JavaMan
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The Szemeredi-Trotter incidence theorem gives an upper bound on the total number of incidences between a finite set of points and a finite set of lines in the plane. An incidence is a pair $(\ell, x)$...

The question as it stands is not well posed as you are asking the wrong question. In order to prove that the circumference formula $C = 2\pi r$ holds for all circle of radius $r$, we first have to ...

All congruences are done modulo $p$. Assume that $x = \pm a^{n+1}$. Then, \begin{align*} x^2 \equiv a^{2n + 2} = a^{\frac{4n+4}{2}} = a^{\frac{p+1}{2}}. \end{align*} We need to show that $a^{\... View answer 0 votes There is an algorithm for this process called Gauss-Jordan elimination. Applying Gauss-Jordan to the example given: 1) The first step is to get a leading one in the first row of your matrix. The ... View answer 0 votes Try the ratio test: $$a_n = \frac{1}{n!} \quad \implies \quad \left| \frac{a_{n+1}}{a_n} \right| = \frac{n!}{(n+1)!} = \frac{1}{n+1}$$ so that $$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \... View answer 2 votes Hint: The Handshaking Theorem gives the precise relation between the number of edges in a graph and the sum of the degree of the vertices of a graph. If you know the degree of each vertex, then, ... View answer 2 votes You could also note that$$ \lim_{x \to 0} \frac{2 \tan x \sec^2 x}{6x} = \lim_{x \to 0}\frac{2 \sec^2 x}{6} $$since$$ \lim_{x \to 0} \frac{\tan x}{x} = \lim_{x \to 0} \frac{\sin x}{x} \cdot \frac{... View answer 3 votes A few to get you started: The number of hands in a Super 4 of a kind would be calculated as: $${5 \choose 4} \cdot {12 \choose 1} \cdot {13 \choose 1} = 780$$ There are${5 \choose 4}$ways to get ... View answer 2 votes This doesn't answer your question directly, but I can't help mentioning that Generating functions are helpful with these kinds of problems. For the first problem, note that$x_i$can be either$0, 1, ...

Generating functions are helpful here. You need to find the number of solutions to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 + 2x_6 \leq 5$$ where all variables are nonnegative $(x_i \geq 0)$. ...

Hint: If $\sum a_n < \infty$, then $\displaystyle \lim_{n \to \infty} a_n = 0$. In particular for large $n$, we have $|a_n| \leq 1$. Then $$|a_n|^2 = |a_n| |a_n| \leq \dots$$

This is well known. Let $p = 2n + 1$ be an odd prime. Then, it is known that if $\gcd(a,p) = 1$, then $a^{p-1} = a^{2n} \equiv 1 \pmod{p}$, by Fermat's Little Theorem. Hence, $(a^n)^2 \equiv 1 \... View answer 0 votes First, let me say that you need to be more careful with your work. Writing$EQ(1) + EQ(2) = 2a + 6b -10c = 4$is a little sloppy. Rather, you should write what$EQ(1) + EQ(2) \implies 2a + 6b - 10c ...

Note that $$x + y = 2 \iff 2x - 3(2-x) = 3 \iff 5x = 9$$ This equation has a solution when...

It is common in number theory to write $c \gg 1$ to mean that $c$ is a sufficiently large (but finite) constant. The $\gg$ notation is Vinogradov's Notation.

No. A sequence converges to zero if and only if we can eventually make all the terms which are large enough as close to zero as we want. In symbols, $a_n \to a$ if for all $\epsilon > 0$, there ...

If you want to use induction, I assume you have checked the base case $n = 5$. To do the inductive step, assume that the statement holds for some $k$: $k^2 < 2^k$, and then under this assumption, ...

A system approach is to find integers $s$ and $t$ (via the Euclidean algorithm) such that $13s + 17t = 1$ (note that we can do this as $\gcd(13,17) = 1$). Then, $$1 = 13s + 17t \equiv 13s \pmod{17}$$...

Use the fact that $AB = 0 \implies A = 0 \text{ or } B = 0$. You have already found when $\csc(x) = 0$, so $\frac{\pi}{4}$ and $\frac{3\pi}{4}$ are both solutions. Now, when does $\cos(x) = 0$? ...

For completeness, here is the proof by induction: Base Case: The case $n = 1$ is easy to check. Inductive Step: Assume that $$\frac{k!}{2^{k-1} \left( \frac{k-1}{2} \right)!^2} \leq \sqrt{k}.$$ ...
Your work is fine. You then need to find the $t$-values that satisfy both conditions: $-4 \leq t \leq 2$ and $t \in (-\infty, -2) \cup (0, \infty)$. Alternatively, you could consider two cases: $t + ... View answer 7 votes Perhaps the easiest way here (though not the easiest way in general) is to write $$A^{-1} = \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right)$$ ... View answer Accepted answer 3 votes For polynomial and logarithmic comparisons, these should be quite clear by looking at the limit itself. With polynomial, you have$\frac{n^k}{n^{\ln n}}$. Since$k$is fixed and$\ln n \to \infty$... View answer 0 votes To get to the heart of the matter, since these are square roots, you are looking for the largest factor of the radicand that is a perfect square. For example, with$\sqrt{32ab^2}$you are looking for ... View answer Accepted answer 0 votes If$f(x^*)$is the maximum of the function over$[0,1]$, then$f(x^*)\geq f(x)$for all$x\in[0,1]$. In particular,$f(x^∗)≥f(x_i)$for$i=1,…,n$. View answer 1 votes The way that mathematics is actually done and the way that mathematics books are written are completely different. No mathematician would ever just sit down and guess a formula to be proven later. ... View answer 1 votes Tom Wolff published a collection of notes titled "Lectures on Harmonic Analysis" which he made freely available. Since his passing, the notes have been kept on Izabella Laba's website. The notes are ... View answer Accepted answer 2 votes Note that if$\alpha \in (-1,1)$, then$1 + \alpha^k \leq 2$so that $$\frac{1}{1 + \alpha^k} \geq \frac{1}{2}.$$ For the second one, since$\frac{1}{3} \leq \frac{1}{2}$, then:$$\frac{1}{2} + \... View answer 2 votes Hint: What is the probability that you will get exactly$0$heads? What is the probability that you will get exactly one head? If it helps, there are$2^4$possibilities for the sequence of four ... View answer Accepted answer 2 votes It is helpful to first draw a picture: Notice that$x = x^2$implies that$x^2 - x = 0$so that$x = 0$or$x = 1$. So we are rotating the area around the line$y = 2\$. We think of the volume we ...