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# February 13

If no question is asked then any answer written here is correct. Count Iblis (talk) 12:32, 18 February 2014 (UTC)

And you didn't even link to vacuous truth! —Quondum 15:26, 18 February 2014 (UTC)
Thanks, I didn't know that this is called that way (English is not my mother tongue) :). You also encounter this when taking the intersection of a collection of sets. If the collection of the sets is empty then the intersection is the whole space. In some statements in topology or measure theory you then don't have to explicitly address the exceptional cases when a collection of sets is empty. Count Iblis (talk) 15:59, 18 February 2014 (UTC)

# March 5

## Where are the constraints violated?

It is well known that, using only a compass and an unmarked straightedge, it is impossible to trisect an arbitrary angle.

Here are three methods to trisect an arbitrary angle - where are the constraints violated?

Let DCE, or gamma, be an arbitrary acute angle. (An arbitrary obtuse angle can be made acute by lopping off one or multiples of 90 degrees, and an angle of 90 degrees is trisectable using a compass and unmarked straightedge.)

Let line segments CB and BA have the same length as DC, both CB and BA lie on the same side of EC as does point D, and point A is constrained to lie on the extension of EC.

One method is to move point A (which moves point B along the arc of a circle) along the extension of EC, until the extension of BA strikes point D. The angle BAC (=DAC), or alpha, is 1/3 of gamma (proof shown below).

A second method is to draw the extension DBA' (where A' is also on the extension of EC), and to move point A (not A' !!) along the extension of EC, until line BA is colinear with DBA'.

A third method is to initially let segment BA=DC (as above) but to let CB have arbitrary length, and to change the length of CB by moving point B (which moves point A along the extension of EC) until CB=BA (or CB=DC).

(Proof of trisection (for all three methods):

Angle BAC of isosceles triangle ABC is equal to alpha, so that the other isosceles angle ACB is equal to alpha, so that the third angle ABC is 180-2*alpha.

Angle ABC (=180-2*alpha) plus angle CBD is a straight angle, so that angle CBD is 2*alpha, as is the other isosceles angle BDC of triangle BCD, thus the third angle BCD of triangle BCD is 180-4*alpha.

Angle DCE plus angle BCD plus angle ACB = straight angle, or gamma+(180-4*alpha)+alpha=180, or gamma-3*alpha=0, or alpha=gamma/3.)

My question: Where are the constraints (of using only a compass and an unmarked straightedge) violated?Bh12 (talk) 00:05, 5 March 2014 (UTC)

First, compass and straightedge constructions don't involve moving points. You're restricted to a very specific set of operations: given two points, you can draw the line containing them; given two points, you can draw the circle centered at one and containing the other; given intersecting lines/circles, you can draw the point of their intersection.
Also, while I don't understand what you're saying with your second method, in your first and third method, B = D. This causes your later argument to break down.
Ignoring that, I don't see why you claim DCE + BCD + ACB = 180. It seems that DCE - (BCD + ACB) = 0. Or perhaps BCD - (DCE + ACB) = 0, depending on the relative positions of B and D.--80.109.80.78 (talk) 06:23, 5 March 2014 (UTC)
You might be interested in our article on Neusis construction which allows methods giving accurate trisections. Dbfirs 08:33, 5 March 2014 (UTC)

## Constant second quotients property

We know that some functions have special properties on how $x$ and $f(x)$ are related:

Linear functions have an add-add property. This means that adding a constant to $x$ will add a constant to $f(x)$.

Exponential functions have an add-multiply property. Adding a constant to $x$ will multiply $f(x)$ by a constant.

Logarithmic functions have a multiply-add property. Multiplying $x$ by a constant will add a constant to $f(x)$.

Multiply-multiply property

Power functions have a multiply-multiply property. Multiplying $x$ by a constant will multiply $f(x)$ by a constant.

Constant second difference property Quadratic functions have a kind of property called constant second difference. Is there a similar kind of function that has a power called constant second quotient?? Please reveal a general equation for this kind of function. Georgia guy (talk) 20:52, 5 March 2014 (UTC)

When you say "constant second difference" do you mean as in (x+1)^2 - x^2 = 2x + 1, 2(x + 1) + 1 - (2x + 1) = 2? I'm not sure what exactly "constant second quotient" is as a property, could you elaborate? Are these your terms, or are they from some source?Phoenixia1177 (talk) 05:55, 6 March 2014 (UTC)
Constant second difference is a property of sequences. You make the difference sequence, and then the difference sequence of that, and you get a constant sequence.--80.109.80.78 (talk) 06:03, 6 March 2014 (UTC)
We can make constant second quotient easily enough by working backwards. Let's start with the quotient sequence: $a_{i+1}/a_i = c$. So $a_i = c^i a_0$. Then let's make the original sequence: $b_{i+1}/b_i = a_i = c^ia_0$. So $b_{i+1} = c^ia_0b_i$. So $b_i = c^{i(i-1)/2} a_0^i b_0$. There's your general form.--80.109.80.78 (talk) 06:03, 6 March 2014 (UTC)
A slight modification of above. Note $a_0=c^\alpha$ for some $\alpha$ and $b_0=c^\beta$ for some $\beta$. Hence
$b_i = c^{\,i^2/2 - i/2} c^{\alpha i} c^{\beta}=c^{\,\frac{1}{2}i^2 + (\alpha-\frac{1}{2}) i + \beta}=e^{A i^2+Bi+C}$
For some A, B, C. Basically the exponential of any quadratic function will have the constant second quotient property. (I've reforatted the question to remove sub-headings, hope thats OK)--Salix alba (talk): 06:45, 6 March 2014 (UTC)
Perhaps the OP is thinking of Finite differences? A math-wiki (talk) 14:49, 9 March 2014 (UTC)

Our article on the turning radius of a vehicle doesn't give a formula :( I'm trying to work out the turning circle (i.e. the turning diameter) of a car. This website (http://www.ehow.co.uk/how_7225784_calculate-turning-circle.html) does:

(TrackLength/2)+(Wheelbase/sin(TurningAngle))


Other websites say it is

2*(Wheelbase/sin(TurningAngle))


Which do you think is correct? I just need an approximate. Thanks! 143.210.123.60 (talk) 21:02, 5 March 2014 (UTC)

Probably the former is intended to give the radius and the latter the diameter of the circle. The website refers to the average steer angle which I take to mean the average between that for the wheel on the inside of the bend and for the wheel on the outside (the wheel on the outside travels around a bigger circle of course). So (wheelbase / sin(average steer angle)) is an estimate of the turning radius for the mid-point of the axle, and adding half the axle length gives the turning radius for the outside wheel. For the latter expression, the turning angle is not precisely defined (whether for the inside wheel, the outside or the average), but in any case, the factor of 2 means it gives a diameter rather than a radius. --catslash (talk) 01:37, 6 March 2014 (UTC)
You are simply AWESOME! Thanks for explaining that to me; it makes so much more sense when someone explains it. 143.210.123.113 (talk) 14:08, 6 March 2014 (UTC)
Also note that, as a practical matter, you may not want to turn at the tightest radius possible. Many cars will make a squealing noise at the sharpest turn possible, and that sure doesn't sound like it's good for the car. I will mark this Q resolved. StuRat (talk) 14:29, 6 March 2014 (UTC)
Resolved

StuRat (talk) 14:30, 6 March 2014 (UTC)

See Ackermann steering geometry (also steering) --catslash (talk) 17:46, 6 March 2014 (UTC)

# March 6

## derivative of (y=x^(2x+1))

If I want to find d/dx of y=x^(2x+1), the normal way is taking ln of both sides:

ln(y) = ln(x^(2x+1))

ln(y) = (2x+1)(ln(x))

d/dx (ln(y)) = d/dx ((2x+1)(ln(x)))

(1/y)(dy/dx) = (2x+1)(1/x) + (ln(x))(2)

dy/dx = y((2xln(x) + 2x + 1) / x)

dy/dx = (x^(2x+1))((2xln(x) + 2x + 1) / x)

But I can also write the original equation as log base x of y = 2x+1

d/dx (log base x of y) = d/dx (2x+1)

( - (lny) / (x ((lnx)^2 )) (dy/dx) = 2

dy/dx = 2x((lnx)^2) / (-(ln(x^(2x+1))))

The two answers are different. What gives? 75.80.145.53 (talk) 07:38, 6 March 2014 (UTC)

Your derivative of $\log_x y$ is wrong. It should be $[(dy/dx) (1/y)(\ln x) - (\ln y)/x]/(\ln^2 x)$.--149.148.255.161 (talk) 09:13, 6 March 2014 (UTC)

## quintic polynomial

I know that the general quintic is insoluble in radicals. But I have a specific quintic polynomial, with integer coefficients which neither I nor mathematica can solve except numerically. Is it possible to prove that a specific integer coefficient quintic polynomial has no solutions in radicals? If so, how to go about it? thanks, 164.11.203.58 (talk) 08:50, 6 March 2014 (UTC)

See Quintic_function#Solvable_quintics, which gives (rather messy) criteria for a quintic to be solvable by radicals. AndrewWTaylor (talk) 09:18, 6 March 2014 (UTC)
(OP) perfect, thanks. It never occurred to me to look at knowledge specific to the quintic.
Resolved
164.11.203.58 (talk) 09:37, 6 March 2014 (UTC)

## Tryes slipping upon deceleration equation

For those that did A-level maths, this is Mechanics 1, but what I have got just doesn't feel right.

(1) Force = mass * acceleration i.e. F=ma

(2) Force = coefficient of friction * mass * acceleration due to gravity i.e. F=µmg ∴ a=µg

In other words, the maximum acceleration (or deceleration) of a car before the tyres slip is only ten times (approx.) the coefficient of friction. That's what the maths seems to say, but it just doesn't feel right; that feels to low. I suspect I might be misusing equation (2). Could you please confirm if that makes sense or not. If not, could you please explain what does make sense. Thank you! 143.210.123.73 (talk) 17:01, 6 March 2014 (UTC)

Note that the coefficient of friction can be more than one. This can happen if the two surfaces aren't completely flat, so a piece of tire tread extends down into dip and pushes against a ridge. Also, if the tire partially melts you can get molten rubber acting as a adhesive. StuRat (talk) 17:20, 6 March 2014 (UTC)
Yes, I was guesstimating a dry friction coefficient of 1.5 for my Hoosier 19.5X6.5-10 R25B tyres. But still a deceleratn of 15ms^-2 before the car skids seems low, so I think maybe I can't use that second equation in that way, but I'm not sure. For a wet friction coefficient (really worst case scenario), I'm guessing a friction coefficient of 0.3. This is for Formula Student by the way 143.210.123.73 (talk) 17:53, 6 March 2014 (UTC)
For a race car there is quite a bit of down force due to aerodynamics. This can amount to a force equal to the weight of the car. It's this enhanced force on the tires that you must multiply by the coef. of friction to get the maximum horizontal force before skidding. Combine this with the fact that racing tires have increased coef. of friction compared to street tires (which last much longer) and it may account for higher acceleration before skidding that you're thinking of. Of course all I know about racing comes from Gran Turismo so take my explanation with a grain of salt. --RDBury (talk) 19:11, 6 March 2014 (UTC)
The car my team are building doesn't have any aerodynamics (it's hard to do) so it's just the weight of the car (320kg). But, how does that fit in mathematically, as the mass cancels out when equating equation (1) and (2). (I know my IP address will change now, but I'm still the OP) 81.101.120.9 (talk) 21:34, 6 March 2014 (UTC)
The calculation feels about right to me. On normal roads, I would expect my tyres to skid with a horizontal acceleration or deceleration approaching one "g", and I notice that my seat belts lock at less than a quarter of this. On wet leaves, I've noticed that tyres skid at about one tenth "g", corresponding to a coefficient of friction of about 0.1
I haven't tested out the deceleration required to trigger air bags. Does anyone have any figures? I would expect them to operate at about 2g because this cannot be achieved by normal braking. Dbfirs 09:00, 7 March 2014 (UTC)
I think you're underestimating how much acceleration you're getting. 1 ms-2 = 2.237 mph/s, so at 15 ms-2, you're doing 0 to 60 (or vice versa) in a little under 2 seconds. That seems like a decent upper bound for acceleration/deceleration to me, at least for any car with no downforce effects. (0-60 time = 60/(2.237*acceleration in ms-2) MChesterMC (talk) 09:12, 7 March 2014 (UTC)
Yes, I've seen drivers trying to achieve that on British road surfaces, and their tyres skid and smoke impressively, but they don't achieve that acceleration because many British roads have a coefficient of friction of around 0.5 see Road slipperiness#Reduction Dbfirs 10:21, 7 March 2014 (UTC)
Note that the coefficient of friction isn't solely a property of the road, but of the road/tire combo, as well as anything between them, like grains of sand and moisture, with even air temperature playing a role. There's also a static and dynamic coefficient of friction and the tires may exhibit a different coefficient when moving forward or sideways, too. StuRat (talk) 16:37, 7 March 2014 (UTC)
True, but I think you'd need special tyres to get more than 0.5 on many British roads. Dbfirs 16:57, 7 March 2014 (UTC)
Thanks for the info! Maybe I'm overestimating my friction coefficient then - I'll reduce it :) 143.210.123.106 (talk) 21:05, 7 March 2014 (UTC)

# March 7

## Cos(pi/2)

My own sketch of a cosine wave and Wolfram Alpha both agree that cos(pi/2) is equal to zero. My TI-83 calculator tells me that it's 0.999624... Why is there a difference? Thanks, Dismas|(talk) 16:51, 7 March 2014 (UTC)

You are working in degrees. Wolfram Alpha works in radians for functions of pi (as does everyone else). Dbfirs 16:55, 7 March 2014 (UTC)
Ah! Yeah, I forgot about that... Thanks, Dismas|(talk) 17:06, 7 March 2014 (UTC)

## Finding two unknowns in one equation

e=(Ae^(-3))+Be

My teacher tells me that just from that we can find that A=0 and B=1. To demonstrate this, she gave me the following example:

3apples = Aoranges + Bapples

Since there are no oranges on the left side of the equal sign, A must equal 0 and B must equal 3. I see that but then apples and oranges are simply variables. e is a number. Isn't it possible that somewhere along the number line with its infinite values that there is another answer that will give you something that you can multiply times e in order to get Ae^-3?

And sorry for the formatting. I deal with the math functions so rarely that I've forgotten how to use them and don't have much time right now to re-learn them for this Q.

Thanks, Dismas|(talk) 17:16, 7 March 2014 (UTC)

It's trivial to show that for any A value, $B = -A\,e^{-4} + 1$, so yes of course there are infinite solutions. Though since $e^{-4}$ is irrational, the only solution with both A and B as integers is the one you already gave. Dragons flight (talk) 17:33, 7 March 2014 (UTC)
(WP:EC)Sure, check $A= e^4, \; B=0$. That gives us right hand side = $Ae^{-3} +Be=e^4\cdot e^{-3} + 0 \cdot e= e+0 = e$, which is the left hand side. If you isolate ("solve for") B, then you get $B = \frac{e^4-A}{e^4} = 1-Ae^{-4}$, which is just the equation of a line (if you write A->x, B->y, you get something of the form y=mx+b). So there are infinitely many points (A,B) that satisfy the equation. Pick any A you want, and the last equation will tell you what B has to be. Unless there's some transcription error, or other bits of the problem, something is very wrong here... your teacher might just be showing you how to "guess and check", but that's no excuse to ignore the uncountably many solutions to the equation. Also, here's what Wolfram alpha shows [1]. Perhaps the interest is in integer solutions, but if that's the case, proving/finding integer solutions is a much more complex subject, e.g. Diophantine equations. SemanticMantis (talk) 17:41, 7 March 2014 (UTC)

I asked another math teacher and she agrees with me that there are infinitely many solutions. I hope that I misunderstood the first teacher in that we were trying to find a solution and not necessarily the only one. Thank you for the responses! Dismas|(talk) 20:59, 7 March 2014 (UTC)

As hinted by Dragon's flight, the question probably meant to ask to find integer A and B. And then the only answer is indeed $A=0,\ B=1$, since $e^4$ is irrational. -- Meni Rosenfeld (talk) 19:08, 9 March 2014 (UTC)

## What's the average change in cents for a purchase in US$2 or more? What's the average change in cents for a purchase in US$2 or more? Does it differ if I say above $10 for various different items? Thanks. μηδείς (talk) 23:23, 7 March 2014 (UTC) Averaged over what? —Tamfang (talk) 01:38, 8 March 2014 (UTC) Erm I'd guess about 1 cent. So many things are sold at$5.99 or $5.95.--Salix alba (talk): 06:20, 8 March 2014 (UTC) There is a related observation that does not suffer from the skewing of psychological factors that lead to strange pricing patterns like this: the wear of keys on a keyboard – the '1' gets used substantially more than the '9'. If you were to take the prices in a store and adjust them so that they produced a smooth distribution, I guess you'd get an expected average of more than$0.50 in change in coins from low prices, tending to exactly $0.50 as the range of the prices under consideration moves to larger values. —Quondum 06:46, 8 March 2014 (UTC) Psychological pricing has a nice plot of the distribution of the final digit of prices--its 9 about 60% of the time. --Mark viking (talk) 07:05, 8 March 2014 (UTC) Yeah, I do remember reading a decade ago that a little more than about 1/3 numerical expressions begin with the figure 1. This seemed to be an artifact of the expressions 1, 10-19, 100-199, 1000-1999, etc., That's not at all what I am asking, since change based on those amounts would be 8-99, 81-999, etc.... μηδείς (talk) 07:27, 8 March 2014 (UTC) That would be Benford's Law, which would be skewed in this case because of psychological pricing. Prices are more likely to be 999.99 than 1000.00, so the first digit 9 will be overrepresented. 7 and 8 will likely still be rare. Staecker (talk) 12:37, 8 March 2014 (UTC) To answer the question, you have to have a probability distribution of all prices. Perhaps this could be checked empirically in certain situations. E.g. you could save all your receipts and calculate and answer. Or do similar if you were e.g a store owner. You could assume a uniform distribution of all whole-cent prices, but that is almost certainly wrong. SemanticMantis (talk) 22:23, 8 March 2014 (UTC) • I don't think this is something the reference desk can answer, given that it would be mainly speculation. I'm sure such a distribution would be highly clustered, which isn't ideal for statistical methods.--Jasper Deng (talk) 09:22, 9 March 2014 (UTC) I agree that it can't be answered without further specification. And even if specified, it's unclear if volunteers here would have access to the right kinds of data. However, I don't agree that clustering is a problem. Say OP was interested in "what is the average (e.g. mean or median) value of change in cents of a WalMart purchase over$2 (paid in cash), in the 2000-2010 period?" -- then, regardless of clustering in the distribution, WalMart could easily answer the question, based on their data. SemanticMantis (talk) 17:09, 9 March 2014 (UTC)
Most states and localities have sales tax, but the rates are quite variable -- so that's an additional complication. If you start with "x.99" or "x.95" pricing and then add variable sales taxes, the story will get very complicated very quickly. Looie496 (talk) 17:32, 9 March 2014 (UTC)
The reason why I say statistical methods wouldn't be desirable is that, since a census is impossible, you will have to take a (simple random) sample and make a confidence interval for the mean. But a clustered, irregular distribution without a well-behaved pdf (let alone a cdf) would make that almost impossible to do except by numerically estimating.--Jasper Deng (talk) 19:24, 9 March 2014 (UTC)
Ah, I see what you mean now, thanks for clarifying. I was mostly thinking of an empirical approach. And even then, you're right that tight clusters and e.g. multi-modality in the pdf will make that hard to sample effectively. SemanticMantis (talk) 21:16, 9 March 2014 (UTC)

# March 10

## Calculating average ranking

I have a number of lists that rank Chinese characters by frequency of occurrence in various types of text. I only have the ranking order, not the actual frequencies. The lists are all different lengths. The most common characters appear in all lists, but many others appear only in a subset of lists. Characters that do not appear in a list are assumed to be less common (by some unknown amount) than the lowest-ranked character that does appear. I want to calculate an average ranking for each character, but I don't know what to do when a character does not appear in one or more lists. If I just ignore those lists, and take an average among the lists in which the character does appear, then the calculated average could be higher than if the character actually did appear, with a low ranking, in an ignored file. This is illogical. What is the sensible way to handle this, to make maximum use of the information that I do have, while not creating illogical results? 86.160.86.139 (talk) 14:13, 10 March 2014 (UTC)

There is no general purpose way to do this- basically you are asking for a voting procedure which will aggregate a set of rankings into a single average ranking. There are serious mathematical obstacles to this- see voting system and specifically voting paradox for some idea of what problems can arise. The specific issue of some characters not appearing on some lists is yet another major problem which as far as I can see doesn't have a clear resolution. You'll have to think about what sort of "average" your ranking is meant to represent, and choose a method based on that. For example if you only want an average of the most common or well-known characters, it might make sense to disallow any characters that fail to appear on one of your lists. This is a bit arbitrary, but whatever you choose to do will be similarly arbitrary. Staecker (talk) 15:02, 10 March 2014 (UTC)
There is no generally consistent way to perform rank aggregation (see Arrow's impossibility theorem), but folks do use heuristic methods that work well enough for their purposes. This paper discusses some methods, based on minimizing distance from the aggregate to the individual lists, and has an associated R package for computations. --Mark viking (talk) 15:39, 10 March 2014 (UTC)
Above answers are good and correct, but don't get too discouraged. Just because there is no mathematically perfect system doesn't mean there isn't one that will work well for your purposes. Since there are a very large number of characters, I suspect Borda count would be a decent place to start. SemanticMantis (talk) 15:45, 10 March 2014 (UTC)
Thanks for the great answers everyone. 86.160.86.139 (talk) 18:35, 10 March 2014 (UTC)

## Why is arcsinh(1) so close to √π/2 ?

Is there some deeper reason, or meaning, or explanation, for the fact that $\text{arcsinh }1\simeq\frac{\sqrt\pi}2 ?$ I ask this in the light of $\sinh x=\frac{e^x-e^{-x}}2,$ and $\frac{\sqrt\pi}2=\int_0^\infty e^{-x^2}dx$ . — 86.125.196.90 (talk) 18:25, 10 March 2014 (UTC)

It must have something to do with the fact that $\text{arcsinh}{(x)} = \ln{(x+\sqrt{x^2+1})}$. It follows that this is a question of showing that $\int_1^{(1+\sqrt{2})} \frac {dx}{x} \simeq \frac{\sqrt{\pi}}{2}$. I suppose there is a complex analysis method to evaluate the integral. I will revisit this later.--Jasper Deng (talk) 19:46, 10 March 2014 (UTC)
Speaking of integrals, plotting $e^{-x^2}$ and $\frac1{\sqrt{1+x^2}}$ simultaneously on the positive semiaxis reveals some very interesting results. There's a symmetry there with regards to the vertical line x = 1, and one can easily see how the area of $\frac1{\sqrt{1+x^2}}$ in between 0 and 1 that lies above that of the Gaussian function nicely mirrors that of $e^{-x^2}$ from 1 to $\infty.$ Also, I was perhaps thinking of expanding $\ln(1+\sqrt2)$ into Taylor series, as well as writing the binomial series of $\frac1{\sqrt{1+x^2}}$, and then integrating it in between 0 and 1. — 86.125.196.90 (talk) 20:18, 10 March 2014 (UTC)
My personal opinion is that expanding it into a series is not going to be particularly helpful. You're right that I want to reduce it to a Gaussian integral or something.
It actually seems to be simpler than going into the complications of the natural logarithm, since the derivative of arcsine is $\frac{1}{\sqrt{x^2+1}}$. Then we're considering $\int_0^1\frac{dx}{\sqrt{x^2+1}}$. The inverse function of the integrand is of course $\frac{1}{\sqrt{\frac{1}{x^2}-1}}$. But I'm not sure about this integral's relation with the Gaussian integral. More later.--Jasper Deng (talk) 23:02, 10 March 2014 (UTC)
Actually, it's a completely moot point - the two quantities are not equal, i.e. the hyperbolic arcsine of 1 is not equal to the square root of pi divided by two. The former is about .881, the latter about .886.--Jasper Deng (talk) 00:15, 11 March 2014 (UTC)
I never said they were... I was just surprised by their being so close to one another... (Maybe that approximation sign is to blame, it looks too close to an equality sign). — 86.125.196.90 (talk) 01:23, 11 March 2014 (UTC)
It's probably a coincidence, then. Here's where Taylor series get into the picture, I'd think. But even that is probably a pure coincidence. Basically, we're not going to be able to answer the question using the integral methods I really like to use.--Jasper Deng (talk) 02:00, 11 March 2014 (UTC)
We have an article on Mathematical coincidences, but this one doesn't seen to be mentioned (and it's not nearly as close an approximation as some that are). AndrewWTaylor (talk) 11:12, 11 March 2014 (UTC)