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What are the most mindblowing things in mathematics? (lemmy.world)
submitted 1 year ago* (last edited 1 year ago) by cll7793@lemmy.world to c/nostupidquestions@lemmy.world
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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) > 10^10^10^10^10^10^10^10^10^10^10^10^10^10^10 (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:

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[-] Jordan117@lemmy.world 34 points 1 year ago

Euler's identity, which elegantly unites some of the most fundamental constants in a single equation:

e^(iπ)+1=0

Euler's identity is often cited as an example of deep mathematical beauty. Three of the basic arithmetic operations occur exactly once each: addition, multiplication, and exponentiation. The identity also links five fundamental mathematical constants:

  • The number 0, the additive identity.
  • The number 1, the multiplicative identity.
  • The number π (π = 3.1415...), the fundamental circle constant.
  • The number e (e = 2.718...), also known as Euler's number, which occurs widely in mathematical analysis.
  • The number i, the imaginary unit of the complex numbers.

Furthermore, the equation is given in the form of an expression set equal to zero, which is common practice in several areas of mathematics.

Stanford University mathematics professor Keith Devlin has said, "like a Shakespearean sonnet that captures the very essence of love, or a painting that brings out the beauty of the human form that is far more than just skin deep, Euler's equation reaches down into the very depths of existence". And Paul Nahin, a professor emeritus at the University of New Hampshire, who has written a book dedicated to Euler's formula and its applications in Fourier analysis, describes Euler's identity as being "of exquisite beauty".

Mathematics writer Constance Reid has opined that Euler's identity is "the most famous formula in all mathematics". And Benjamin Peirce, a 19th-century American philosopher, mathematician, and professor at Harvard University, after proving Euler's identity during a lecture, stated that the identity "is absolutely paradoxical; we cannot understand it, and we don't know what it means, but we have proved it, and therefore we know it must be the truth".

[-] dandroid@dandroid.app 6 points 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@programming.dev 2 points 1 year ago* (last edited 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.

[-] aja@lemm.ee 4 points 1 year ago

Time for a deep dive. Wish me luck, lads.

[-] sadbehr@lemmy.nz 2 points 1 year ago

Let us know what you find in the depths.

this post was submitted on 03 Aug 2023
630 points (100.0% liked)

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