math fun

Lets have some math fun The way this works is we'll ask each other math questions and try to solve them. I'll start it off Prove that e ≠ a/b, basically prove e is irrational.

e = \bigsum_{i=1}^\infty 1/i! Suppose e = a / b. Then be = a \in N. This means that \bigsum_{i=1}^\infty b/i! \in N. However, i! will eventually dominate any positive integer, so this cannot possibly be a natural number. Contradiction, therefore e =/= a/b for any a,b \in N

A function f from the vertices of a graph to R is harmonic if f(v) = 1/deg(v) \sum_{v' adj v} f(v') = v, that is, the function at every vertex is the sum of the function at all adjacent vertices. Prove that f is harmonic iff f is constant on each connected component of the graph.

>>13637 *the function at every vertex is the average of the function at all adjacent vertices sorry bout that

>>13636 >i! will eventually dominate any positive integer Can you explain more how this means that the infinite sum will not approach a natural number? I'm not sure what dominate means here. How is it different from $$ \sum 1/2^n = 2 $$? Does 2^n dominate 2?

>>13637 Assume it attains a maximum somewhere in a given connected component, show that the adjacent vertices all have the same value, iterate, conclude it's constant on the connected component. This only works for graphs with a finite number of vertices, tho. Consider the graph whose vertices are the integers, and connect n with n-1 and n+1, for all n. The identity function is trivially harmonic, but it's nonconstant. I'm sorry, but I don't really have any appropriate questions.

Here's a fun little puzzle I came across today, the math behind it is far from complex. It took me a bit to solve it though. A bat and ball cost $1.10. The bat costs one dollar more than the ball. How much does the ball cost?

>>15962 Too simple

>>15963 Sorry, couldn't think of a better one. Perhaps you should propose one, lainon.

>>13636 >However, i! will eventually dominate any positive integer, so this cannot possibly be a natural number. That's not a complete proof. It could be that the sum of terms where i! dominates b sums to a natural number. Complex analysis proof: assume e = p/q for positive integers p and q, so qe = p. Then (qe)^(i 2\pi) = p^(i 2\pi). Thus q^(i 2\pi) = p^(i 2\pi). By the definition of complex exponentiation, for x > 0 \in R, x^(i 2\pi) = e^(i 2\pi ln(x)) = x. Therefore q^(i 2\pi) = p^(i 2\pi) = q = p, so e = 1, a contradiction.

>>15968 I'm sure I made a mistake with this proof but I'm not sure how to fix it... it shows that *any* expression re = s for *any* real numbers r,s implies r = s, but this is obviously not true: plugging in r = 1 and s = e obviously gives the right result. So the proof must be circular. I want to give a proof that e is transcendental but I don't really remember how it goes ;_;

>>15969 the error is here: >for x > 0 \in R, x^(i 2\pi) = e^(i 2\pi ln(x)) = x e is an example of a number where x^{i 2\pi} = x fails. exponentiation of a real number by a pure imaginary number always ends you up on the unit circle, and you only end on a real if x is a power of e (i.e. log(x) is an integer). >I want to give a proof that e is transcendental but I don't really remember how it goes ;_; I think the only proof I learned uses Galois theory or something.

>>13636 >However, i! will eventually dominate any positive integer, If sum_{i=1}^infty (1/i!) converges (or in fact, is bounded) then so does sum_{i=1}^infty (b/i!).

>>15962 This one is often used to showcase how our brain solves problems like these. Many people report they initially think the bat costs $1 and the ball $0.10, that's the intuitive answer that the brain takes as an initial guess that isn't too costly to compute. But eventually people get the right answer by way of another process in the brain that works a lot more rationally but is costly and long-winded.

I'm fuarrrking retard ;_;.

>>20245 You are not anon, you just have to practice and find things you are good at.

>>20246 Thanks anon.

>>20225, 11/10 - x = x - 1 <=> 21/20 = x, is straightforward.