这是一份Caltech加州理工学院的Numerical analysis数值分析代考案例

加州理工的这门数值分析课属于入门课程,exam考察的内容主要也是数学分析的基本功和数值分析中的一些基本概念和方法,比如absolute error, polynomial interpolation.

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数学代写|数值分析代写Numerical analysis final exam
Problem 1.

TRUE or FALSE? Prove or disprove the following statements (no points for correct answer without a valid proof):
a) (2 points) If $f \in C^{1}[a, b]$ and $\left|f^{\prime}(x)\right| \leq 1$ for all $x \in[a, b]$, then there can be at most one number $p \in[a, b]$ for which $f(p)=2 p$.
b) (2 points) If $f(x)$ is Lipschitz on $(-\infty, \infty)$, then so is $f(x)^{2}$ on $(-\infty, \infty)$.

Proof .

a) TRUE – Suppose there are two values $p, q$ with $p \neq q$ s.t. $f(p)=2 p$ and $f(q)=2 q$. Then by the MVT, there exists $\xi \in[a, b]$ s.t.
$$
f^{\prime}(\xi)=\frac{f(p)-f(q)}{p-q}=\frac{2 p-2 q}{p-q}=2 .
$$
But $\left|f^{\prime}(x)\right| \leq 1$ for $x \in[a, b]$, so by contradiction $p=q$.
b) FALSE – Consider e.g. $f(x)=x$, which is Lipschitz with $L=1$. But $f(x)=x^{2}$ is not, since $f^{\prime}(x)=2 x$ which is unbounded.

Problem 2.

a) (3 points) Show that the fixed point iteration
$$
p_{n}=\frac{p_{n-1}^{2}+3}{5}, \quad n=1,2, \ldots
$$
converges for any initial $p_{0} \in[0,1]$.
b) (1 point) Estimate how many iterations $n$ are required to obtain an absolute error $\left|p_{n}-p\right|$ less than $10^{-4}$ when $p_{0}=1$. No numerical value needed, just give an expression for $n$.

Proof .

a) Show $g(x) \in[0,1]$ for $x \in[0,1]$ :
$$
\begin{aligned}
&g(0)=3 / 5 \
&g(1)=4 / 5
\end{aligned}
$$
increasing function
Show $\left|g^{\prime}(x)\right| \leq k<1$ :
$$
g^{\prime}(x)=\frac{2 x}{5} \leq \frac{2}{5}=k<1
$$
b)
$$
10^{-4}=\left|p_{n}-p\right| \leq k^{n} \max {1,0}=\left(\frac{2}{5}\right)^{n} \Rightarrow n \approx \frac{-4}{\log _{10} \frac{2}{5}}
$$

Problem 3.

Find the fourth degree polynomial $f(x)$ which satisfies the conditions
$$
f(0)=1, f(1)=2, f^{\prime}(1)=\alpha, f(2)=2 \alpha, f^{\prime}(2)=2 \alpha
$$

Proof .

Divided differences

$\Longrightarrow f(x)=1+x+(\alpha-1) x(x-1)-\frac{1}{2} x(x-1)^{2}+\left(\frac{9}{4}-\frac{\alpha}{2}\right) x(x-1)^{2}(x-2)$

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