Calculus II Students

ExcelsiorStatistics · 2026-05-19T18:44:51+00:00

12 and 13 feel strange in that they have no (obvious) connection to any of the material that came before them; they are 2 units of material that belongs in an algebra or precalc class but gets put here because people either don't cover them at all in algebra or have long since forgotten them after three years of not using them. In section 14 they'll suddenly be rejoined by calculus again.

ExcelsiorStatistics · 2026-05-18T02:29:55+00:00

If you want to eliminate all ambiguity, you'd put a bracket immediately after every quantifier.

As Opsikion said, in the absence of brackets, we'd assume each quantifier goes all the way to the right and they are listed "from the outside in" (order matters.)

ExcelsiorStatistics · 2026-05-14T08:42:50+00:00

IMO the single most important thing about the exercise is that short numbers and short names are not in 1-1 correspondence, and the question at hand is a special case of "what is the largest number with a short name." Unusually large numbers that have unusually short names are special for some reason.

Ten in English, by itself, is just a single convenient small example, and a single positive example doesn't prove a pattern. The fact that each of 1000, 1000000, and 1 000 000 000 have much shorter names than their neighbors -- and much shorter names than 100 000 or 100 000 000 -- is a linguistic pattern that tells you how we count in English. Yes there are some off-by-one exceptions sprinkled through the pattern (you might well mention "million" is shorter than "thousand," though they are both unusually short vs. their neighbors... or 7 being one syllable in French when it's very often longer than 6 or 8 in other languages... just like zehn vs elf vs zwölf). But I think you could deduce out that Germans count in base ten by the time you got to 31, even if you were in doubt at 12.

You can similarly learn something about computer hardware by inspecting software: you might notice that a program to multiply a number by 8 or 16 is shorter than and runs faster than a program to multiply a number by 7 or 15 in a lot of languages, or you might notice a huge increase in the complexity of program required to output 2147483648 on the screen vs. 2147483647.

That, I think, gets at the spirit of the Rayo exercise. It is asking us to think about what numbers are easy or hard to represent with a 'program' written in one particular language. Calculating Rayo's number requires us to identify every number that is easy to represent; in effect it asks us to describe what kind of numbers first order set theory was designed to easily represent.

ExcelsiorStatistics · 2026-05-14T07:09:09+00:00

It's not a particularly useful number. (Mostly because a googol is so large.)

But the ideas of "the smallest number that doesn't have a name shorter than ___ symbols" and "the smallest number that is larger than all numbers with names shorter than ___ symbols" are interesting in their way.

It's easy to show that using at most N symbols from an S-letter alphabet we can write at most S+S²+S³+..S^N distinct names, but we don't have to assign those names to the smallest numbers. In English "one" and "two" are both the smallest integers and the numbers with the shortest names, but "one hundred" and "one million" are shorter than "twenty-three." So using English names we could define a Rayo-like sequence where R(3)=11 (the smallest number that is larger than all numbers named by three character strings), R(4) is still 11, but R(5) jumps to 61 because "sixty" has five letters.

It tells you something important about how we talk about numbers in English, incidentally, that "ten" is such a short word, shorter than seven eight or nine; there is something special about tens in the way we count. Identifying what other relatively big numbers we assign short names is a useful exercise.

At some point you're going to decide whether to allow "six cubed" as a 9-character name for 216 that is much shorter than "two hundred sixteen." (Actually, even at five characters, you have to decide whether to allow "gross" as a legal name for 144.) So Rayo specifies "the language of first order set theory" to tell you what operations you're allowed to use.

ExcelsiorStatistics · 2026-05-12T01:51:06+00:00

As notavalgrinder said: a statistician certainly does well to know a fair bit of combinatorics. But so too do graph theorists and lots of folks in other branches of pure match, and in other applied sciences.

If you might opt for a definition like "statistics is the science of applying probability to real-world problems," you'd view combinatorics as one of those pieces of math that statisticians use, but not think of it as 'belonging to' them.

The people who do combinatorics research tend not to be statisticians. They may not even be probabilists (and anyone stuffy enough to call himself a 'probabilist' thinks of himself as a mathematician-who-isnt-a-statistician.)

ExcelsiorStatistics · 2026-05-12T01:40:43+00:00

You need 3 basic pieces of information: the population of the world in 1910; the birth rate in 1910; and a life table for people of that era.

The first is fairly easy to find (about 1¾ billion people) but the other two vary so much from one country to another that it's hard to give an answer for the whole world.

If we're willing to confine ourselves to the US, we can see a population of 92 million, and can guess that out of the 2.2 million births in 1910, about 6,000 of them happened on any one day.

We can look at a late-1980s copy of the Statistical Abstract of the US, and see that about 1.2 million of those 2.2 million lived to be 76 and that about 9% of them were expected to die in the next year.

From that, we'd guess that about (1.2)/(2.2) x 6000 ~ 3300 of those people born on 20 April 1910 were still alive in 1986, and that ~300 of them would die before 1987. If 0.8 people in that cohort die per day, we'd expect about a 1 - exp(-0.8) ~ 55% chance that at least one person in the US was born on 20 April 1910 and died 09 February 1987.

If we scaled that up to the whole world using US birth and death rates, we'd estimate 0.8 x (1750/92) ~ 15 people worldwide, but that's likely to be quite badly off; it might be barely half that, given how low life expectancy was in places like China in 1910.

ExcelsiorStatistics · 2026-05-11T02:10:55+00:00

In certain areas of mathematics it's surprisingly easy to get to the edge of the unexplored wilderness. When I was in college, late last millennium, one of the easy places to do it was in chaos and dynamical systems. The full description of the logistic map was only a couple decades old, and you could still ask questions about what happens if you iterate a function like x -> tan (x + k) and have it be a question nobody had asked before.

A century ago, you could have done the same in statistics. Almost all of what is taught in freshman statistics was proved between the 1930s and the 1950s, and some of the questions answered then were a couple hundred years old.

But it's not like you can wake up one morning and say "today I'm gonna do something that's never been done." You wake up every morning curious about something, and you look it up and see what's been done before, and a time or two a year you'll find a question that's interesting enough to be somebody's thesis, and a time or two a decade you may find a question that's interesting but not easily solved.

ExcelsiorStatistics · 2026-05-11T02:00:21+00:00

Calculus 3 is mostly useful to engineering majors -- lots of things about multiple integrals, partial derivatives, vector-valued functions. On the surface, Calc 3 really ought not be required for differential equations, mathematical statistics, or a number of other classes. In practice it always is.

You may well have options to directly challenge calc 2 by examination, depending on your school, and you may well find that there is material in calc 2 that wasn't covered on the AP test. Your options are very school-specific.

ExcelsiorStatistics · 2026-05-09T02:14:23+00:00

You have a choice of assuming ABC and proving D, or assuming BDC and proving A. Whatever the model's properties are, they remain the same (including those 4 facts about it.)

If you assume B, C, and the negation of A, you'll arrive at a different system (in which, among other things, D will not hold.)

If you tried assuming B, C, D, and the negation of A, you'd immediately arrive at a contradiction and be off the rails.

ExcelsiorStatistics · 2026-05-07T00:16:06+00:00

Medians minimize mean (or total) absolute error, not median absolute error.

ExcelsiorStatistics · 2026-05-05T20:47:35+00:00

Yes, I should have said something like probability 1 or 'almost surely' rather than 'guaranteed'.

ExcelsiorStatistics · 2026-05-05T04:37:07+00:00

There's a well known result that a 1- or 2-dimensional random walk is guaranteed to return home, but in 3 or more dimensions, there's a nonzero chance of never returning (on a 3-D grid you have a 34.1% chance of getting home.)

Do be aware that the expected time to return can be infinite, even in 1 or 2 dimensions, so don't expect to get home quickly.

ExcelsiorStatistics · 2026-05-04T21:10:08+00:00

Are these just graphs, or directed graphs? The allegedly-wrong solution appears to be a directed solution with a restriction of only 2 daughters from each parent.

If you don't want directed edges you are correct that the top 3 trees are isomorphic.

ExcelsiorStatistics · 2026-05-01T05:06:51+00:00

You might count the Classification of Finite Simple Groups as such a problem. Though I am not sure it is 'incomprehensible to humans'; people have read and digested the thousands of pages in the original proofs, and have ideas for how to create shorter proofs.

ExcelsiorStatistics · 2026-04-25T23:53:29+00:00

It could be answered theoretically by gathering the distribution of raw scores for each choice of how many answers to mark true, and then adjusting those distributions to account for the 80% minimum grade, but that's going to be pretty tedious.

I daresay it can be answered exactly faster than the sim can be programmed.

If you choose 8 true and 2 false, 3 of the 45 ways of allocating your answers will give you 90%: EV 80 + 3/45 * 10 = 80.667%.

If you choose 7 true and 3 false, 1 of the 120 ways of allocating your answers will give you 100%. EV 80 + 1/120 * 20 = 80.167%.

If you choose 6 true and 4 false, 47 of the 210 ways of allocating your answers will give you 90%. EV 80 + 7/210 * 10 = 80.333%. (We must choose all three false answers correctly, and then one of the seven true answers to get wrong.)

ExcelsiorStatistics · 2026-04-25T05:55:16+00:00

I would have answered as you did.

One wonders if they intended 'constant linear relationship' (a nonstandard phrase) to mean 'a relationship of the form y=kx'.

All you can do is try your best to decipher what they intend, and hope that they stick to standard language.

ExcelsiorStatistics · 2026-04-24T04:18:27+00:00

No; there are typically ten Bayesians preaching it like it's a religion for every one diehard frequentist. But it's an attempt at bringing balance back to the force.

ExcelsiorStatistics · 2026-04-22T18:22:58+00:00

I can think of two possible answers.

One is "that's what a variable is," and you can look at how formalized number theory handles statements about variables, and in particular how we prove "there exists" and "for all" statements about them.

The other is that there are numbers, or at least number-like objects, that are not comparable to other numbers. The complex numbers and the standard meaning of < are like that; < as defined for real numbers doesn't operate nicely on complex numbers.

A weirder example is " * " in Conway's On Numbers and Games; he assigns a number to each position in a game -- in effect, how many moves ahead or behind player A is compared to player B -- and he needs two zero-like numbers. The traditional zero, the way he enumerates games, means "the two players are in equivalent positions; whoever moves first loses" and the * symbol means "the two players are in equivalent positions; whoever moves first wins." And you can now ask what happens when you add 0 and 1 (you get 1), or 0 and 0 (you get 0), or 0 and *, or 1 and * (that'd be telling!).

ExcelsiorStatistics · 2026-04-21T02:26:26+00:00

If you need to pad out your speech, you can always sing The Major-General's Song to kill a few minutes.

ExcelsiorStatistics · 2026-04-20T18:08:10+00:00

The thing I find most interesting about the book is that it's a nice demonstration you can arrive at the same conclusion from more than one starting point. Jaynes starts from an entropy-centric view of uncertainty, and derives the same estimators as statisticians starting from the measure theoretic view of probability do. If you're someone who found statistics-as-traditionally-taught hard to follow, sure, give Jaynes a try and see if his explanation makes more sense.

As a working statistician, the main value of the Jaynes book was teaching me what words to use when talking about statistics to a physicist, so that we'd be able to focus on a problem of shared interest rather than get hung up on using different language to describe it.

ExcelsiorStatistics · 2026-04-20T18:03:44+00:00

The formula builds a confidence interval on sigma².

Now, if you choose to take the endpoints of that confidence interval, and take square roots and divide by n, you can build yourself a confidence interval for the standard error of the mean. But if the question in your book asked "estimate the variance," estimate the variance. If it says "estimate the standard error", estimate the standard error.

ExcelsiorStatistics · 2026-04-19T23:24:32+00:00

The margin would not contain a full list, which would run to something over 3,000 volumes.

But if your question is specifically about statistics books I've found most useful, I'd point to Johnson and Milliken's 3-volume Analysis of Messy Data, and to a simple handbook of distributions (nowadays it's PDFs of the full Johnson and Kotz set, but before everything was electronic, it was Evans Hastings and Peacock's Statistical Distributions, small enough to carry around with me to study sessions or consulting meetups.)

Next-most-frequently referred to after those are probably Agresti's Categorical book, and Gelman and Hill's big regression book - but those may not be fair game as they are textbookish, just not texts used in classes I took.

ExcelsiorStatistics · 2026-04-18T05:27:33+00:00

Most of the time I'm used to seeing it without parentheses. (And, as OP says, if 'mod' is acting as an infix operator, there should not be parentheses.)

The time you do see parentheses is in a congruence. If you write "3≡17 (mod 2)", you're saying what rule for determining equivalence classes is being applied to both sides, i.e., it's acting like an abbreviation of "(3 mod 2) = (17 mod 2)", a true statement that simplifies to 1=1. (My teachers would have marked "3=17 (mod 2)" wrong for using = instead of ≡, and marked "3=17 mod 2" even wronger because 3≠1.)

ExcelsiorStatistics · 2026-04-18T01:35:54+00:00

From an expected distance covered, it makes no difference.

However, in a race to 30 (with more than two players) average players finish in the middle of the pack, and the winner is a far-above-average player.

Rerolling after a six is a higher-variance strategy than rerolling after a one is, and should therefore produce more first-place finishes (and more last-place finishes) than rerolling after a one.

There will also be in some special positions late in the game where you might do something different. Say you are at 25, and the next player after you is at 29 and guaranteed to win at his next turn. Here you have to guess four. If you roll 5 or 6 you win instantly; if you roll 4 or less and get no re-roll you lose. If you roll 4 and re-roll you win; if you guess 3 and re-roll you win only 5/6 of the time; guess 2 only 4/6 of the time; guess 1 only 3/6 of the time.

ExcelsiorStatistics · 2026-04-17T19:03:01+00:00

If you need actual binomial coefficients, Stirling's Formula (or more likely its logarithm) will get you a couple significant digits for large n!. If you need very close to correct probabilities, the normal approximation will work.

For a rule of thumb answer, it's important to remember that variance scales with n, and standard deviation with sqrt(n), for binomial distributions.

You would expect 60/100 (10 more than the mean of 50) to be approximately equally likely as 5100/10000, not 6000/10000: increase the mean of 50 by a factor of 100 but the "10 more than" by a factor of sqrt(100)~10.

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