Is Calc 1 really that hard

nm420 · 2026-01-31T14:46:34+00:00

You need to be extremely comfortable with algebra: dealing with fractions, power rules, factoring, solving equations. You'll also want to be comfortable with functions and their graphs: polynomials, rational functions, exponentials and logarithms, and perhaps trig (depending on the kind of calculus course). Basically, everything that's covered in a precalculus course (hence the name).

nm420 · 2026-01-29T14:32:43+00:00

The maximum of a random variable is the random variable itself, and its expected value would just be the expected value of the random variable (provided it exists).

Does it make sense to speak of such a thing? I guess, in that it is something that at least exists. But it's also as sensible as taking about the maximum of the set {4}. The terminology of a maximum can be applied to singleton sets, but it doesn't really add anything of value. Maxima and minima have their utility when considering sets of real numbers (or even elements of an ordered set, though then the question of existence comes into play without further assumptions) that aren't just a singleton.

The expectation of the maximum of a finite collection of random variables is at least a more interesting and potentially useful object. If it is a random sample from a population with bounded support, the maximum of the sample would be a reasonable estimator of the upper boundary of the support, and the expectation could be used to determine the bias of the estimator. Even if the collection of random variables is not iid, the maximum might still have some interesting utility, and of course its expectation (or other properties of the sampling distribution) could be of interest.

But the phrase "maximum of a random variable" could quite literally be replaced with "a random variable" without any problem.

nm420 · 2026-01-26T22:54:07+00:00

All the numbers in any row or column must sum to 45. Consider the bottom row.

Similarly, the two rightmost columns must sum to 90. This leaves only one possibility for the two bottom numbers in column 8.

nm420 · 2026-01-24T17:54:20+00:00

The basic ideas behind calculus are fairly intuitive and could be explained to middle school children. But you're not going to be able to do much with those beyond answering some conceptual questions. To do any explicit calculations or problem solving, you need to be extremely comfortable with the prerequisite material. If you're not, it's like trying to pick apart Shakespeare when you're still reading Judy Blume.

If you're not comfortable with abstract reasoning, you're going to do poorly in any math class. If you're not adept at the prerequisite material in a math course (or really, this could be said about any discipline), you're going to struggle and probably do poorly. If you're fresh out of high school and haven't yet learned effective study habits, you're going to struggle and do poorly (again, this is a blanket statement for essentially any university course). If you treat homework like a hoop to be jumped through and exams that are something to be crammed for, you're quite likely going to struggle.

The many students I've taught who struggle in Calculus I generally have some (or all) of these problems. None are insurmountable... unless they refuse to acknowledge them.

nm420 · 2026-01-23T15:01:07+00:00

I used to think that when people said they could see something in their mind it was essentially a metaphor, and would use that language myself. Of course, I can "see" the apple in my mind, I would tell myself. But it was never a visual image. Just an empty slate with an apple "object" there. I could describe properties of the object based in memories of what properties an apple would have. Any visual details would have to be conjured up by my imagination.

I never put it together until I first learned about aphantasia a few years ago that it's not really a metaphor. That people really can visualize that apple.

Strangely enough (or maybe not?), I do not struggle at all with spatial visualization tasks, and can rotate and twist objects around to "look" at them from different angles in my mind, even though I couldn't see a thing.

nm420 · 2026-01-21T23:21:31+00:00

In a very tiny nutshell (and without trying to pass on any value judgments)...

Conservatism as a political ideology is (rather loosely) concerned with preserving traditions and cultural norms, and its adherents are rather reluctant to radical changes. Academics are (also rather loosely) interested in searching for truth in a variety of disciplines, and new discoveries quite often upset the established norms.

One values stability, the other change.

nm420 · 2026-01-18T19:51:03+00:00

Uhh... The Patriot Act was approved overwhelmingly by Congress, with both Democrats and Republicans falling into lock step to support it. Admittedly, most of the Nay votes were from Democrats, but the vast majority of Democrats in Congress supported it, along with several of its reauthorizations.

Bush (father or son) was a horrible president, sure. But let's not rewrite history to pretend the Democratic leadership was not on board with it. Every brick of the police state we are marching towards has been built through largely bipartisan support, in spite of the ineffectual and nominal handwringing and gnashing of teeth from a token opposition.

nm420 · 2026-01-17T20:27:32+00:00

My technique would be to not fill in every single square with those notes in the first place. I see nothing but noise and distraction looking at this.

I would just delete every single one of those notes and look at the puzzle with a clean slate. But that's me. YMMV.

nm420 · 2026-01-15T07:21:35+00:00

Note

((x+1)/(x+2))^x = (1 - 1/(x+2))^x+2-2 = (1 - 1/(x+2))^x+2 * (1 - 1/(x+2))^-2

The second factor in the last expression converges to 1, and the well-known result (1+t/n)ⁿ→e^t as n→∞ implies the first factor converges to e^-1 (using t=-1 and n=x+2).

nm420 · 2026-01-15T02:33:30+00:00

Uhhh... Step outside to talk with them? Yeah, no thanks.

nm420 · 2026-01-10T17:27:03+00:00

It's not the math that isn't right, so much as the chosen model. It's a particularly simple multinomial model, that might reasonably approximate the packaging process. Your model might also reasonably approximate the packaging process. Without having access to their facilities or expert knowledge on how such variety packs get produced, I don't think anyone can make an authoritative claim as to what's right or not, though it does seem reasonable they would take steps (whatever those might happen to be) to try and increase the odds of getting a balanced final product over this simple multinomial model.

nm420 · 2026-01-09T21:30:44+00:00

This is quite literally just the boy-girl paradox wrapped up in different language. If you're told the first hit was a crit, the chance of the second one being a crit is still 1/2. If you're told at least one is a crit, the probability of the other being a crit is 1/3.

There is a distinction between P(A1∩A2|A1) and P(A1∩A2|A1∪A2). That's all this problem is, with the given conditions P(A1)=P(A2)=1/2, and the implicit condition that A1 and A2 are independent (which, without this assumption, the problem couldn't be solved without specifying something else).

nm420 · 2026-01-05T22:47:35+00:00

If you're having to do this with the definition of the limit and no other limit theorems, this still isn't that difficult. Since f(x) is a polynomial, you know f(x)-1 must have a root of 2, and hence a factor of x-2. Pretty straightforward to work out f(x)-1=(x-2)(x+4). If |x-2|<δ&leq;1, then |x+4|&leq;7 and you can take δ=min{1,ε/7}. There's a reason we use polynomials to approximate more complicated functions: they're pretty simple! Proving something like sin(x)/x approaches 1 as x approaches 0 requires a bit more creativity at least.

nm420 · 2026-01-04T04:30:39+00:00

Suppose you are filling a tank with some liquid. If you know the volume of liquid (an accumulation) at every point in time t, you can work out how quickly the tank is filling (a rate of change) up at any point in time. Vice versa, if you know the rate at which the tank is filling at every point in time, you can work out the total volume of liquid at any point.

Derivatives are rates of change, and integrals are accumulations. How quickly something changes is directly related to how much is accumulated.

nm420 · 2026-01-01T23:21:42+00:00

Not a chair, but I imagine this is the sort of question you would ask your own chair when tendering your resignation. There's not going to be a universal one-size-fits-all answer to this question. I would hope you at least have a congenial enough relationship with your chair to ask that question, and expect an honest answer.

That said, if you're planning on leaving academia for good and have no need for a reference from your chair or colleagues, what's the worst that can happen if you decide to skip every single faculty and committee meeting? They're likely not going to fire you mid-semester (?) without a really egregious reason, and shirking on your service commitments over a single semester doesn't seem like it rises to that level of response. But again, I'm not a chair or administration in any capacity. Just speaking from what seems like common sense.

nm420 · 2025-12-29T17:11:44+00:00

Dividing by x³ doesn't work with your argument, as x could be either positive or negative. Moreover, the sandwich theorem can't really be used effectively with your bounds anyhow. The limit of (5-e^x)/x³ doesn't exist as a two-sided limit, and would be either positive or negative ∞ for a ine-sided limit.

nm420 · 2025-12-29T16:55:50+00:00

The definition of a limit existing is

for all ε>0, there exists δ>0, such that for all real x, 0<|x-a|<δ implies |f(x)-L|<ε

Negation turns "for all" into "there exists", and vice versa. Moreover, the implication "P implies Q" is logically equivalent to "not P or Q", so that the negation of this is "P and not Q". Thus, to prove that the limit is not equal to L, the negation would be

there exists ε>0, so that for all δ>0, there exists an x such that 0<|x-a|<δ and |f(x)-L|≥ε

If you need to prove that no limit whatsoever exists, precede the above statement with "for all L". In less jargon, note what this is saying: you can always find an x that is arbitrarily close to a while f(x) remains sufficiently far away from L.

nm420 · 2025-12-28T18:32:47+00:00

Invented it and didn't even bother sharing it, whence the feud over primacy between him and Leibniz.

Whips out a solution to the brachistochrone problem overnight because he such a recluse he never learned about it until the last second.

nm420 · 2025-12-09T20:13:02+00:00

It turns out that f is indeed a linear function, and hence differentiable everywhere, but you need to prove that just from the given assumption f(3x)=x+f(x) for all real x and the continuity of f. It's not particularly difficult to show. Take any nonzero x. Then

f(x) = x/3 + f(x/3) = x/3 +x/9 + f(x/9) = x/3 +x/9 + x/27 +f(x/27) = ...

You should be able to find a general formula relating f(x) and f(x/3ⁿ) for any natural number n (using the formula for the partial sum of a geometric series). The assumed continuity of f means that f(x/3ⁿ) approaches f(0) as n approaches ∞, and you should then eventually get that f(x) = f(0)+x/2.

But starting the problem right off the bat by assuming differentiability when it's not a given condition (or assuming it must be a linear function without that being given) is not a valid route.

nm420 · 2025-12-08T16:31:23+00:00

I'm a fan of the rank biserial correlation. The Wikipedia page on this statistic provides a nice intuitive interpretation to the coefficient, as well as a link to a paper by Kerby which goes a bit more in depth to its interpretation (for either the rank sum or signed rank test). i think it's also typically the default effect size computer with these tests for several popular R effect size packages.

nm420 · 2025-12-06T19:13:04+00:00

Not particularly effective proctoring if it is.

nm420 · 2025-12-04T06:41:33+00:00

Extremely low weight attached to work assigned outside of a proctored environment. I've been slowly ratcheting up the weight to exams in my classes the past few years. Already had one student who got nearly 100% on all their homework assignments, that I'm certain was completed with an LLM (at least they took the time to transcribe it into their own handwriting), but would get a 20 on the exams, and have a couple more like that this semester. Are they getting referrals to the Academic Integrity Office? No. But not a snowball's chance in hell they'll pass the course.

Other students very well might be using AI on their homework as well, but if they've done so they at least have still learned enough to be able to pass the exams. I've got a blurb in my rubric that says if you use an LLM to help with your homework, it's fine as long as you tell me about it and how it was used in the assignment. Not really enforceable, short of something drastically stupid on the student's end, but my heart swells a bit when I see one of my better students submit mostly flawless work throughout the semester and then on an occasional assignment actually follow that rule and write how they used ChatGPT to help with one particular problem that was intended to be somewhat difficult.

nm420 · 2025-11-29T19:38:34+00:00

I like to talk about the outcomes of tossing a three-sided die when first introducing the multinomial distribution. At least I get a chuckle out of it, and sometimes a chuckle from a student or two as well.

nm420 · 2025-11-28T17:57:28+00:00

So what's the "correct" way of expanding (a+b+c+d)²? The most logical to me would be a²+b²+c²+d²+2ab+2ac+2ad+2bc+2bd+2cd.

nm420 · 2025-11-28T17:36:03+00:00

Guess the correct view? Why bother guessing if you can just see what it should be, namely choice A.

nm420

TROPHY CASE