Connor Harris
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This is a well-known result analogous to the formula $$\frac{d^k}{dx^k} x^k = k!$$for differentiation of polynomials. I'll try to state it clearly: Definition: Let the discrete derivative of a ...

Not an answer, but a quick table with the number of turns until the recurrence of the initial position on rectangular grids small enough to be testable by computer. The ant is initially facing up (...

For all integers $n \geq 2$, note that $$n(n+1) < e^n. \tag{*}$$ We may show this by expanding $e^n > 1 + n + \frac{n^2}{2} + \frac{n^3}{6}$, which implies: $$e^n- n(n+1) > 1 - \frac{n^2}{2} +... View answer Accepted answer 8 votes Reduce the problem to showing that \mathbb{R}^2 minus a countable set of points is path-connected, using one of these methods. Let S^2 \subset \mathbb{R}^n be a 2-sphere that contains p and q ... View answer 7 votes The arrangement is indeed impossible. As the answer to the question you linked mentions, the sums of each segment of three adjacent numbers all themselves have to add up to 3 (1 + 2 + \cdots + 12) = ... View answer Accepted answer 7 votes Subadditivity is implied by a requirement that g(x) := \frac{f(x)}{x} be monotonically decreasing. If g(x) \geq g(x+y) and g(y) \geq g(x+y), then$$f(x) + f(y) = x g(x) + y g(y) \geq x g(x+y) + ...

There's no solution. Consider reduction modulo 3.

Hint: suppose some number $n$ is not prime, but $n$ has no prime factors less than 100. That is, $n$ is the product of at least two prime numbers greater than 100. What's the smallest that $n$ could ...

Recall that a fixed point of a function $f$ is a solution to $f(x) = x$. The only fixed point of $g \circ g$ is $1$, so the only fixed point of $g$, if any, is also $1$, as any fixed point of $g$ is ...

Here's a less orthodox answer. Consider just the interval $[0, 1)$, and divide it into intervals $[0, 1/2), [1/2, 2/3), \ldots, [1 - \frac{1}{n}, 1 - \frac{1}{n+1}), \ldots$, each of width $\frac{1}{n(... View answer 5 votes Multiply by$2and rearrange to \begin{align*}(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2ad - 2d + \frac{4}{5} = 0. \tag{\star}\end{align*} For fixeda$and$d$, the minimum value of$(a-b)^2 + (b-c)...

Denote $f(n) = n + 3n^3 + 7n^7 + 9n^9$. The terms in $f(n)$ are either all odd or all even, so $f(n)$ is even. By Fermat's little theorem, $9n^9 \equiv 4n \bmod 5$ and $7n^7 \equiv 2n^3 \bmod 5$. ...

The main branches of $\sqrt{\cdot}$ and $\cos^{-1}$ have nonnegative ranges, so their sum is zero only if they are both zero themselves. This forces $y = x$.
If we replace the indefinite integral with a definite one so as to keep one point (say, $(1, 0)$) fixed as we vary the exponent, then there's no jump at all. For all $s \neq 0$, define $$f_s(x) := \... View answer Accepted answer 5 votes The only triples of positive integers (a, b, c) for which a \neq b and c^a + c^b = a^c + b^c, up to interchange of a and b, are (1, 3, 2), (2, 4, 2), and (2, 4, 4). Allowing a = b ... View answer Accepted answer 5 votes The series diverges, because$$\sum_{n=1}^\infty \frac{1}{\sqrt{4n^2 - 1}} > \sum_{n=1}^\infty \frac{1}{2n}.$$View answer 5 votes Hint: the formula$$\int x^c\, dx = \frac{x^{c+1}}{c+1} + C$$applies to all real numbers c \neq -1, not just integers. View answer Accepted answer 4 votes f can only be the 2-adic valuation on \mathbb{Q}. That is, f(0) = 0 (a simple deduction from f(0 \times 2) = f(0) f (2)), and if q = \pm 2^k \frac{a}{b} with a and b both odd, then f(q) ... View answer 4 votes It's completely possible that x \in g(x) for every x \in A. But if this is the case, then the set B is just the empty set \emptyset, which is a perfectly good element of \mathcal{P}A that ... View answer 4 votes If f(n) = k_n n, then (k_n n)^2 = n f(k_n n), so f(k_n n) = k_n^2 n = k_n (k_n n). Inductively, therefore, f(k_n^r n) = k_n^{r+1} n. And k_n has to be an integer: if it weren't, then there ... View answer Accepted answer 4 votes A brief sketch of an argument: Continuous functions are determined by their values on any dense subset of their domain, such as the rationals. The rationals are countable, so pick some enumeration ... View answer 4 votes As user113102 hints, if n is odd, then n^2 \equiv 1 \bmod 8. Thus, in the equation$$p_3^2 + p_4^2 + p_5^2 + p_6^2 + p_7^2 = p_8^2 - 8$$where all p_i are odd primes, the LHS is congruent to 5 ... View answer Accepted answer 4 votes Following up on Daniel Schepler's comment. Let$$P_n(x) = \frac{1}{1 + \frac{1}{2 + \ddots \frac{1}{n+x}}}.This is basically the RHS of the recurrence equation for K(n). Then: \begin{align*} P_1(... View answer Accepted answer 4 votes Here's what I think you're asking, generalized slightly. There are n balls numbered 1, \ldots, n. Lottery players choose k distinct numbers with k < n, then k balls are drawn. Tickets are ... View answer Accepted answer 4 votes Method 1: Recall that [L:K][K:F] = [L:F], note that \mathbb{Q}(2^{1/3}, 2^{1/2}) = \mathbb{Q}(2^{1/6}), and show that X^6 - 2 is irreducible over \mathbb{Q} using Eisenstein or similar. ... View answer Accepted answer 4 votes Note that\frac{d}{dx} \frac{\log(x+1)}{\log x} = \frac{\frac{\log x}{x+1} - \frac{\log (x+1)}{x}}{\log^2 x} < 0,$$so \log_{999} 1000 is larger. View answer Accepted answer 4 votes They're not the same. In your solution, for example, x could be a teacher; in that case, for all y, S(x) \wedge F(y) is false, so the implication (S(x) \wedge F(y)) \Rightarrow \neg A(x, y) is ... View answer Accepted answer 4 votes The number of graphs on n distinct vertices is 2^{(n^2 - n)/2}, giving the upper bound. To show the lower bound, consider a set \mathcal{S} of non-isomorphic graphs on n vertices. Construct ... View answer Accepted answer 4 votes If two adjacent squares on an edge are different colors, they can only be made the same color (this is what they mean by "fixed") if one square is in the rectangle being inverted and one of them isn't.... View answer Accepted answer 4 votes For point (2), if you can find quantities a, b, c, d, not all zero, such that$$ a m^2 + b(m+1)^2 + c(m+2)^2 + d(m+3)^2 = 0 for any integer $1 \leq m \leq n$, then this gives you a nontrivial ...