# 数学代写|数值分析代写numerical analysis代考|MATH3003

## 数学代写|数值分析代写numerical analysis代考|Inner products for function spaces

To facilitate the development of least squares approximation theory, we introduce a formal structure for $C[a, b]$. First, recognize that $C[a, b]$ is a linear space: any linear combination of continuous functions on $[a, b]$ must itself be continuous on $[a, b]$.

Definition 2.2. The inner product of the functions $f, g \in C[a, b]$ is
$$\langle f, g\rangle=\int_a^b f(x) g(x) \mathrm{d} x .$$
The inner product satisfies the following basic axioms:

• $\langle\alpha f+g, h\rangle=\alpha\langle f, h\rangle+\langle g, h\rangle$ for all $f, g, h \in C[a, b]$ and all $\alpha \in \mathbb{R}$;
• $\langle f, g\rangle=\langle g, f\rangle$ for all $f, g \in C[a, b] ;$
• $\langle f, f\rangle \geq 0$ for all $f \in C[a, b]$.
With this inner product we associate the norm
$$|f|_2:=\langle f, f\rangle^{1 / 2}=\left(\int_a^b f(x)^2 \mathrm{~d} x\right)^{1 / 2} .$$
This is often called the ‘ $L^2$ norm,’ where the superscript ‘ 2 ‘ in $L^2$ refers to the fact that the integrand involves the square of the function $f$; the $L$ stands for Lebesgue, coming from the fact that this inner product can be generalized from $C[a, b]$ to the set of all functions that are square-integrable, in the sense of Lebesgue integration. By restricting our attention to continuous functions, we dodge the measuretheoretic complexities.

## 数学代写|数值分析代写numerical analysis代考|Least squares minimization via calculus

We are now ready to solve the least squares problem. We shall call the optimal polynomial $P_* \in \mathcal{P}n$, i.e., $$\left|f-P\right|_2=\min {p \in \mathcal{P}_n}|f-p|_2 .$$ We can solve this minimization problem using basic calculus. Consider this example for $n=1$, where we optimize the error over polynomials of the form $p(x)=c_0+c_1 x$. The polynomial that minimizes $|f-p|_2$ will also minimize its square, $|f-p|_2^2$. For any given $p \in \mathcal{P}_1$, define the error function \begin{aligned} E\left(c_0, c_1\right):=\left|f(x)-\left(c_0+c_1 x\right)\right|{L^2}^2 &=\int_a^b\left(f(x)-c_0-c_1 x\right)^2 \mathrm{~d} x \ &=\int_a^b\left(f(x)^2-2 f(x)\left(c_0+c_1 x\right)+\left(c_0^2+2 c_0 c_1 x+c_1^2\right.\right.\ &=\int_a^b f(x)^2 \mathrm{~d} x-2 c_0 \int_a^b f(x) \mathrm{d} x-2 c_1 \int_a^b x f(x) \mathrm{d} x \ & \quad+c_0^2(b-a)+c_0 c_1\left(b^2-a^2\right)+\frac{1}{3} c_1^2\left(b^3\right. \end{aligned}
To find the optimal polynomial, $P_$, optimize $E$ over $c_0$ and $c_1$, i.e., find the values of $c_0$ and $c_1$ for which
$$\frac{\partial E}{\partial c_0}=\frac{\partial E}{\partial c_1}=0 .$$
First, compute
\begin{aligned} &\frac{\partial E}{\partial c_0}=-2 \int_a^b f(x) \mathrm{d} x+2 c_0(b-a)+c_1\left(b^2-a^2\right) \ &\frac{\partial E}{\partial c_1}=-2 \int_a^b x f(x) \mathrm{d} x+c_0\left(b^2-a^2\right)+c_1 \frac{2}{3}\left(b^3-a^3\right) . \end{aligned}
Setting these partial derivatives equal to zero yields
\begin{aligned} 2 c_0(b-a)+c_1\left(b^2-a^2\right) &=2 \int_a^b f(x) \mathrm{d} x \ c_0\left(b^2-a^2\right)+c_1 \frac{2}{3}\left(b^3-a^3\right) &=2 \int_a^b x f(x) \mathrm{d} x \end{aligned}

# 数值分析代考

## 数学代写|数值分析代写numerical analysis代考|Inner products for function spaces

$$\langle f, g\rangle=\int_a^b f(x) g(x) \mathrm{d} x$$

• $\langle\alpha f+g, h\rangle=\alpha\langle f, h\rangle+\langle g, h\rangle$ 对所有人 $f, g, h \in C[a, b]$ 和所有 $\alpha \in \mathbb{R}$;
• $\langle f, g\rangle=\langle g, f\rangle$ 对所有人 $f, g \in C[a, b] ;$
• $\langle f, f\rangle \geq 0$ 对所有人 $f \in C[a, b]$.
有了这个内积，我们将规范联系起来
$$|f|_2:=\langle f, f\rangle^{1 / 2}=\left(\int_a^b f(x)^2 \mathrm{~d} x\right)^{1 / 2}$$
这通常被称为 ‘ $L^2$ 范数，’其中上标’ 2 ‘在 $L^2$ 指的是被积函数涉及函数的平方 $f$; 这 $L$ 代表 Lebesgue，因 为这个内积可以从 $C[a, b]$ 在勒贝格积分的意义上，所有平方可积函数的集合。通过将我们的注意力限 制在连续函数上，我们避开了测度论的复杂性。

## 数学代写|数值分析代写numerical analysis代考|Least squares minimization via calculus

$$|f-P|_2=\min p \in \mathcal{P}_n|f-p|_2 .$$

$$E\left(c_0, c_1\right):=\left|f(x)-\left(c_0+c_1 x\right)\right|{L^2}^2=\int_a^b\left(f(x)-c_0-c_1 x\right)^2 \mathrm{~d} x \quad=\int_a^b\left(f(x)^2-2 f(x)\left(c_0+c_1 x\right)\right.$$

$$\frac{\partial E}{\partial c_0}=\frac{\partial E}{\partial c_1}=0 .$$

$$\frac{\partial E}{\partial c_0}=-2 \int_a^b f(x) \mathrm{d} x+2 c_0(b-a)+c_1\left(b^2-a^2\right) \quad \frac{\partial E}{\partial c_1}=-2 \int_a^b x f(x) \mathrm{d} x+c_0\left(b^2-a^2\right)+c_1 \frac{2}{3}(b$$

$$2 c_0(b-a)+c_1\left(b^2-a^2\right)=2 \int_a^b f(x) \mathrm{d} x c_0\left(b^2-a^2\right)+c_1 \frac{2}{3}\left(b^3-a^3\right) \quad=2 \int_a^b x f(x) \mathrm{d} x$$

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