### 澳洲代写｜MATH1003｜Algebra and Calculus Methods代数与微积分方法 澳洲国立大学

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The course will discuss the three main classes of equations, elliptic, parabolic and hyperbolic. It is intended both for mathematics students continuing to honours work and for other students using mathematics at a high level in theoretical physics, engineering and information technology, and mathematical economics.

Topics to be covered will include fundamental solutions, maximum principles, regularity (smoothness) of solutions, variational problems, Holder and Sobolev spaces.

## Algebra and Calculus Methods代数与微积分方法问题集

Consider the expression
$$\left(\cdots\left(\left((x-2)^2-2\right)^2-2\right)^2-\cdots-2\right)^2 \quad \text { (n squares) } .$$
Find the coefficient of $x^2$.

Solution. Let $a_n$ be the coefficient of $x^2$, and $b_n$ the coefficient of $x$. We observe that for any $n$, the term not containing $x$ in the above development is 4 . It follows that
$$a_n=4 a_{n-1}+b_{n-1}^2 \quad \text { and } b_n=4 b_{n-1},$$
where $a_1=1$ and $b_1=-4$. We first obtain that $b_n=-4^n$. Substituting in the recurrence relation corresponding to $a_n$, we obtain $a_n=4 a_{n-1}+4^{2 n-2}$, which implies $a_n=4^{n-1}\left(4^n-1\right) / 3$.

An interesting additive decomposition of positive integers in terms of Fibonacci numbers is presented in what follows.

Prove that any positive integer $N$ may be written as the sum of distinct and nonconsecutive terms of the Fibonacci sequence.

Solution. Let $\left(F_n\right){n \geq 1}$ be the Fibonacci sequence. Then $F_1=F_2=1$ and $F{n+2}=$ $F_{n+1}+F_n$, for all $n \geq 1$. Let us assume that $F_n \leq N<F_{n+1}$. So, $0 \leq N-F_n<$ $F_{n-1}$. It follows that there exists $s<n-1$ such that $F_s \leq N-F_n<F_{s+1}$. Hence $0 \leq N-F_n-F_s<F_{s-1}$ and $s-1<n-2$. We thus obtain that $N$ may be written
1.3 Recurrent Sequences
as $N=F_n+F_s+F_p+\cdots+F_r$, where the consecutive subscripts $n, s, p, \ldots, r$ are nonconsecutive numbers.

We need in what follows elementary differentiability properties of polynomials.

For any real number $a$ and for any positive integer $n$ define the sequence $\left(x_k\right){k \geq 0}$ by $x_0=0, x_1=1$, and $$x{k+2}=\frac{c x_{k+1}-(n-k) x_k}{k+1}, \quad \text { for all } k \geq 0 .$$
Fix $n$ and let $c$ be the largest real number such that $x_{n+1}=0$. Find $x_k$ in terms of $n$ and $k, 1 \leq k \leq n$.

Solution. We first observe that $x_{n+1}$ is a polynomial of degree $n$ in $c$. Thus, it is enough to find $n$ values of $c$ such that $X_{n+1}=0$. We will prove that these values are $c=n-1-2 r$, for $r=0,1, \ldots, n-1$. In this case, $x_k$ is the coefficient of $t^{k-1}$ in the polynomial $f(t)=(1-t)^r(1+t)^{n-1-r}$. This property follows after observing that $f$ satisfies the identity
$$\frac{f^{\prime}(t)}{f(t)}=\frac{n-1-r}{1+t}-\frac{r}{1-t},$$
that is,
$$\left(1-t^2\right) f^{\prime}(t)=f(t)[(n-1-r)(1-t)-r(1+t)]=f(t)[(n-1-2 r)-(n-1) t] .$$
Identifying the coefficients of $t^k$ in both sides, we obtain
$$(k+1) x_{k+2}-(k-1) x_k=(n-1-2 r) x_{k+1}-(n-1) x_k .$$
In particular, the largest $c$ is $n-1$, and $x_k=C_{n-1}^{k-1}$, for $k=1,2, \ldots, n$.
The next problem is devoted to the study of the normalized logistic equation, which is a successful model of many phenomena arising in genetics and mathematical biology. In its simplest form, the logistic equation is a formula for approximating the evolution of an animal population over time. The unknown $a_n$ in the following recurrent sequence represents the number of animals after the $n t h$ year. It is easy to observe that the sequence $\left(a_n\right)_{n \geq 1}$ converges to zero. The interesting part of the problem is to deduce the first-and second-order decay terms of this sequence.

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