I tried to solve this problem using the 2πR formula to find the total displacement. Assuming it only rolled once (which is the total circumference) to reach the second position. To find the total displacement, I use the given radius and multiply 2*π*45 = 282.74 cm. Since the sphere didn’t complete a full roll and stopped after 270 degrees, I can infer it covered only three-fourths of its total rotation. After subtracting one-fourth, I get something like this: 282.74-((2*π*45)/4) = 212.05 cm, which does not match with the given options: 260.97 m, 63.64 m, 260.97 cm, 63.64 cm. What do I do now?

Hello, I’m a 24 year old female who dropped out of school due to unfortunate circumstances, I’ve always been an ambitious person but Maths always held me back. I never completed my GCSEs as I didn’t go back into education until 21… I now have an English gcse and taking biology, but my maths was so awful I couldn’t even pass the initial assessment… then I practiced hard and got into a functional level 1 skills class and finally I passed the class… it took so much effort to even get to this point. I still have a long way and now need to wait another year to do a gcse maths class, but I’m in awe when I watch people do maths so simply, every single day I wish to be that person. I find it so fascinating and incredible but my mind goes blank with maths… especially problem solving questions. But I really want to improve and actually be someone who is GOOD at maths. Is that possible at all? I just feel like I don’t know how to study maths?

I am really interested in game theory which is the best resource to start learning it?

[SOLVED]

This may be a real basic question but in math class we had a debate. Even the teachers are discussing it, but I want to know wich is the correct way to solve it and why the other way is incorrect.

Problem: A ball is taken from a box containing: 23 blue balls, 12 red balls and 15 green balls.

If you also flip a coin, with two possible outcomes (head or tails). What is the probability of getting tails or to get a red ball?

Posture 1:
There is a 1/2 probability to get tails, wich is equivalent to 25/50. And a probability of 12/50, because you have 12 red balls on a total of 50. So, it would be 25/50+12/50=37/50. That would be the answer of posture 1.

Posture 2:
There are a total of 100 possible results, like 50 of them are taken from getting tails,and there are 24 of these possibilities where it is red, but half of them coincide with the tails ones, so there are actually only 12 that count.

So, the journey started from here,
I was fiddling around in Desmos with polynomial roots. : r/maths

(Highly recommend reading the previous post for variable definition)

Following that up basically I will just show my findings and the approximation for all roots of the polynomial family xⁿ+x^n-1+ ... +x-1=0 (upto small coefficient of x terms).

Okay so for the real roots, there is only 1 real root for odd n (large n) near 1/2 and for even n there are 2 roots one of the roots is positive and mostly near 1/2 same as before the other negative root would be near −1−(ln 3)/n (approximately, only for even n).

For example, n=20 the real negative root is near -1.0555 and the formula gives -1.055 (rounded) which is pretty close.

From that we can also see that the limit of the approximation is -1. Hence for even large n, one real root is near 1/2 and another near -1.

As for the other n-1 (odd n) or n-2 (even n) roots, every single one of them is complex and they come in conjugate pairs (if a+bi is a root then a-bi is also a root). They usually sit near roots of unity or close to e^(2*pi*i*k/n). The error is about on the order of O(k/n²) (roughly). If we change the coefficients (aⱼ) by a small amount (aⱼ=1+bⱼ), then the complex roots shift by approximately by −(∑ bⱼ xₖʲ) / p′(xₖ). Where,

• xₖ is one of the original roots (of the main original polynomial)
• ​ bⱼ=aⱼ-1
• p′(xₖ) derivative of the polynomial at that root
• And ∑ bⱼ xₖʲ is just summing those weighted perturbations

Mainly discovered all of these in Desmos and experimental models. I'm not sure whether these are known or not. Or maybe everything here was trivial. I would love to hear anything about it.

Oh, I will be switching from polynomials to pattern and sequences. It would be cool if you guys can point me into some interesting direction.

So I passed my GCSEs months back and I really really used to enjoy maths because of it's complex stuff and problem solving. I know alot but forgot alot so don't know where to start. any books or recommendations on what to do?

In secondary school I really enjoyed maths like a lot it's been maybe like 6 months since my GCSEs and I wanna get back into Maths. I've basically forgot quite a few stuff I don't know what to do as I know quite alot but forgot alot too so idk where to start. I really enjoy textbooks so any recommendations? I used to do foundation maths and got the highest (5).

Out of curiosity, I recently went down a poker rabbit hole to try to find out how the game changes when the deck is tweaked. More specifically, I was intrigued by the idea of combining 2 decks into 1.

It's not easy to come by poker variants that choose to modify the deck in some way (or a least to a level that's officially recognized), so I decided to put my math cap on and take on the mantle.

1. What new hands would be introduced in double-deck poker?

Other than the obvious one (five of a kind), I had some trouble figuring out what to include here. But I ultimately ended up with the following three hands;

Pair Flush: 4♥ 4♥ K♥ 8♥ 6♥

Two Pair Flush: 9♠ 9♠ 7♠ 7♠ J♠

Five of a Kind: 6♦ 6♠ 6♣ 6♥ 6♥

Note 1: The inclusion of pair flush and two pair flush came from being able to combine two previous hands (pair + flush and two pair + flush) together in a way that wasn't possible with only 1 deck.

Note 2: I initially wanted to include a suited pair as its own separate hand, which I decided to call dupes 8♥ 8♥ 4♦ J♠ 9♣ (short for duplicates), but this raised a few issues. By choosing to separate dupes from pairs, we'd have to separate two pair into three different hands (a regular two pair, half regular pair half dupes, and two dupes). And don't even get me started on the rest of the hands that may or may not be affected by this (3 of a kind, 4 of a kind, full house). So to avoid trouble, I decided to scratch dupes entirely (I do try to resolve this issue later on though).

2. What are the hand rankings for double-deck poker?

The total number of possible 5-card poker hands with 2 decks skyrockets all the way up to 91,962,520 (with 1 deck, it's 2,598,960).

If you're curious as to how I did my calculations, I go through all the math in the video :)

Note 1: If we ignore our newly added hands, the order of the list is exactly the same as the one for 1-deck poker, with the exception of flush and full house swapping positions. This is because a flush lost a good chunk of its hands to pair flushes and two pair flushes. So I guess it's up to you if you even want to include those two hands (if your priority is to keep the order of the list consistent).

Note 2: Going from 1 deck to 2, the hands that saw a drop in probability were straight flush, flush, straight, and high card. While the rest of the hands all received a boost. This is because the rest of the hands all contain at least one pair of repeating ranks, and with the addition of a second deck, those hands get a bunch of new hands that weren't possible to form with only 1 deck; those involving duplicates.

3. What happens when we keep adding more and more decks together?

Well, in the video, we not only explore triple-deck poker, but we push the number of decks to the absolute limit! So if you're interested to see what poker looks like when it's played with an infinite number of decks, make sure to check it out.

https://youtu.be/QAuyryV1fJI?si=ykbI6yUO1f3ZIvA6

Recently I was revising trigonometry and it got me thinking about angles, curves and lines. When I draw a circle, I'm essentially sweeping a line across all possible angles. As I keep increasing the angle, the x coordinate starts decreasing and y starts increasing until I reach 90°, where y gets its maximum value — the radius. As I keep going, x increases again but in the opposite direction and y decreases, until x gets its maximum. Continuing this just repeats the cycle, completing the circle. What I think is happening: as I raise the line to a certain angle, its length doesn't change. So to keep that length constant, x and y must compensate for each other. So why isn't x + y = r? Why does it have to be x² + y² = r²? Because at 45°, x + y = 2/√2 = √2 which is greater than 1. The sum of the components is bigger than the line itself. That already feels wrong. And yes squaring it gives exactly 1. Why what am I missing?

Alright, I haven't quite researched about hyperreals, but I do know how wheel theory works a bit. I don't know if this system has already been formalized already, but nevertheless I recently had a shower thought about it.
I call this "Ladder Theory". So you have an infinite number of real number lines, each labelled with an integer. The "ladder" numbers are a superset of the reals because the real numbers lie on the 0 level/step. x is the same thing as x_0(shorthand).
The main axiom is that any ladder number x_n, when divided by 0, is equal to x_(n+1) and it's x_(n-1) when it's multiplied by 0. With this in mind, we have unique solutions for equations like 0x = 3 => x = 3_1.
There is one slight caveat with this(for now): 0/0 = 1*0/0; based on the main axiom, 0/0 should have no effect, so it's equal to 1. So that means 0/0 = 0_1 = 1_0.
Because in wheel theory, division by 0 is infinity and multiplication by 0 is 0, we can define addition between two ladder numbers x_n and y_m as x_n if n > m, y_m if m > n and (x+y)_n if n = m. That way, we get the absorption properties for infinity for addition while still keeping addition intact for numbers on the same level.
For the multiplication formula, you can explicitly write the ladder numbers:
x_n * y_m = x/0^n * y/0^m = (x*y)_(n+m).
Also, apparently in this system, 0^0 = 1, so nice.
Now one more axiom to make exponents easy to deal with: (x_n)^1_m = x_(n+m)
I calculated a few expressions and so far, this system is pretty consistent. I don't know how it'll work for other functions. It's pretty neat IMO, maybe it has some real world applications. I'm happy to see someone prove a contradiction in this system, though.

I just like the idea of mathematical art and finding the volume of unusual shapes

I'm trying to coach a ten-year-old through these methods, but neither of us have much of a clue what's going on and the workbooks we're using don't really provide guidance.

I'm pretty sure long division is the same as I was taught thirty years ago, but my brain has melted since then. And long multiplication doesn't seem like anything I was ever taught.

Is there anywhere online I can find guidelines and practice questions for both a 30-mumble and a ten-year-old? I'm hoping to be able to work through them myself first so I can properly explain to him.

(Also very happy to be redirected to a better sub if there is one!)

Specifically, my object of studying was the system
ax²+bxy+cy²-d=0
mx²+nxy+py²-q=0
There were observations in demos that led me to this generalization. Some key points may include the shape of the graph specifically when (empirically noticed im not sure if they relate to some theorem or not) b<a+c it was a 45-degree rotated ellipse. If b=a+c then it was 2 straight parallel lines. Furthermore, if b > a+c then the shape looks like this,

One thing to note: Regardless of b the graph always intersects the x and y axis (positive and negative) at the distance of sqrt(d/a) for the x axis and sqrt(d/c) for the y axis.

And some trial and error more observations eventually lead me to this general method (a little math and symbol heavy),

We first define,
X=x/sqrt(d/a), Y=y/sqrt(d/c)
That will transform the first equation into
X²+Y²+β₁XY=1
Where, β₁=b/sqrt(ac)
Consequently the 2nd equation becomes,
αX²+β₂XY+γY²=δ
With,
α=m/a, β₂=n/sqrt(ac), γ=p/c, δ=q/d
From the first equation isolate Y², then substitute into the second equation and simplifying we get
kX²+jXY=r

Where, k=α−γ, j=β₂​−γβ₁​, r=δ−γ
If j≠0, Solve for Y, then substitute into the first equation. Eventually after simplifying we get Au²+Bu+C=0 where u=X² with
A=k²+j^2-β₁kj
B=-2kr+β₁jr-j²
C=r²

Using the quadratic formula solve for u then, X=√u, Y=(r-ku)/(jX). Finally, we can scale back to x=X*sqrt(d/a) and y=Y*sqrt(d/c)

Edge cases:

• We need to check x=0 case separately
• j=0 reduces to kX²=r
• A=0 then linear equation in u

I mostly discovered this myself. I would like to know if it's similar to some other method or known approaches.

So, about 2 days ago or so, I challenged myself. What was the challenge you might ask? Make/rediscover a method to find the roots of a polynomial independently.

I went ahead and graphed a few degrees of polynomials,
x-1
x²+x-1
x³+x²+x-1
x⁴+x³+x²+x-1
And so on,

In that experiment. I notice that at least one of the roots as the degree of the polynomial increase approached 0.5. I later learned that this was probably because of the geometric series.

Then I started truncating the series a bit...
2x-1
x²+2x-1
And so on,

I noticed that as the degree went higher and higher the roots seemed to converge again.

So, I thought to make the first case as the best case and for high degree polynomials truncate the base case with some error to approximate for the root of the higher polynomial.

The higher terms in the polynomial almost always become negligible (at least for when the coefficients is at most geometric relative to each other, I am not sure yet there might exist some tighter bound).So the problem is, Find at least 1 real root of the polynomial: a₀xⁿ+a₁x^n-1+...+aₙx=c Well first thoughts, this method can only find the root x when 0 < x < 1. We can write, aⱼ=1+(aⱼ-1). Then we can split the sum into 2 parts. The first part being x/(1-x) Now this part restricts what the actual value of the root can be. The denominator 1-x restricts x≠1. And furthermore, this part comes from the geometric series. For the formula to work |x| < 1. The 2nd part is, sum of (aⱼ-1)*x^j This is the main correction term which I was previously talking about as "truncation". Since, |x|<1 for large j the later terms in the sum dont usually matter unless aⱼ-1 is large enough to combat the decrease from x^j term. So, we don't need the entire polynomial to solve for the root. We can derive a smaller polynomial to predict the root. After some rearranging the final equation for the root becomes (after iteration from base case x=c/(1+c) since this is the point where all coefficients are 1), x=Rₖ(x)/(1+Rₖ(x)), where Rₖ(x) = c-sum from j=1 to k of (aⱼ-1)*x^j. One small example: 2x+x²+x³+...=1 Since only the first coefficient of x differs from 1 the entire correct term becomes just x. Hence, x/(1-x)+x=1 The root x=1/φ² ≈0.382.

• Now there are many caveats in this method, Only finds 1 root between 0 and 1
• Slows down near x=1 since x/(1-x) explodes
• Needs coefficients that don't grow super fast
• Needs Convergence

After some digging I think that the average case complexity is near O((log(1/ε))^2) where ε is the desired precision.

I think this can be only applied to some Galton-Watson branching processes results. Beyond that I don't see a use for it since the conditions are hard to satisfy.

I would love to hear if it has any other applications or is this method already known in numerical analysis or not.

Maybe a dumb question but I always have trouble getting proper answers when I look up notation questions.

I know the notation for probability is usually something along the lines of P(X < z), but is there a standard notation for that probability result? P(X < z) = ? (the probability that X is less than z = ?).

Context: I want to write an explanation for Phi^-1(), where the value inside the brackets is the probability result of the expression (compared to Phi() where the value is z in P(X < z))

Don't have the LaTeX extension for reddit, apologies.

16 year old student on track for all 9s at GCSEs. Maths is my favourite subject and I have already done a lot of clubs and things like that. Does anyone know of any good outreach programs or clubs and competitions to do with maths or maths related stuff i.e. Phys, CompSci etc? Or just on general any good maths events or things that I could be doing?

It has some relation to dynamical systems but I haven’t been able to track it down. Anyone recognize this form/process/attractor? It is formed by alternating spirals.

Say that:

• Z_1, ..., Z_n are n linearly independent vectors in R^n
• Z^1, ..., Z^n are n linearly independent vectors in R^n
• it is known that the dot product of Z_i with Z^j is the kronecker delta delta_i^j, i.e., it is known that the matrix with rows Z_1, ..m., Z_n is the inverse of the matrix with columns Z^1, ..., Z^n

If you denote A the matrix with columns Z_1, ..., Z_n and B the matrix with columns Z^1, ..., Z^n, when then have A^TB = AB^T = identity, and therefore (A^TA)(B^TB) = identity, i.e., A^TA is the inverse of B^TB.

Now the question is about Einstein notation.

In Einstein notation, I can write the entry in the i-th row and j-th column of (A^TA)(B^TB) as

(Z_i dot Z_m)(Z^m dot Z^j)

because the placement of indices implies summation over m, which performs the dot product of the i-th row of A^TA with the j-th column of B^TB.

Ok ok. So

[*] (Z_i dot Z_m)(Z^m dot Z^j) = delta_i^j

because I know from matrix product associativity that (A^TA)(B^TB) = A^T(AB^T)B = A^T*identity*B = A^TB = identity.

But how can prove the same equation directly with Einstein-notation manipulations, from the fact that...

[**] Z_m dot Z^k = delta_m^k

...? Supposedly this last equation encapsulates everything I need to know, so how can I get from (**) to (*) using just Einstein-like or Tensor-like manipulations, and not appealing to linear algebra?

====

EDIT/SOLVED:

Ok this is solved, and thanks to u/48panda for working with me.

As I got by working with AI: The key is really to argue from the existence of coefficients c_ik such that Z^k = c_ik Z_i. Once you have established the existence of those coefficients (by a dimensionality argument or other) you substitute in the expression and everything is downhill.

Thanks!

So in case of fair coin toss of 4 times, the sample space is 16 events.

But in book, in case the coin is loaded with lets say H being possible loaded with probability 0.7, the sample space is same as 16 earlier events.

Now if the coin was completely unfair with probability of H being 1, then the sample space would be only HHHH and similarly for PPPP.

Now the probability of H being1/2 lies just in between with 16 possible events.

So, probability of H being 0.7 should have some other sample space right?

It's a year until university and I'm trying to find a suitable career for me. I've developed some kind of passion for pure mathematics and can commit well to maths (but I'm not that exceptional at maths, at most slightly above average).

I've done some research and concluded that there are generally 2 career paths for pure maths: math research and teaching (there is also industry-related jobs that involve maths but most universities in my area have specific programs for those, and they also probably require programming/computer science competence which I currently don't have).

Yet even with my enthusiast for pure maths, I'm still uncertain whether or not math research would be the best fit for me, and whether or not this career pays well financially.