What concepts or facts do you know from math that is mind blowing, awesome, or simply fascinating?

Here are some I would like to share:

  • Gödel’s incompleteness theorems: There are some problems in math so difficult that it can never be solved no matter how much time you put into it.
  • Halting problem: It is impossible to write a program that can figure out whether or not any input program loops forever or finishes running. (Undecidablity)

The Busy Beaver function

Now this is the mind blowing one. What is the largest non-infinite number you know? Graham’s Number? TREE(3)? TREE(TREE(3))? This one will beat it easily.

  • The Busy Beaver function produces the fastest growing number that is theoretically possible. These numbers are so large we don’t even know if you can compute the function to get the value even with an infinitely powerful PC.
  • In fact, just the mere act of being able to compute the value would mean solving the hardest problems in mathematics.
  • Σ(1) = 1
  • Σ(4) = 13
  • Σ(6) > 101010101010101010101010101010 (10s are stacked on each other)
  • Σ(17) > Graham’s Number
  • Σ(27) If you can compute this function the Goldbach conjecture is false.
  • Σ(744) If you can compute this function the Riemann hypothesis is false.

Sources:

  • Dandroid@dandroid.app
    6·
    1 year ago

    This is the one that made me say out loud, “math is fucking weird”

    I started trying to read the explanations, and it just got more and more complicated. I minored in math. But the stuff I learned seems trivial by comparison. I have a friend who is about a year away from getting his PhD in math. I don’t even understand what he’s saying when he talks about math.

    • Kogasa
      2·
      1 year ago

      Recall the existence and uniqueness theorem(s) for initial value problems. With this, we conclude that e^(kx) is the unique function f such that f’(x) = k f(x) and f(0) = 1. Similarly, any solution to f’’ = -k^(2)f has the form f(x) = acos(kx) + bsin(kx). Now consider e^(ix). Differentiating it is the same as multiplying by i, so differentiating twice is the same as multiplying by i^(2) = -1. In other words, e^(ix) is a solution to f’’ = -f. Therefore, e^(ix) = a cos(x) + b sin(x) for some a, b. Plugging in x = 0 tells us a = 1. Differentiating both sides and plugging in x = 0 again tells us b = i. So e^(ix) = cos(x) + i sin(x).

      We take for granted that the basic rules of calculus work for complex numbers: the chain rule, and the derivative of the exponential function, and the existence/uniqueness theorem, and so on. But these are all proved in much the same way as for real numbers, there’s nothing special behind the scenes.