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!

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

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.

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.

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.

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.

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?

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).

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.

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.

That reads like the set-up of a good graph theory problem 😉

The answer is 200… in base 3 😉

Right 😅 lucky for me, 100*2^100 still lower than 100!

(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.

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.

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.

Oof 😮‍💨 on a good day I can’t even get to the second 3 😅

For pi day, a “convenient” expression 😉

#MathSky

16

A 10$ rebate for a preselected offer vs a customized order.

Again some care must be taken with 0, when ending the numbers in the 100s: … 103102101 we can just continue with the number already defined starting from 2 digit numbers only.

I think we can extend that idea to numbers with 10 digits, of which there are 10! that have different digits to build from.