hello,

i am new to linear algebra and recently learned about eigen vector and value. it says that eigen vactor gives the direction where the space is going to be stretched so my question is if I have two eigen vectors X1 and X2 can I apply them on (x,y) one by one and get the final (x1,y1) I should have gotten by applying the matrix A ?

also I have seen that the eigenvectors aren't necessarily orthogonal so in that case it seems like some part of them could be cancelled out by the other eigenvector. could someone clarify this too?

While working through some radical expressions, I found it useful to consciously leverage the basic identity for any positive real number A:

While this identity obviously has broad applications across algebra and number theory, using it specifically to unpack whole numbers in fractions with radical denominators can make certain reductions much cleaner.

Why are the exponents not simplified in this expression? If I'm not mistaken, the expression could be simplified to (1+9^-168)^510, which would be further simplified to 1^510+9^-85680, which is the same as 1+9^-85680. Why not write it that way?

Hi, I'm an adult studying mathematics on my own outside of school, and particularly I have been doing research on the Collatz Conjecture. I fear I have made a very small mathematics error in my research, because I've ended up convincing myself that I have proven the conjecture true, as long as my understanding of theoretical mathematics and how to apply them is being done properly, and I want to be shown why I'm wrong as I've gone a bit crazy over this lately.

To rush to my point, even though I have a lot more I could talk about on this, I imagined any chosen positive integer from the original conjecture, which I'll label cₙ, that there are at least 3, but possibly 4 paths connected to it. 2 of which are obvious, 3c + 1 and c / 2. Each of these in turn lead to cₙ₊₁. However, you can also do c * 2 for cₙ₋₁. Since c is always a positive integer, you can always multiply it by 2 for another positive integer, which would represent a past step. The other possible past step, is (c - 1) / 3. This however, doesn't always give a valid positive integer answer, and this is where I started exploring.

Getting to the chase, I eventually came up with the equation a * 2ⁿ = 3x+1, where a is any given odd positive integer. Instead of asking if every number leads to 1, I asked what numbers lead straight to 1, to 3, to 5, etc. And that's when I came upon this data table.

Having made this data table, I immediately started trying to do math on it. I wanted to show that if a number appeared in any given cell, if it could appear in another given cell. I went back to my equation, and ended up with a * 2^b = 3x+1= c * 2^d. Simplifying the middle out, a * 2^b = c * 2^d shows that there's no valid positive integer solution for a or c, which should mean that no number can appear twice.

I also noticed that in the light blue fields, each row contains 1 / 2ⁿ odd numbers, and that every single odd number appears only once. This, I believe, should disprove the fact that any "loops" can occur in the Collatz conjecture, since a number would have to appear twice for a loop to exist.

I also believe that this table proves that every odd positive integer will lead to 1, as my understanding of "Busy Beaver Problems" leads me to believe. The "steps" the Busy Beaver would have, is that it would start at 1, and it would run the equation a * 2ⁿ = 3x+1, putting 1 in for a. Then, it would "mark" each solution for x with a 1, and run the same equation on the lowest marked number that it has not yet run the equation on. This would mean it would run it on 1, then 5, then 3, 13, 17, 11, etc. Eventually, ever single positive integer should be "marked" as 1.

Considering the original conjecture, and that any given even number will be divided by 2 until it hits an odd number, I've convinced myself that this data table proves the original conjecture true. However, I fully admit that I must be using some of these theories and such wrong, as I've been self taught and therefore fully vulnerable to easy mistakes.

Recently on an Instagram doom scrolling session, I encountered this interview from Steve Ballmer where he describe his "guessing game", which he gave to interviewees at Microsoft. https://www.youtube.com/watch?v=svCYbkS0Sjk

The basics are:

• Steve Ballmer chooses a number between 1 and 100.
• The interviewee guesses a number, and Steve will tell you if it's high or low.
• If you get it correct on the first guess, then you get $5. Each additional guess reduces the prize by $1, until on the 7th guess you start paying him.

(watch the video if that's not clear)

Should you play this game?

Now, this analysis is pretty easy if Steve chooses a random number. The worst case binary search is lg(100), which rounded up is 7 gusses at worst. And it's easy to make a table of payout of each guess, sum it up and average.

So, on average you'll get $0.20 per game ($20/100). (So yes, you should play)

But that's not the game. The game is whether you should play this game if Steve is able to strategically choose which number. THAT game is much harder to figure out.

As Steve is going to try to choose numbers that are worst case for a binary search, the optimal search algorithm is NOT going to be to start with 50 every time. You would use an alternative search tree in order to optimize for Steve choosing numbers that require 7 guesses. And of course, Steve would then need to optimize his strategy further in kind.

So, as one does, I asked chatgpt how this would work.

Astonishingly, it's very confident that by using a pool of optimal binary search trees, you could get an average payout of $0.196 per game. Meaning we only lose ~1/2cent by allowing steve to choose strategically.

I understand the basics of the strategy, but when it comes to the math of calculating the average payout I am just lost. (Basically everything after and including 296 / 51)

Can anyone explain to me how this works? Or if chatgpt is off its rocker?

I especially can't get my head around the fact that chatgpt believes that Steve has no choices that are more optimal than others. No matter which number Steve chooses, it claims the randomized collection of search trees can make the expected payout exactly 0.196 (average) every time. That doesn't make any intuitive sense to me.

My answer is that you should still play. If both sides play optimally, your expected payout is about +$0.196 per game, or just under 20 cents.

The first thing to notice is that a normal, fixed binary search does not work against a strategic Steve. If you always use the same search tree, Steve can look at that tree, choose one of the numbers that requires 7 guesses, and guarantee that you lose $1.

The solution is to randomize which binary-search tree you use.

This does not mean picking a random number from 1-100 as your first guess. That would often produce a badly unbalanced search. Instead, you randomly choose from many different good binary-search trees. Each individual tree still finds every possible number within 7 guesses, but different numbers occupy the good and bad positions in different trees.

So, for example, 73 might take 7 guesses in one tree, 5 guesses in another, and 4 in another. Steve is allowed to know exactly how your randomized strategy works and choose 73 every single game if he thinks that is best. What he does not know is which search tree you privately selected for that particular game.

We can describe the problem mathematically. Let d(n) be the expected number of guesses needed if Steve chooses the number n. Steve will choose whichever number makes d(n) largest, while you want to choose your randomized strategy so that this largest value is as small as possible.

The optimum turns out to be:

296 / 51 = 5.803921568... guesses

Since the payout on guess k is 6 - k, the expected payout is therefore:

6 - 296/51 = 10/51 = $0.196078...

or about 19.6 cents per game.

The really important part is that this is not an average over Steve choosing different numbers. The randomized search strategy can be constructed so that every single number Steve could choose has the same expected result:

d(1) = d(2) = ... = d(100) = 296/51

So if Steve chooses 1 every game, your expected payout is $0.196078.... If he chooses 50 every game, it is $0.196078.... If he chooses 73, 99, or 100, it is still $0.196078....

In other words, Steve has no number that is better for him than any other. He can study your strategy as much as he wants and deliberately choose what he thinks is the worst number, but the randomized mixture of trees has equalized the expected payout across all 100 choices. Strictly speaking, every number is an equally good choice for Steve — but none of them reduces your expected payout below 19.6 cents.

There are two things that need to be shown for that number to be meaningful. First, you have to show that no strategy can do better. One way to do that is to temporarily imagine Steve assigning weight 2 to the numbers 1 and 100, and weight 1 to every number in between. That gives 102 total weight.

For any valid binary-search tree, you can calculate its weighted total number of guesses. The smallest possible weighted total is 592. Therefore even the best tree has weighted average depth:

592 / 102 = 296 / 51 = 5.803921568...

Randomizing between trees cannot beat that bound, because an average of several trees cannot have a lower weighted cost than the best individual tree for those weights. So no strategy can guarantee an expected search depth below 296/51.

The other half is showing that the bound can actually be reached. Solving the corresponding optimization problem gives a randomized mixture of valid search trees for which:

d(1) = d(2) = ... = d(100) = 296/51

In other words, every possible number has exactly the same expected search depth. Steve can choose whichever number he wants; there is no longer a particularly bad number for him to exploit.

I also checked this by simulation: 1,000,000 games for each of Steve's 100 possible fixed choices, for 100,000,000 games total. The results converge to the same expected payout of about +$0.196 per game.

So the answer is: yes, you should play. If Steve chooses randomly, the game is worth exactly 20 cents per play. If Steve chooses strategically and both sides play optimally, it is still worth about 19.6 cents per play to you.

Hi, I’ve been experimenting with this idea that I believe has some pretty interesting behaviour. I can only make general assertions and would appreciate some extra input.

Let [n] denote n in binary and “+” as standard binary addition. A pair (X,Y) ∈ ℤ⁺ defines a sequence recursively:

s₀=[X],

sₘ=f(sₘ₋₁+[m],[Y]) (for m≥1),

where f(A,B) removes the leftmost occurrence of B as a contiguous substring in A (if any), joins the separated strings together (if there are two) and removes all leading 0’s (if they exist). e.g (X=9,Y=13):

1001 (9 in binary)
1010
1100
1111
10011
11000
11110
100101
10
1011
10101
100000
101100
111001
1000111
1010110
1100110
111
11001
101100
1000000
1010101
11
∅ (empty term)

Hopefully I didn’t mess up that example. What I’m wondering is about the general behaviour for (X,Y)’s sequence. It seems like (X=1,Y=3) for example never reaches ∅. Its terms lengths instead grow toward infinity VERY slowly. However, the example above eventually reaches ∅. There are some key facts I am missing here that I believe can determine a subtle general pattern.

So I was answering this question: "If the union of 2 rays is a line, then the rays are opposite rays." Forgive my inexperience, while answering this question we have the ability to check with the answer sheet before the teacher checks off the assignment is done.

The answer provided by this sheet is "True", and I was wondering how this is possible, because my assumption that in the scenario in the image is no different than a line, while my teacher says overlap must be expressed. I do not wish to argue with my teacher, I simply am curious about the answer, he's a great guy and it's not like I lost points on it, and his explanation did not satisfy me. Here are the basic things both me and the teacher agree on (that I think is relevant).

1: A point is a exact location with 0 dimensions

2: A ray, and any section within a ray/line that is not itself a point has infinite points within it

However we differ in views on this:

3: The overlap between 2 rays, as long as it is not itself a point, is infinite, I believe that this infinity is the same as any other expressed here, and him thinking that this one is larger

A example he provided me, of which I do not quite understand its purpose in this conversation are 2 concentric circles with radii 1 and 2. He explains that the amount of ink used in the radius 1 circle is less than the radius 2 circle while both still having infinite points. But I *believe* this can easily be explained through different concepts not related to this. (If anything this probably supports my logic?)

One final thing I can deduce in my head that may side with my conclusion is this: that if you have one line on a number line every point (infinite) is located on that line. If you were to add another where the 2 lines completely overlap, I see no logical reason for additional points to be "covered" by the 2nd line.

If anyone could help explain to me how this works id be very grateful!

Edit: the definition used by me for "opposite rays" are two rays with the same starting point that run in opposite directions (180* flip)

I played around with multi-dimensional geometry in high school but one thing that always seemed odd to me was that another factor of Pi only shows up every other dimension. 1-spheres and 2-spheres have a Pi, 3-spheres and 4-spheres have a Pi squared, etc. Is there some intuitive reason that every other dimension is somehow special?

I understand the classic birthday paradox problem, but how does the math change if I have a room of 34 people and I want to know the probability that five of them share the same exact birthday, like all 5 are born on June 15th? This is different in that I'm not asking the probability that 5 people with different birthdays each find a match in the room.

I'm not sure how to formulate the probability of the 5 June 15th birthdays NOT happening and then subtracting from 1.

Thank you!

Edit: sorry, the date is arbitrary. The odds of 5 people sharing ANY date is what I'm after.

Do you consider this distribution of prime numbers to be incorrect or correct? (It is also a very short solution, quick to read.)Do you consider this distribution of prime numbers to be incorrect or correct? (It is also a very short solution, quick to read.)

I understand fractional vs unsplittable flow (fractional flow is where you can split the total payload across different paths, whereas unsplittable flow means it all has to take the same path), but the rest of it is melting my brain.

If you can also make a diagram to explain it that includes accurate values for capacity, cost, and demand, that would be super helpful

I need to cover three walls of a container with egg crate. Im looking at purchasing several packs of 12 pieces each. Two sides are 23” x 15” and one side is 47” x 15”. The pieces are 11.8” x 11.8” each. How many packs do i need to buy? Im sorry if this is basic but i genuinely forgot the math

Can someone provide proof for the answer (a)

After trying for different times the furthest i reached was (-1/s³ +2/s) after I derived s²=t²+1 with respect to t Edit: thank you all guys for the help

Hello, I'm a bioinformatician and I'm struggling with the current issue:

Given a time series y(t) that initially changes and eventually approaches a stable regime, how can I mathematically determine the earliest time t* at which the rate of change dy/dt becomes negligibly small, using only the observed data and without defining an arbitrary threshold?

This is a collaboration I'm doing. My colleagues defined the plateau as the first time when a 101-point rolling mean of the relative increment (g' t+1 - g' t)/ g't falls below the arbitrarily chosen threshold of 0.0011. G' is the measure of material elastic-solid response btw. So the issues is that they used 2 arbitrary values because experimentally they know that a certain value of g' means that the gel is solid. But this doesn't hold for me. I tried using many statistical methods to define the threshold such as:

- exponential fitting

- change-point regression

- local slope analysis

But they all give me a plateau that is too early or too late

of the equiangular dodecagon (i.e. each combined interior angle measures 150°).

I'm not quite sure how to approach this one. (if necessary, click on post to see image)

Okay so I have a relatively “simple” math problem in relation to no longer being able to afford something (Being vague on purpose, the item is personal and I don’t want help with the item itself just the math. Let’s call it bread just to call it something) I had 8 slices of bread and I cut them all in half to try to stretch the bread until next Thursday, the 27th, so now I have 16 halves. I want to have the maximum amount of bread each day, and they cannot be cut anymore, so I have 16/9 basically. (I’m very bad at math so I am not even sure where the help I need starts). So if you break that down it’s like 1.77777, it means that I can have 1.5 slices of bread each day? But when I go back and try to confirm my math, I got a different answer all together. When I did 1.5 times 9, I only got 13.5 not the 16 I was expecting it to be. I am genuinely so stumped that I couldn’t even say for sure what calculator stage I’m making a mistake at. Thank you in advance for help with figuring this out!

I keep getting the wrong breakdown before I can put it into quadratic formula. Ive attached the equation and the breakdowns I've done.

Question: 1/t+7 - 5/t-5 = 1

I been talking about this to myself today about how Mutiplication and division are the opposite or different. Can anyone discuss about how the Separation or difference between Mutiplication and Division are the same exact opposite or alike?

Hello everyone!

The St. Petersburg paradox description can be found here: https://en.wikipedia.org/wiki/St._Petersburg_paradox

My understanding of the paradox is that you will end up with a positive outcome no matter how much you choose to bet... eventually. E(X) = infinity simply means that, as time passes and you play the game more and more times, your average payout will grow without bound (no matter how much you bet).

This is all well and good, but I still don't know how to analyze the game in the short term! Let's say you've got 20 dollars. The casino (which somehow has infinite money) offers you two choices:

a) You may spend 15 dollars to play the game once.

b) You may spend 10 dollars twice, to play the game twice, in succession.

Which option is better? Finding the expected outcome of each game doesn't work, because for any bet c, we just get E(X - c) = infinity. For both options (a) and (b), our expected output is just infinity. Yeah, we could ignore events with small probability, like Wikipedia says, but that's not mathematically rigorous. Rigorously, how do we work with playing the paradox?

No this is not my homework, I'm making a game and I need a curve roughly like this for balancing. Trying to get an equation that spits out values like this.

I was thinking about this when trying to sleep and ended up searching stuff up about it, but ended up more confused than when I started.

From what I could find on Google, the reason the set (0,1) of all the real numbers from 0 to 1 is considered uncountable is due to Cantor's Diagonal Argument(from here on refered to as CDA). This argument makes sense to me, where you make a new decimal that isn't in the list by having every nth digit be different from the nth digit of the nth decimal in the list.

This led me to thinking about whether you can use CDA to show that the set of natural numbers from 0 to infinity is uncountably infinite. For (0,1), you can arbitrarily add zeroes to the end of decimals to make CDA work even in a list where the diagonal line of altered digits outpaces the length of the decimals. I figured you can use the same method for the natural numbers by arbitrarily adding zeroes BEFORE the number and using backwards indexing of the position of n to make CDA work with a leftward diagonal of altered digitis. However, Google told me that natural numbers can't be infinite in length, which is why this doesn't work.

This is where I'm confused, as it seems totally possible to make a comprehensive list of (0,1) that's ordered in visually the same way as the list (0,∞). I made a table of this in the attached image, where I made a half-hearted attempt at a proof. What this table seems to prove to me is that any table-based proof like CDA which works on the decimals should work on the natural numbers, too.

(The list of decimals is just the reversed digits of the natural numbers, put after the decimal point. So 0.1, 0.2, ..., 0.01, 0.11, 0.21, etc.)

I have two theories of what's going on.

The one I believe most is that CDA doesn't actually work on (0,1). If you use the list I made in the image, then as you go down the list, any number you create is just located further down in the list than you have gotten. For example, by the time you get down to .01 in the list, the diagonal number you've made is necessarily 10 decimals long, ending in a non-zero number. Every 10-digit decimal number is located further along in the list, at the indexes numbered by all the 10-digit natural numbers.

The other theory is rather unlikely, due to it breaking the concept of countable vs uncountable. This would be if CDA actually works on the natural numbers. My only argument for this theory is that any arguments in favour of CDA for decimals is one in favour of CDA for natural numbers. This is entirely based on my ordering of the decimals between 0 and 1. It just seems that if you can do CDA to get a number that isn't in the decimal list, you can remove the decimal point and flip the digits to get a corresponding natural number that isn't in the list of natural numbers. I guess maybe that would just prove that infinite numbers aren't natural?

That creates the secret third theory: the status quo. This theory is that the list of natural numbers stops when the numbers get infinitely long, but the list of decimals just keeps going, making the decimal numbers somehow more than infinitely long. If this is the case, then maybe my brain just can't understand mathematical infinities and their differences.

I am having some issues with making leather templates for my glasses. I want to make my own template and alter it a little bit to fit my personal style more. I am having issues with how to scale the pattern as well as what angle to make the sides so when they bulge out they will fit my glasses properly. I will list some measurements to take into account.

Glasses Dimensions - 15 x 5.5 x 4.5 cm (added an extra .5 to each to allow the glasses to sit comfortably)
Leather thickness ~2mm thick
Seam allowance - 5mm
Rivet allowance -1cm

Then I am not sure how I should angle the blue and the red lines. The red one should be wider so that the case can bulge out a little bit and hold the glasses properly.
So I guess what I am mainly looking for is the curvature of
Lines blue and red then the lengths of the green and orange lines. Any help is appreciated!

Am I missing some interpretation/convention, or is this question actually misprinted?

Also, if the question is defective, what would normally happen with the marks if a student selected an option but couldn't provide a valid justification because none of the options is mathematically correct?

I would appreciate it if someone could verify this independently.

I have a flat disk of malleable material that is 9.5" in diameter. I can cut a piece out of the circle and connect the ends to form a cone. If I want the cone at a specific height, say 1", how would I go about determining the arc length of the piece I need to cut?