# Orthonormal basis for the poynomials of degree maximum 2

• MHB
• mathmari
In summary, the orthonormal basis for the poynomials of degree maximum $2$ is given by $\tilde{q}_1=1, q_1=\frac{\tilde{q}_1}{\|\tilde{q}_1\|}, q_2=x-\langle x^2, q_1\rangle q_1, q_3=x^2-\frac{3}{20}, q_4=x^2-\frac{3}{20}, q_5=x^2-\frac{9}{16}$.
mathmari
Gold Member
MHB
Hey!

We consider the inner product $$\langle f,g\rangle:=\int_{-1}^1(1-x^2)f(x)g(x)\, dx$$ Calculate an orthonormal basis for the poynomials of degree maximum $2$.

I have applied the Gram-Schmidt algorithm as follows:

\begin{align*}\tilde{q}_1:=&1 \\ q_1:=&\frac{\tilde{q}_1}{\|\tilde{q}_1\|}=\frac{1}{\langle \tilde{q}_1, \tilde{q}_1\rangle}=\frac{1}{\int_{-1}^1(1-x^2)\cdot 1\cdot 1\, dx}=\frac{1}{\int_{-1}^1(1-x^2)\, dx}=\frac{1}{\left [x-\frac{x^3}{3}\right]_{-1}^1}=\frac{1}{\left [1-\frac{1^3}{3}\right]-\left [(-1)-\frac{(-1)^3}{3}\right ]}\\ =&\frac{1}{1-\frac{1}{3}+1-\frac{1}{3}}=\frac{1}{2-\frac{2}{3}}=\frac{1}{\frac{4}{3}}=\frac{3}{4}\end{align*}

\begin{align*}\tilde{q}_2:=&x-\langle x, q_1\rangle q_1=x-\left (\int_{-1}^1(1-x^2)\cdot x\cdot \frac{3}{4}\, dx\right )\cdot \frac{3}{4}=x-\frac{3}{4}\cdot \left (\int_{-1}^1(x-x^3)\, dx\right )\cdot \frac{3}{4}=x-\frac{9}{16}\cdot \left [\frac{x^2}{2}-\frac{x^4}{4}\right ]_{-1}^1 \\ = & x-\frac{9}{16}\cdot 0=x\\ q_2:=&\frac{\tilde{q}_2}{\|\tilde{q}_2\|}=\frac{x}{\langle \tilde{q}_2, \tilde{q}_2\rangle}=\frac{x}{\int_{-1}^1(1-x^2)\cdot x\cdot x\, dx}=\frac{x}{\int_{-1}^1(x^2-x^4)\, dx}=\frac{x}{\left [\frac{x^3}{3}-\frac{x^5}{5}\right]_{-1}^1}=\frac{x}{\left [\frac{1}{3}-\frac{1}{5}\right]-\left [-\frac{1}{3}+\frac{1}{5}\right ]}\\ =&\frac{x}{\frac{1}{3}-\frac{1}{5}+\frac{1}{3}-\frac{1}{5}}=\frac{x}{\frac{2}{3}-\frac{2}{5}}=\frac{x}{\frac{4}{15}}=\frac{15x}{4}\end{align*}

\begin{align*}\tilde{q}_3:=&x^2-\langle x^2, q_1\rangle q_1-\langle x^2, q_2\rangle q_2=x^2-\left (\int_{-1}^1(1-x^2)\cdot x^2\cdot \frac{3}{4}\, dx\right )\cdot \frac{3}{4}-\left (\int_{-1}^1(1-x^2)\cdot x^2\cdot \frac{15x}{4}\, dx\right )\cdot \frac{15x}{4}\\ =&x^2-\frac{9}{16}\cdot \int_{-1}^1(x^2-x^4)\, dx-\frac{225x}{16}\cdot \int_{-1}^1(x^3-x^5)\, dx =x^2-\frac{9}{16}\cdot \frac{4}{15}-\frac{225x}{16}\cdot 0=x^2-\frac{3}{20}\\ q_3:=&\frac{\tilde{q}_3}{\|\tilde{q}_3\|}=\frac{x^2-\frac{3}{20}}{\langle \tilde{q}_3, \tilde{q}_3\rangle}=\frac{x^2-\frac{3}{20}}{\int_{-1}^1(1-x^2)\cdot \left (x^2-\frac{3}{20}\right )\cdot \left (x^2-\frac{3}{20}\right )\, dx}=\frac{x^2-\frac{3}{20}}{\int_{-1}^1\left (-x^6+\frac{13x^4}{10}-\frac{129x^2}{400}+\frac{9}{400}\right )\, dx}=\frac{x^2-\frac{3}{20}}{\frac{9}{140}}\\ =&\frac{140}{9}\left (x^2-\frac{3}{20}\right )\end{align*} I wanted to check if I have the correct answers and I noticed (if I am not mistaken) that the polynomials that I found are not orthonormal in respect to the other terms. Have I applied a wrong formula? :unsure:

Hey mathmari!

I see you wrote $\|\tilde q_1\|=\langle\tilde q_1,\tilde q_1\rangle$.
Shouldn't that be $\|\tilde q_1\|=\sqrt{\langle\tilde q_1,\tilde q_1\rangle}$?

Klaas van Aarsen said:
I see you wrote $\|\tilde q_1\|=\langle\tilde q_1,\tilde q_1\rangle$.
Shouldn't that be $\|\tilde q_1\|=\sqrt{\langle\tilde q_1,\tilde q_1\rangle}$?

Oh yes, you 're right! Now I get the correct result! :giggle:

## 1. What is an orthonormal basis for polynomials of degree maximum 2?

An orthonormal basis for polynomials of degree maximum 2 is a set of three polynomials that are orthogonal to each other and have unit length. This means that the inner product of any two polynomials in the set is equal to 0, and the norm (length) of each polynomial is equal to 1.

## 2. Why is it important to have an orthonormal basis for polynomials of degree maximum 2?

An orthonormal basis allows us to represent any polynomial of degree maximum 2 as a linear combination of the basis polynomials. This makes it easier to manipulate and analyze polynomials, as well as solve problems involving them.

## 3. How do you find an orthonormal basis for polynomials of degree maximum 2?

To find an orthonormal basis for polynomials of degree maximum 2, we first start with a set of three linearly independent polynomials of degree 2 or less. Then, we use the Gram-Schmidt process to orthogonalize and normalize the polynomials, resulting in an orthonormal set.

## 4. Can an orthonormal basis for polynomials of degree maximum 2 be used for higher degree polynomials?

No, an orthonormal basis for polynomials of degree maximum 2 is specific to polynomials of degree 2 or less. To find an orthonormal basis for higher degree polynomials, we would need to use a different set of basis polynomials.

## 5. What are some applications of using an orthonormal basis for polynomials of degree maximum 2?

An orthonormal basis for polynomials of degree maximum 2 is commonly used in fields such as signal processing, image processing, and data analysis. It can also be used in solving differential equations and in numerical methods for solving problems involving polynomials.

• General Math
Replies
6
Views
1K
• General Math
Replies
4
Views
803
• General Math
Replies
18
Views
3K
• General Math
Replies
6
Views
2K
• General Math
Replies
3
Views
1K
• General Math
Replies
7
Views
1K
• General Math
Replies
4
Views
1K
• General Math
Replies
2
Views
907
• General Math
Replies
2
Views
1K
• General Math
Replies
4
Views
1K