Edgy Monist
@edgymonist.bsky.social
Discrete anarchist mathematician.
Mathématicien anarchiste discret.
Mathématicien anarchiste discret.
So they finally found x after all those years?
May 2, 2025 at 5:45 PM
So they finally found x after all those years?
I am sure they didn’t say that just because the class was only on every other day 😉 great job!
May 1, 2025 at 11:16 AM
I am sure they didn’t say that just because the class was only on every other day 😉 great job!
Woah there! Did you just assume a and b to even exist? I want my existence proof first! Uniqueness can then be left as an exercise to the reader.
April 28, 2025 at 1:06 PM
Woah there! Did you just assume a and b to even exist? I want my existence proof first! Uniqueness can then be left as an exercise to the reader.
My_book_revised_final-version_v2_final-for-real-this-time_v3(4).pdf
April 27, 2025 at 4:11 PM
My_book_revised_final-version_v2_final-for-real-this-time_v3(4).pdf
First they = the reference. Second they = the administration.
April 8, 2025 at 11:52 AM
First they = the reference. Second they = the administration.
The reference is somewhat wrong I think? They say epsilon and phi cancel each other out. But didn’t they take the ratio and divide by 2? So it’s like if they decided that epsilon*phi = 2, which is not canceling out.
And also there is a minimum of 10%, so there is a condition somewhere.
And also there is a minimum of 10%, so there is a condition somewhere.
April 8, 2025 at 11:51 AM
The reference is somewhat wrong I think? They say epsilon and phi cancel each other out. But didn’t they take the ratio and divide by 2? So it’s like if they decided that epsilon*phi = 2, which is not canceling out.
And also there is a minimum of 10%, so there is a condition somewhere.
And also there is a minimum of 10%, so there is a condition somewhere.
That’s just a circle… if it’s me drawing on the blackboard 😅
March 27, 2025 at 10:26 AM
That’s just a circle… if it’s me drawing on the blackboard 😅
Even if it is not especially about doing research, I think Ian Stewart’s “Letters to a young mathematician” is a worthwhile read, to complement more straight to the point resource.
March 25, 2025 at 12:04 PM
Even if it is not especially about doing research, I think Ian Stewart’s “Letters to a young mathematician” is a worthwhile read, to complement more straight to the point resource.
Anyway, the line of equality could be defined in the limit (with e.g. 10, 10+epsilon, 10+2*epsilon, …) it’s just the interpretation can be tricky and not “pass to the limits”, so to speak, when talking about the end result of the limit process.
March 25, 2025 at 11:37 AM
Anyway, the line of equality could be defined in the limit (with e.g. 10, 10+epsilon, 10+2*epsilon, …) it’s just the interpretation can be tricky and not “pass to the limits”, so to speak, when talking about the end result of the limit process.
So if we ask ourselves of every household, if it is in the bottom x% or not (x>0), to answer consistently, they would all be included, so they would have 100% of the income/wealth.
March 25, 2025 at 11:06 AM
So if we ask ourselves of every household, if it is in the bottom x% or not (x>0), to answer consistently, they would all be included, so they would have 100% of the income/wealth.
Here’s a citation from the Wikipedia source: “it shows for the bottom x% of households, what percentage (y%) of the total income they have”.
If everyone has the same amount of income/wealth, the “bottom x%” does not uniquely exist.
If everyone has the same amount of income/wealth, the “bottom x%” does not uniquely exist.
March 25, 2025 at 11:06 AM
Here’s a citation from the Wikipedia source: “it shows for the bottom x% of households, what percentage (y%) of the total income they have”.
If everyone has the same amount of income/wealth, the “bottom x%” does not uniquely exist.
If everyone has the same amount of income/wealth, the “bottom x%” does not uniquely exist.
I have a problem with the term “line of equality” we usually can find with that type of graph. If everyone has the same wealth/income, there are no population quantiles anymore, so no line…
March 25, 2025 at 12:40 AM
I have a problem with the term “line of equality” we usually can find with that type of graph. If everyone has the same wealth/income, there are no population quantiles anymore, so no line…
How do we test if they have been scraped?
March 24, 2025 at 11:58 AM
How do we test if they have been scraped?
Yes, with a = 2 and b = n + sqrt(n^2+4).
March 21, 2025 at 7:05 PM
Yes, with a = 2 and b = n + sqrt(n^2+4).
Yes, starting from a rectangle with sides of length 1 and sqrt(2), the process will yield rectangles that alternate between scale factors of sqrt(2) and 1+sqrt(2).
March 20, 2025 at 4:49 PM
Yes, starting from a rectangle with sides of length 1 and sqrt(2), the process will yield rectangles that alternate between scale factors of sqrt(2) and 1+sqrt(2).
That rectangle is similar to the previous rectangle with a scale factor of (sqrt(5)+1)/2, i.e. the golden ratio!
So if we continue the process, we always obtain rectangles that are similar to the starting rectangle, and thus we never terminate on a final square.
So if we continue the process, we always obtain rectangles that are similar to the starting rectangle, and thus we never terminate on a final square.
March 20, 2025 at 3:34 PM
That rectangle is similar to the previous rectangle with a scale factor of (sqrt(5)+1)/2, i.e. the golden ratio!
So if we continue the process, we always obtain rectangles that are similar to the starting rectangle, and thus we never terminate on a final square.
So if we continue the process, we always obtain rectangles that are similar to the starting rectangle, and thus we never terminate on a final square.
No, it might not terminate. Let‘s start with a rectangle with the short side having length 2 and the long side having length sqrt(5) + 1.
Removing a 2x2 square, we are left with a rectangle of sides measuring 2 and sqrt(5) - 1.
Removing a 2x2 square, we are left with a rectangle of sides measuring 2 and sqrt(5) - 1.
March 20, 2025 at 3:34 PM
No, it might not terminate. Let‘s start with a rectangle with the short side having length 2 and the long side having length sqrt(5) + 1.
Removing a 2x2 square, we are left with a rectangle of sides measuring 2 and sqrt(5) - 1.
Removing a 2x2 square, we are left with a rectangle of sides measuring 2 and sqrt(5) - 1.
That reads like the set-up of a good graph theory problem 😉
March 20, 2025 at 10:39 AM
That reads like the set-up of a good graph theory problem 😉
The answer is 200… in base 3 😉
March 18, 2025 at 3:26 PM
The answer is 200… in base 3 😉
Right 😅 lucky for me, 100*2^100 still lower than 100!
March 18, 2025 at 1:03 AM
Right 😅 lucky for me, 100*2^100 still lower than 100!
(2^100)! < (2^100)^100 = 2^10000 so 2^(100!) much bigger
March 16, 2025 at 10:27 PM
(2^100)! < (2^100)^100 = 2^10000 so 2^(100!) much bigger
Congrats on polishing your work! It’s something we learn over time, that not everything needs to be laid out in full detail. In my case, I had to sort out, in long derivations of various formulas, which steps were trivial and could be skipped vs tricky ones that needed to be spelled out.
March 16, 2025 at 2:51 PM
Congrats on polishing your work! It’s something we learn over time, that not everything needs to be laid out in full detail. In my case, I had to sort out, in long derivations of various formulas, which steps were trivial and could be skipped vs tricky ones that needed to be spelled out.
So a simple strategy for the first player could be : always pick the largest available number.
I am not 100% sure about the way to solve the recurrence equations that this approach yield without assuming a kind of minimax principle.
I am not 100% sure about the way to solve the recurrence equations that this approach yield without assuming a kind of minimax principle.
March 16, 2025 at 2:14 PM
So a simple strategy for the first player could be : always pick the largest available number.
I am not 100% sure about the way to solve the recurrence equations that this approach yield without assuming a kind of minimax principle.
I am not 100% sure about the way to solve the recurrence equations that this approach yield without assuming a kind of minimax principle.
If n is odd (n=2m+1), the first player can win with probability (m+1)/(2m+1) if he picks his guess such that the second player, whatever the result if the guess is wrong, is left with an even amount of numbers to guess in.
March 16, 2025 at 2:14 PM
If n is odd (n=2m+1), the first player can win with probability (m+1)/(2m+1) if he picks his guess such that the second player, whatever the result if the guess is wrong, is left with an even amount of numbers to guess in.
I think it‘s something along these lines.
If n is even, the probability that the first player wins is 1/2 whatever his choice. So no strategy here.
If n is even, the probability that the first player wins is 1/2 whatever his choice. So no strategy here.
March 16, 2025 at 2:14 PM
I think it‘s something along these lines.
If n is even, the probability that the first player wins is 1/2 whatever his choice. So no strategy here.
If n is even, the probability that the first player wins is 1/2 whatever his choice. So no strategy here.