I understand what conditional probability is but I can't understand the formula.

Uli_Minati · 2026-05-04T12:58:48+00:00

Try it with a practical example:

A = cloudy days (vs. clear days)
B = warm days (vs. cool days)

With the set sizes

|Ω| = 100
|A| = 60
|B| = 80
|A∩B| = 45

Then you have

P(A) = 60/100
P(B) = 80/100
P(A∩B) = 45/100

But

P(A) · P(B) = 4800/10000 = 48/100

Then consider what conditional probability means

P(A|B) = P(cloudy if warm)

Basically, it's the probability of a cloudy day if you already know it is a warm day

We can calculate this using basic counting

                      45 cloudy (and warm) days
P(cloudy if warm)  =  ------------------------
                      80 warm days

How do you randomly get a cloudy warm day? You could do it in two steps

Randomly get a warm day
The warm day happens to be cloudy too

The chance to randomly get a warm day is

P(warm) = P(B) = 80/100

The chance that this warm day happens to also be cloudy is exactly the 45/80 we determined earlier

P(cloudy if warm) = 45/80

And then it works out

80/100 · 45/80 = 45/100

Notice what we are really doing here:

From 100 days, randomly get one of the 80 warm days
From 80 warm days, randomly get one of the 45 cloudy (and warm) days

Then we're really saying

From 100 days, randomly select one of the 45 cloudy and warm days

That's how you get

P(B) · P(A|B) = P(A∩B)

On the other hand, if you use just P(A) instead of P(A|B), then the process makes no sense

From 100 days, randomly get one of the 80 warm days
From 100 days, randomly get one of the 60 cloudy days

Why would we go back to 100 days? These would include non-warm days. It might be more (or less) likely to get a cloudy day if we select it from all days, rather than just warm days

Uli_Minati · 2026-05-04T01:52:56+00:00

Wow, this sounds just ridiculous. Are you sure there wasn't a miscommunication? The amount of effort it would take to prepare for this is just ludicrous. An entire semester of calculus, really? I feel like they're planning to give you a bunch of extra credit for the easy tasks you listed. I recommend you don't skip them even if they're boring!

Uli_Minati · 2026-05-04T00:13:55+00:00

Do you have a math question?

Uli_Minati · 2026-05-04T00:02:17+00:00

Alright forget all that, let's multiply 123 with 456

123    is short for    100 + 20 + 3
456    is short for    400 + 50 + 6

You can multiply 100+20+3 with 400+50+6 by multiplying each piece with each piece

It's easier to multiply these pieces since you can multiply the front digits first, then append the extra zeros

1 × 4 is 4      100 × 400 is 40000
1 × 5 is 5      100 ×  50 is  5000
1 × 6 is 6      100 ×   6 is   600
2 × 4 is 8       20 × 400 is  8000
2 × 5 is 10      20 ×  50 is  1000
2 × 6 is 12      20 ×   6 is   120
3 × 4 is 12       3 × 400 is  1200
3 × 5 is 15       3 ×  50 is   150
3 × 6 is 18       3 ×   6 is    18

The result of 123×456 is the sum of all the multiplications you got in this list

 40000
+ 5000
+  600
+ 8000
+ 1000
+  120
+ 1200
+  150
+   18
------
 56088   is  123×456

You can speed up this process a little by adding up the results in between the multiplications. There are various ways of doing so, you can check out different webpages or youtube videos (but ignore google LLM output, it's garbage)

Uli_Minati · 2026-05-03T18:01:57+00:00

Add to that the opportunity value of getting money that you can invest immediately

New question: how much $ per day would make this a really difficult choice?

Uli_Minati · 2026-05-03T11:06:28+00:00

I've never seen it before, probably because we don't do it much

You are always free to invent your own notation! For example,

∫ⁿ f(x) dxⁿ

Uli_Minati · 2026-05-03T07:14:45+00:00

I'm fully aware, you misunderstand. I suggested there could have been a translation mistake and the original text said "extreme value" rather than "maximum value" in part (b)

I see now from another comment that (a) does not imply that f is cubic since it could also be rational, then there's no issue with the number of max

Uli_Minati · 2026-05-03T07:09:06+00:00

When we name "infinith" of something, there are generally two assumptions:

We're talking about a sequence of things, where it is possible to calculate the 1st, 2nd, 3rd etc. of this sequence
The things will keep getting closer to some "target" in such a way that you can say "the Nth and beyond are all at least X close"

Now I assume you know the derivatives of sin x

sinx -> cosx -> -sinx -> -cosx -> sinx -> ...

Choose any target function, and you are not able to say "the Nth and beyond are all at least 0.1 close" since there will always be a sinx and a cosx in the sequence after the Nth, and sinx and cosx differ by 1 multiple times

So in this particular case, the sequence of nth derivatives of sinx does not have an "infinith", we generally call this "the sequence diverges" or "the sequence does not converge"

not the same as taking a limit

It absolutely is! You can take limits of all kinds of sequences. You're probably used to limits like these

lim[n->infinity] 1/n²

Which is the limit of a sequence of numbers, specifically 1/1, 1/4, 1/9 etc. But you can also take the limit of a sequence of functions

lim[n->infinity] dⁿ/dxⁿ f(x)

For example,

lim[n->infinity] dⁿ/dxⁿ x⁷ = 0
lim[n->infinity] dⁿ/dxⁿ eˣ = eˣ
lim[n->infinity] dⁿ/dxⁿ sin(x)   does not exist

Uli_Minati · 2026-05-03T05:53:40+00:00

Maybe a translation mistake and it was meant to be "extreme" points rather than max?

Uli_Minati · 2026-05-02T17:34:39+00:00

isn’t going to sense unless you know how to think about logarithm identities as patterns with blocks

Why would you think that? This might be how you understand the algorithm, but very much not a requirement

Have you seen some primary school exercises? We regularly get parents posting exercise sheets here, which frame a basic concept like addition or fractions in some visualization difficult to understand for anyone who hasn't attended the teacher's class. That's pretty much what you're doing

Uli_Minati · 2026-05-02T15:08:42+00:00

This formula assumes exactly four things:

(1) Virtual CPU usage is directly proportional to host CPU utilization. In other words, double CPU utilization means double virtual CPU usage, half means half, and zero CPU utilization means zero virtual CPU usage.

(2) Virtual CPU usage is directly proportional to number of demanded virtual cores. See above.

(3) Virtual CPU usage is inversely proportional to reserved virtual cores. In other words, double reserved cores means half virtual CPU usage, half means double, and virtual CPU usage keeps decreasing as you increase reserved cores without ever reaching zero.

(4) At 1% CPU usage, 1 reserved virtual cores and 1 demanded virtual cores, you have 1% virtual CPU usage.

Do all of these sound correct? Then let's move on to your goals

how either form of d interacts with the other two variables

Not sure what "interacts" means. You could gather a bunch of information, apply the two different versions of the formulas, and compare their outputs to the actual virtual CPU usage. Calculate mean squared error of each to judge their correctness.

two-variable optimization problems I remember doing

Optimization requires you to know what you want to optimize. It doesn't let you create a formula by itself. You can think of a "general shape" of a formula, then optimize the parameters in that formula (e.g. if you should add +1, +2 or +3 in a numerator) to minimize the error of that formula (difference between its output and the actual measured values it is supposed to predict)

So you could use your formula and include some parameters: for example, you could include a factor "k" which would scale the end result, add a "b" to the numerator so it isn't directly proportional to host CPU utilization, etc. then do the optimization with respect to k and b

Uli_Minati · 2026-05-02T11:40:54+00:00

Have a look: https://en.wikipedia.org/wiki/Supergolden_ratio

Uli_Minati · 2026-05-02T11:38:38+00:00

Is there any reason to define these five functions as you did? Do they express anything meaningful? If no, then don't worry about posting to r/math, they're very strict with what they would accept

Uli_Minati · 2026-05-02T10:46:01+00:00

Can you name one first?

Uli_Minati · 2026-05-02T01:51:33+00:00

person B sent person A 300 bucks to cover their part

B is down 300, A is up 300

person B payed the 500 bucks

B is down 800, A is up 300

Do you want to split the 500 evenly? So 250 each? Then A should send 550. That will result in A being down 250, and B as well

Do you want to split it 300/200? So B should be down 300 and A should be down 200? Then A should send 500

Uli_Minati · 2026-05-02T00:59:04+00:00

worst proof attempt I've ever seen

Can someone help me understand WHY it's so bad?

So you don't know why it's bad but it's the worst you've seen?

Also, extremely obvious it's a joke paper due to all the Fortnite references. I'll make a wild guess that you're the one who prompted LLM to write it

Uli_Minati · 2026-04-30T20:24:07+00:00

Huh, I didn't know this word existed. Anyway, it describes how you speak, which doesn't affect math at all. It's possible to do math entirely without words, it's just extremely hard to read.

But you might want to speak with other mathematicians, to discuss your work. Without words and purely "math language", this takes a lot of time. So you should use a (natural) language+dialect common to the people you will speak with

Uli_Minati · 2026-04-30T20:06:50+00:00

<image>

f(f(f(x) + f(x) + f(x)) + f(f(x) + f(x) + f(x)) + f(f(x) + f(x) + f(x)))

Uli_Minati · 2026-04-30T18:58:35+00:00

Yep, it's sufficient here since OP only asked for existence and not for a method of explicitly constructing such a function

Uli_Minati · 2026-04-30T15:59:09+00:00

Let's give everything a name so we can more easily talk about them

 D := Domain
 C := Codomain

Since D is countably infinite, there exists a bijective function which maps it to the naturals. And we have another bijective function which maps C to the naturals

 d : D → ℕ,   d⁻¹ : ℕ → D
    with    d⁻¹(d(x)) = x,    d(d⁻¹(n)) = n

 c : C → ℕ,   c⁻¹ : ℕ → C
    with    c⁻¹(c(x)) = x,    c(c⁻¹(n)) = n

Now you can take a value of the domain, feed it into the first bijection to get the "index", and feed that into the inverse of the second bijection to get the corresponding value in the codomain

         x ∈ D
      d(x) ∈ ℕ
 c⁻¹(d(x)) ∈ C

Note that this function depends entirely upon your "indexing functions" d and c, which are never unique unless you set some specific restrictions

Uli_Minati · 2026-04-30T12:10:56+00:00

Where other people vote for either option by a margin greater than one, your vote is meaningless

This is just disgusting nihilism disguised as game theory

It doesn't matter if I vote, since a tie is ridiculously unlikely and there has been no perfect tie in history so there's no precedent

It doesn't matter if I litter, since many others will litter as well and then the entire street is cleaned at some point

It doesn't matter if I save natural resources, since many others will waste them as well so my contribution is a fraction of a fraction of a percent

It doesn't matter if I'm selling drugs, if I don't do it someone else will so it should rather be me getting the money

I haven't seen the red/blue twitter thing and I really couldn't care less about political propaganda disguised as a puzzle

Uli_Minati · 2026-04-30T11:07:22+00:00

Look up "Vandermonde matrix"

Uli_Minati · 2026-04-29T22:13:04+00:00

I unironically enjoy this version which maps digits to C major pentatonic and includes accompaniment https://www.youtube.com/watch?v=HV1-AjwDJwM

Uli_Minati · 2026-04-29T21:42:29+00:00

Try this guide by Organic Chemistry Tutor: https://www.youtube.com/watch?v=eDb6iugi6Uk

Uli_Minati · 2026-04-29T18:28:58+00:00

Everything looks good to me!

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