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May 10

Finding a bijection between a permutation of n digits and the range [1,n!]

Hello,

I am given a permutation of the digits 1-9, representing an ID number. I try to find a bijection that would link each number to an integer in the range 1-9!, according to its relative order. For example:

f(123456789) = 1, f(123456798) = 2, ....,f(987654312) = 9!-1 f(987654321) = 9!

Any hints or suggestions regarding the finding of such a function?

Thanks! — Preceding unsigned comment added by 212.179.21.194 (talk) 07:44, 10 May 2016 (UTC)

I think it is slightly easier to do this for g(N)=f(N)-1. You can write

$g(N)=a_{1}8!+a_{2}7!+a_{3}6!+\dots +a_{9}$,

where $a_{1}$ is one less than the first digit of N (and it may be easier to do this with digits 0..8 instead of 1..9). To find $a_{2}$, you map the remaining digits to 1..8 and continue recursively. But I have no idea whether this is the most efficient way of going about this. —Kusma (t·c) 09:54, 10 May 2016 (UTC)

Its not particularly pretty, but you can construct the answer recursively by noting that the first 8! numbers start with 1, the next 8! with 2 etc... and applying the same logic to the remaining digits. — Preceding unsigned comment added by 128.40.61.82 (talk) 10:08, 10 May 2016 (UTC)

Thank you both for your answers, why did you choose to represents the number as a linear combination of factorial terms? 212.179.21.194 (talk) 11:11, 10 May 2016 (UTC)

It seems natural, as the first digit is 1 for 8! times, the second digit is 2 for 7! times, the third digit is 3 for 6! times, ... (and similar things with other digits). Really, as 128.X has said, you just look at how many numbers start with what digits. —Kusma (t·c) 14:48, 10 May 2016 (UTC)

It doesn't add much to what's already here but you might be interested to read the section of the article Permutation#Permutations in computing which discusses this problem. Dmcq (talk) 15:05, 10 May 2016 (UTC)

Relation vocabulary

WP:RDL#Fill in the blanks includes a question on mathematical vocabulary. (Answers would be best placed there, on the Language Desk.) -- ToE 14:14, 10 May 2016 (UTC)

Railroad tracks

I stand on a bridge over railroad tracks, directly above the symmetry axis between the two rails, and take a picture of the (totally straight) rails all the way to the horizon, while pointing the camera along the symmetry axis. The images of the rails get closer and closer the higher you look on the photo. What is the equation for the curve of the image of one of the rails? And how about in the special case where the camera is no higher than the rails? Loraof (talk) 17:03, 10 May 2016 (UTC)

It is a straight line. The articles Graphical projection and vanishing point have a few details. Sławomir Biały (talk) 17:09, 10 May 2016 (UTC)

Sławomir Biały's answer is correct for a rectilinear projection, which is what you get with a typical camera and lens (modulo some distortion). In the case of a vertical cylindrical panorama, like this one, the image is a sinusoid. (If the camera is in the plane of the rails, then it's a straight line, which in this case is a sinusoid of amplitude 0.) -- BenRG (talk) 19:02, 10 May 2016 (UTC)

Thanks. Now if I'm on a small asteroid with a very near horizon, presumably both rails seem to just end at the height of the horizon, without merging. But what if I'm on a flat planet of infinite extent? Do the two straight lines intersect at a place on the photo representing the same elevation as the camera, with neither continuing beyond there? Loraof (talk) 19:31, 10 May 2016 (UTC)

Assuming a flat planet and an ideal pinhole camera, the horizon is the intersection of the film with the plane parallel to the ground and containing the aperture (so any object at the same height as the camera will also project to the horizon). The image of a track is the intersection of the film with the plane containing the track and the aperture. The track images intersect at a point on the horizon, namely the intersection of the film with the line parallel to the tracks and containing the aperture.

On a curved asteroid there are no straight lines (unless you mean great circles on a sphere, in which case there are no parallel lines), so you'd have to be more precise about how the tracks are laid out, but most likely they will not intersect anywhere in the image. -- BenRG (talk) 20:24, 10 May 2016 (UTC)

The same effect can be seen close taken on photos from buildings. See original[1] and modified[2] picture. The modification is an stretch of the original photo from |__| to \__/. The original picture appears like /__\. The reason is the photographer on the flor and the relative distance form the camera lens to each corner of the building. Such fixing can make a fat tower appear more fat. In Jim Button and Luke the Engine Driver, Michael Ende created the fictional Mr. Tur Tur, who against physical law, appered even more tall from the distance he was seen, described as a Scheinriese in fiction. During the Cold War, the borders of nations were shown on a globe or map to make the enenmy apper smaller. Statistic sometims is being displayed linear or scaled in a logarithm to make the smaller values appear bigger and in a smaller difference. --Hans Haase (有问题吗) 12:06, 12 May 2016 (UTC)

May 11

Maximum coverage of plane by identical regular polygons...

Looking for the maximum coverage of the plane by each regular identical N-polygon. For N=3,4,6 the answer is 1.00. For 8, I think it is the Truncated square tiling where the part covered = 2sqrt(2)-2. Any ideas on 5 ,7 and other numbers? The limit as N gets large is equal to the $\eta _{h}={\frac {\pi }{2{\sqrt {3}}}}\approx 0.9069.$. Which is worst?Naraht (talk) 18:26, 11 May 2016 (UTC)

I suggest starting here and following the links backwards. (An earlier version of this post had a bunch of other links, but I think this is the best one and will answer most of your questions.) JBL (talk) 19:43, 11 May 2016 (UTC)

May 12

System of Linear Eequations

Is the following statement true? This seems to me wrong, but I can't find any counterexample...

Statement: Let $k>1$, and let a system of linear equations such that each equation has exactly k terms $y_{i}\in \{x_{i},1-x_{i}\}$ where $\{x_{i}\}$ are the variables of these equations . That is, we have a system of linear equations, each of the form: $\sum _{y_{i}\in \{x_{i},1-x_{i}\}}{y_{i}}=1$. For example, the system may consist of the equation $x_{1}+(1-x_{2})=1$ (here, $y_{1}=x_{1},y_{2}=1-x_{2}$).

This system of equations has some solution over $\mathbb {Z}$ iff it has a 0-1 solution (i.e, it has some solution over $\{0,1\}\subset \mathbb {Z}$). 213.8.204.30 (talk) 06:43, 12 May 2016 (UTC)

I bet it's true for k=2. For k=3, here is a counter-example:

a+b+c=1

(1-a)+d+(1-d)=1

(1-b)+d+(1-d)=1.

If you don't like having the same variable twice in a clause, use

a+b+c=1

(1-a)+d+(1-e)=1

(1-b)+d+(1-e)=1

d+f+g=1

e+f+g=1

--JBL (talk) 04:02, 13 May 2016 (UTC)

It's a great counterexample! Thank you!

What about the following statement: for $k=3$, if the system of equations has some solution over $\mathbb {Z}$, and it doesn't have any 0-1 solution, then there exists some variable $x_{i}$, such that $x_{i}=-1$ in all the solutions over $\mathbb {Z}$ of the system of equations? (that is, if a solution is an assignment to the variables, then in every solution, it holds that xi is assigned the value -1) 213.8.204.65 (talk) 05:07, 13 May 2016 (UTC)

That is certainly false, since I could switch c for 1-c everywhere. If you replace "-1" by "-1 or 2" then I don't have any intuition about it, unfortunately. JBL (talk) 12:59, 13 May 2016 (UTC)

Thank you! 213.8.204.65 (talk) 06:43, 14 May 2016 (UTC)

Ok, here's another example that maybe will make you give up hope ;):

a+b+w=1

(1-a)+b+x=1

a+(1-b)+y=1

(1-a)+(1-b)+z=1

JBL (talk) 18:52, 14 May 2016 (UTC)

Numerical recipes for fixed point convergence

I have a smooth function $f:\mathbb {R} ^{N}\rightarrow \mathbb {R} ^{N}$ with at least one fixed point ${\vec {x_{*}}}$ such that $f\left({\vec {x_{*}}}\right)={\vec {x_{*}}}$. It also has the desirable property that if you take ${\vec {x_{i+1}}}=f\left({\vec {x_{i}}}\right)$ then ${\vec {x_{i}}}\rightarrow {\vec {x_{*}}}$ as $i\rightarrow \infty$ for most starting guesses. In other words, it is an attractive fixed point.

My problem is that the function $f$ is very computationally expensive to evaluate, requiring about 5 minutes per iteration (using parallel processing on a 16-core workstation). Currently I need to follow the sequence 40 or 50 steps to get a value for ${\vec {x_{i+1}}}$ within the tolerance that I need. Having this process take a few hours would be "okay" except that I'd like to repeat the process many more times for other variants of $f$. Hence, I'd like to find a way to accelerate the process of finding ${\vec {x_{*}}}$ while hopefully only calling $f$ 10 or 15 times. I tried using a canned root finding system (on $\lVert f\left({\vec {x}}\right)-{\vec {x}}\rVert$) but that converged even more slowly than just following the sequence.

Does anyone have suggestions for numerical recipes to accelerate the process of finding such a fixed point? Though $f$ is smooth, it is also safe to say that I don't have any useful prior knowledge of its derivatives, so gradient type methods will have to rely on whatever information can be gained from evaluating $f$. Dragons flight (talk) 15:01, 12 May 2016 (UTC)

Can you get anything by looking at first differences ${\vec {x_{i+1}}}-x_{i}$ (or second differences or third...)? Robinh (talk) 19:48, 12 May 2016 (UTC)

As the function is slow to compute, approximate it by some other function which is fast to compute. Bo Jacoby (talk) 08:40, 13 May 2016 (UTC).

Have you tried something from the article about series acceleration? 213.8.204.65 (talk) 09:26, 13 May 2016 (UTC)

I've since tried implementing Broyden's method. However, I'm not sure if I have done it correctly as it doesn't seem to be improving the convergence right now. Dragons flight (talk) 12:42, 13 May 2016 (UTC)

I think it depends on how big your N is. If you're talking about R⁵ then it certainly should be possible to accelerate convergence since after 10 iterations you should be able to estimate the Jacobian pretty accurately. If it's R⁵⁰ then even if f was linear you wouldn't know the Jacobian until you're already done, but it might be possible to get some useful information about it before then. If R⁵⁰⁰ then the Jacobian is basically unknown and I think acceleration techniques pretty much rely on it. It also depends on how rapidly the Jacobian changes, otherwise no estimate is going to be useful. Another question is how fast do the iterates converge now; if convergence is already quadratic then you're already doing the best you can with linear methods. One additional thought though, since you're doing this with multiple f's, if it's possible to gauge how similar they are then you might try sequencing them so that similar f's are together, then use the ending point of each f as the starting point for the next one. --RDBury (talk) 14:33, 13 May 2016 (UTC)

N is roughly 3,000. I haven't systematically tried to characterize the speed of convergence, but it doesn't feel very fast. I often observe values move 1 unit at step n, and then 0.8 to 0.9 units in the same direction at step n+1, which would suggest a fairly slow rate of convergence. As you suggest, I am hopeful that the endpoint of similar f's can be useful for acceleration, but I haven't tried that yet. Dragons flight (talk) 15:18, 13 May 2016 (UTC)

Infinite product

Let $f(a)=\prod _{k=2}^{\infty }{\frac {k^{a}-1}{k^{a}+1}},a\geq 1$. Does $f(a)$ have a closed form? What is $f'(a)$? 24.255.17.182 (talk) 23:22, 12 May 2016 (UTC)

$\prod _{k=2}^{\infty }{\frac {k^{a}-1}{k^{a}+1}}$

$=\exp \left(\sum _{k=2}^{\infty }\log({\frac {1-k^{-a}}{1+k^{-a}}})\right)$

$=\exp \left(\sum _{n=1}^{\infty }n^{-1}\sum _{k=2}^{\infty }\left((-k^{-a})^{n}-(k^{-a})^{n}\right)\right)$

$=\exp \left(-2\sum _{n=0}^{\infty }(2n+1)^{-1}\sum _{k=2}^{\infty }k^{-a(2n+1)}\right)$

$=\exp \left(-2\sum _{n=0}^{\infty }(2n+1)^{-1}(\zeta (2an+a)-1)\right)$

Bo Jacoby (talk) 09:19, 13 May 2016 (UTC).

Some limits, $f(1)=0;f(3)=2/3;f(\infty )=1;f'(1)=0;f'(\infty )=0$ Dragons flight (talk) 13:39, 13 May 2016 (UTC)

May 13

About Mazarae and de:Mazarä

Hello Wikipedia:Reference desk/Mathematics folks,
Article WP:PRODed and then un-PROD-ed
I realio, trulio have only the faintest clue what this article about.
Your thoughts about this?
Pete "Math class is tough - let's go shopping!" AU aka --Shirt58 (talk) 12:24, 13 May 2016 (UTC)

I also have only the faintest clue what the article about. That is not because the topic is difficult; it's the kind of topic I enjoy. It's because the article is very poorly written, probably by some who already knows what a mazarae is and can't appreciate that most readers don't know. Maproom (talk) 22:08, 15 May 2016 (UTC)

I agree with Maproom. It seems to me like Wikipedia_talk:WikiProject_Mathematics might be a better venue, though. --JBL (talk) 01:12, 16 May 2016 (UTC)

I have made the article shorter, and, I hope, more comprehensible. What it was trying to say was really quite simple. Maproom (talk) 14:30, 16 May 2016 (UTC)

predicting grade distributions?

Hey,

I was wondering how accurately one could predict the grade distributions for a class of n students if you know what the lower and upper quartiles and the median are after 60% of the points have been awarded. (The last 40% being a final which is assumed to be of equal difficulty as the other exams).

Thanks — Preceding unsigned comment added by 140.233.174.42 (talk) 18:30, 13 May 2016 (UTC)

See https://www.academia.edu/25108263/Statistical_Induction_and_Prediction Bo Jacoby (talk) 22:02, 13 May 2016 (UTC).

Thanks Bo, but I honestly don't understand most of that. Im sorry — Preceding unsigned comment added by 140.233.174.42 (talk) 12:42, 14 May 2016 (UTC)

May 14

Neighborly polytopes

Let P be the Cartesian product of two simplices (not necessarily the same dimension) and let D be the dual of P. When is D neighborly? For example the dual of the product of two triangles is neighborly, but the dual of the product of a line segment and a tetrahedron is not. --RDBury (talk) 05:46, 14 May 2016 (UTC)

When does a "multiple comparisons problems" occur (statistical inference / t-tests)

Hello!

I am wondering whether I am facing a Multiple comparisons problem in my analysis. I set up a "global" (compound/intersection) hypothesis for 2 independent groups, each testet with a t-test. I want to accept the global hypothesis in case both (indepedent) groups are improving on one variable (outcome).

$H_{AB}=H_{A}\cap H_{B}:\mu _{A}>0\quad {\text{and}}\quad \mu _{B}>0$

So the way I see this: both single t-test are highly significant. To accept the global hypothesis I don't have to add anything up. Can I infer that the intersection hypothesis is accepted since each of the two independent tests are significant?

Thanks for advice, --WissensDürster (talk) 08:40, 14 May 2016 (UTC)

I don't think you can infer that. See Simpson paradox. Loraof (talk) 14:21, 14 May 2016 (UTC)

Stupid complex number question

Hi,

I apologize in advance for this probably being a stupid question:

I want to convert (1-i) into polar form. Now, I believe the magnitude of this number is sqrt(1^2 - i^2) = 0??? How is this possible? Also, I get from trig that the angle would be arctan(-i)? What's going on?? — Preceding unsigned comment added by 140.233.174.42 (talk) 14:25, 14 May 2016 (UTC)

Not a stupid question! The point (1-i) in the complex plane is one unit to the right (in the real direction) and one unit down (in the imaginary direction).So its angle is minus 45°, or equivalently 360° - 45° = 315°. Its magnitude is found as the square root of the sum of squares of the coefficients (of 1 and -1, not of 1 and i). So the magnitude is ${\sqrt {1^{2}+(-1)^{2}}}={\sqrt {1+1}}={\sqrt {2}}.$ Loraof (talk) 14:36, 14 May 2016 (UTC)

And the angle can be written as arctan(-1/1) = arctan(-1). Loraof (talk) 14:41, 14 May 2016 (UTC)

oh! thank you! — Preceding unsigned comment added by 140.233.174.42 (talk) 14:45, 14 May 2016 (UTC)

Are complex numbers in polar coords really meaningful ? While it's useful for regular old X and Y coords, to tell us the distance and direction from the starting point, using it for complex numbers seems like saying 1 apple and 1 orange equals 1.41 fruit at a 45 degree angle. StuRat (talk) 14:50, 14 May 2016 (UTC)

They most certainly are, which is most obvious when you multiply them: multiplying a complex number by (1+i) multiplies its magnitude by the square root of 2, and adds 45 degrees to the angle it makes with the x-axis. Double sharp (talk) 14:57, 14 May 2016 (UTC)

There is also no denying that it is useful, as can be seen if you try to work out (1+i)¹⁷ both ways. Double sharp (talk) 15:00, 14 May 2016 (UTC)

So is it just useful as a calculation method, or do complex numbers in polar coords relate to anything in the real world ? If so, what does the angle and magnitude represent ? StuRat (talk) 15:02, 14 May 2016 (UTC)

you're getting philosophical, which will be dismissed by mathematicians..the idea being that all math is just a calculating method, so to speak..and none of it directly relates to the real world (though it can be used for applications in the real world)...68.48.241.158 (talk) 16:19, 14 May 2016 (UTC)

Very much yes! Putting a complex number in its modulus-argument form is frequently very useful, and is no or less meaningful than representing it in its real and complex components. For an example, look at complex number § electromagnetism and electrical engineering, where the argument is often used to represent phase. —  crh ₂₃  (Talk) 16:38, 14 May 2016 (UTC)

In linear difference equations, the characteristic roots may be complex. If so, the magnitude of the largest pair determines whether the real-world quantity being modeled converges to a steady state (magnitude less than 1) or diverges (magnitude greater than 1). The departure of the magnitude from 1 determines how fast the convergence or divergence is. The additive contribution of this pair of complex conjugate roots to the solution for the dynamic variable is $2m^{t}\cdot \cos(\theta t+\delta )$ where m is the magnitude of the roots, t is time, $\theta$ is the angle of one of the complex numbers, and $\delta$ depends on the real and imaginary parts of the complex numbers scaled by their magnitude. Something similar happens with linear differential equations. So the magnitude and angle appear in a solution equation that contains only real elements. Loraof (talk) 16:54, 14 May 2016 (UTC)

Thanks all, for providing examples. StuRat (talk) 22:42, 14 May 2016 (UTC)

A formalism rigorous

When is a formalism rigorous? Is a rigorous formalism just another word for exhaustive or explicit? --Llaanngg (talk) 20:51, 14 May 2016 (UTC)

Presumably this question comes with context. You will get better answers if you provide the context than if you obscure it. --JBL (talk) 22:27, 14 May 2016 (UTC)

It depends with time. I'm sure the Egyptians and Babylonians and Greeks thought what they did was rigorous. However new axioms have had to be added to even Euclids Geometry to fix things missed out in it. Nowadays a formal system would be one that can be checked by a proof checker on a computer. In the future, and I for one welcome our new AI overlords, ;-) artificial intelligences will presumably think our idea of rigor is naive and primitive. Dmcq (talk) 22:52, 14 May 2016 (UTC)

For example: "The objective of this work is to present a rigorous formalism for the solution of engineering problems on vibrations in which the vibrating structure has a discrete distribution of loads." But there are plenty of mathematical or technical texts setting their objective to outline/present/develop a "rigorous formalism" for some problem/field/issue. What would be the difference if their formalism were not rigorous?--Llaanngg (talk) 23:07, 14 May 2016 (UTC)

From formalism (philosophy of mathematics):

Formalism is associated with rigorous method. In common use, a formalism means the out-turn of the effort towards formalisation of a given limited area.

Assuming this is the concept of formalism that is relevant to your sources, it seems to me that this quote implies that "rigorous formalism" is a redundancy meaning "formalism". Loraof (talk) 23:55, 14 May 2016 (UTC)

I think your question is more about the language being used as opposed to what "formalism" is/means...I think the "rigorous" word is just tagged on here to emphasize the formal nature of a formalism...68.48.241.158 (talk) 14:18, 15 May 2016 (UTC)

Couldn't it be that "rigorous" means from basic principles, explicitly describing each step? Llaanngg (talk) 17:52, 15 May 2016 (UTC)

I think that's more along the lines of what "formal" means here...the "rigorous" just meaning here that it's carefully formalized..68.48.241.158 (talk) 17:56, 15 May 2016 (UTC)

May 15

infinity/infinity

If Infinity is divided by infinity, is the answer 1? Or can the answer be anything you want depending upon how the infiniteies were derived?--178.106.99.31 (talk) 17:54, 15 May 2016 (UTC)

Infinity is not a number, so dividing infinity by infinity is undefinable. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:59, 15 May 2016 (UTC)

No, that's not an answer. Infinity can be a number in some contexts — see for example Riemann sphere, aka the "extended complex numbers". But even in the Riemann sphere, you can't divide infinity by infinity, though you can add it to anything except infinity, and multiply it by anything except zero. --Trovatore (talk) 20:12, 15 May 2016 (UTC)

some infinite sets can be placed in one to one correspondence with other infinite sets, so they can be divided out in a sense along these lines (ie..one to one) but other infinite sets are strictly larger than others (real numbers vs integers) so cannot be divided out one to one)...68.48.241.158 (talk) 18:03, 15 May 2016 (UTC)

I think a case can be made that the Dirac delta function is one special case. I believe also that 0/0 is the same as infinity/infinity and it's apparent that x^2/x is 0 for x=0. These relate to the series expansion above. --DHeyward (talk) 18:25, 15 May 2016 (UTC)

0 divided by 0 is also undefined. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:28, 15 May 2016 (UTC)

But for continuous functions, it's important to point out that it's not indeterminate just because it evaluates numerically to 0/0. x^2/x is not indeterminant undefined for any value of x even though straight substitution is 0/0. The answer is "0". it's trivial as to why but is the starting point for evaluating series and functions that converge to 0 or infinity at different rates. --DHeyward (talk) 21:37, 15 May 2016 (UTC)

Literally speaking, you cannot evaluate x²/x at x=0. You can take the limit as x approaches 0, and that limit is 0. But that is not the same thing. --Trovatore (talk) 21:45, 15 May 2016 (UTC)

It trivially reduces to f(x)=x^/x=x . It seems odd that f(0)=x^2/x is only 0 in the limit but f(0)=x=0 is not. That seems to defy laws of equivalence/identity. --DHeyward (talk) 22:29, 15 May 2016 (UTC)

It doesn't reduce to x at x=0. The identity ax/bx=a/b assumes x≠0. --Trovatore (talk) 22:38, 15 May 2016 (UTC)

More precisely it can be evaluated in the limit and shown to be 0 and limits are tools for resolving it. That the function f(0)=x/x=1 and f(0)=x^2/x=0 shows that evaluating to 0/0 is indeterminant but not undetermined for any function. They are substitutional identities. The identity ax/bx=a/b does not have any discontinuities and is a/b when x=0. --DHeyward (talk) 22:49, 15 May 2016 (UTC)

No, sorry, that is incorrect. The identity does not hold when x=0. --Trovatore (talk) 22:55, 15 May 2016 (UTC)

I should say, there are things you could correctly mean by what you say. For example, it's true if you mean it to be interpreted in the ring of rational functions over x. In that case, you're not dividing the number ax by the number bx, but rather the function λx.ax by the function λx.bx. (Does anyone know what the html is for $\mapsto$?) But this is not the usual interpretation of the claim; if you want to be correct you have to make that clear.) --Trovatore (talk) 23:11, 15 May 2016 (UTC)

In the cases I provided, they demonstrate "indeterminant" but the different functions are differentiable. It's my understanding that in those cases, the solution is exact and the functions are continuous without exception for x=0. L'Hospital's rule seems to apply. Am I missing something? --DHeyward (talk) 08:08, 16 May 2016 (UTC)

Yes. You're missing what x²/x, evaluated at x=0, actually means. What it means is, you take 0 and square it. Then you take 0. Then you divide the first by the second; that is, 0/0. Not the "form" 0/0, but literally zero divided by zero, which is undefined.

By default, that is what it means. Period, end of discussion. There is no opportunity to apply L'Hospital's rule; it is completely irrelevant. --Trovatore (talk) 08:12, 16 May 2016 (UTC)

But i would not say 0/0 is undefined, rather it is indeterminate. It can have multiple solutions depending on function ans space, including undefined but it can also have very defined solutions such as 1,0,infinty, etc. x²/x doesn't have a discontinuity at x=0. It is 0. When faced with indeterminant substitutions, there are first principle derivations that lead to a solution. Plugging in numbers demonstrates "indeterminant" but it is not the same as "undetermined" or "undefined." Continuous functions are continuous even through values that substitute into an indeterminant form. f(x)=x^2/x is continuous with solutions throughout real values of x including 0. --DHeyward (talk)

See Indeterminate form. --Kinu ^t/_c 19:34, 15 May 2016 (UTC)

Indeterminate forms are quite common with +-infinity. With only real numbers (i.e. no infinities) there are only 4 indeterminate forms; 0/0, 0 to the 0, the zeroth root of 1, and the logarithm of 1 in base 1. Georgia guy (talk) 20:41, 15 May 2016 (UTC)

Your latter two are not actually traditional "indeterminate forms". There's a temptation to want to come up with an abstract notion of "indeterminate form" and figure out what else it includes besides the traditional ones, but personally I think it's a waste of time. "Indeterminate form" is a historical notion, pretty much superseded now that we have rigorous notions of continuity and so on. It's still useful for teaching calculus, because it serves as a sort of warning system for where common expressions are discontinouous, but I don't believe it's useful to extend it.

So there are exactly seven indeterminate forms: 0/0, 0⁰, ∞−∞, ∞/∞, 0×∞, 1^∞, ∞⁰. Its a closed-ended list and will never be extended, not because there aren't other expressions that are arguably similar, but because there's just not much point. --Trovatore (talk) 20:59, 15 May 2016 (UTC)

Actually, I looked up the dates, and it appears as though the notion of continuous function actually predates the notion of "indeterminate form", so I have to backpedal a little bit on the historical sequence I suggested above. I still think the basic thrust of what I said is correct, though — once you understand continuity, you don't really need "indeterminate forms", but they're still useful as a sort of reminder of things to watch out for, but not useful to extend them. --Trovatore (talk) 21:43, 15 May 2016 (UTC)

Sort of playing the devil's advocate here, but the "indeterminate forms" are precisely those indeterminate operations that can be constructed by the usual concepts of addition, multiplication, and division – with exponentiation thrown in for good measure, although I think there are probably some reasons for regarding that as less fundamental. The often cringeworthy "L'Hopital's rule" tells us that one way of making sense of expressions like $\infty /\infty$ is to change what one means by ∞ (or, for the other forms, 0). It is not the "number" infinity that one is dividing, but rather an order of growth. Thus, I believe, to make sense of limits such as these, one is in effect thinking of the "number" $\infty$ as a point of the Stone-Cech compactification of the real line. (Maybe the "Hewitt compactum"?) Sławomir Biały (talk) 21:49, 15 May 2016 (UTC)

Well, in this context, though, I don't see any good reason to exclude the complex numbers. So if you want to generalize, you probably ought to include e^i∞, or equivalently sin(∞), or i^∞. But my point is that I don't think it's a good idea to generalize. The notion is probably just barely worth learning in calculus, but to take it beyond that is to focus on the wrong things. --Trovatore (talk) 22:07, 15 May 2016 (UTC)

Oh, to be clear, I'm not saying there might not be interesting math to do along these lines. But I would avoid calling it "indeterminate forms", which is a name for a specific pedagogical aid that people need to first understand, then understand the limitations of. --Trovatore (talk) 22:27, 15 May 2016 (UTC)

This is a perennial question, and I think it is worth attempting to give a real answer. I am not trying to generalize it to complex numbers because, in most cases when someone is tempted to write ∞/∞, they are certain that they mean something, and that something usually does not involve complex numbers. For example, when a calculus student writes $\infty /\infty =1$, they need to be told gently that $\infty$ is a concept that does not behave as other numerical quantities do. The "two infinities" are not "the same infinity". Instead, the relevant concept of infinity is really that of an order of growth. For instance, although it is true that the polynomials $n+1$ and $n^{100}+1$ both tend to infinity, they do not go at the same rate, and therefore they occupy "different infinities". This is, of course, merely an intuitive explanation. Part of our task as post-modern mathematicians is to assign meaning to this sort of sensical nonsense. What is the sense in which we really mean that there are different infinities? What is the space of all infinities? Does it have a topology? What algebraic structures does it support? Etc. Also, I disagree about indeterminate forms as merely a pedagogical crutch. And in any case it is not very relevant to the original question whether indeterminate forms as such are a complete list or no.

I guess this is related to the ultrafilter construction of the hyperreals. Sławomir Biały (talk) 23:17, 15 May 2016 (UTC)

I take issue with Trovatore. Infinity can be added to infinity and the answer is infinity. This video explains how [3]. 80.44.167.65 (talk) 14:34, 16 May 2016 (UTC)