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$\noindent Hint: All you need to do is show that the unique closest point on the circle to $A$ is the point $B$. This can be done by showing that $g\left(\alpha\right) := 1+r^{2}-2r\cos\left(\theta - \alpha\right)$ has a \emph{unique} point $\alpha$ in $\left[0,2\pi)$ where it is minimised...

$\noindent I assume you want the coefficients $a_n$ in the Maclaurin series for $e^{2x}\cdot (1-x)^{-1}$. Since $e^{2x} = \sum _{k = 0}^{\infty}b_{k}x^{k}$, where $b_{k} = \frac{2^{k}}{k!}$, and for $(1-x)^{-1}$, it's $\sum_{k=0}^\infty c_k x^k$, where $c_{k} \equiv 1$, the desired $a_{n}$ is...

$\noindent The kernel is just the set of solutions to $y'' - y' = 0$. The characteristic equation is $r^{2} - r = 0$, which has roots $0,1$. So the solutions to this ODE (hence the kernel) are the span of $1$ and $e^{x}$ as required (note that $\sinh x + \cosh x = e^{x}$).$ $\noindent For the...

$\noindent Given $\varepsilon > 0$, take $N= \frac{1}{\varepsilon^{2}}$, then $|f(n) - 0| < \varepsilon$ for all $n >N$, by your working.$ $\noindent Don't need floor or ceiling functions because there's no need for the $N$ above to be an integer, it suffices for it to just be a positive real...

$\noindent Note that for $\mathbf{x} = \begin{bmatrix}x_{1}\\ x_{2}\\ x_{3}\end{bmatrix} \in \mathbb{R}^{3}$, we have $T\left(\mathbf{x}\right) = x_{1} \begin{bmatrix}1 & 0 \\ -1 & 0\end{bmatrix} + x_{2} \begin{bmatrix}-1 & 1 \\ 0 & 0\end{bmatrix} + x_{3} \begin{bmatrix}0 & -1 \\ 1 &...

Can't just use m + 1, since could have a factors be included in the lower sets in general. Instead, use n = a prime number bigger than m. I assume you can see why this'll work (also there infinitely many primes so this is well-defined).

$\noindent $S_{n} \not\subset S_2 \cup S_3 \cup \cdots \cup S_m$$

For less trivial ones, they'll be precisely those 'arrow diagrams' where only loops from a dot to itself exist. So there are four possibilities. E.g. you could do one with a loop from 1 to itself and a loop from 2 to itself (and no other arrows).

Here's one (essentially a trivial one): just draw 1 and 2 (represented by dots for example) with no arrows whatsoever.

Re: Discrete Maths Sem 2 2016 $\noindent (This is pretty much the foundation of the Euclidean Algorithm.) Let $g = \gcd (a,b)$ and $h = gcd(a, a-b)$, assuming $(a,b) \neq (0,0)$. Note $g$ divides both $a$ and $b$, so it divides $a-b$ too. Therefore, $g \leq h$, since $h$ is the \emph{greatest}...

Yeah in the integral test, the sum and integral have the same convergence/divergence status. This is because when proving the integral test, we bound the sum between two integral estimates using the integral in question, and then take the limit and use the squeeze law. Besides, if you believe...

For the 3 | (x^2 + y^2), just look at all possible squares modulo 3: they are 0^2 = 0, 1^2 = 1, 2^2 = 1. The only way to get two numbers from {0,1,1} to add to 0 (mod 3) is if both are 0. In other words, if two squares add to a multiple of 3, they are each 0 mod 3, as required.

Given any non-zero complex number r*cis(theta), it has two square roots (+/-)sqrt(r)*cis(theta/2). (These are clearly distinct if z =/= 0, i.e. r =/= 0.)

What necessary condition? Q3 is just asking for an example. An example would be f:R->R, f(x) = sin(x), with A = [0, 2*pi] and B = [2*pi, 4*pi]. Then f(A) = [-1,1] = f(B), but A∩B = {2*pi}. So f(A∩B) = {sin(2*pi)} = {0}, whereas f(A)∩f(B) = [-1,1].

What was the reasoning for your answer? You can use the inclusion/exclusion principle to get the answer.

Re: Discrete Maths Sem 2 2016 Well if all the letters are distinct, the answer is by symmetry just (1/2)*N, where N is the no. of possible eight (distinct) letter strings of which two of the letters and x and y. I.e. N = (24C6)*(8!). (Can of course also write N as (8*7*(24P6).)

Re: Discrete Maths Sem 2 2016 Main idea is that exp(ix) will be onto because clearly any complex number on the unit circle can be written in the form e^{ix}, and it'll be one-to-one because two "essentially different" x's clearly yield two different points on the unit circle (as they'll have...

It's decreasing because the denominator is increasing. The denominator is increasing because (n+1)^{(n+1) + 1/(n+1)} > n^{(n+1) + 1/(n+1)} (*) > n^{n + 1/n} (**). Note (*) follows since the function x^{(c+1) + 1/(c+1)} is increasing on the positive reals for any given positive integer c, and...

The progression is the sequence of terms, series is the sum of them. • Geometric progression: https://en.wikipedia.org/wiki/Geometric_progression • Geometric series: https://en.wikipedia.org/wiki/Geometric_series

$\noindent Q3) Suppose $p > n^{\frac{1}{3}}$. Assume for the purposes of contradiction that $m$ is not prime. Then the number $m$ (which has a prime factorisation (fundamental theorem of algebra)) can be written as $m = q_{1}l$, for some integer $l >1$, where $q_{1}$ is prime. (Note if $l$ were...