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The characteristic equation is $x^2-7x+12=0$, which factors as $(x-3)(x-4)=0$, yielding two roots, 3 and 4. So $f(n)=a\cdot 3^n+b\cdot 4^n$ for some constants $a$ and $b$. Now use the values of $f(1)$ and $f(2)$ to solve for $a$ and $b$.

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Write $a_n = f(n)$ instead. Step 1 You can note that $$a_{n+1}-4a_n = 3(a_n-4a_{n-1})$$ so putting $b_n=a_n-4a_{n-1}$ you get $$b_{n+1} = 3b_n$$ so $b_n$ is geometric progression, with $b_2=3$ so $b_1=1$ and thus $$b_n = 3^{n-1}$$ so $$\boxed{a_{n+1}-4a_n =3^n}$$ Step 2 You can also note that $$a_{n+1}-3a_n = 4(a_n-3a_{n-1})$$ so putting $c_n=a_n-3a_{n-... 4 If$\ b+5\ $and$\ b^2+7\ $are not coprime there must be a prime number$p$with $$b\equiv -5\mod p$$ This gives $$b^2\equiv 25\mod p$$ and because of $$b^2\equiv -7\mod p$$ we have$\ p\mid 32\ $, which implies$\ p=2\ $. Hence$\ 2\ $is the only possible common prime factor. But$\ b+5\ $is odd if$\ b\ $is a multiple of$\ 32\ $. Hence$\ b+5\ and ... 4 Unfortunately I don't know what your mathematical background is to know if this is a useful answer, but I'll post it for the sake of completeness. What you have is a linear constant-coefficient difference equation. There are lots of ways to solve them, some specialized, but the usual generic one is linear algebra: \begin{align*} \overbrace{\begin{... 3 Let S(i) = \sum_{x_1+...+x_k = n}x_i $. Note that$S(i) = S(j)$for every$i,j \in \{1,...,k\} \sum_{i=1}^k S(i) = \sum_{i=1}^k \sum_{x_1+...+x_k = n}x_i = \sum_{x_1+...+x_k = n}\sum_{i=1}^k x_i =\sum_{x_1+...+x_k = n}n = n{ n-1 \choose k-1 } $On the other hand$\sum_{i=1}^k S(i) = k\cdot S(1) $So:$S(1) = \frac{n}{k}{ n-1 \choose k-1 } $3 The first person avoid his own hat. Hence$N-1$options. What about the second person person? His number of option relies on whether the first person has chosen the hat of the second person. He might have$N-2$options (if his hat wasn't chosen earlier) and$N-1$options (if his hat has been chosen earlier). Hence the formula for$P(A^C)$is not correct. ... 3 Let's look first for real numbers$a>b>0$with$a^b=b^a$. Write$a=tb$with$t>1$. Then the equation becomes $$(tb)^b=b^{tb}$$ which is equivalent to $$tb=b^t$$ and so to $$t=b^{t-1}$$ and finally to $$b=t^{1/(t-1)}.$$ So $$(a,b)=(t^{t/(t-1)},t^{1/(t-1)})$$ for$t>1$is the parametric solution of$a^b=b^a$with$a>b>0. Next question: when ... 3 Well, countability of a set means that one can tabulate the elements of the set in a list like this: $$\begin{array}{cccc} 0 & 1 & 2 & 3 & \ldots \\ a_0 & a_1 & a_2 & a_3 &\ldots \end{array}$$ This holds for the natural numbers, $$\begin{array}{cccc} 0 & 1 & 2 & 3 & \ldots \\ 0 & 1 & 2 & 3 &\... 3 Let f(x) = ax+b and let g(x) = cx + d. Then let h(x) denote the function$$h(x) = (f\circ g)(x) - (g\circ f)(x) = a(cx+d) + b - c(ax+b) - d = ad + b - bc - d.$$So you're right that this is a constant, and hence can't be surjective (since the codomain \mathbb{R} has more than one point). You're confused about the definition of injectivity though. A ... 3 The torturous chain of comments has reached a point where an attempt at an Answer seems appropriate. Note again that the incompleteness theorem, in the guise of the unsolvability of the "decision problem", shows that it's impossible to give a perfect answer, in the form "To decide whether \phi is valid do the following", where "the following" can be done ... 3 What you have is precisely Pascal's triangle, but with the n^{\text{th}} row multiplied by n+1 for each n \ge 0; that is, the k^{\text{th}} entry in row n is (n+1) \dbinom{n}{k}. To see this, note that$$ \begin{align*} & \frac{d}{dx} \sum_{i=0}^n x(1-x)^i \\ &= \frac{d}{dx} \sum_{i=0}^n \sum_{r=0}^i (-1)^r \dbinom{i}{r} x^{r+1} &&... 2 The relationR$can be described as follows:$(A,B) \in R$iff either both$A$and$B$contain$3$or neither contains$3$. Thus, for any$A$you just have to consider two cases: a)$3 \in A$and b)$3 \notin A$In case a) the equivalence class consists of all sets that contain$3$and in case b) the equivalence class consists of all sets that do not ... 2 Choose 8 cells from 32 to be one color, then choose next 8 from 24 to be second color and so on... So we have $${32\choose 8}{24\choose 8}{16\choose 8}{8\choose 8}= {32!\over 8!^4}$$ 2 I agree with your answer. In particular, your term for$x,y \in A$exceeds the whole book answer and your logic is sound. 2 I also think that you are correct. I get the same answer by just partitioning the remaining eight people ... $$N(A)= \binom8{3,3,2}+\binom8{5,1,2}+\binom8{5,3,0} \\= \binom{8}{3}\binom{5}{3} +\binom{8}{1}\binom{7}{5}+\binom{8}{0}\binom{8}{5} \\=784$$ 2 There is a straightforward bijection, though - contra your last sentence - gotten by "interleaving:" $$0, -1, 1, -2, 2, -3, 3, ...$$ I think this is readily understandable. It's when we look at the rationals that things get difficult. 2 Well, one can define ''component'' as follows. Define a binary relation on the nodes of an undirected graph by saying that two nodes$u,v$are connected if there a path in the graph between$u$and$v$. This is an equivalence relation (transitive, reflexive, symmetric) on the node set. The equivalence class containing the node$v$is the set of all nodes ... 2 We have $$a_n-3=2(a_{n-1}-3),$$ which says that$a_n-3$is a geometric progression, which gives $$a_n-3=(-1-3)\cdot2^n$$ or $$a_n=3-2^{n+2}.$$ 2 Added on request of Mehrdad. Say we have $$\boxed{a_{n+1} = (x+y)a_n-xya_{n-1}}$$ then we can do: $$a_{n+1}-xa_n = y(a_n-xa_{n-1})$$ and $$a_{n+1}-ya_n = x(a_n-ya_{n-1})$$ Putting$\boxed{b_n =a_n-xa_{n-1}}$and$\boxed{c_n = a_n-ya_{n-1}}$we can finish as before. In general$x,y$are solution of quadratic (characteristic) equation$t^2-pt-q=0$of ... 1 The group axioms are:$(1)$Closure under binary operation of the group i.e.$(a \star b) = c \in S \ \forall \ a,b \in S$.$(2)$Associativity of the operation '$\star$' i.e. for any$a,b,c \in S , (a \star b) \star c = a \star (b \star c)$.$(3)$Existence of identity of the group under$\star$i.e. an element$e$such that$(e \star a) = a = (a \star ...

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The theorem of Lagrange says that subgroup order divides group order. Since 6 does not divide 13, a group of order 13 cannot have a subgroup of order 6. Subgroups can only have order 1 or 13 (trivial subgroups) as 13 is a prime.

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The key fact is that since $\psi$ has no quantifiers, its meaning does not depend on the domain. Now, notice that the statement asserts the existence of three elements of the domain (which may be equal to one another) that satisfy the universal formula. What if you restrict your seven element model to only consist of these three (or possibly one or two) ...

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Let $b=32a$ and so $b+5=2^5a+5$ and $b^2+7=2^{10}a^2+7$. The $1^{st}$ term is a linear polynomial in $a$ with coefficient in $\mathbb{R}$ and its root is in $\mathbb{R}$. But the $2^{nd}$ polynomial does not have any real root. Hence it cannot be factorized as a product of linear polynomial with real coefficient. So they are coprime.

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$b^2+7 = (b+5)(b-5) + 32$ So any common factor of $b^2+7$ and $b+5$ must also be a factor of $32$. Since $b$ is a multiple of $32$, both $b+5$ and $b^2+7$ are odd, so common factor is not a multiple of $2$. Therefore the only common factor of $b+5$ and $b^2+7$ is $1$ i.e. they are coprime.

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By the Euclidean algorithm $\:\overbrace{(b^{\phantom{|}}\!\!\!+\!5,\,\color{#c00}{b^2\!+\!7})\, =\, (b\!+\!5,\,\color{#c00}{32})}^{\!\!\!\!\!\!\!\!\!\large \bmod b+5:\ \ \ b\ \equiv\ -5\ \ \Rightarrow\ \ \color{#c00}{b^{\Large 2}+7\ \equiv\ 32_{\phantom{|}}}}\, =\, \overbrace{(\underbrace{\color{#0a0}{32n^{\phantom{|}}\!\!\!+\!5}}_{\large b \ =\ 32n},\,32)\,... 1 The logic of mathematics works as follows: There are a number of axioms, which are defined to be true. From these axioms, we deduce other properties and results. From these, we then get more and more results. Taking all the results that are already proven to be true (assuming the axioms to hold) and deducing a new result from that is what is called "to ... 1 Let's try an easier question: what is the expected number of rooms with$K$or more occupants? Considering a single room, we can use a binomial distribution: the probability of exactly$K$occupants of that room is${P \choose K}\frac{(N-1)^{P-K}}{N^P}$so the probability of at least$K$occupants of that room is$\sum\limits_{j=K}^N{P \choose j}\frac{(N-...

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Consider the difference between these two statements: $$\lnot\forall x\in\mathbf{R}(x^2>0)$$ $$\lnot\exists x\in\mathbf{R}(x^2>0)$$ The first way to understand the difference between $\lnot\forall$ and $\lnot\exists$ is to read directly what these statements mean. The first statement is equivalent to "it is not true that for all real $x$ that $x^2$ is ...

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Both statements are very different. There is no human who can fly means $\neg\exists x [f(x)]$ or if one explicitly refers to humans $\neg\exists x[ h(x)\wedge f(x)]$. The statement $\neg \forall x[f(x)]$ or more explicitly referring to humans, $\neg \forall x[h(x)\Rightarrow f(x)]$, means that it is not true that all humans can fly.

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You have copied the definition wrongly. The correct definition of $c \otimes x$ is $c \otimes x=(\sqrt [3] c)x$.

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