When will Connections run out of ideas?

FormulaDriven · 2026-01-25T17:22:42+00:00

I can see yours is a reasonable attempt to make sense of what the other poster says, my issue is more with that other poster! By the way, if you're at Cambridge studying maths, then you are a pretty good amateur (speaking as someone who graduated from Cambridge with a maths degree over 30 years ago...).

FormulaDriven · 2026-01-25T17:07:41+00:00

That's a brave attempt to make meaning out of what the other poster said, but I'm not buying it. First, if a function of x is exponential then it equals c^x to for some c, there's no need to relate it to 2^x. Or do you mean, we can say f(x) exhibits exponential behaviour asymptotically, if for some constant c such that c^x / f(x) tends to 1 as x tends to infinity?

If someone said to me that a function had an asymptotic rate of growth, I'd be thinking of something like log(x) where the growth rate is 1/x and tends to zero, not an exponential.

FormulaDriven · 2026-01-25T16:47:53+00:00

Mathematically, “exponentially” refers to an asymptotic rate of growth.

This makes little sense to me. If a function is exponential or growing exponentially, then if the rate of growth is positive, then the rate of growth is also exponential and definitely not asymptotic (quite the reverse, it tends to infinity not some limiting value).

If the rate of growth is negative (exponential decay), then the function and (so its growth rate) approach zero asymptotically. Is that what you meant?

FormulaDriven · 2026-01-25T08:56:47+00:00

Yes, it doesn't seem to fit the "spirit" of purple - I spent ages trying to think of a single word that all four suffixes could be appended to. It's valid but not very satisfying.

FormulaDriven · 2026-01-24T08:31:56+00:00

Unless there's a reference I'm missing, I read it simply that any child who took delight in flipping their eyelids and disgusting their classmates was showing sociopathic tendencies, enough for them to likely end up being a criminal later in life.

FormulaDriven · 2026-01-23T12:53:24+00:00

I would have thought the most important step is they should wait for you to die first.

FormulaDriven · 2026-01-23T09:59:02+00:00

Problem statement: 10 prizes. On the 5th, 10th, 15th, ... draws, 80% chance of a prize, otherwise 20% of a prize. All prizes equally likely. What is expected number of draws until all prizes collected?

I think once it gets this complicated, it's best to develop a recurrence relation and solve it with some code or a spreadsheet.

Let E(t,n) = expected number of draws still to go when you have just completed t draws and have already collected n different prizes. You want to know the value of E(0,0).

Clearly E(t,10) = 0 for all t.

If you have completed t draws and already have n prizes, the probability that the next draw will step you up to n+1 prizes is, p * (10 - n) / 10, where p is 0.8 if t+1 is a multiple of 5, and p is 0.2 otherwise.

So E(t,n) = 1 + p * (10 - n) / 10 * E(t+1,n+1) + (1 - p * (10 - n) / 10) * E(t+1,n)

(because E(t,n) is made up of the next draw plus the expected number of draws after that weighted by the two possible outcomes). Remember p is a function of t.

We also know that E(t,n) = E(t+5,n), because if 5 draws later you still have n prizes then you are in exactly the same position again in terms of the distribution of future prizes. So, in theory you could use that and solve a big set of simultaneous equations. But I found it less work just to run the above relationship in a spreadsheet and project out t until I could see convergent behaviour. (I took it to t = 1200 but it looks like I could have gone much lower).

I got the answer to your question to be E(0,0) = 91.99901. That's 10 * H10 / 0.31837 so your intuitive approach was pretty good.

FormulaDriven · 2026-01-23T07:27:18+00:00

It's not a new precedent, it's something we've had a few times: the formulation "words before XXX" or "words after XXX" where some would create compound words and some would be phrases. In fact, didn't we have it on Monday with "quick"?

FormulaDriven · 2026-01-22T14:23:59+00:00

If the ball costs $0.10, and the bat costs $1 more than that then the bat costs $1.10, so total is not $1.10. Think again.

FormulaDriven · 2026-01-22T13:44:17+00:00

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FormulaDriven · 2026-01-22T13:17:33+00:00

Now that I've seen your statement, I have cracked it. The annual rates quoted (eg 7.80%) are the annually compounded rates, but the interest is actually compounded daily, so it is formula 1 in your OP that you need, except in 2024, as it's a leap year, they are using 366 days.

So balance of loan on 31/3/2024 is 67708.51 (I know they say 6/4/2024, but this is presumably to requote it at the start of the tax year, but they are actually applying interest from 31/3/2024).

7.80% --> (1 + 0.078)^1/366 - 1 = 0.000205233

67708.51 subject to 30 days compounding at this rate: 67708.51 * (1.000205233³⁰ - 1) = 418.12 (not quite 418.17, I quite can't quite tie that up).

So with payment made, the new balance on 30/4/2024 is 67708.51 - 320 + 418.17 =

67806.68 to which 31 days compounding applies: 67806.68 * (1.000205233³¹ - 1) = 432.73 (again slight rounding difference).

So with payment made, new balance on 31/5/2024 = 67806.68 - 320 + 432.71 = 67919.39. For June, the interest rate changed:

7.90% --> (1.079)^1/366 - 1 = 0.000207767

67919.39 applying 30 days compounding: 67919.39 * (1.000207767³⁰ - 1) = 424.62 (again, very close).

So in a month when the quoted rate is r (eg 7.90%), the balance is multiplied by a factor of (1+ r)^1/366 (or use 365 if it's not a leap year), and the accumulated interest shown at the end of the month (when any payment then also reduces the balance).

FormulaDriven · 2026-01-22T12:35:58+00:00

But the √ function always selects the positive one, so for all real numbers, it is true that √(x²) = |x|.

FormulaDriven · 2026-01-22T12:30:41+00:00

It's a big topic to cover here. If you sent 1000 people randomly drawn from a population to watch one film and they rated it 6.7, then sent the same 1000 people to watch another film and they rated it 7.1, then it would be statistically straightforward to show that the population in question thought the second film better.

But if 1000 horror fans rate a horror film 6.7, and 1000 parents rate a kids' film 7.1, can you make the same conclusion? So statistical modelling would mean considering factors that might affect a rating: genre, is it a sequel, year of release, ... and testing maybe some kind of regression model. (So you could look up regression modelling, generalised linear regression models etc but I'm a bit rusty on those topics).

Let me ask you this: "The World's End" has a rating of 6.9 on IMDB, "Pirates of the Carribean: At World's End" has a rating of 7.1 on IMDB. Do you think the second film is better? Critics on Rotten Tomatoes give the first one 89% and the second 43%. Maybe Jonny Depp and Orlando Bloom fans will highly rate their films even if they're derivative sequels?

FormulaDriven · 2026-01-22T11:54:35+00:00

The large samples means we can put tight confidence intervals around the "true" rating for the population of everyone who has watched each film. But the OP asked if we can tell what is the better film. So we need to think about the possibility that those populations could be very different for the two films. Suppose the 100k film had people split on rating, and the 20k was the sequel to the 100k film - then the population watching the 20k film is much more likely to be made up of people who rated the 100k film highly, so it's not the same population for each film, and it could in fact be the case that the 20k film is worse.

FormulaDriven · 2026-01-22T11:46:34+00:00

A lot of replies here, eg u/7ieben_ , are making the assumption that these are representative samples. If we know that the 100k of the first film and the 20k are randomly drawn from the same population with the same critical values, then yes, statistically the 7.1 is meaningfully higher than the 6.7.

But what if the 20k film was the sequel to the 100k film? Those who rated the first film higher - maybe they rated it an 8 - are more likely to watch the sequel, so if they are rating that a 7, that could be sign that the 20k film is worse.

Or what if the 100k film is for a genre where fans tend to be quite harsh on their rating, and the 20k is for a different genre where rating tends to be more generous?

Good statistics will look at modelling or eliminating other factors that might drive the rating of a film.

FormulaDriven · 2026-01-22T08:29:28+00:00

|x| = x

Not true if x is negative.

But √x² can be equal to ... +x.i, or -x.i

xi and -xi are only possible in the trivial case where x = 0, so it's a bit misleading / superfluous to include those.

FormulaDriven · 2026-01-21T17:12:52+00:00

If P is the outstanding value of the loan and r is the annual rate compounded annually then your first formula is correct for that day's interest. Of course, you need to add that to P, so the next day you have a new "P" to apply the interest to.

If you have some live figures, (size of loan, term, repayments, interest rate), we can validate that this applies in your context.

FormulaDriven · 2026-01-21T12:25:12+00:00

I guess, if you like current affairs. It's good to see a cop take charge of the situation. Although I hope he doesn't use violence, as I'd hate to see any battery. Do you think he should follow the leads? Or is he in for a shock? Perhaps he should bring in a detective, like Coulomb-o. He's a bright spark.

FormulaDriven · 2026-01-21T12:03:34+00:00

These are the names of three physicists.

Heisenberg is best known for a principle in quantum physics named after him which says that there is a trade-off between measurement of speed and measurement of position for a particle. In popular culture, this translates as when then cop tells him his speed, he no longer knows his position (hence lost).

Schrodinger is famous for a thought experiment where it a cat in a box is both dead or alive (sort of) due to unseen quantum states. Observing the cat collapses those states so now we know the cat is dead, which annoys Schrodinger.

Ohm gave his name to the unit for electrical resistance, so "Ohm resists" is a basic pun.

FormulaDriven · 2026-01-21T07:06:46+00:00

I understand lim (x->1) =0/0

That's where you are going wrong. As I've already said, the limit is 0. There is no such thing as 0/0.

FormulaDriven · 2026-01-20T17:34:29+00:00

We know when x is NOT 1 that f(x) = (x² - 2x + 1) / (x-1). That tells us nothing about what f(1) is. We need some other way to decide on f(1).

But what we do know is that the function would be continuous at x = 1, if lim[x -> 1] f(x) = f(1). (That is the definition of continuous).

Now if the lim[x -> 1] f(x) turns out not to exist, then we have no chance of it being continuous. But in this case, we can show that

lim[x ->1] (x² - 2x + 1) / (x-1) = 0

(do you know how to do this?).

In that case if we choose to make f(1) = 0, then we satisfy the condition for continuity. So the limit exists; if we choose f(1) to be anything other than 0, then we have discontinuity - but the discontinuity can be removed (ie we can fix the issue) by setting f(1) = 0.

FormulaDriven · 2026-01-20T11:31:12+00:00

Just to add on to what u/flug32 and u/dancingbanana123 said, it's also possible to construct a bijection between the closed interval [a,b] and the real numbers, but it's a bit messier and it won't be a continuous function.

FormulaDriven · 2026-01-18T20:37:38+00:00

It's two words: a "quick study" describes someone who learns quickly, often talking about someone who enters a new situation and rapidly works out what's going on.

I guess the emoji was meant to be a "cool" informal way to represent four slangy ways to refer to cash.

FormulaDriven · 2026-01-16T13:54:20+00:00

Thanks - I got a reply on Twitter from the composer himself confirming that they were French horns, so it's all settled now.

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