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# July 14

## Slope–intercept form

Hey,

Can I wrote a slope intercept form for y=8x+5 like this: y-8x+5 (without equality)?

--141.226.146.224 (talk) 05:48, 14 July 2017 (UTC)

You can write y-8x=5, and this means the same, but without an equals sign you don't have "verb" in your "sentence" so there is no meaning, it is just an expression, and cannot represent a line. Dbfirs 06:02, 14 July 2017 (UTC)
thank you. I wasn't sure. --141.226.146.224 (talk) 06:31, 14 July 2017 (UTC)
Note that it should be ${\displaystyle y-8x-5=0}$, not ${\displaystyle y-8x+5=0}$. -- Meni Rosenfeld (talk) 08:46, 14 July 2017 (UTC)
... yes, that's an alternative way to write the equation, but it does need an equals sign. Dbfirs 20:48, 14 July 2017 (UTC)
Yes, but that's a separate problem. My understanding is that the OP tried to inquire if he can write the expression ${\displaystyle y-(8x+5)}$ to refer to the line, with the ${\displaystyle =0}$ part being implicit. You explained that this is not a thing and the equality should be included. I explained the unrelated issue that the OP had a calculation error and that ${\displaystyle y-(8x+5)}$ is really ${\displaystyle y-8x-5}$. -- Meni Rosenfeld (talk) 00:54, 16 July 2017 (UTC)

But the Slope–intercept form is specifically y=mx+b. Bubba73 You talkin' to me? 02:59, 15 July 2017 (UTC)

... or y=mx+c in the UK. Dbfirs 08:41, 15 July 2017 (UTC)

## Algebraic numbers such as those in casus irreducibilis

If a cubic equation over the rationals has three real roots, none of which are rational, the only way to express those roots algebraically is in terms of non-real complex numbers. This case is called casus irreducibilis.

What do we know about the set of all real algebraic numbers that are expressible algebraically only in terms of non-real complex expressions? Loraof (talk) 15:30, 14 July 2017 (UTC)

For example, are "almost all" real algebraic numbers in this category? Loraof (talk) 00:00, 16 July 2017 (UTC)
Since there are countably many algebraic numbers and the subset expressible in terms of real radical expressions is infinite, and so is its complement, it follows they have the same cardinality.--Jasper Deng (talk) 16:58, 17 July 2017 (UTC)

# July 15

## Trig function derivative (degrees)

I think I asked this question a few years ago:

What is the derivative of ${\displaystyle \arctan \left({\frac {\sin(x)}{\cos(x)+c}}\right)}$ while ${\displaystyle x}$ is in degrees? יהודה שמחה ולדמן (talk) 20:51, 15 July 2017 (UTC)

It is the derivative of the function with the argument expressed in radians multiplied by ${\displaystyle \pi /180}$. Ruslik_Zero 20:56, 15 July 2017 (UTC)
${\displaystyle {\frac {d}{dx}}\arctan \left({\frac {\sin({\frac {\pi }{180}}x)}{\cos({\frac {\pi }{180}}x)+c}}\right)={\frac {\pi }{180}}\left({\frac {c\cos({\frac {\pi }{180}}x)+1}{c^{2}+2c\cos({\frac {\pi }{180}}x)+1}}\right)}$ ? יהודה שמחה ולדמן (talk) 21:37, 15 July 2017 (UTC)
This is an easy consequence of the chain rule. Let ${\displaystyle u={\frac {\pi }{180}}x}$ and calculate ${\displaystyle {\frac {dy}{dx}}={\frac {dy}{du}}{\frac {du}{dx}}={\frac {dy}{du}}{\frac {\pi }{180}}}$.--Jasper Deng (talk) 04:09, 16 July 2017 (UTC)
• But if you want the result to also be in degrees, don't you have to multiply this result by ${\displaystyle {\frac {180}{\pi }}}$? Then the result is just the unadjusted formula. Loraof (talk) 15:25, 16 July 2017 (UTC)
The function he has is dimensionless, so the units of the derivative should be inverse degrees, which agrees with the expression given above.--Jasper Deng (talk) 17:41, 16 July 2017 (UTC)
Actually the OP's function is the arctan function ("the angle whose tangent is ..."), whose dimensions are degrees or radians. Loraof (talk) 19:17, 16 July 2017 (UTC)
Right, I got the order of functions reversed. The conversion factor you mentioned should be applied at the beginning; then it drops out as you mentioned.--Jasper Deng (talk) 19:52, 16 July 2017 (UTC)

# July 17

## Inverse Function

Please what is the inverse function of y= x×arcsinx41.58.81.54 (talk) 20:32, 17 July 2017 (UTC)

First you have to restrict x to positive numbers, or an inverse does not exist. Even then, the inverse will not be an elementary function, so the best you can get would probably be a power series expansion using the Lagrange inversion theorem.--Jasper Deng (talk) 06:14, 18 July 2017 (UTC)

# July 18

## Achi (game), a possibly solved game

Hi there. Is Achi (game) easily proven to be a solved game? It is related to the "trivially solvable" game called Three Men's Morris, so I thought it might be.

I'm not sure if this should be asked here, but I will anyway:

Should we create Category:Solved games and Category:Partially solved games and add items listed at solved game, including, of course, article solved game?

Cheers, Anna Frodesiak (talk) 19:23, 18 July 2017 (UTC)

Ah, a user has pointed out here that it appears to be solved. I couldn't see the ref that said so.

So, many of the games at solved game do not mention "solved", and probably should, right? Again, this may not be the right place to discuss this, but what the heck. Shall we boldly get all this sorted? Anna Frodesiak (talk) 19:27, 18 July 2017 (UTC)

Looking at Three Men's Morris, it seems very badly written. Someone needs to have a look at it. I may have the time later this week... -- SGBailey (talk) 15:00, 19 July 2017 (UTC)

### Wild tic-tac-toe

Is Wild tic-tac-toe solved? Anna Frodesiak (talk) 11:31, 20 July 2017 (UTC)

The article says it is. Any game with such a small number of possible states is trivial to just brute-force with a computer anyway. --Deacon Vorbis (talk) 12:17, 20 July 2017 (UTC)

# July 19

## Integrating over countable set

As a note, this is not an assigned problem, just something I was thinking about. Suppose ${\displaystyle f(x)}$ is strictly positive and continuous over an open interval ${\displaystyle S=(a,b)}$, so that ${\displaystyle \int _{S}f(x)dx>0}$. We commonly encounter integration over these uncountably infinite sets like ${\displaystyle S}$ in evaluating integrals. Note that if we create the finite set ${\displaystyle D=\{x_{k}:x_{k}\in S,k=1,...,n\}}$, for example, some ${\displaystyle n}$ integers lying in ${\displaystyle (a,b)}$, it appears that ${\displaystyle \int _{D}f(x)dx=\sum _{k=1}^{n}\int _{x_{k}}^{x_{k}}f(x)dx=0}$. So my question is, is it ever possible to have a finite, or countably infinite, subset ${\displaystyle S^{*}\subset S}$, such that ${\displaystyle \int _{S^{*}}f(x)dx>0}$? 50.101.93.214 (talk) 00:09, 19 July 2017 (UTC)

Not in the usual sense. The Lebesgue measure of any countable (including any finite) set is zero.--Jasper Deng (talk) 05:51, 19 July 2017 (UTC)
But counting measure will do the trick.
By the way, the formula for computation you suggest (the sum of integrals) does not make sense. YohanN7 (talk) 07:29, 19 July 2017 (UTC)
Not for infinite sets it wouldn't.--Jasper Deng (talk) 07:36, 19 July 2017 (UTC)
Why not? Integrals are generally allowed to assume the values ${\displaystyle \pm \infty }$. In the present case, since strictly positive functions are considered, only finite values or ${\displaystyle +\infty }$ can occur, and the integral is never ill-defined. YohanN7 (talk) 08:25, 19 July 2017 (UTC)
I (personally) usually don't tend to consider an integral to exist with infinite value, rather that it exists or doesn't.--Jasper Deng (talk) 08:39, 19 July 2017 (UTC)
I have no problem with that. But even if the set is infinite, the integral may be finite. In the present case one has
${\displaystyle \int _{S^{*}}fd\mu =\Sigma _{x\in \mathbb {R} }x\cdot \mu (f^{-1}(x)).}$
(So ${\displaystyle f}$ is actually a step function or simple function according to some definitions (e.g. Herbert Federer's Geometric measure theory) of the term. The seemingly uncountable sum needs its proper definition of course, see Federer again.) The r h s is a countable sum, with strictly postitive terms. It may well be finite. YohanN7 (talk) 09:02, 19 July 2017 (UTC)
Yes, it can indeed be finite since for some functions, ${\displaystyle S^{*}}$ can be chosen to produce an (absolutely) convergent series. But it won't work for every function and every ${\displaystyle S^{*}}$ meeting the OP's definition.--Jasper Deng (talk) 09:11, 19 July 2017 (UTC)
There's no point in arguing over conventions and terminology, but I dare say that in measure and integration theory, infinite values are allowed more often than not, both for the range of functions and the value of integrals. This is probably because it simpliefies the exposition. The only things not allowed (i.e. left undefined) are expressions of the type ${\displaystyle \infty -\infty }$. Whenever ${\displaystyle 0\cdot \infty }$ occurs, it is defined to be zero (at least, I can't think of an exception atm). (Division may also cause problems, but rarely pops up.) YohanN7 (talk) 09:17, 19 July 2017 (UTC)

# July 20

## Goldman and G-domain

I've seen definitions of both Goldman domain and G-domain, but do those two pages describe formally equivalent structures? I understand that G-domain has been used as a synonym for Goldman domain, but is this the standard use? Klbrain (talk) 21:23, 20 July 2017 (UTC)