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February 10

How to prove

How to prove that tan20⋅tan30=tan10⋅tan50 ? 117.227.40.234 (talk) 03:32, 10 February 2013 (UTC)

Are these in degrees ? If so, why not just multiply them out ? StuRat (talk) 03:34, 10 February 2013 (UTC)

Ya, they are in degrees. How I can I do it using tanA = cot(90-A)? 117.227.145.29 (talk) 03:47, 10 February 2013 (UTC)

Or some any other basic identities. 117.227.18.193 (talk) 06:06, 10 February 2013 (UTC)

((( My browser doesn't show the symbol used between the tans here, can someone say what it is please (* / + - ???) -- SGBailey (talk) 09:49, 10 February 2013 (UTC) )))

Hello, I think the symbol is meant to show multiplication, like latex's '\cdot' (in a hex editor it comes up as 0xB7 which is Unicode (UTF-16)'s 'MIDDLE DOT') so it's

$\tan(20)\cdot \tan(30)=\tan(10)\cdot \tan(50)$.77.86.3.26 (talk) 10:44, 10 February 2013 (UTC)

There may be simpler ways, but here's an outline of mine. Multiply both sides by cos 20 cos 30 cos 10 cos 50 so that the entire identity deals only with sin and cos. Now using the product-to-sum formulas at Trigonometric_identity#Product-to-sum_and_sum-to-product_identities, you can get rid of the 20's and 30's entirely and only be left with angles of 10 and 50. Using the double-angle formulas, you can then replace some of the 50's with 100's. The 100's can in turn be replaced with 10's using the formulas for shifting by 90. Now it's not too hard to prove the identity. You'll still need an angle addition formula, though. 96.46.195.35 (talk) 13:01, 10 February 2013 (UTC)

The identity $\tan(10)=\tan(20)\cdot \tan(30)\cdot \tan(40)$ is mentioned at Trigonometric identities#Identities without variables (though not with a clear explanation, unfortunately) and this identity is identical to the one above by recognising that $\tan(40)=\cot(90-40)=\cot(50)={\frac {1}{\tan(50)}}$. So, the identity the IP mentions is certainly true, provided that WP page is correct. EdChem (talk) 13:03, 10 February 2013 (UTC)

Yes, I just noticed that. See also Morrie's law, which relates directly to the OP's identity in the same way. 96.46.195.35 (talk) 13:08, 10 February 2013 (UTC)

How about this: we can write $\tan(30^{\circ })=1/{\sqrt {(}}3)$, then we have:

$\tan(20)/{\sqrt {(}}3)=\tan(10)\cdot \tan(50)$.

Using your "tanA = cot(90-A)" we have

$\tan(20)\cdot \tan(40)/{\sqrt {(}}3)=\tan(10)$.

From List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae we have:

$\tan(2A)={\frac {2\tan(A)}{1-\tan ^{2}(A)}}$, so we have:

$\tan(20)\cdot {\frac {\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {(}}3)=\tan(10)$.

Using the half-angle formula:

$\tan(A/2)={\frac {\tan(A)}{1+{\sqrt {1+\tan ^{2}(A)}}}}$ we have

$\tan(20)\cdot {\frac {\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {(}}3)={\frac {\tan(20)}{1+{\sqrt {1+\tan ^{2}(20)}}}}$.

What do people think?77.86.3.26 (talk) 13:15, 10 February 2013 (UTC)

You dropped a factor of 2 along the way. Also, although if you restore the 2 the calculations are correct, the argument is still incomplete. It's something special about the number tan(20) that makes the last equation true - for tan(21) it would be false.96.46.195.35 (talk) 14:42, 10 February 2013 (UTC)

Thanks for spotting my mistake. One possible way to go is by expanding the above into an equation for $\tan(20^{\circ })$ and noting that:

$\tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}$, so

$\tan(60^{\circ })={\frac {3\tan(20^{\circ })-\tan ^{3}(20^{\circ })}{1-3\tan ^{2}(20^{\circ })}}={\sqrt {3}}$

and comparing? 77.86.3.26 (talk) 16:28, 10 February 2013 (UTC)

Sorry that's no help, one equation is quartic and the other cubic. 77.86.3.26 (talk) 16:35, 10 February 2013 (UTC)

Complete proof

Recalling the double angle formula for the sine function and rearranging, we can state that, for any angle α:

$\cos(\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}$     and similarly that     $\cos(2\alpha )={\frac {\sin(4\alpha )}{2\sin(2\alpha )}}$     and     $\cos(4\alpha )={\frac {\sin(8\alpha )}{2\sin(4\alpha )}}$

On multiplying these together, we get that

$\cos(\alpha )\cdot \cos(2\alpha )\cdot \cos(4\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}\cdot {\frac {\sin(4\alpha )}{2\sin(2\alpha )}}\cdot {\frac {\sin(8\alpha )}{2\sin(4\alpha )}}={\frac {1}{2\sin(\alpha )}}\cdot {\frac {1}{2}}\cdot {\frac {\sin(8\alpha )}{2}}={\frac {\sin(8\alpha )}{8\sin(\alpha )}}$

Setting α = 20°, we get

$\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {\sin(160^{\circ })}{8\sin(20^{\circ })}}$

and since $\sin(20^{\circ })=\sin(180^{\circ }-20^{\circ })=\sin(160^{\circ })$

$\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {1}{8}}$     . .     . .     . .     (1)

This is the so-called Morrie's law.

Now, recalling the sums-to-products formula for the sine function we can state that, for any angles α and β:

$\sin(\alpha )\cdot \sin(\beta )={\frac {\cos(\alpha -\beta )-\cos(\alpha +\beta )}{2}}$

Taking α = 80° and β = 40°:

$\sin(80^{\circ })\cdot \sin(40^{\circ })={\frac {\cos(80^{\circ }-40^{\circ })-\cos(80^{\circ }+40^{\circ })}{2}}={\frac {\cos(40^{\circ })-\cos(120^{\circ })}{2}}$

Then, nothing that   $\cos(120^{\circ })=-\cos(180^{\circ }-120^{\circ })=-\cos(60^{\circ })={\frac {-1}{2}}$:

$\sin(80^{\circ })\cdot \sin(40^{\circ })={\frac {\cos(40^{\circ })+{\frac {1}{2}}}{2}}$

On multiplying by sin(20°), we get:

$\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })=(\cos(40^{\circ })+{\frac {1}{2}})\cdot {\frac {\sin(20^{\circ })}{2}}={\frac {\cos(40^{\circ })\cdot \sin(20^{\circ })}{2}}+{\frac {\sin(20^{\circ })}{4}}$

Now, invoking the sums-to-products formula for the product of a sine and a cosine function, for any angles α and β:

$\sin(\alpha )\cdot \cos(\beta )={\frac {\sin(\alpha +\beta )+\sin(\alpha -\beta )}{2}}$

with α = 20° and β = 40°:

$\sin(20^{\circ })\cdot \cos(40^{\circ })={\frac {\sin(20^{\circ }+40^{\circ })+\sin(20^{\circ }-40^{\circ })}{2}}={\frac {\sin(60^{\circ })+\sin(-20^{\circ })}{2}}$

Now, as $\sin(60^{\circ })={\frac {\sqrt {3}}{2}}$     and     $\sin(-20^{\circ })=-\sin(20^{\circ })$

$\sin(20^{\circ })\cdot \cos(40^{\circ })={\frac {{\frac {\sqrt {3}}{2}}-\sin(20^{\circ })}{2}}$

and it follows that:

$\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })={\frac {\frac {{\frac {\sqrt {3}}{2}}-\sin(20^{\circ })}{2}}{2}}+{\frac {\sin(20^{\circ })}{4}}={\frac {\sqrt {3}}{8}}-{\frac {\sin(20^{\circ })}{4}}+{\frac {\sin(20^{\circ })}{4}}={\frac {\sqrt {3}}{8}}$     . .     . .     . .     (2)

Now, divind equation (2) by equation (1), we get:

${\begin{aligned}{\frac {\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })}{\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })}}={\frac {\frac {\sqrt {3}}{8}}{\frac {1}{8}}}\\\tan(20^{\circ })\cdot \tan(40^{\circ })\cdot \tan(80^{\circ })={\sqrt {3}}\\\end{aligned}}$

Recalling that one of the properties of the tangent function is that:

$\tan(90^{\circ }-\theta )=\cot \theta$     and applying it at θ = 50° and θ = 10° and recalling that     $\cot(30^{\circ })={\sqrt {3}}$     and substituting, we find:

$\tan(20^{\circ })\cdot \tan(90^{\circ }-50^{\circ })\cdot \tan(90^{\circ }-10^{\circ })=\cot(30^{\circ })$

$\tan(20^{\circ })\cdot \cot(50^{\circ })\cdot \cot(10^{\circ })=\cot(30^{\circ })$

$\tan(20^{\circ })\cdot {\frac {1}{\tan(50^{\circ })}}\cdot {\frac {1}{\tan(10^{\circ })}}={\frac {1}{\tan(30^{\circ })}}$

$\tan(20^{\circ })\cdot \tan(30^{\circ })=\tan(10^{\circ })\cdot \tan(50^{\circ })$

as required. EdChem (talk) 06:22, 11 February 2013 (UTC)

Alternate proof

Here is an alternate proof, loosely modelled on the outline above. We compute cos² 50 + tan 10 sin 50 cos 50 = (1 + cos 100)/2 + tan 10 (sin 100 /2 ) = (1 - sin 10)/2 + tan 10 (cos 10 / 2) = 1/2 - (sin 10)/2 + (sin 10)/2 = 1/2 = cos 60 = cos 10 cos 50 - sin 10 sin 50.

Dividing the result by cos 50, we find cos 50 + tan 10 sin 50 = cos 10 - sin 10 tan 50.

We rearrange to obtain cos 10 - cos 50 = tan 10 sin 50 + sin 10 tan 50 = tan 10 tan 50 cos 10 + tan 10 cos 10 tan 50 = tan 10 tan 50 (cos 10 + cos 50).

Now we divide by cos 10 + cos 50 and use the identity for a product of tangents to find tan 10 tan 50 = (cos 10 - cos 50)/(cos 10 + cos 50) = tan 20 tan 30. 64.140.122.50 (talk) 08:14, 12 February 2013 (UTC)

Extension

Above, an IP noted that the result is equivalent (after re-adding a lost 2) to the statement that:

$\tan(20)\cdot {\frac {2\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {3}}={\frac {\tan(20)}{1+{\sqrt {1+\tan ^{2}(20)}}}}$

which is the same as the polynominalin t:

${\frac {2t}{1-t^{2}}}={\frac {\sqrt {3}}{1+{\sqrt {1+t^{2}}}}}$

Solving this for t in the range 0 < t < 1 will give the exact value for tan 20°. EdChem (talk) 06:22, 11 February 2013 (UTC)

Letting t = tan 20, we have sqrt(3) = tan 60 = tan 3t = (t^3 - 3t)/(3t^2 - 1), hence t^3 -3sqrt(3)t^2 - 3t + sqrt(3) = 0.

Multiplying this equation by t^3 + 3sqrt(3)t^2 - 3t -sqrt(3), we find that t^6 - 33t^4 + 27t^2 - 3 = 0.

By Eisenstein's criterion, the polynomial on the left is irreducible over the rational numbers, so it is not possible to find a polynomial equation of lower degree with rational coefficients of which t is a root. 64.140.122.50 (talk) 08:13, 12 February 2013 (UTC)

Multiplying by a uniformly-distributed random variable

If I multiply random variable A by random variable B, and random variable B has a uniform distribution, will random variable A*B then have the same type of distribution as what random variable A had? Thorstein90 (talk) 22:29, 10 February 2013 (UTC)

No. I don't think it will ever be the same, and it can be very different, as for example if you multiply a narrow Gaussian r.v. by a broad uniform r.v. Looie496 (talk) 00:40, 11 February 2013 (UTC)

But what I mean is if you multiply a narrow Gaussian by a broad uniform, would it still be (maybe a broader) Gaussian? Thorstein90 (talk) 04:51, 11 February 2013 (UTC)

Clearly not, consider the Gaussian N(1,0). (That's the constant 1.) McKay (talk) 05:54, 11 February 2013 (UTC)

A zero-variance Gaussian? Thorstein90 (talk) 06:48, 11 February 2013 (UTC)

Yes, a zero-variance Gaussian is just a limiting case of a Gaussian as the variance goes to zero (it has all its probability mass at the mean). If something (such as preservation of the class of distributions) were true for all members of the class, it would be true in the limiting case as well. But here, with a unit-mean, zero-variance Gaussian (or any other distribution of unit mean and zero variance), the product with a uniform distribution is simply the same uniform distribution. Duoduoduo (talk) 15:20, 11 February 2013 (UTC)

See Normal distribution#Zero-variance limit. Duoduoduo (talk) 15:50, 11 February 2013 (UTC)

So how do you get the answer to be "yes, it will be random", the same as xoring by a random one time pad? You have a good source of randomness so is there some way to "xor" by it in the same way, to get an equally random result as the "OTP" what you're xoring by? 91.120.48.242 (talk) 07:42, 11 February 2013 (UTC)

My interpretation if 91.120's question is this: A one-time pad (OTP) is uniformly distributed over some fixed range of integers. Is there any random variable A, somehow distributed, that we could multiply by the OTP distribution to get another distribution that is also uniformly distributed? This is a different question from the original one, which wanted the product to be distributed like A rather than like the uniform. This question is also different because one-time pads are discretely distributed. My own answer is that I doubt it, but I'm not sure. Duoduoduo (talk) 15:32, 11 February 2013 (UTC)

What you probably want is (modular) addition, not multiplication. If A is a random variable in [0,1], and B is independent and uniformly distributed in [0,1], then $C=(A+B)\ \mod 1$ is uniformly distributed in [0,1], independent of A, and using B and C you can reconstruct A. This can be extended to (bounded) ranges which are not [0,1]. -- Meni Rosenfeld (talk) 13:44, 12 February 2013 (UTC)

February 11

Variance of the product of two normals

I know that for independent random variables, $Var(XY)=Var(X)Var(Y)+Var(X)[E(Y)]^{2}+Var(Y)[E(X)]^{2}$.

Does this mean that the variance of the product of $X\sim N(0,\sigma _{X}^{2})$ and $Y\sim N(0,\sigma _{Y}^{2})$ is simply $\sigma _{X}^{2}\sigma _{Y}^{2}$?

Thorstein90 (talk) 01:22, 11 February 2013 (UTC)

With all respect, a person who has the sophistication to ask that question should have the sophistication to answer it, since the product of two zero-mean Gaussians is a zero-mean Gaussian. Looie496 (talk) 04:36, 11 February 2013 (UTC)

Oh really? Then what is this: http://mathworld.wolfram.com/NormalProductDistribution.html Thorstein90 (talk) 04:47, 11 February 2013 (UTC)

At a glance that looks okay to me. In equation 2 of your wolfram reference, the expression $(|u|)/(\sigma _{x}\sigma _{y})$, looks to me like the ratio of the centered random variable to the standard deviation, so the variance would be $\sigma _{x}^{2}\sigma _{y}^{2}$. Duoduoduo (talk) 14:54, 11 February 2013 (UTC)

However, the product of two zero mean Gaussians is not Gaussian. That's true for sums not products. Duoduoduo (talk) 14:59, 11 February 2013 (UTC)

The answer to your (original) question is "yes". McKay (talk) 05:52, 11 February 2013 (UTC)

No: It's not Gaussian. Michael Hardy (talk) 02:07, 16 February 2013 (UTC)

The original question didn't say anything about Gaussians. -- Meni Rosenfeld (talk) 17:59, 21 February 2013 (UTC)

The number of subsets of {1,2,...,n} with a total sum k

What's the generation function $f_{n}$ of the series $a_{k}=\left|\left\{A\subseteq \{1,2,\dots ,n\}:\sum _{i\in A}i=k\right\}\right|$? Thanks, 79.179.188.147 (talk) 11:56, 11 February 2013 (UTC)

See Restricted partition generating functions in partition (number theory). Gandalf61 (talk) 13:19, 11 February 2013 (UTC)

Still can't solve it. The function $q(k)$ is similar but doesn't restrict the subsets $A$ to be contained in $\{1,2,\dots ,n\}$. 79.179.188.147 (talk) 14:17, 11 February 2013 (UTC)

The restriction to subsets of $\{1,2,\dots ,n\}$ means that the upper bound of the product is not infinity but is ... what ? Gandalf61 (talk) 14:31, 11 February 2013 (UTC)

Oh, thanks! 79.179.188.147 (talk) 17:11, 11 February 2013 (UTC)

The number of ways to color N balls with I colors is. ${\binom {N+I-1}{I-1}}$. Bo Jacoby (talk) 02:51, 12 February 2013 (UTC).

Wouldn't the answer be (1 + x)(1 + x²)⋅⋅⋅(1 + xⁿ)? If you expand the product, the coefficient of x^k is a_k. 64.140.122.50 (talk) 04:56, 13 February 2013 (UTC)

Squared triangular number representative polychora.

Not sure if this belongs here or on WP:MATH. In the article Squared triangular number, is the following text:

These numbers can be viewed as figurate numbers, a four-dimensional hyperpyramidal generalization of the triangular numbers and square pyramidal numbers.

. Now this makes it sound like there is a single 4 dimensional figure that if sliced differently into two dimensional pieces would make it hypergeometically obvious. While the sum of cubes is obviously a cubical hyperpyramid (shape made of a cube and 6 square pyramids), it is unclear to me how the hyper pyramid would be cut to give the square of the triangular number. Can someone please help with explaining this to me here, which might help in changing the explanation in the article?Naraht (talk) 15:08, 11 February 2013 (UTC)

Don't know of any, but there is a nice proof without words at [1] Dmcq (talk) 01:04, 13 February 2013 (UTC)

Not doubting the proof (though some more information as to which proof used by each of the discoverers might be nice. Part of the problem with the concept is that the cubical hyperpyramid has a single vertex which is unique so figuring out which of the 6 it lines up with.Naraht (talk) 19:01, 13 February 2013 (UTC)

Have you tried examining the cross-sections involving planes involving the w-axis? They are hard to see. Double sharp (talk) 15:50, 18 February 2013 (UTC)

There are two types of cross section on the cubical hyperpyramid. The ones from the unique vertex to the largest cube are essentially from the point to the largest cube with a cube getting larger. In the other three cross sections, it starts with a square, builds up to a square pyramid and then back down to a square. Not sure how that helps. :(Naraht (talk) 15:43, 20 February 2013 (UTC)

February 12

Cosets and normal subgroup equality

Hello. I recently came across an exercise, and would be very grateful if someone could help.

Given a normal subgroup H in G, with g in G, if $gH=H$, then is $g\in H$?

Neuroxic (talk) 11:31, 12 February 2013 (UTC)

Any subgroup H contains the identity. Think what happens with that. Dmcq (talk) 12:11, 12 February 2013 (UTC)

That question is a special case of the following question, which may actually be easier (intuitively) to solve: if $gH=L$, then is $g\in L$? Basically you're just asking "Is $g\in gH$ for some subgroup $H$?" and it is a very well known fact that the answer is yes, and the proof is also very well known and straightforward: $1\in H\implies g\cdot 1\in gH\implies g\in gH$ --AnalysisAlgebra (talk) 19:21, 12 February 2013 (UTC)

There's no need for $H$ to be normal, by the way. --AnalysisAlgebra (talk) 19:23, 12 February 2013 (UTC)

Estimating parameters of a distribution from censored data

Suppose you hypothesize that your data are drawn from, say, a lognormal distribution or a gamma distribution. Your data are right-censored at the value x*. That is, you have exact data whenever x≤x*, but whenever x>x* all you know is that x>x*.

(1) How do you estimate the parameters of the hypothesized distribution? I know it would be by maximum likelihood, but how do you handle the censored data?

(2) Is there a command in, say, SAS that will do this?

(3) How do you test the hypothesis that the data did in fact come from that distribution? Duoduoduo (talk) 18:37, 12 February 2013 (UTC)

(1) If your data consists of uncensored points $x_{1},\ldots ,x_{m}$ and n censored points, and your pdf with parameters $\theta$ is $f(x|\theta )$, then the likelihood of $\theta$ is $\left(\int _{x^{*}}^{\infty }f(t|\theta )\ dt\right)^{n}\prod _{i=1}^{m}f(x_{i}|\theta )$. There's ostensibly an inconsistency in that you use a density for the contribution to likelihood of some points and a probability for others, but since likelihood functions are only meaningful up to a constant factor it doesn't matter. -- Meni Rosenfeld (talk) 19:49, 12 February 2013 (UTC)

Thanks, Meni! Duoduoduo (talk) 14:19, 14 February 2013 (UTC)

February 15

How do you solve ${\frac {n!}{(n-r)!n^{r}}}\geq 1-\epsilon$ for n?

How do you solve ${\frac {n!}{(n-r)!n^{r}}}\geq 1-\epsilon$ for $n$, given $r$ and $\epsilon$? The parameter values of interest are of the order $r\sim 10^{9}$ and $\epsilon \sim 10^{-5}$. Can someone give a good approximation formula that can be evaluated without overflows or gross loss of precision during computation? Thanks. --173.49.13.216 (talk) 03:56, 15 February 2013 (UTC)

I'm not a specialist of this kind of thing, but this can be rewritten (exactly) as

$\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)\cdots \left(1-{\frac {r-1}{n}}\right)\geq 1-\epsilon$

$\log \left(1-{\frac {1}{n}}\right)+\log \left(1-{\frac {2}{n}}\right)+\cdots +\log \left(1-{\frac {r-1}{n}}\right)\geq \log(1-\epsilon ).$

In approximate terms, we can write this as

$-{\frac {1}{n}}-{\frac {2}{n}}-\cdots -{\frac {r-1}{n}}\geq \log(1-\epsilon )$

$-{\frac {r(r-1)}{2n}}\geq \log(1-\epsilon )$

$n\geq {\frac {r(r-1)}{-2\log(1-\epsilon )}}\approx {\frac {r^{2}}{2\epsilon }}$

In linearizing the logarithms, the error in the terms on the left is at most about ${\frac {r}{2n}}$ times their value. Given the approximate value of n, the error on the left is, in proportion, about ${\frac {\epsilon }{r}}$. In other words, in using the formula $n\geq {\frac {r(r-1)}{-2\log(1-\epsilon )}}$ you commit an error of at most about ${\frac {\epsilon }{r}}\approx 10^{-14}$ times the vaue of n. (In fact, a more careful calculation shows that the error is about 2/3 of this, and that it results in an underestimation of the minimum possible n.) If you replace r - 1 with r, this rises to about 1/r, or 10^-9. If you replace $\log(1-\epsilon )$ with $\epsilon$, the error rises to about $\epsilon /2=5\times 10^{-6}$ times the value of n. 64.140.122.50 (talk) 05:58, 15 February 2013 (UTC)

I forgot to mention this - I'm using natural logarithms here. 64.140.122.50 (talk) 06:08, 15 February 2013 (UTC)

Thanks. Your formula is more than good enough. --173.49.13.216 (talk) 12:15, 15 February 2013 (UTC)

February 16

Quadratic roots

Is is possible to have two equal complex roots of a quadratic equation, where a,b,c are real numbers is a quadratic equation? — Preceding unsigned comment added by 117.227.197.26 (talk) 13:06, 16 February 2013 (UTC)

No - only complex conjugate, if they are complex. Using the standard form of the quadratic equation:

$ax^{2}+bx+c=0$

where as usual a, b, c are real numbers, think about the quadratic formula solution:

$x={\frac {-b\pm {\sqrt {b^{2}-4ac}}}{2a}}$

so for negative discriminant,

$b^{2}-4ac<0$

the roots are complex conjugate. M∧Ŝc²ħεИ_τlk 13:25, 16 February 2013 (UTC)

Altenative method - if the roots are equal and both equal to k then

$ax^{2}+bx+c=a(x-k)^{2}$

$\Rightarrow ax^{2}+bx+c=ax^{2}-2akx+ak^{2}$

$\Rightarrow b=-2ak$

$\Rightarrow k=-{\frac {b}{2a}}$

and so k is real since a and b are real. Gandalf61 (talk) 14:10, 16 February 2013 (UTC)

Notation for representation theory of the Lorentz group

See also: Talk:Representation theory of the Lorentz group § Feedback

The notations D(Λ) = "representation of the Lorentz group" and (m, n) = "finite dimensional irreducible representations" seem clear enough, however the notation (say) "D^{(1/2, 0)}" (i.e. including the superscript) is confusing...

My burning questions are...

• is D^(2m) = (m, 0) for half-integer m, or equivalently D^(m) = (m/2, 0) for integer m? Or...
• is D^(m) = (m, 0) for integer or half-integer m?

Just alternative notations/conventions?...

• In this paper by Gábor Zsolt Tóth (see appendix A4); letting m and n be integers:

D^(m) = (m/2, 0) ⊕ (0, m/2),

D^(m) = (m/2, m/2),

D^{(m, n)} = (m/2, n/2) ⊕ (n/2, m/2),

what does the last expression have for equivalent notation:

D^{(m, n)} = D^(?) ⊕ D^(?) ?

while...

• The WP article takes m, n to be half-integers in (m, n), so does that translate to

"D^{(2m, 2n)} = (m, n) ⊕ (n, m)" ?

and this has what notation:

"D^{(m, n)} = D^(?) ⊕ D^(?) ?

• In all... is the statement

D^{(m, n)} = (m/2, n/2) ⊕ (n/2, m/2) = "(2m  +  1)(2n  +  1)-dimensional irreducible representations of D(Λ)"

true?

• Are any of the differing conventions, where to put the 1/2 factor, or just the choice of what is integer and half-integer, is "the" standard?

Thanks in advance for any and all replies. Best, M∧Ŝc²ħεИ_τlk 13:23, 16 February 2013 (UTC)

Don't get lost in the notation. The finite dimensional real irreducible representations of the spin group of the Lorentz group all have one of the following forms (for n and m integers):

$(n/2,m/2)\oplus (m/2,n/2),(n/2,n/2).$

These are irreducible as real representations, which means you cannot decompose them further without breaking invariance under complex conjugation. However those of the first kind do decompose as a sum of two complex conjugate represnetations. The source you cite calls the first kind of representation $D^{(n,m)}$ and the second kind ${\widetilde {D}}^{(n)}$. The fact that these are irreducible implies that there is no decomposition of $D^{(n,m)}$ into a direct sum of two real representations D's. Sławomir Biały (talk) 14:40, 16 February 2013 (UTC)

It might be added that the above exhaustive list applies to O⁺(1;3) which includes parity inversions (the orthochronous Lorentz group). In physics SO⁺(1;3) (the proper orthochronous Lorentz group) is often of interest because parity is not a symmetry in every theory (see e.g. weak interactions). For SO⁺(1;3) the irreducible representations are of the form

$(m/2,n/2)$

for n and m integers.YohanN7 (talk) 16:18, 16 February 2013 (UTC)

Excellent explanations of the facts - if that's all there is to it (the notation and it's meaning), that's fine. Thanks (again)! M∧Ŝc²ħεИ_τlk 23:43, 16 February 2013 (UTC)

February 17

probability and habit.

Lets say that someone wants to break a habit , what is the best way of modeling the probabilities?

If I take avoiding Facebook as an example where I have 1/3 chance of logging in in 1 day and 2/3 chance of not logging in:

Using a binomial distribution, if I avoid Facebook for 365 days then I have (2/3)^365 which is 5 x 10^-65 which is smaller than the Planck length.

I am guessing that the binomial is not working for this case but would the modelling of breaking a simple habit cause such small numbers?

Ap-uk (talk) 12:35, 17 February 2013 (UTC)ap

Your analysis would be right if the probability were to remain constant day after day for the whole year. But in reality it would not. If you've made it for say 200 days without logging in, it's probably not by chance but rather it's that you have adjusted your behavioral tendencies -- that is, you've adjusted your log-in probability down to near zero. Duoduoduo (talk) 14:53, 17 February 2013 (UTC)

Agreed. In math-speak, the probabilities from one day to the next are not independent, which is required for the formula you used. So, you need to use P = P_day1P_day2P_day3P_day4... instead. Of course, actually determining P_day1, P_day2, P_day3, P_day4 ... is not easy. StuRat (talk) 17:14, 17 February 2013 (UTC)

You'd really need to use conditional probabilities:

$\Pr[{\hbox{365 days}}]=\Pr[{\hbox{day 1}}]\cdot \Pr[{\hbox{day 2}}\mid {\hbox{day 1}}]\cdot \Pr[{\hbox{day 3}}\mid {\hbox{day 1 and day 2}}]\cdots \Pr[{\hbox{day 365}}\mid {\hbox{day 1 and day 2 and }}\ldots {\hbox{ and day 364}}]$

The probabilities near the end are probably really close to 1; given that you have successfully avoided Facebook for the first 364 days, the probability that you'll avoid it on the 365th day is probably almost 100%.

Also, comparing a probability to the Planck length is meaningless; a probability is not a length. —Bkell (talk) 13:15, 19 February 2013 (UTC)

Analytical geometry question

I'm stuck trying to solve the following problem. Any help would be appreciated!

Which lines crossing point (2, 6) together with the x-axis and the y-axis border a triangle with an area of 25? I'm trying to solve both rising and falling lines.

I know that the equation for the line is $y-6=m(x-2)$ where m is the slope of the line. And I also know that I need to find the points where y=0 or x=0. Then the area would be the product of those two points divided by two.

So when the line crosses the x-axis y=0 and $mx-2m+6=0$. And when it crosses the y-axis x=0 and $y-6+2m=0$. But I think I'm doing something wrong here, any ideas? 82.181.210.44 (talk) 15:32, 17 February 2013 (UTC)

No hang on, I got right two of the lines.

$x={\frac {2m-6}{m}}$ and $y=6-2m$. So ${\frac {2m-6}{m}}*(6-2m)=50$ and we can see that m is either $-2$ or ${\frac {-9}{2}}$. And the equations of the two lines are $y=-2x+10$ and $y={\frac {-9x}{2}}+15$.

But I still can't figure out the two remaining lines.82.181.210.44 (talk) 17:05, 17 February 2013 (UTC)

If the triangle is in the second quadrant its area is negative, so the corresponding value in your analysis is -50. This will give a second quadratic equation in m, with two positive roots.←86.186.142.172 (talk) 17:22, 17 February 2013 (UTC)

You found two triangles in the first quadrant, but there is also a triangle in the 2nd quadrant and one in the fourth quadrant. As 86.186.142.172 says, both those areas can be thought of as negative, so plug that into your above equation to find the two lines. BTW, your initial statement of the problem was quite hard to follow. Try this: "I need to find 4 lines in the XY plane, each of which pass through (2,6), and, in conjunction with the X and Y-axes, form a triangle with an area of 25". The "4" and "each" are the critical bits that were missing from your Q. StuRat (talk) 17:51, 17 February 2013 (UTC)

Yes, thanks to both of you I finally figured it out. And sorry about the unclear question. The problem was in different language and I had to translate it myself and I'm not good at that. And sorry about the math tags that were missing at first. I was too busy working with this problem so I couldn't be bothered to learn how to use them. But anyways, I think the issue was that I was thinking that area can't be negative, and didn't realise that the roots would be positive. Thanks again! 82.181.210.44 (talk) 18:13, 17 February 2013 (UTC)

You're quite welcome, and I will mark this Q resolved. StuRat (talk) 18:19, 17 February 2013 (UTC)

Resolved

February 19

Tensors

Why is tensor a very tough topic? No one is explaining tensor in a simple language. I tried to read Wikipedia article about Tensor, but it is also very tough. Could someone make me understand about tensors in a simple language? 27.62.227.241 (talk) 14:11, 19 February 2013 (UTC)

Tensors are geometric objects. Triangles, rectangles, and parallelograms are also geometric objects. 140.254.226.230 (talk) 14:34, 19 February 2013 (UTC)

I don't agree that tensors are geometric objects in the way that triangles and rectangles are. That seems to suggest that tensors are "objects" in a naive sense, which is not true. Rather tensors express geometrical information. Sławomir Biały (talk) 16:37, 19 February 2013 (UTC)

As the picture of a tensor shows on the article Tensor, it looks like tensors look like the vectors used in physics to describe magnitude and direction. But they are not just vectors. Tensors would probably be the geometrical objects that may use vectors to specify some linear relationship between two objects. They may be vectors themselves. Perhaps, the force vectors in physics would count as tensors. If that is the case, then see Kinematics. 140.254.226.230 (talk) 16:55, 19 February 2013 (UTC)

Right. Vectors are a special case of tensors. The illustration there shows, more generally, that certain kinds of tensors express information that require specifying one or more directions in a space. Strain, for instance, is an example. Sławomir Biały (talk) 17:25, 19 February 2013 (UTC)

A similar question was asked recently on the Science Desk, and got some good answers. AndrewWTaylor (talk) 18:22, 19 February 2013 (UTC)

Tensors can be confusing because there are several (equivalent) ways of defining them. Here is one way of looking at a tensor: It is a function that takes a number of vectors as arguments and produces a number as output. This function is required to be linear in each of its arguments. Such a function is called multi-linear. To figure out the components of the tensor one takes combinations of basis vectors as arguments and computes the output. This description is basis dependent, because if one uses another basis for the vectors, one will, in general, get a different multi-linear function describing the same tensor! YohanN7 (talk) 21:11, 19 February 2013 (UTC)

I found the Dan Fleisch video mentioned in the previous post illuminating. It was mostly a rehash ("Tensors are N-dimensional vectors"), but the gem at the end for me was the realization that a tensor is almost always accompanied by a (potentially implicit) basis vector set, and that the basis vector set of a tensor can combine basis vectors from potentially distinct vector spaces. The prevalent use of the implicit x/y/z basis vectors for most usage of (1-dimensional) vectors can make it difficult to realize that there can be more complex sets of basis vectors that accompany a tensor. -- 205.175.124.30 (talk) 22:38, 19 February 2013 (UTC)