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April 2
Angle condition for equilateral polygons
By equiangular polygon and its source 5, there is a polynomial condition for a sequence of side lengths to be consistent with an equiangular polygon: namely that be divisible by Is there an analogous condition for a sequence of angles to be consistent with an equilateral polygon? Loraof (talk) 17:06, 2 April 2016 (UTC)
- Take n=5; it's easy to generalize from this. If the angles are α1, α2, α3, α4, α5 then first α1+α2+α3+α4+α5 = 3π by properties of pentagons in general. Eliminating α5, there are two other conditions: 1 - cosα1+cos(α1+α2)-cos(α1+α2+α3)+cos(α1+α2+α3+α4)=0 and sinα1-sin(α1+α2)+sin(α1+α2+α3)-sin(α1+α2+α3+α4)=0. To prove this, let β1, β2, β3, β4, β5 be the corresponding exterior angles. Take the first side to be from 0 to 1 in the complex plane; from which the remaining vertices must be 1+eiβ1, 1+eiβ1+ei(β1+β2), etc. The 5th vertex will be at the start again iff 1+eiβ1+ei(β1+β2)+ei(β1+β2+β3)+ei(β1+β2+β3+β4)=0. Note, this assumes that there isn't any self intersection or similar weirdness going on, otherwise the first condition isn't always true. --RDBury (talk) 23:06, 2 April 2016 (UTC)
Thanks! Loraof (talk) 16:22, 3 April 2016 (UTC) Is there by any chance a source I could use in order to put this into the equilateral polygon article? Loraof (talk) 16:24, 3 April 2016 (UTC)
April 3
Inhomogeneous heat equation on a finite interval
Could I get an example of solving the 1D heat equation ? The article has an example over the entire real axis (Heat_equation#Inhomogeneous_heat_equation)- how can I apply Green's function over a finite interval (I'm guessing not just changing the limits of integration from to )? 24.255.17.182 (talk) 20:56, 3 April 2016 (UTC)
- If there are no boundary conditions on and , then you can get a solution of this equation just by extending the forcing term by zero outside of (so, effectively, yes, changing some limits of integration). I'm not sure how to get general boundary conditions. Some special cases can be gotten using Jacobi theta functions (for zero boundary conditions) or the method of images. Sławomir
Biały 00:22, 4 April 2016 (UTC)
April 4
How do i calculate this special permutation?
I have a permutation that regards two sets to choose from. This will act almost like a combination (order doesnt matter), but in a restricted and unexpected sense. Lets say the sets are as follows:
Set A = {A1, A2, A3, A4) Set B = (B1, B2, B3)
The GENERAL ORDER does matter, but which element is chosen from the same set does not. For instance:
A1, B1, A2, B2, A3, B3, A4 is NOT the same as
A1, B1, A2, B2, A3, A4, B3
But.... A1, B1, A2, B2, A3, B3, A4 IS the same as
A4, B2, A2, B1, A3, B3, A1
.... This is because what matters is the group that is in each position, not the specific element. In other words, you can simplify the permutations to read like this: A B A B A B A, and in cases like that, the elements are indistinguishable. The order still matters to some degree, so this isn't a combinations situation, but how do i eliminate repeats that are counted if i calculate nPr for this special case? It seems hard to talk about when two elements in A are swapped or etc.
.... should i take the permutations of the union of sets a and b, then divide by the permutations of a, and also by permutations of b? This seems reasonable but i am unsure.
Thanks in advance for any assistance! 216.173.144.188 (talk) 18:06, 4 April 2016 (UTC)
- The 4 A's are put in 7 positions in ways. See combination. Bo Jacoby (talk) 18:20, 4 April 2016 (UTC).
OH! So this IS a combination, but of the "places in the list". In other words, we are taking "7 positions in an ordered list, choose 4", where it doesn't matter what position is picked first, second, etc. What threw me off is i was thinking of the binomial coefficient as choosing ITEMS, not positions. Is this a correct way to see things?
Thanks! 216.173.144.188 (talk) 18:41, 4 April 2016 (UTC)
- To answer your last question, yes: to count combinations, you count the permutations of the sum and divide by the permutations of the subsets; thus for your problem , because the 4! permutations of A and the 3! permutations of B don't matter. —Tamfang (talk) 08:50, 5 April 2016 (UTC)
Right, i was thinking along those lines at one point Tamfang, but i was unsure of myself. Thanks everyone! :D
216.173.144.188 (talk) 13:05, 5 April 2016 (UTC)
April 5
Vector proof
Given a Quadrilateral and are midpoints of respectively,
Prove: . יהודה שמחה ולדמן (talk) 09:44, 5 April 2016 (UTC)
- Please see our note at the top of the page regarding homework. What have you done so far and where are you stuck?
- To start you out:
- Have you drawn a diagram of the problem?
- Do you know and understand why, given the points , then ?
- How many ways can you represent in terms of vectors not utilizing the points ?
- Can you think of a way to combine those equations so that you can solve for in the desired terms?
- Our relevant article is Vector addition. -- ToE 14:07, 5 April 2016 (UTC)
April 6
Sums using multiple variables: A question about sigma notation
Hello, I have a question about sigma notation. One could write the following, having the understood meaning shown afterward in an equality:
However, what if we have two variables, i and j, with i going UP from 1 to 3 and j going down from 4 to 2, simultaneously? (NOT a double sigma where all possibilities of i,j are summed!) I want to express the following sum:
I am tempted to write as follows, or similar: OR etc...
... but this seems silly! How can i PROPERLY express what i am trying to, using sigma notation? (Or a better way, if what im trying to do as short hand isnt appropriate)
216.173.144.188 (talk) 04:41, 6 April 2016 (UTC)
- One option is . -- ToE 06:04, 6 April 2016 (UTC)
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- Yes, I think that's the way it would usually be done. If both i and j were used a lot of times in the expression, I suppose you could write something like
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- and it would be understood. --69.159.61.172 (talk) 06:47, 6 April 2016 (UTC)
- Yes, I think that's the way it would usually be done. If both i and j were used a lot of times in the expression, I suppose you could write something like
Okay, thank you, this makes sense! However, it is a simplification based on what i guess now i see as a poor substitute example. Let me ask about my actual situation and see what the correct way is to write this concisely as a mathematician. We have the following:
Given , Consider ...
Does this make sense? Is there any reason why the index of a sum cant decrease instead of increase?
(It took me a while to realize i could express i in terms of n and j together, which helped in this)
216.173.144.188 (talk) 13:20, 6 April 2016 (UTC)
- In the sigma notation the index can't decrease because Summation#Formal_definition says that , for b < a. The fix is simple though, just write . Egnau (talk) 16:02, 6 April 2016 (UTC)
Indeed,
Thank you all very much for your input. I consider this matter closed. 216.173.144.188 (talk) 16:35, 6 April 2016 (UTC)
- The Iverson bracket is useful.
- Bo Jacoby (talk) 22:31, 6 April 2016 (UTC).
What are the odds that a person will die on his birthday?
What are the odds that a person will die on his birthday? Is it 1 out of 365? Or is it more complicated than that? I was surprised to see that this happened with Merle Haggard today. Thanks. Joseph A. Spadaro (talk) 19:49, 6 April 2016 (UTC)
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- yes, would have to be, if looking at whole population...or whatever it end up being if include the extra leap days....UNLESS it can be determined that more people tend to be born in the spring or summer and more people tend to die in the fall or winter.....68.48.241.158 (talk) 20:26, 6 April 2016 (UTC)
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- But once you have a birthday, you have a birthday. That's that. So, it's simply a matter of "matching" the death day. No? Joseph A. Spadaro (talk) 20:28, 6 April 2016 (UTC)
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- Close to 365, unless they were born on leap day. I suppose a surprise and big birthday meal might put the death rate ever so slightly higher that day. StuRat (talk) 20:27, 6 April 2016 (UTC)
if 65% of population is born in spring or summer and 65% of population dies in fall or winter, then looking at the whole population...a human being has a smaller chance than one 1/365(or so) of dying on her birthday... 68.48.241.158 (talk) 20:34, 6 April 2016 (UTC)
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- I am asking about a specific person, not just any generic person. And a specific person, once born, already has a set birth date (whatever that date might be). So, the question is not so much: "What are the odds that any one of us human beings will die on our birthday?" The question is more like: "What are the odds that Brad Pitt will die on his birthday?" Which is another way of saying: "What are the odds that Brad Pitt will die on December 18?". Joseph A. Spadaro (talk) 21:14, 6 April 2016 (UTC)
- if asking specific person than depends on individual factors..so unanswerable...person might be dying of cancer, only have 6 weeks to live...birthday 6 months away.....?????68.48.241.158 (talk) 21:21, 6 April 2016 (UTC)
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- but if brad pitt is expected to live many, many more years (and human being are more likely to die in fall or winter) then he has a smaller chance of dying on his birthday....68.48.241.158 (talk) 21:23, 6 April 2016 (UTC)
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- if asking specific person than depends on individual factors..so unanswerable...person might be dying of cancer, only have 6 weeks to live...birthday 6 months away.....?????68.48.241.158 (talk) 21:21, 6 April 2016 (UTC)
- I am asking about a specific person, not just any generic person. And a specific person, once born, already has a set birth date (whatever that date might be). So, the question is not so much: "What are the odds that any one of us human beings will die on our birthday?" The question is more like: "What are the odds that Brad Pitt will die on his birthday?" Which is another way of saying: "What are the odds that Brad Pitt will die on December 18?". Joseph A. Spadaro (talk) 21:14, 6 April 2016 (UTC)
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- Actually, is there a mistake? December 18 is still technically fall, I believe (not winter). I think winter starts on December 21 or so. No? Joseph A. Spadaro (talk) 22:29, 6 April 2016 (UTC)
- There are many definitions of when seasons start. Don't be fooled by news people (especially in North America) who like to talk about "the official first day of winter". Besides, everyone knows the official first day of winter is in October. --69.159.61.172 (talk) 23:07, 6 April 2016 (UTC)
- Summer starts on December 1st in almost all of the temperate southern hemisphere, though I expect someone will be able to find an odd exception. Dbfirs 07:46, 7 April 2016 (UTC)
- There are many definitions of when seasons start. Don't be fooled by news people (especially in North America) who like to talk about "the official first day of winter". Besides, everyone knows the official first day of winter is in October. --69.159.61.172 (talk) 23:07, 6 April 2016 (UTC)
- Actually, is there a mistake? December 18 is still technically fall, I believe (not winter). I think winter starts on December 21 or so. No? Joseph A. Spadaro (talk) 22:29, 6 April 2016 (UTC)
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- Agreed. There are slightly more deaths in winter where I live and also where you live, but that will not be the case in some other parts of the world. Dbfirs 21:50, 6 April 2016 (UTC)
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- That would be true if seasonal variation were as great in humans as it is in some other animals. The answer will be approximately "364:1 against" (the question asked for odds, not probability), but with slight variations as mentioned above. Dbfirs 21:06, 6 April 2016 (UTC)
- certainly there's some variation so determining the actual odds or probability (is there a difference in this context?) would in practice be complicated...68.48.241.158 (talk) 21:14, 6 April 2016 (UTC)
- Yes, I wasn't disagreeing, just commenting that the effect would be small. It would also vary by region and culture. I agree that the fine detail would be complicated to calculate. In the future, it might be possible to predict, from an individual's DNA, address, and lifestyle, the most likely time of year of death, but the probability is unlikely to change significantly from 1/365 (the small correction for leap years possibly cancels out the small birthday surprise effect or birthday parachute jump). The answer for Brad Pitt will be very close to the answer for any other individual. Dbfirs 21:26, 6 April 2016 (UTC)
- certainly there's some variation so determining the actual odds or probability (is there a difference in this context?) would in practice be complicated...68.48.241.158 (talk) 21:14, 6 April 2016 (UTC)
- (Not to be morbid, but it's a morbid question anyway) Infant mortality may dominate other effects. 11000 / 4000000 ≈ 1/350 of all babies born in the U.S. die within 24 hours of birth. Probably more than 1/700 die on the same date (more births are in the morning, and probably more deaths are earlier in the 24-hour period too). That's quite large compared to 1/365. -- BenRG (talk) 04:55, 7 April 2016 (UTC)
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- Good point. We should probably restate the problem as "What is the probability of dying on your birthday, but not your birth date". StuRat (talk) 05:17, 7 April 2016 (UTC)
- In that case, mortality is highest on the day after the birthday, and still high the days after that, from infant deaths with a small delay after birth. Then, deaths on the birthday itself are probably even lower than average since it is the last date to appear after we start counting, and therefore least affected by infant mortality. Unless some other intrinsic birthday effect offsets this, of course. If all effects related to infant mortality are to be excluded from the question, one should exclude all data below age 5 or so. Gap9551 (talk) 16:18, 8 April 2016 (UTC)
- Good point. We should probably restate the problem as "What is the probability of dying on your birthday, but not your birth date". StuRat (talk) 05:17, 7 April 2016 (UTC)
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- Yes, I apologise to all new-born babies for not considering them as "person"s. Dbfirs 07:56, 7 April 2016 (UTC)
- Check out "Variation of mortality rate during the individual annual cycle", which investigates all births/deaths in the city of Kiev over the period 1990-2000. Kiev men showed a 44.4% excess in deaths on their birthday compared to the expected value (women slightly less at 36.2% excess), and it looks like other studies have found similar results. So working out the exact odds would take a lot of actuary work, but it looks like it's something more like 1:300 than 1:365. Smurrayinchester 08:46, 7 April 2016 (UTC)
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- I've started a page Birthday effect which collects the studies. Please feel free to improve it. Smurrayinchester 11:53, 7 April 2016 (UTC)
- Thank you for that excellent summary of the research. The effect is much bigger than I expected. Dbfirs 14:14, 7 April 2016 (UTC)
- I've started a page Birthday effect which collects the studies. Please feel free to improve it. Smurrayinchester 11:53, 7 April 2016 (UTC)
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- Very interesting. I will have to read that new article. Thanks. Joseph A. Spadaro (talk) 18:04, 7 April 2016 (UTC)
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I downloaded the first third of the Social Security Death Index (2.9 GB). From the 28,607,398 recorded birth / death pairs, I found 45,707 individuals who were recorded as dying on their birthday (0.1597%) giving a chance of approximately 1 in 625 of dying on your birthday (well below the expected 1 in 365). It should be noted that the SSDI only records deaths of individuals who were issued US social security numbers. As SSNs are not issued to individuals who die prior to the birth being officially registered, none of the records include infants who died on the day they were born. Dragons flight (talk) 18:50, 7 April 2016 (UTC)
As for seasonal differences:
January | February | March | April | May | June | July | August | September | October | November | December | |
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Births | 2420302 | 2269663 | 2525845 | 2330376 | 2386245 | 2296193 | 2442604 | 2487487 | 2457619 | 2387806 | 2226876 | 2376382 |
Deaths | 2708214 | 2407497 | 2539914 | 2348925 | 2331869 | 2217327 | 2290327 | 2248835 | 2197259 | 2391209 | 2362526 | 2563496 |
Dragons flight (talk) 18:59, 7 April 2016 (UTC)
Same dataset has 2,393,596 individuals (8.37%) dying in the same month that they were born, or a chance of 1 in 11.952. Dragons flight (talk) 19:10, 7 April 2016 (UTC)
- Why would there be such a disparity between what we would expect (statistically) and the actual results? That does not seem to make sense. I am referring to this comment: I found 45,707 individuals who were recorded as dying on their birthday (0.1597%) giving a chance of approximately 1 in 625 of dying on your birthday (well below the expected 1 in 365). Thanks. Joseph A. Spadaro (talk) 21:00, 7 April 2016 (UTC)
- What is the distribution of death dates? E.g., how many days (out of 365) have frequency less than 0.16% ? --JBL (talk) 21:09, 7 April 2016 (UTC)
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- Hhmmm, I've now taken a closer look at the data and it appears that about half of the SSDI death dates are recorded without a real day of death, e.g. March 0, 1979. Some of the birth dates also are also missing detail (but a much smaller fraction than the death dates). If I exclude all of the birth / death pairs that don't involve fully-formed dates, I am left with 14,879,058 individuals. Of these 44,665 were recorded as dying on their birthday (0.3002%; a chance of 1 in 330). That is now slightly more than the expected 1 in 365, though much closer than the initial estimate. Dragons flight (talk) 07:19, 8 April 2016 (UTC)
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- Looking a bit further:
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Birthday + X days Chance of dying (according to SSDI records) -40 days 1 in 364 -30 days 1 in 366 -20 days 1 in 366 -10 days 1 in 366 -5 days 1 in 366 -4 days 1 in 366 -3 days 1 in 363 -2 days 1 in 370 -1 days 1 in 365 Birthday 1 in 330 +1 days 1 in 361 +2 days 1 in 361 +3 days 1 in 363 +4 days 1 in 364 +5 days 1 in 365 +10 days 1 in 363 +20 days 1 in 363 +30 days 1 in 365 +40 days 1 in 366
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- Dragons flight (talk) 08:08, 8 April 2016 (UTC)
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- that seems more in line with common sense...the previous estimate seemed weird..68.48.241.158 (talk) 11:54, 8 April 2016 (UTC)
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- confused about the chart though...saying only 1/364 chance dying the time frame of 40 days before birthday?? that can't be right...??68.48.241.158 (talk) 16:33, 8 April 2016 (UTC)
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- It's the chance of dying exactly that many days before or after your birthday. Dragons flight (talk) 17:00, 8 April 2016 (UTC)
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- That chart seems strange. All of the time intervals are listed with odds of very close to 1 in 365. They are all 362, 363, 364, 366, etc. Why would the exact birthday be the only one with a "weird" amount listed for odds as 330? That (the birthday) is the only one on that chart that really deviates from the expected 365 value. Why? Joseph A. Spadaro (talk) 18:25, 8 April 2016 (UTC)
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- Or does that strange value (330) include babies that die on the exact day they are born? But even that theory wouldn't make sense, I don't think? Joseph A. Spadaro (talk) 18:31, 8 April 2016 (UTC)
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April 7
Area, vertex, and boundary centroids
Two related questions: For what polygons does the area centroid (centroid of the entire enclosed region) coincide with (a) the centroid of the vertices, or (b) the centroid of the boundary points? For (a), it's true for all triangles and of course for all regular polygons. I'm guessing it's true for all parallelograms too. But what other polygons? As for (b), it does not hold for all triangles or even for all isosceles triangles; it holds for all regular polygons, and I'm guessing for all parallelograms, but what else? Loraof (talk) 19:04, 7 April 2016 (UTC)
- Trivial remarks: obviously for any region with a line of symmetry, all three centroids must lie on that line. If there are two lines of symmetry, all three lie at the intersection and so coincide. (In this case there is a rotational symmetry, and so it is reasonable to speak of "the center" of the region.) --JBL (talk) 20:50, 7 April 2016 (UTC)
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- Case (a) is much easier since the area centroid and vertex centroid are both rational functions of the vertex coordinates. For the edge centroid you need to weight by the edge lengths, which gives you √'s in the expression and everything gets messier. For quadrilaterals I'm pretty sure (meaning I have most of a proof worked out but it's too much trouble to write out the deatials) that (a) is true iff it's a parallelogram. The crucial point is, and correct me if I'm wrong about this, that as operators on quadrilaterals, the point centroid and area centroid both commute with affine transformations. So you can simplify by taking three points of your quadrilateral to be (0, 1), (0, 0), and (1, 0). This idea won't work for the edge centroid because a linear transformation does not change length by a constant factor. --RDBury (talk) 03:30, 8 April 2016 (UTC)
Interpolation and extrapolation
I have been told, and told to teach, that interpolation is generally more reliable than extrapolation. Obviously, everything in science and statistics rests, at least tacitly, on a bed of assumptions; for example, in the case of science, we usually assume that induction works. What are the "minimal" statistical assumptions required for extrapolation to be generally less reliable than interpolation?--Leon (talk) 20:07, 7 April 2016 (UTC)
- I don't think "interpolation is more reliable than extrapolation" is a formalizable statement in the sense you're hoping for. --JBL (talk) 20:52, 7 April 2016 (UTC)
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- From the lede of our article Extrapolation: "It is similar to interpolation, which produces estimates between known observations, but extrapolation is subject to greater uncertainty and a higher risk of producing meaningless results." -- ToE 21:14, 7 April 2016 (UTC)
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- @Thinking of England: But that just tells me what I've already been told, with no explanation of why.
- @Joel B. Lewis: In that case, is it in the same category as induction? By this I mean that it is a ill-defined heuristic that is important to science.--Leon (talk) 09:44, 8 April 2016 (UTC)
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- @Thinking of England: If this is meant as a response to me then I don't understand its relevance to my comment. If it is meant only as a response to Leon, please feel free to remove or ignore this comment. --JBL (talk) 21:40, 7 April 2016 (UTC)
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- (ec) I think it's something like this: if you have a simple regression line, I think you can calculate the variance of the predicted value (i.e., its variance over a large number of replications of the regression) as a function of the independent variable's value x. I think this variance is greater if you are outside the range of values of x that you used in the regression. The reason is this: draw a linear regression line through some data points. Suppose it correctly goes through a point in the middle, which I'll call the pivot point. Now except by coincidence, the estimated slope coefficient will not be exactly right; the correct line goes through the pivot point with a somewhat different slope. The farther away from the pivot point you look, the more the true line and your estimated line diverge from each other.
- Another less formal way in which the maxim is true is this: a linear (or any other) regression may have its functional form misspecified. This might not matter much in the range that the x data used for the regression were in, but matters more the farther outside that range that you try to extrapolate. For instance, draw a bunch of data points almost exactly satisfying y=x2. Pick a narrow range and run a linear regression. It may fit reasonably well for the data points included in the regression, but fits worse the farther outside that range you look. Loraof (talk) 21:20, 7 April 2016 (UTC)
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- Assuming only that uncertainty increases monotonically with distance outside a range of samples, there is no limit to the error that an excessive extrapolation can incur. See Extrapolation#Quality of extrapolation. AllBestFaith (talk) 02:04, 8 April 2016 (UTC)
April 8
Goat problem
Several centuries ... I have been searching (and got stuck) in the 19th century:
- Here (beautiful poem!)
- ... and here (3 horses).
- (a) Is someone able to find older descriptions of this problem?
- (b) Is the term "goat problem" the official/ultimate/best description - or are there better ones? THX 213.169.163.106 (talk) 11:45, 8 April 2016 (UTC)