# 代数代写 Algebra II|MATH 8807 Boston College Assignment

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## Instructions:

Algebra is a branch of mathematics that deals with the study of mathematical symbols and the rules for manipulating these symbols. In algebra, letters and symbols are used to represent unknown quantities, and mathematical operations such as addition, subtraction, multiplication, division, and exponentiation are used to solve equations and find solutions to problems.

Algebra is used extensively in many areas of mathematics, science, engineering, and economics, among other fields. It is essential in solving problems related to geometry, calculus, and statistics, and is also used in everyday life, such as calculating mortgage payments or determining the best price for a product.

Algebraic expressions can take many forms, such as linear equations, quadratic equations, polynomial equations, and systems of equations. These expressions can be solved using various methods, including factoring, completing the square, and the quadratic formula.

Some fundamental concepts in algebra include variables, coefficients, constants, and equations. Variables are letters or symbols that represent unknown quantities, while coefficients are numbers that multiply the variables in an equation. Constants are fixed numbers, and equations are mathematical expressions that use an equal sign to show that two expressions are equal.

Consider the random walk of a man on the integer line $\mathbb{Z}$ such that, at each step, that the probability to go from position $i$ to position $i+1$ is $p$, and the probability to go from $i$ to $i-1$ is $1-p$. The man “falls off the cliff” if he reaches the position 0. Suppose that the man starts at the initial position $i_0 \geq 1$. Find the probability that he falls off the cliff.

Let $P_i$ be the probability that the man falls off the cliff starting from position $i$. We want to find $P_{i_0}$ given that $i_0 \geq 1$. Note that $P_0 = 1$ since the man has already fallen off the cliff when he starts at position $0$. Also note that $P_i = 0$ for all $i \leq 0$ since the man has already fallen off the cliff.

We can now set up a recurrence relation for $P_i$. If the man is currently at position $i$, then he has two options for his next step: he can either move to position $i+1$ with probability $p$, or he can move to position $i-1$ with probability $1-p$. Thus, we have

$$P_i = p P_{i+1} + (1-p) P_{i-1}$$

for all $i \geq 1$. This is a second-order linear recurrence relation with constant coefficients. We can solve it using the characteristic equation:

$$r^2 – (1-p)r – p = 0.$$

The roots of this equation are

$$r_1 = \frac{1-p + \sqrt{(1-p)^2 + 4p}}{2} \quad \text{and} \quad r_2 = \frac{1-p – \sqrt{(1-p)^2 + 4p}}{2}.$$

Note that $r_1 > 1$ and $r_2 < 0$. Therefore, the general solution to the recurrence relation is

$$P_i = A r_1^i + B r_2^i$$

for some constants $A$ and $B$. We can find $A$ and $B$ by using the boundary conditions $P_0 = 1$ and $P_i = 0$ for all $i \leq 0$. This gives us the system of equations

\begin{align*} A + B &= 1, \ Ar_1^i + Br_2^i &= 0 \quad \text{for all } i \leq 0. \end{align*}

Solving for $A$ and $B$, we get

$$A = \frac{r_2}{r_2 – r_1} \quad \text{and} \quad B = -\frac{r_1}{r_2 – r_1}.$$

Therefore, the probability that the man falls off the cliff starting from position $i_0 \geq 1$ is

$$P_{i_0} = \frac{r_2^{i_0}}{r_2^{i_0} – r_1^{i_0}}.$$

Note that this expression makes sense since $r_1 > 1$ and $r_2 < 0$, so $r_1^{i_0}$ grows exponentially while $r_2^{i_0}$ decays exponentially as $i_0$ gets large. This means that $P_{i_0}$ approaches 1 as $i_0$ increases, which makes sense since the man is more likely to fall off the cliff the further he is from it.

For $1 \leq k \leq n / 2$, find a bijection $f$ between $k$-element subsets of $\{1, \ldots, n\}$ and $(n-k)$-element subsets of $\{1, \ldots, n\}$ such that $f(I) \supseteq I$, for any $k$-element subset $I$.

Let $S$ be a set of size $n$ and let $k$ be an integer such that $1 \leq k \leq n/2$. We will define a bijection $f$ between the set of $k$-element subsets of $S$ and the set of $(n-k)$-element subsets of $S$ that satisfies the given property.

Let $I$ be a $k$-element subset of $S$. We will define $f(I)$ as follows: first, let $J = S \setminus I$ be the complement of $I$ in $S$. Since $|I| = k$, we have $|J| = n-k$. Next, let $T$ be the $k$-element subset of $J$ that contains the $k$ smallest elements of $J$ (with respect to some fixed total ordering of $S$). Note that $T$ is well-defined since $|J| = n-k \geq k$.

Finally, we define $f(I) = T \cup I$. We claim that $f(I)$ is an $(n-k)$-element subset of $S$, and that $f$ is a bijection between the set of $k$-element subsets of $S$ and the set of $(n-k)$-element subsets of $S$.

To see that $f(I)$ is an $(n-k)$-element subset of $S$, note that $|f(I)| = |T \cup I| = |T| + |I| – |T \cap I|$, and $|T| = k$, $|I| = k$, and $|T \cap I| = 0$ (since $T$ and $I$ are disjoint). Therefore, $|f(I)| = 2k \leq 2(n/2) = n$, so $f(I)$ is a subset of $S$ with at most $n$ elements. Moreover, $|f(I)| = n-k$ since $|T| = k$ and $|J| = n-k$, so $f(I)$ is indeed an $(n-k)$-element subset of $S$.

To see that $f$ is a bijection, we will define its inverse. Let $A$ be an $(n-k)$-element subset of $S$, and let $J = S \setminus A$. Since $|A| = n-k$, we have $|J| = k$. Let $T$ be the $k$-element subset of $J$ that contains the $k$ smallest elements of $J$ (with respect to the same fixed total ordering of $S$ as before). Note that $T$ is well-defined since $|J| = k \geq k$. Finally, we define $f^{-1}(A) = T \cup A$. We claim that $f 问题 3. Prove that the number of set-partitions$\pi$of the set$[n]:=\{1, \ldots, n\}$such that, for any$i=1, \ldots, n-1$, the consecutive numbers$i$and$i+1$do not belong to the same block of$\pi$equals the number of set-partitions of the set$[n-1]$. 证明 . To prove the given statement, let us first denote by$a_n$the number of set-partitions of$[n]$such that no consecutive numbers belong to the same block, and by$b_n$the number of set-partitions of$[n-1]$. Now, let us consider a set-partition$\pi$of$[n]$such that no consecutive numbers belong to the same block. We will show that we can obtain a set-partition$\pi’$of$[n-1]$in a one-to-one fashion by removing the element$n$from$\pi$and merging the block containing$n-1$with the block containing$n$. Formally, let$B_n$be the block of$\pi$containing$n$, and let$B_{n-1}$be the block of$\pi$containing$n-1$. Since$n$and$n-1$do not belong to the same block, we have$B_n \neq B_{n-1}$. We can define a new set-partition$\pi’$of$[n-1]$by setting$\pi'(i) = \pi(i)$for$i=1,\ldots,n-2$, and by merging$B_n$and$B_{n-1}$into a single block in$\pi’$. Conversely, given a set-partition$\pi’$of$[n-1]$, we can obtain a set-partition$\pi$of$[n]$such that no consecutive numbers belong to the same block by adding the element$n$to$\pi’$as a new block. Formally, we can define a new set-partition$\pi$of$[n]$by setting$\pi(i) = \pi'(i)$for$i=1,\ldots,n-1$, and by setting$\pi(n) = {n}$. It is easy to see that these constructions are one-to-one, so the number of set-partitions of$[n]$such that no consecutive numbers belong to the same block is equal to the number of set-partitions of$[n-1]$. Thus, we have$a_n = b_{n-1}\$, as required.