IS Math-2014-02

题目来源：Problem 2 日期：2024-08-09 题目主题：Math-Calculus-Orthogonal Polynomials

解题思路

在这个问题中，我们需要处理的主要任务是计算给定多项式之间的内积，以及利用这些内积构建正交归一的多项式集合。解题步骤可以大致分为以下几个部分：

1. 计算多项式的内积：

   • 首先，通过积分计算给定多项式 $q_0(x)$、$q_1(x)$ 和 $q_2(x)$ 之间的内积 $(q_i,q_j)$。这里，利用偶函数和奇函数的积分特性来简化计算。例如，任何偶函数在对称区间 $[-1,1]$ 上的积分等于两倍的从 $0$ 到 $1$ 的积分，而奇函数在对称区间 $[-1,1]$ 上的积分为零。

2. 构建正交归一的多项式集合：

   • 利用已经计算出的内积，通过归一化（使内积为 1）和正交化（使不同多项式之间的内积为 0）的过程，来构建一组正交归一的多项式。具体来说，我们使用格拉姆-施密特正交化过程（Gram-Schmidt Process）来逐步构建一组满足正交归一条件的多项式集合。

3. 构造更高次的正交多项式：

   • 在构造第 $p_3(x)$ 多项式时，需要确保它不仅与前面的多项式正交，还必须是一个标准化多项式（即归一化的）。通过格拉姆-施密特正交化过程，可以找到 $p_3(x)$ 的形式，并确保满足所有条件。

4. 证明正交多项式的唯一性：

   • 通过假设存在另一组不同的多项式集合，并分析其与已知正交归一多项式集合的关系，证明这种多项式必须是已知多项式的相反数，从而证明了正交多项式在符号上的唯一性。

整个解题过程依赖于对多项式内积、正交化过程及其几何意义的理解，同时需要熟练掌握积分计算和函数的奇偶性分析。在解决这些问题时，关注每一步的逻辑推导和计算细节，以确保最终构造的多项式集合确实满足正交归一条件。

Solution

Question 1

To compute the integrals, we need to evaluate the inner product $(q_i,q_j)={\int }_{-1}^1{q}_i(x)q_j(x)\mathrm{d}x$ for the given polynomials:

$$q_0(x)=1,q_1(x)=x,q_2(x)=3x^2-1.$$

1.1. Compute $(q_0,q_1)$

$$(q_0,q_1)={\int }_{-1}^1{q}_0(x)q_1(x)\mathrm{d}x={\int }_{-1}^{1}1\cdot x\mathrm{d}x$$

Since $x$ is an odd function and the integral is over a symmetric interval $[-1,1]$, we have:

$$(q_0,q_1)=0.$$

1.2. Compute $(q_0,q_2)$

$$(q_0,q_2)={\int }_{-1}^1{q}_0(x)q_2(x)\mathrm{d}x={\int }_{-1}^{1}1\cdot (3x^2-1)\mathrm{d}x$$

This can be split into two integrals:

$$(q_0,q_2)=3{\int }_{-1}^1{x}^2\mathrm{d}x-{\int }_{-1}^{1}1\mathrm{d}x$$

Since $x^2$ is even:

$${\int }_{-1}^1{x}^2\mathrm{d}x=2{\int }_0^1{x}^2\mathrm{d}x=2{\left[\frac{x^3}{3}\right]}_0^1=\frac{2}{3},$$

and:

$${\int }_{-1}^{1}1\mathrm{d}x=2.$$

Thus:

$$(q_0,q_2)=3\cdot \frac{2}{3}-2=2-2=0.$$

1.3. Compute $(q_1,q_2)$

$$(q_1,q_2)={\int }_{-1}^1{q}_1(x)q_2(x)\mathrm{d}x={\int }_{-1}^{1}x\cdot (3x^2-1)\mathrm{d}x$$

This can be split into two integrals:

$$(q_1,q_2)=3{\int }_{-1}^1{x}^3\mathrm{d}x-{\int }_{-1}^{1}x\mathrm{d}x.$$

Both $x^3$ and $x$ are odd functions, so:

$$(q_1,q_2)=0.$$

1.4. Compute $(q_0,q_0)$

$$(q_0,q_0)={\int }_{-1}^1{q}_0(x)q_0(x)\mathrm{d}x={\int }_{-1}^{1}1\cdot 1\mathrm{d}x=2.$$

1.5. Compute $(q_1,q_1)$

$$(q_1,q_1)={\int }_{-1}^1{q}_1(x)q_1(x)\mathrm{d}x={\int }_{-1}^1{x}^2\mathrm{d}x=\frac{2}{3}.$$

1.6. Compute $(q_2,q_2)$

$$(q_2,q_2)={\int }_{-1}^1{q}_2(x)q_2(x)\mathrm{d}x={\int }_{-1}^1(3x^2-1{)}^2\mathrm{d}x.$$

Expanding the square:

$$q_2(x{)}^2=9x^4-6x^2+1.$$

Thus:

$$(q_2,q_2)=9{\int }_{-1}^1{x}^4\mathrm{d}x-6{\int }_{-1}^1{x}^2\mathrm{d}x+{\int }_{-1}^{1}1\mathrm{d}x.$$

Compute each integral:

$${\int }_{-1}^1{x}^4\mathrm{d}x=2{\int }_0^1{x}^4\mathrm{d}x=2{\left[\frac{x^5}{5}\right]}_0^1=\frac{2}{5},$$ $${\int }_{-1}^1{x}^2\mathrm{d}x=\frac{2}{3},$$ $${\int }_{-1}^{1}1\mathrm{d}x=2.$$

Thus:

$$(q_2,q_2)=9\cdot \frac{2}{5}-6\cdot \frac{2}{3}+2=\frac{18}{5}-4+2=\frac{18}{5}-\frac{20}{5}+\frac{10}{5}=\frac{8}{5}.$$

Question 2

2-1. Orthonormal Set $\{p_0(x),p_1(x),p_2(x)\}$

We need to normalize $q_0(x)$, $q_1(x)$, and $q_2(x)$ and orthogonalize them to form an orthonormal set.

$$p_0(x)=\frac{q_0(x)}{\sqrt{(q_0,q_0)}}=\frac{1}{\sqrt{2}}.$$ $$p_1(x)=\frac{q_1(x)}{\sqrt{(q_1,q_1)}}=\frac{x}{\sqrt{\frac{2}{3}}}=\sqrt{\frac{3}{2}}x.$$ $$p_2(x)=\frac{q_2(x)}{\sqrt{(q_2,q_2)}}=\frac{3x^2-1}{\sqrt{\frac{8}{5}}}=\sqrt{\frac{5}{8}}(3x^2-1).$$

So, the orthonormal set is:

$$p_0(x)=\frac{1}{\sqrt{2}},p_1(x)=\sqrt{\frac{3}{2}}x,p_2(x)=\sqrt{\frac{5}{8}}(3x^2-1).$$

2-2. Finding $p_3(x)$

To find $p_3(x)$ using the Gram-Schmidt process, we begin by expressing $p_3(x)$ as a linear combination of a cubic polynomial and the previously established orthogonal polynomials $q_0(x)=1$, $q_1(x)=x$, and $q_2(x)=3x^2-1$. Specifically, we can express $p_3(x)$ as:

$$p_3(x)=a_3{x}^3+a_2(3x^2-1)+a_{1}x+a_0.$$

Given that $q_0(x)$, $q_1(x)$, and $q_2(x)$ are mutually orthogonal and their norms (self-inner products) are already computed, we aim to ensure that $p_3(x)$ is orthogonal to all of them.

Step 1: Applying Orthogonality Conditions

We leverage the orthogonality conditions that $p_3(x)$ must satisfy:

1. Orthogonality to $q_0(x)$:

$$(p_3,q_0)={\int }_{-1}^1{p}_3(x)q_0(x)\mathrm{d}x=a_0\cdot {\int }_{-1}^{1}1\mathrm{d}x=2a_0=0.$$

This implies $a_0=0$.

2. Orthogonality to $q_1(x)$:

$$(p_3,q_1)={\int }_{-1}^1{p}_3(x)q_1(x)\mathrm{d}x.$$

Given the symmetry and oddness of certain terms, the significant contributions come from:

$$(p_3,q_1)=a_3\cdot \frac{2}{5}+a_1\cdot \frac{2}{3}=0\Rightarrow a_1=-\frac{3}{5}{a}_3.$$

3. Orthogonality to $q_2(x)$:

$$(p_3,q_2)={\int }_{-1}^1{p}_3(x)q_2(x)\mathrm{d}x.$$

Given the specific form of $p_3(x)$ and the orthogonality:

$$(p_3,q_2)=9a_2\cdot \frac{2}{5}-6a_2\cdot \frac{2}{3}+2a_2=0\Rightarrow a_2=0.$$

Step 2: Normalize $p_3(x)$

Given the orthogonality conditions, the polynomial $p_3(x)$ simplifies to:

$$p_3(x)=a_3\left(x^3-\frac{3}{5}x\right).$$

The next step is to normalize $p_3(x)$. The norm of $p_3(x)$ is:

$$\Vert p_3{\Vert }^2={\int }_{-1}^1{\left(a_3\left(x^3-\frac{3}{5}x\right)\right)}^2\mathrm{d}x=a_3^2\left(\frac{2}{7}-\frac{6}{5}\cdot \frac{2}{5}+\frac{9}{25}\cdot \frac{2}{3}\right).$$

Simplifying this:

$$\Vert p_3{\Vert }^2=a_3^2\cdot \frac{8}{175}.$$

To normalize $p_3(x)$, we set $a_3=\sqrt{\frac{175}{8}}$, so the normalized polynomial is:

$$p_3(x)=\frac{5\sqrt{14}}{4}\left(x^3-\frac{3}{5}x\right).$$

This final expression for $p_3(x)$ ensures that the polynomial is both orthogonal to the previous polynomials and normalized, thus satisfying the orthonormality conditions.

Question 3

3-1. Expressing Coefficients $c_k$

Given:

$$f_N(x)=\sum _{i=0}^N{c}_i{p}_i(x),$$

where $p_i(x)$ are orthonormal, we have:

$$(f_N,p_k)=\sum _{i=0}^N{c}_i(p_i,p_k)=c_k.$$

Thus:

$$c_k=(f_N,p_k).$$

3-2. Uniqueness of $p_N(x)$

Suppose ${\tilde{p}}_N(x)$ is another polynomial of degree $N$ such that the set $\{p_0(x),p_1(x),\dots ,p_{N-1}(x),{\tilde{p}}_N(x)\}$ is orthonormal. Since ${\tilde{p}}_N(x)$ is also a polynomial of degree $N$, it can be expressed as a linear combination of the orthonormal polynomials $p_0(x),p_1(x),\dots ,p_N(x)$:

$${\tilde{p}}_N(x)=\sum _{i=0}^N{\tilde{c}}_i{p}_i(x),$$

where the coefficients ${\tilde{c}}_j$ are given by the inner products:

$${\tilde{c}}_j=({\tilde{p}}_N(x),p_j(x)).$$

Since ${\tilde{p}}_N(x)$ is orthogonal to $p_0(x),p_1(x),\dots ,p_{N-1}(x)$, we have:

$${\tilde{c}}_j=0\text{for}j\le N-1.$$

Thus, ${\tilde{p}}_N(x)$ simplifies to:

$${\tilde{p}}_N(x)={\tilde{c}}_N{p}_N(x).$$

Now, consider the fact that both $p_N(x)$ and ${\tilde{p}}_N(x)$ are normalized, meaning that:

$${\int }_{-1}^1{p}_N(x{)}^2\mathrm{d}x=1\text{and}{\int }_{-1}^1{\tilde{p}}_N(x{)}^2\mathrm{d}x=1.$$

Substituting ${\tilde{p}}_N(x)={\tilde{c}}_N{p}_N(x)$ into the normalization condition for ${\tilde{p}}_N(x)$, we get:

$${\int }_{-1}^1{\tilde{p}}_N(x{)}^2\mathrm{d}x={\int }_{-1}^1({\tilde{c}}_N{p}_N(x){)}^2\mathrm{d}x={\tilde{c}}_N^2{\int }_{-1}^1{p}_N(x{)}^2\mathrm{d}x={\tilde{c}}_N^2.$$

Given that ${\int }_{-1}^1{p}_N(x{)}^2\mathrm{d}x=1$, it follows that:

$${\tilde{c}}_N^2=1\Rightarrow {\tilde{c}}_N=\pm 1.$$

Therefore, ${\tilde{p}}_N(x)=\pm p_N(x)$, proving that $p_N(x)$ is unique except for its sign.

知识点

正交多项式Gram-Schmidt正交化线性代数内积空间正交归一条件定积分

难点思路

本题的难点在于证明正交多项式的唯一性，即对于给定的正交归一化条件，如何证明在除了符号之外，不存在其他的正交多项式满足条件。通过将多项式的表达形式带入归一化条件，并结合积分的性质，我们能够严格地证明多项式的唯一性。这需要对多项式展开的系数进行细致的分析，并使用平方积分的方法来消除系数的不确定性。

解题技巧和信息

1. Gram-Schmidt 正交化：当构造正交多项式时，Gram-Schmidt 正交化过程是标准方法。该方法通过逐步消除各个多项式与之前所有多项式的相关性，最终获得正交多项式。

2. 唯一性证明技巧：在证明多项式唯一性时，考虑到多项式的展开形式及其在内积空间中的正交性，通过平方和积分的方法可以直接得出唯一性。这种方法能够有效消除多项式展开系数的不确定性。

3. 计算简化：在计算过程中，利用已知的正交条件和内积结果可以大大简化计算。这意味着你可以直接利用之前的计算结果，避免重复积分计算。

重点词汇

orthonormal condition 正交归一条件

inner product 内积

Gram-Schmidt process 格拉姆-施密特正交化过程

polynomial 多项式

uniqueness 唯一性

linear combination 线性组合

normalization 归一化

参考资料

1. Arfken, G.B., Weber, H.J., and Harris, F.E. “Mathematical Methods for Physicists,” 7th ed. Academic Press, Chapter 12.
2. Kreyszig, E. “Advanced Engineering Mathematics,” 10th ed. Wiley, Chapters on Linear Algebra and Inner Product Spaces.
3. Strang, G. “Introduction to Linear Algebra,” 5th ed. Wellesley-Cambridge Press, Chapters on Orthogonal Projections and Orthogonality.