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3

Yes. Just solve the equation $f(x) = f(y)$: $$2x^2 + 4x + 8 = 2y^2 + 4y + 8 \\ 2(x^2-y^2) + 4(x-y) = 0 \\ (x-y) \big( 2(x+y) + 4 \big) = 0 \\ x = y \vee x+y = -2$$ So $(x, y) \in E \iff x = y \text{ or } y = -x-2$.

0

The function is just $2(x+1)^2 + 6$. Clearly, $f(x)=f(y)$ iff $|x+1|=|y+1|$. In other words, the distance from $-1$ is the same for $x$ and $y$.

0

I would have thought that this issue is related to the possibility that two elements of $D_n$ $$\left[\begin{array} \, & c& a_1 & a_2 & \ldots \\ \, & b_1 & b_2 & \ldots\end{array}\right]$$ and $$\left[\begin{array} \, & a_1 & a_2 & \ldots \\ \, &c & b_1 & b_2 & \ldots\end{array}\right]$$ could ...

2

Whenever we have a function $f\colon X\to Y$, we can define a relation $\sim_f$ over $X$ by declaring $$a\sim_f b\qquad\text{if and only if}\qquad f(a)=f(b)$$ This is readily seen to be an equivalence relation: for all $a\in X$, $a\sim_f a$: indeed $f(a)=f(a)$ for all $a,b\in X$, if $a\sim_f b$, then $b\sim_f a$: indeed $f(a)=f(b)$ implies $f(b)=f(a)$ ...

2

Compute $$\det\pmatrix{a&0\\0&1}$$ What do you notice? The partition property is inherent because $\det$ is a function and $x \sim_f y :\Leftrightarrow f(x) = f(y)$ is an equivalence relation for any function $f$.

4

Hints: It suffices to show that $\sim$ is an equivalence relation on $M_2(\mathbb R)$. So simply show that it's reflexive, symmetric, and transitive. This follows immediately, since $\sim$ is defined in terms of $=$. Try some examples. Pick a matrix at random, and find its equivalence class. Then try going backwards: pick a determinant at random, then try ...

1

For (d), OEIS A000930 delights in the name of Narayana's cows sequence. Starting with $n=0$: $1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745$

4

This is essentially an explanation of Jack D'Aurizio's answer. Let $k$ be the number of twos, then we have $n-2k$ ones. There are then $$\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n-k}{k}=\sum_{k=0}^\infty\binom{n-k}{n-2k}$$ ways to write $n$ as a sum of ones and twos. One way to compute this number is to look at the generating function. \begin{align} ... 0 This problem could be formulated as follows: A rectangle 133\times1 is given and dominoes 1\times1 and 2\times1. In how many ways can one cover the rectangle with dominoes? The result can be obtained by induction like shown in other answers or there are other methods like cutting rectangle approximately in half. However, the beauty of the ... 0 I present a proof using PET (Polya Enumeration Theorem) for future reference. We will be using Z(S_k), the cycle index of the symmetric group on k elements. The partition into at most k parts is given by[z^n] Z(S_k)\left(\frac{1}{1-z}\right)$$where the fraction includes the empty term to account for slots being left blank, which represents at most ... 1 Ok, Here is a simple answer that you can explain to an elementary school student. The answer is 67. First you make 133 by using as many twos as possible, which is 133/2 = 66. That gives you one way of making 133. Now you can look at the number of twos in your answer, and you can add another way by splitting one of those twos into ones. Since you have 66 ... 4 I'll prove that this holds in general, not just for 133 Let P_i be the solution to this problem for number i. That is, the answer to the exact question is P_{133} Let F_i be the ith Fibonacci number. Proof by induction that \forall i\geq1:P_i = F_{i+1} Base: i=1 Can make with only a single 1. So P_1 = 1 = F_2 Hypothesis: Assume holds ... 2 Let some of the pieces in the first case contain zero. Then add 1 to all of them. 42 For 133 ones there is 1 outcome. For 131 ones and 1 two there are 132 outcomes. For 129 ones and 2 twos there are {131 \choose 2} outcome. Thus there are {133 \choose 0}+{133-1 \choose 1}+{133-2 \choose 2}+...+{133-66 \choose 66} outcomes, which is, as noted by Jack D'Aurizio, the 134th Fibonacci number (You may refer to Fibonacci number ... 3 Hint:$$[x^{133}]\,\frac{1}{1-(x+x^2)}=F_{134}=4517090495650391871408712937.$$More information and context can be found in the questions here and here. Just to be clear, the number we want to compute is given by the coefficient of x^{133} in the sum:$$1+(x+x^2)+(x+x^2)^2+(x+x^2)^3+\ldots = \frac{1}{1-(x+x^2)}$$and since the Taylor coefficients a_n of ... 2 Consider for instance how many ways to make a sum of 1? Then of 2? To make a sum of 3, you can add a 2 to all the ways to make a sum of 1, and a 1 to all the ways to make a sum of 2 (so the number of ways of making 3 is the sum of the ways of making 2 plus the number of ways of making 1). For a total of 4, you can add a 2 to the ways to make 2, and a 1 to ... 10 Hint: Note that the first digit in the sum is either a 1 or a 2, so the number of ways of getting a sum of 133 is the number of ways of getting a sum of 132 (add one at the beginning) plus the number of ways of getting a sum of 131 (add two at the beginning). Let's do it with the number of ways of getting to 5. If the first number in the sum ... 1 You can do this using generating functions. This is analogous to the classic coin change problem. In your case, you want to look at the term for x^{133} in the expansion of$$ f(x) = \frac{1}{(1-x^1)(1-x^2)(1-x^3)}.$$We may rewrite f as$$ f(x) = \frac{1/6}{(1-x)^3} + \frac{1/4}{(1-x)^2} + \frac{17/72}{1-x} + \frac{1/8}{1+x} + \frac{-j/9}{x-j} + ...

1

I'm not sure what the question is, but you're 100% correct. The way I look at it: There's a category $\mathbf{CMon}$ of commutative monoids. The underlying set functor $U:\mathbf{CMon} \rightarrow \mathbf{Set}$ has a left-adjoint $F:\mathbf{Set} \rightarrow \mathbf{CMon}$. Let $\Phi$ denote an object of $F \dashv U$, where I write $F \dashv U$ for the ...

1

You can assume that $x_N\ge N/2$ and so you want to prove that $$\frac{\sum \limits_{i = 1}^N x_i^4}{\left(\sum \limits_{i = 1}^N x_i^2\right)^2}\ge \frac{x_N^4 + (N-x_N)^4}{(x_N^2 + (N-x_N)^2)^2}$$ Set $\lambda:=N-x_N$, and assume $\lambda>0$ (else the inequality is trivially true), and set also $$x:= \frac{x_N}{\lambda},\quad\text{and}\quad ... 2 Hint: if x\in\bar{0}\cap \bar{1}, then in particular x\in\bar{0} and x\in\bar{1}. Now, x\in\bar{0} means that xR0, so that x-0=x is divisible by 2 so that x is even. Perform a similar computation assuming that x\in\bar{1} and derive a contradiction. 2 You can partition any set X as X=\displaystyle\bigcup_{x\in X}\{x\}. -1 The trick here is to consider the following "switching operation":$$\sigma_{ij}(\lambda_1,\ldots,\lambda_N) = (\lambda_1,\ldots,\lambda_i+1,\ldots,\lambda_j-1,\ldots,\lambda_N). Now the question becomes: what can we say about $f(\sigma_{ij}(\lambda))$? Assume without loss of generality that $\lambda_j\ge1$, and that all $\lambda_k\ge0$. A short ...

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