Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For this question, dene an inner product on the vector space of $P_2$ of polynomials, through the formula

$$p(x)q(x) = \int p(x)q(x)dx$$

What are the lengths $$\|\ 1\ \|, \|\ x\ \|,\|\ x^2\ \|$$

is it just $\sqrt{1}, x \ and \ x^2 $ respectively?

also, Find constants $a$, $b$ so that $1, 2x - 1, ax^2+ bx + 1$ is an orthogonal basis of $P_2$. (`or- thogonal basis' means each basis vector has inner product $0$ with every other basis vector).

share|improve this question
6  
You need some integration bounds... –  1015 Mar 25 '13 at 21:15
1  
Also, note that lengths are numbers, so asking whether the length of the polynomial $x$ is just $x$ betrays a fundamental misunderstanding of what's going on. –  Gerry Myerson Mar 26 '13 at 12:21

1 Answer 1

As @julien notes in a comment, you need upper and lower limits of integration for your "definition" of the inner product to actually define an inner product. Solely for the purpose of writing some kind of answer that may help you, I will assume that the definition of the inner product is $$\int_{17}^{42}p(x)q(x)\,dx$$ Now if $g(x)$ is any polynomial, then the length of $g(x)$, denoted $\|g\|$, is the nonnegative number defined by $$\|g\|^2=\int_{17}^{42}(g(x))^2\,dx$$ Now can you work out $\|1\|,\|x\|,\|x^2\|$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.