JavaMan
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It is well known that the number of ways to get to the lattice point $(x,y)$ (supposing $x, y \geq 0$) by taking steps of one unit each either in the eastward or northward direction is exactly $${x +... View answer 64 votes Note that$$\begin{align} 99^{100} > 100^{99} &\iff 99 \cdot 99^{99} > 100^{99} \\ &\iff 99 > (100/99)^{99} \\ &\iff 99 > \left( 1 + \frac{1}{99}\right)^{99} \end{align}... View answer Accepted answer 0 votes Hint: \begin{align} t_1 &= 1 \\ t_2 &= 1 + 2 \\ t_3 &= 1 + 2 + 3 \\ t_4 &= 1 + 2 + 3 + 4 \end{align} Can you guess a formula fort_n$? Once you have a formula for$t_n$,... View answer 1 votes Hint: $$\begin{array} &\sin(\pi) = 0 \\ \sin(2 \pi) = 0 \\ \sin(3\pi) = 0 \\ \qquad\qquad\vdots \end{array}$$ View answer 6 votes This is not a particularly elegant solution, but an alternative route is to simply note that$(1 \pm i)^2 = \pm 2i$. I will show a lot of steps, but this method involves only very easy calculations. ... View answer Accepted answer 6 votes The number of solutions is always infinity or a nonnegative integer. Let$z = re^{i \theta}$, to see that $$z^{\pi} = r^{\pi}e^{i \pi\theta} = 1.$$ The solutions are then$r = 1$and$\theta = ...

Hint: \begin{align} \sin(x) = &x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 + \dots. \\ &\Downarrow \\ \frac{\sin(x)}{x} = &1 - \frac{1}{3!}x^2 + \frac{1}{5!}x^4 + \dots \\ &\... View answer 1 votes Euler showed that the Dirichlet series of a multiplicative function f can be written as a product (now called an Euler product): L(s, f) := \sum_{n \geq 1} \frac{f(n)}{n^s} = \prod_p \left(1 + ...

The complex number $z = c$ is a solution to the equation $f(z) = 0$ (also called a zero or root of $f(z)$) if and only if $(z-c)$ is a factor of $f(z)$. The proof of this statement follows easily by ...

If $f(x)$ and $f^{-1}(x)$ are inverse functions (meaning that $(f \circ f^{-1})(x) = (f^{-1} \circ f)(x) = x$ on the respective domains), then the domain of $f$ is the range of $f^{-1}$ and the domain ...

Hint: It is enough to prove that $A \times B$ is countable whenever $A$ and $B$ are countable. To do this, write $$A = \{a_0,a_1,a_2, \dots\}$$ and $$B = \{b_0,b_1,b_2,\dots\}.$$ To find a ...

Hint: $$f(-2) = (-2)^5 -2(-2) + 10 = -32 + 4 + 10 = -18 < 0$$ while $$f(2) = 2^5 - 2(2) + 10 = 38 > 0.$$
The Cross Product $v \times w$ is always orthogonal to both $v$ and $w$. This is easy to see by a direct calculation: Write $v = \langle v_1, v_2 , v_3\rangle$ and $w = \langle w_1, w_2, w_3\rangle ... View answer Accepted answer 4 votes They do not provide a derivation, but this is actually written up in Wikipedia. I use the standard notation$e(x) = \exp(2 \pi i x)$. Assuming$\gcd(k,n) = 1$, we have $$\sum_{x \in \mathbb{Z}/n \... View answer 1 votes Hint \# 2: tards answer might be a simpler way to go, but here is an alternative route: Let d(n) denote the number of divisors of the integer n. Show that$$ \gcd(m,n) = 1 \quad \implies \... View answer 5 votes Rough sketch: Have you seen the Wikipedia section on the distribution of squarefree integers? It essentially gives your solution. All you need to notice is that for large$n$, the "probability" ... View answer 1 votes If$\gcd(x,m) = d > 1$, then the equation$y \equiv ax \pmod{m}$has$d$times the solutions of$y \equiv a\frac{x}{d} \equiv \frac{m}{d}$. In other words, we may assume$\gcd(x,m) = 1$. In ... View answer 3 votes Fix$k \geq 0$. Show that $${n+1 \choose k} = \frac{(n+1)!}{k!(n+1-k)!} \geq \frac{n!}{k!(n-k)!} = {n \choose k}.$$ But this follows whenever$n+1 \geq n+1 - k$, so... View answer 4 votes Aside from it being unlikely that$\gcd(x,n) \neq 1$, note that the only possibilities are$\gcd(x,n) \in \{1,p,q,n\}$. Therefore, if$x$and$n$are not coprime, the one can decipher the text ... View answer 1 votes For the Inductive Step: Suppose that we have $$\frac{n^n}{3^n} < n!$$ and $$n! < \frac{n^n}{2^n}.$$ We want to show that this implies that $$\frac{(n+1)^{n+1}}{3^{n+1}} < (n+1)! ... View answer Accepted answer 7 votes Hint: Assuming "sum" means "union", let A = \{a_1, a_2, \dots\} and B = \{b_1 , b_2, \dots\} be two countable sets. Consider the map$$ f(n) = \left\{ \begin{array}{ccc} b_{n/2} & & n ... View answer 0 votes You are right. The easiest way to see this is to recognize the probability that you roll a$1$or a$2$as $$1 - Pr(\text{not rolling a }1 \text{ or } 2) = 1 - (4/6)^2 = 5/9.$$ View answer 4 votes Let$D_f(s)$denote the Dirichlet generating series: $$D_f(s) = \sum_{n =1}^{\infty} \frac{f(n)}{n^s}.$$ When the series convergences absolutely, we can write$D_f(s)as an Euler product: $$... View answer 1 votes First and foremost, \sqrt{1} is not 1 or -1. When we discuss the square root of a number \sqrt{n} this denotes the positive or principal square root. Therefore, \sqrt{1} = 1 and 1 alone. ... View answer 2 votes Since \xi \in [0,1), then \xi^n \to 0. Hence, f(\xi^n) \to f(0) as f is continuous. An alternative proof: Since f is continuous on [0,1], then f is uniformly continuous on [0,1]. ... View answer Accepted answer 2 votes They are indeed equal. This follows since$$\begin{align} \sqrt[n]{x^n} = \left\{ \begin{array}{ccc} |x| & & n \text{ is even} \\ x & & n \text{ is odd} \end{array} \right. \... View answer 7 votes You were indeed almost there. All that's left to do is just switch the order of summation on the last sum. I won't fill in all the details but here's a start: \begin{align} \sum_{1 \leq n \leq x}... View answer 2 votes This is just an exercise in persistence. First note that \lim_{h \to 0}\frac{\frac{1}{h} - \frac{1}{e^h - 1} - \frac{1}{2}}{h} = \lim_{h \to 0} \frac{2(e^h - 1)-2h - h(e^h - 1)}{2h^2 (e^h - 1)} ... View answer 4 votes Just note that\begin{align} \frac{8}{\sqrt{4 + 3x}} - 2 &= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}} \\ &= \frac{8 - 2 \sqrt{4 + 3x}}{\sqrt{4 + 3x}} \cdot \frac{8 + 2\sqrt{4 + 3x}}{8 + ... View answer 0 votes You can explicitly takeq = \left\| \frac{m}{n}\right\|$and$r = m - n \left\| \frac{m}{n}\right\|$, where$\| \alpha \|$denotes the nearest integer to$\alpha \in \mathbb{R}\$. Note that for any ...