Someone told me that anyone can calculate the values of Sine and Cosine in degrees using nothing but a single complex number.

Define a complex number value "onedeg" as

onedeg = 1 * (Cos[1 deg] + i Sin[1 deg]) = 0.999847695156391 + 0.0174524064372835 i

Next calculate Sine[23.45 deg] and Cos[23.45 deg]

onedeg^23.45 = 0.917408 + 0.397949 i

voila! Cos[23.45 deg] = 0.917408 and Sin[23.45 deg] = 0.397949

It feels like magic, a single complex number is capable of calculating Sine and Cosine in degrees! 110.22.20.252 (talk) 13:25, 1 October 2016 (UTC)[reply]

See Euler's formula or maybe better De Moivre's formula which is based on it for more about this. Dmcq (talk) 13:57, 1 October 2016 (UTC)[reply]

You may also enjoy reading our Circle group and Root of unity articles. -- ToE 19:41, 1 October 2016 (UTC)[reply]

And how exactly do you calculate onedeg^23.45...?

You have to be able to calculate sin and cos in the first place in order to calculate this power.

So while what you say is true (for the reasons explained by Dmcq), it in no way actually helps in the computation.

At best, for a whole number of degrees, you can use the complex representation as a mnemonic to avoid having to remember the formulas for sine/cosine of sums of angles.

And of course - if it wasn't obvious by now - there's nothing special about the particular complex number you've picked. -- Meni Rosenfeld (talk) 18:41, 4 October 2016 (UTC)[reply]

October 4

October 5

Median range

Is there such a thing as the "median range" of a set of numbers? I don't recall encountering the term before, but the article Case Western Reserve University#Undergraduate profile says

Median SAT scores (25% - 75%) were between 1280 and 1450. The median range for ACT scores was 29 to 33.

I have not come across the term before either, but I'm not a statistician. When I was in school, the statistic in your median SAT scores example was called interquartile range. --100.34.204.4 (talk) 01:47, 5 October 2016 (UTC)[reply]

Whittier Law School#Admission ststistics says

Median Range of LSAT: 150-154

Median Range of GPA: 2.75-3.28

Kleinbrook, Texas says

the subdivision had a median price range of $95,000-$151,000.

Traces, Texas says

subdivision had a median price range of $77,000-$124,000 U.S. dollars.

Williamsville, New York says

The median age range was 45–49 years.

Is median range accepted mathematical terminology? Loraof (talk) 01:01, 5 October 2016 (UTC)[reply]