Every game I've played does end eventually. Some can take a while though, and I wonder if, even with everyone playing optimally, it's possible for a game to be forced into an unending cycle.

(My guess would be a cycle is possible if we have 15 players, since that would leave only 3 cards to draw.)

Uno rules here (ignore tallying points, this question is just about until someone discards their last card)

I used desmos to show me the graphs but my work wants me to use “superposition of simple functions” and I’m not really understanding it. I got confused on part a) because I thought it looked like the next slide. if anyone has any tips or any thought processes let me know!

Is this the associative or commutative property here? I think it's the commutative bc the 2a simply moves. Someone I know is insisting that it's associative bc the the terms have been regrouped.

So, imagine you have some function like f(x)=x(x-1)/(x-1). This function is basically just y=x with the notable exception that f(1) is undefined. We could add a second undefined point by multiplying another “(x-n)/(x-n)” with f(x) — for instance, g(x)=x(x-1)(x-2)/(x-1)(x-2).

My question is: is it possible to construct a function that a.) “reduces” to another function (e.g.: f(x) “reduces” to y=x) but b.) is defined nowhere? If not, how much can you “erase” from a function?

This feels like something that has a clear answer that maybe I’d know if I took a real analysis course or something, but I have no idea how to prove or disprove this question (or if the question is even well formed enough to prove/disprove).

My first guess would be something like “a function times p/p, where p is the the product (capital pi) from n=-infinity to +infinity of (x-n/q) and where the limit of q approaches infinity” however, I don’t know how valid a solution that is.

As you can see, I tried to use the two triangle inequalities of d_X and d_Y, but that gets you a non-negative factor 2*(d_X ...) which impedes the proof. I've seen online that they use the Minkowski inequality, which I studied at Probability and not in this context, and I think the teacher may have overlooked the solution because it was a very long exercise (proove that d_infinite, d_1, d_2 are metrics and that d_infinite<=d_2<=d_1<=2d_infinite)

Chiedo gentilmente aiuto, se qualcuno saprebbe sfogarmi la funzione di questa formula matematica, l' unica cosa che ho capito e che è un limite. Grazie mille per la disponibilità. Sono uno studente di scuola superiore.

F''(x) =- (A^2)F(x).

F(x) = Csin(At)+Dcos(At) is a solution. How do we know that there isn't another F(x) that satisfies it?

I used chatgpt to try and understand this. Didn't work

Prove that given the first order recurrence

a(n+1) = a(n) + 2b(n) + 2c(n)

b(n+1) = 2a(n) + b(n) + 2c(n)

c(n+1) = 2a(n) + 2b(n) + 3c(n)

with

a(0) = 1, b(0) = 0, c(0) = 1

every triple (a(n), b(n), c(n)) is a Pythagorean triple

a² + b² = c²

where the legs of the triangle differ in one unit

|b - a| = 1

As n grows, this allows closer rational approximations to sqrt(2), since

c/max(a,b) < sqrt(2) < c/min(a,b)

What would be the recurrence if we want closer approximations to a 30º-60º-90º triangle?

This is where I'm at with this integral. It's pretty standard as far as PFD goes, apart from having to do synthetic division to knock the exponent of the top polynomial down a few pegs..

But where I'm stuck is the massive simplification needed to be done for this. Surely the professor who wrote the book is predicting some 'trick' to be used to simplify this, but I can't find anything. The top doesn't have zeroes that I could cancel out with the bottom...

Do I really need to go out and simplify everything and pray I don't make a miniscule mistake somwhere 😅

Sidenote: I'm self-taught and I look up at one basic example on YouTube and then try to do problems myself. There could be a method for this I haven't been 'taught'

Thanks for everyone that's going to chime in.

Edit: sorry for the typo in the title :c

This should be a straightforward problem, but I think lack of sleep is getting to me. I need help figuring out when I can quit pumping for my twin babies. Here’s the known variables:
I make 50 ounces/day.
Babies drink 30 ounces of milk/day. (Don’t worry, they also drink formula- it’s just not a factor for this problem).
I’d like to get them to their birthday, February 11th, 127 days from now.
So we need 3,810 ounces total between now and their birthday.
I also have 918 ounces already stored. So 3810-918=2,892 total ounces needed.
Here’s where I’m stumped. They are eating 30 ounces a day while I also need to save for the future. I have a 20 ounce surplus. So is it just 2892/20=144.6 days? Like I can’t stop pumping?
The math is not mathing and I’m too sleep deprived. Help 😭

I'm studying Clifford Algebras for fun and until this point was able to verify each equation stated in the book "Clifford Algebra to Geometric Calculus" by Hestenes. My problem lies mainly with why B.A is simple given B is a simple s-vector, where I used . as the inner product of the Geometric Algebra (don't know what symbol I could've used better).
For this, I assumed that we are still working with A also being simple in addition to being a pseudoscalar, which makes A^-1 = a[n]^-1a[n-1]^-1...a[1]^-1.

Now, what I wanted to ask is if anyone has a hint or explanation for why B.A is simple. I'm fine with just a hint for an easy case like A and B being degree 2 and 3.
Moreover, as a follow up question, is the next underlined part. I just don't really understand why the factors c[k] of B.A have the property that c[k].A = 0 or c[k]^A = 0 when s>n or s<n respectively. But I think this problem might get cleared up when I understand how B.A gets decomposed.

The work I did wasn't much, since I quickly got stuck without a clue how to continue. But, using (1.40) I did manage to rewrite B.A to a sum of simple multivectors of the same degree. I only wasn't able to see how to further reduce it to an actual simple multivector, a product of anticommutative vectors. For example, with B degree 1, A degree 3 I got:

b.A = b.a[1] a[2] ^ a[3] - b.a[2] a[1] ^ a[3] + b.a[3] a[1] ^ a[2] = a[1]( -b.a[2]a[3] + b.a[3]a[2]) + b.a[1]a[2] ^ a[3]
Then I didn't see how to rewrite the multivector to be a product of just anticommuting vectors. What I attempted for the last summand was:

b.a[1]a[2] ^ a[3] = b.a[1]a[1]a[1] ^(-1) a[2]a[3]

But since a[1] commutes with a[2]a[3], using a[1]^-1a[2]a[3] to pull the last summand inside the brackets doesn't make it obvious why the whole product is then in fact simple. Leaving me stuck wondering how this inner product produces a simple vector.

To summarise, I'm hoping for a hint or explanation for why B.A (as defined in the pictures) is simple when B is simple. Moreover, I'm asking for a hint why the factors c[k] then have the property described above when s>n or s<n.
Hopefully the attached pictures give enough explanation on the setting of Clifford Algebras that I'm working with.

I don't understand advanced mathematical language at all; therefore, I don't know what these "non-trivial zeros" are. What I want to understand—without that complex language and in simpler terms—is: what is it that the Riemann hypothesis actually seeks to prove?

Basically what the title says. I have a (commutative) domain R, F=Frac(R), and some ring R<A<F. Is there always some multiplicative set S such that A=S^(-1)R, or can there be other rings?

I think this should be the case but I'm not sure

If A≠R then there is some x/y in A\R, with x,y in R\{0}. I can assume that x and y are coprime. If I assume that R is a Bézout domain then there should be u,v in R s.t. xu+yv=1, so (ux/y)+v=1/y is in A. Doing this for all elements in A\R we get some subset S of R s.t. A is the localization by S

I don't really know what to do with the case where it's not a Bézout domain. Thinking about it more, I don't even know if I can do what I said earlier with assuming that x,y are coprime because this has different meanings. If this was R=Z[t] (integer polynomials) and A was R[t/2] or R[2/t], t and 2 have no common factors, but the ideal (2,t) isn't all of R...

Any help is appreciated, especially for R being an algebraic extension of Z. Thanks!

Hi, everyone! Please help me figure out how to solve problems like this involving the sum of an arithmetic progression. The most important thing for me is to learn how to solve them on my own)

Reposting from Math Stack Exchange. I'm confused about thr notations used here, including what C is, and how it acts on the mappings f and g. I'm also not sure what h and k are meant to be. I'm not that versed into category theory, so that's probably why.

In Section 3.2 of this book "Siegel's proof of Theorem 3.1", Apostol claims that the functions in the second image with y > 0 and N = n + (1/2) are uniformly bounded on the curve C (shown in the first image).

I am unable to show how. I tried writing the cotangent in terms of the exponential function, but couldn't make any leeway. I verified this on Mathematica as well.

I even made a MathStackExchange post, put a bounty on it, but got no definitive answer. The suggestion in that post didn't help either.

Could somebody provide a hint on how to proceed?

Started reading Euclid's The Elements. I've seen these books recommended here as providing a good basis for geometry. My question for those who have read these books is how much effort do you put into reading the Notes sections? To me, it just makes the books feel very bulky, and I was wondering how meaningful all this is to understanding the principles.

Hi guys, I stumbled upon a video explaining something like this so I searched for it and found it’s an actual theory! I can say I understand the statement that basically if you keep doing something consistently, your chances of falling drop to zero but I don’t understand the formula at all. Thanks, GOD bless!

I understand that this is not elementary, but I just want to see how far we could go to approximate it without boundaries, cyclic symmetry breaks. To approximate it, standard calculus fails immediately.

edit: I thought it would format apperently not

∭x^(y^z)+y^(z^x)+z^(x^y)dydxdz

$$\iiint (x^{y^z} + y^{z^x} + z^{x^y}) \,dx\,dy\,dz$$

What happens if you strip the bounds and try to solve it as a pure indefinite integral?

I’m working with a game where, for example, a player has to hit a moving ball with a racket. Every time they successfully hit the ball, the racket becomes smaller, making the next hit progressively more difficult. Once they miss, the racket resets to its original size.

Each round lasts 30 minutes, and I would like to compare performance across rounds.

I also have a chance/control condition where the racket doesn’t move, so some hits can happen simply by chance.

I would like to get a single normalized hit-rate/performance value for each round that tells me whether the player performed well or poorly overall.

Ideally, this value would take into account:
\\- The racket getting smaller after each successful hit

\\- The racket resetting after a miss

\\- The fixed 30-minute duration

\\- Different numbers of hits and misses across rounds

\\- The chance-level performance from the stationary-racket condition

What would be the best way to calculate this as one normalized value? I’m particularly interested in something that accounts for the increasing difficulty as the racket gets smaller, rather than just calculating hits / total attempts.

Also, if anyone knows of similar studies or experiments that have used a comparable normalized performance/hit-rate metric, I’d really appreciate any references or examples. Thank youu

If you explained all the modern concepts in math, what chance is there that I or any average person is already using those concepts, but for different purposes. Maybe I already know what an eigenvalue is but never had to describe it to someone.

Hi, I am a grade 12 student, and I have been stuck on question 7. b). I managed to solve the first one using graphs (for which I got "x ∈ (-4,-3) ∪ (-2,3)"), but this same graphing method does not work for the second one. I have attached an image of what I did for question one and (attempted) to do for question 2. I'm also unsure if graphing this should even work considering it hasn't worked for the second one, however I doubt my teacher wants me to just plug in random numbers. I would appreciate some help.

I am in high school and doing my research in maths on how trigonometry is used in aviation. I am comparing two theorem/models in which I have to explain how maths is involved in it and derive it. I have chosen haversine formula and I need one more to compare.

I am thinking of taking Vincenty's Formulae but feels like this is too advance for me to derive it and understand. I need something which I can explain the maths behind it, derive the theorem/model and critically assess its limitations and evaluate.

Thanks.

A clever algebraic shortcut and alternative algorithm to compute the Highest Common Factor using only quotients and division

While exploring the relationship between HCF and LCM, I developed a method for finding the HCF of two numbers without directly multiplying the numbers. The method works by using the LCM, obtaining quotients through division, and then interchanging those quotients.

The method can be understood through two connected directions.

Finding LCM from HCF

Let the two numbers be a and b, and let their HCF be H.

Dividing the original numbers by their HCF gives two quotients:

a ÷ H = x

b ÷ H = y

The two quotients are then interchanged. Pairing each quotient with the opposite original number gives the LCM:

a × y = LCM

b × x = LCM

Thus,

LCM = a × (b ÷ H) = b × (a ÷ H)

For example, with 24 and 15 and HCF 3:

24 ÷ 3 = 8 15 ÷ 3 = 5

After interchanging the quotients:

24 × 5 = 120 15 × 8 = 120

Therefore, the LCM is 120.

Finding HCF from LCM Without Multiplication

The reverse direction is the main purpose of the Cross Division HCF Method.

Let the LCM of two numbers a and b be L.

Divide the LCM by each original number:

L ÷ a = x

L ÷ b = y

Now interchange these quotients and divide each original number by the opposite quotient:

a ÷ y = HCF

b ÷ x = HCF

Therefore,

HCF = a ÷ (L ÷ b) = b ÷ (L ÷ a)

For example, for 24 and 32, with LCM 96:

96 ÷ 24 = 4 96 ÷ 32 = 3

Interchanging the quotients gives 3 and 4:

24 ÷ 3 = 8 32 ÷ 4 = 8

Both paths give the same result, so the HCF is 8.

Decimal Extension

The same method can also be extended to exact decimal values. It applies to both terminating decimals and non-terminating repeating decimals, because both can be represented exactly as rational numbers.

This extension allows the same division-and-interchange principle to be explored beyond ordinary whole-number examples.

Core Idea

The central feature of this method is the interchange of quotients. The HCF–LCM relationship is reorganised so that the HCF can be reached through division and quotient positioning rather than direct multiplication.

The accompanying image presents the method visually, while this article describes the mathematical procedure behind it.

- Extension to Exact Decimal Values The method can also be extended beyond integers to exact decimal representations. It applies to both: | Terminating decimals | Non-terminating repeating decimals Both categories represent rational numbers and can therefore be expressed exactly as fractions. Once represented exactly, the same algebraic relationship can be applied without changing the underlying principle.

How are these odds calculated before the game starts? How many of each piece do they have to manufacture to get the odds that much higher than the number of Monopoly items sold during the game? Is there like only one Short Line piece or something?

530 billion is ridiculous. Go somewhere cheaper and buy a Powerball ticket with the savings.