To be honest I am a bit disappointed…it looked nice… ]]>

Ну тогда вот в чистом виде.

https://www.lix.polytechnique.fr/Labo/Ilan.Vardi/mekh-mat.html ]]>

google books: https://books.google.com/books?id=ho6fMF8ehogC&lpg=PP1&pg=PP1#v=onepage&q&f=false

which is in part about these exams, but also includes essays about Bella Abramovna/Jewish People’s University.

]]>Μια Σοβιετική Εβραία απόφοιτος των Μαθηματικών Ολυμπιάδων, (αργυρό και χρυσό μετάλιο το 1975/1976), περιγράφει τα αδύνατα μαθηματικά προβλήμα…

]]>Since x and y are “switchable”, we are basically given that |F(x)-F(y)|<=(x-y)^2 for all x and y. Pick any two distinct real numbers x and y (assume x<y, without loss of generality), and assume that the distance between them is m.

Now choose any natural number n, and partition the interval [x,y] into n equal subintervals. Let x_0=x, x_1=x+(m/n), x_2=x+2(m/n), and so on, up to x_n=y. We then have:

|F(x)-F(y)|

=|[F(x_0)-F(x_1)]+[F(x_1)-F(x_2)]+…+[F(x_(n-1)-F(x_n)]|

<=|F(x_0-F(x_1)|+|F(x_1)-F(x_2)|+…+|F(x_(n-1)-F(x_n)|

<=(m/n)^2+(m/n)^2+…+(m/n)^2

=m^2/n

But if |F(x)-F(y)|<=m^2/n for ALL natural numbers n, we must have |F(x)-F(y)|=0. So the function values at any two points are equal; the function must be constant.

Oh, and I’m a Jew, by the way. 😀

]]>https://arxiv.org/abs/physics/0605057

🙂

]]>F(x_1) <= F(x_2) + (x_1-x_2)^2

but also

F(x_2) <= F(x_1) + (x_1-x_2)^2

which can only occur if F is the constant function.

]]>