### 数学代写|数值方法作业代写numerical methods代考| Taylor’s Theorem

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## 数学代写|数值方法作业代写numerical methods代考|Taylor’s Theorem

Taylor’s theorem allows us to expand a function as a series involving higher-order derivatives of a function. We take the Cauchy form (with exact remainder):
$f$ is $n$ times differentiable
$$f(b)=\sum_{k=0}^{n-1} \frac{(b-a)^{k}}{k !} f^{(k)}(a)+R_{n}$$
where:
$$R_{n}=\frac{(b-\xi)^{n} f^{(n)}(\xi)}{n !}, a<\xi<b$$

and:
\begin{aligned} &f^{\prime}=f^{(1)}=\frac{d f}{d x}, f^{(2)}=\frac{d^{2} f}{d x^{2}} \ &f^{(n)}(x)=\left(f^{(n-1)}(x)\right)^{\prime}=\frac{d}{d x}\left(f^{(n-1)}(x)\right) \end{aligned}
We conclude with a discussion of the exponential function. It is the only functi that is the same as its derivative. To see this, we use the formal definition (1.7) of derivative (and noting that $e^{x} e^{y}=e^{x+y}, x, y \in \mathbb{R}$ ):
$$\frac{d}{d x} e^{x}=\lim {h \rightarrow 0}\left(\frac{e^{x+h}-e^{x}}{h}\right)=e^{x} \lim {h \rightarrow 0} \frac{e^{h}-1}{h}=e^{x}, x \in \mathbb{R}$$
We summarise some useful properties of the exponential function:
\begin{aligned} &e^{x}=\sum_{n=0}^{\infty} \frac{x^{n}}{n !} \ &e^{x}=\lim {n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n} \ &\frac{d}{d x} e^{x}=e^{x} \ &e^{x+y}=e^{x} e^{y} \ &y=\log x \Longleftrightarrow x=e^{y} \ &\log (a b)=\log a+\log b . \ &\frac{d^{n}}{d x^{n}} e^{x}=e^{x} \forall n \geq 1 \ &e^{x}=\sum{k=0}^{n-1} \frac{x^{k}}{k !}+\mathbb{R}{n} \text { where } \mathbb{R}{n}=\frac{x^{n}}{n !} e^{\xi}, \xi<x . \end{aligned}

## 数学代写|数值方法作业代写numerical methods代考|Big O and Little o Notation

For many applications we need a definition of the asymptotic behaviour of quantities such as functions and series; in particular we wish to find bounds on mathematical expressions and applications in computer science. To this end, we introduce the Landau symbols $\mathrm{O}$ and $\mathrm{o}$.
Definition $1.2$ (O-Notation).
\begin{aligned} &f(x)=O(g(x)) \text { as } x \rightarrow \infty \text { if } \exists M>0, \exists x_{0} \text { s.t. } \ &|f(x)| \leq M|g(x)| \text { for } x>x_{0} \ &f(x)=O(g(x)) \text { as } x \rightarrow a \text { if }|f(x)| \leq M|g(x)| \text { for }|x-a|<\delta \ &\text { Unified definition: } \lim _{x \rightarrow a} \frac{f(x)}{g(x)}<\infty \end{aligned}

An example is:
\begin{aligned} &f(x) \equiv 6 x^{4}-7 x^{2}+2 \ &g(x) \equiv x^{4} \ &f(x)=O(g(x)) \text { as } x \rightarrow \infty \ &f_{n} \equiv 2 n^{3}+6 n^{2}+5(\log n)^{3} \ &f_{n}=O\left(n^{3}\right) \text { as } n \rightarrow \infty \end{aligned}
Definition $1.3$ (o-Notation).
\begin{aligned} &f(x)=o(g(x)) \text { as } x \rightarrow \infty \ &\text { if } \lim _{x \rightarrow \infty} \frac{f(x)}{g(x)}=0 . \end{aligned}
An example is:
\begin{aligned} &2 x=o\left(x^{2}\right) \ &2 x^{2} \neq o\left(x^{2}\right) \ &1 / x=o(1) . \end{aligned}
We note that complexity analysis applies to both continuous and discrete functions.

## 数学代写|数值方法作业代写numerical methods代考|PARTIAL DERIVATIVES

In general, we are interested in functions of two (or more) variables. We consider a function of the form:
$$z=f(x, y) .$$
The variables $x$ and $y$ can take values in a given bounded or unbounded interval. First, we say that $f(x, y)$ is continuous at $(a, b)$ if the limit:
$$\lim _{x \rightarrow a} f(x, y)$$
exists and is equal to $f(a, b)$. We now need definitions for the derivatives of $f$ in the $x$ and $y$ directions.

In general, we calculate the partial derivatives by keeping one variable fixed and differentiating with respect to the other variable; for example:
\begin{aligned} &z=f(x, y)=e^{k x} \cos m y \ &\frac{\partial z}{\partial x}=k e^{k x} \cos m y \ &\frac{\partial z}{\partial y}=-m e^{k x} \sin m y . \end{aligned}

We now discuss the situation when we introduce a change of variables into some problem and then wish to calculate the new partial derivatives. To this end, we start with the variables $(x, y)$, and we define new variables $(u, v)$. We can think of these as ‘original’ and ‘transformed’ coordinate axes, respectively. Now define the function $z(u, v)$ as follows:
$$z=z(u, v), u=u(x, y), v=v(x, y)$$
This can be seen as a function of a function. The result that we are interested in is the following: if $z$ is a differentiable function of $(u, v)$ and $u, v$ are themselves continuous functions of $x, y$, with partial derivatives, then the following rule holds:
\begin{aligned} &\frac{\partial z}{\partial x}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial x}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial x} \ &\frac{\partial z}{\partial y}=\frac{\partial z}{\partial u} \frac{\partial u}{\partial y}+\frac{\partial z}{\partial v} \frac{\partial v}{\partial y} \end{aligned}
This is a fundamental result that we shall apply in this chapter. We take a simple example of Equation (1.11) to show how things work. To this end, consider the Laplace equation in Cartesian geometry:
$$\frac{\partial^{2} u}{\partial x^{2}}+\frac{\partial^{2} u}{\partial y^{2}}=0$$
We now wish to transform this equation into an equation in a circular region defined by the polar coordinates:
$$x=r \cos \theta, y=r \sin \theta$$
The derivative in $r$ is given by:
$$\frac{\partial u}{\partial r}=\frac{\partial u}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial u}{\partial y} \frac{\partial y}{\partial r}=\cos \theta \frac{\partial u}{\partial x}+\sin \theta \frac{\partial u}{\partial y}$$
and you can check that the derivative with respect to $\theta$ is:
$$\frac{\partial u}{\partial \theta}=-r \sin \theta \frac{\partial u}{\partial x}+r \cos \theta \frac{\partial u}{\partial y}$$
hence:
\begin{aligned} &\frac{\partial u}{\partial x}=\cos \theta \frac{\partial u}{\partial r}-\frac{1}{r} \sin \theta \frac{\partial u}{\partial \theta} \ &\frac{\partial u}{\partial y}=\sin \theta \frac{\partial u}{\partial r}+\frac{1}{r} \cos \theta \frac{\partial u}{\partial \theta} \end{aligned}

and:
\begin{aligned} &\frac{\partial^{2} u}{\partial x^{2}}=\cos \theta \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}\right)-\frac{1}{r} \sin \theta \frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \ &\frac{\partial^{2} u}{\partial y^{2}}=\sin \theta \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial y}\right)+\frac{1}{r} \cos \theta \frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial y}\right) . \end{aligned}
Combining these results allows us to write Laplace’s equation in polar coordinates as follows:
$$\frac{\partial^{2} u}{\partial r^{2}}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^{2}} \frac{\partial^{2} u}{\partial \theta^{2}}=0 .$$
Thus, the original heat equation in Cartesian coordinates is transformed to a PDE of convection-diffusion type in polar coordinates.

We can find a solution to this problem using the Separation of Variables method, for example.

## 数学代写|数值方法作业代写numerical methods代考|Taylor’s Theorem

F是n次可微
F(b)=∑ķ=0n−1(b−一种)ķķ!F(ķ)(一种)+Rn

Rn=(b−X)nF(n)(X)n!,一种<X<b

F′=F(1)=dFdX,F(2)=d2FdX2 F(n)(X)=(F(n−1)(X))′=ddX(F(n−1)(X))

ddX和X=林H→0(和X+H−和XH)=和X林H→0和H−1H=和X,X∈R

## 数学代写|数值方法作业代写numerical methods代考|Big O and Little o Notation

F(X)=这(G(X)) 作为 X→∞ 如果 ∃米>0,∃X0 英石  |F(X)|≤米|G(X)| 为了 X>X0 F(X)=这(G(X)) 作为 X→一种 如果 |F(X)|≤米|G(X)| 为了 |X−一种|<d  统一定义： 林X→一种F(X)G(X)<∞

F(X)≡6X4−7X2+2 G(X)≡X4 F(X)=这(G(X)) 作为 X→∞ Fn≡2n3+6n2+5(日志⁡n)3 Fn=这(n3) 作为 n→∞

F(X)=这(G(X)) 作为 X→∞  如果 林X→∞F(X)G(X)=0.

2X=这(X2) 2X2≠这(X2) 1/X=这(1).

## 数学代写|数值方法作业代写numerical methods代考|PARTIAL DERIVATIVES

∂和∂X=∂和∂在∂在∂X+∂和∂在∂在∂X ∂和∂是=∂和∂在∂在∂是+∂和∂在∂在∂是

∂2在∂X2+∂2在∂是2=0

X=r因⁡θ,是=r罪⁡θ

∂在∂r=∂在∂X∂X∂r+∂在∂是∂是∂r=因⁡θ∂在∂X+罪⁡θ∂在∂是

∂在∂θ=−r罪⁡θ∂在∂X+r因⁡θ∂在∂是

∂在∂X=因⁡θ∂在∂r−1r罪⁡θ∂在∂θ ∂在∂是=罪⁡θ∂在∂r+1r因⁡θ∂在∂θ

∂2在∂X2=因⁡θ∂∂r(∂在∂X)−1r罪⁡θ∂∂θ(∂在∂X) ∂2在∂是2=罪⁡θ∂∂r(∂在∂是)+1r因⁡θ∂∂θ(∂在∂是).

∂2在∂r2+1r∂在∂r+1r2∂2在∂θ2=0.

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