1
$\begingroup$

Use orthogonality to write the polynomials $1,x,x^2$ as linear combinations of the orthogonal basis of $p_1(x) =1, p_2(x) = x - \frac{1}{2}, p_3(x) = x^2 - x + \frac{1}{6}$ .

They answer is:

$x = \frac{1}{2}p_1(x) + p_2(x).$

But how did they get that? I know how to do these types of problems in $\mathbb{R^n}$ space but not in function space?

  • 0
    it says ..2012-10-21

4 Answers 4

0

If the inner product is $\langle f,g \rangle=\int_0^1f(x)g(x)dx$ then eveyrthing follows from what Gerry mentioned.

  • 0
    I agree, but without a $s$pecific inner product in mind you won't get a numerical value for these coefficients. In fact, different inner products would even produce different values for these coefficients.2012-10-21
3

Let $v_1,\dots,v_n$ be an orthogonal basis for a vector space $V$, and let $v$ be in $V$, so $v=c_1v_1+\cdots+c_nv_n\tag1$ for some scalars $c_1,\dots,c_n$. You want to know how to find the $c_i$. Take the inner product of (1) with $v_i$; use properties of the inner product, and orthognality of the basis, and $c_i$ should pop out at you.

  • 0
    I see my mistake now. I didn't see that this was a solution for only x. I get how to do it now. Thanks a lot !2012-10-23
2

The polynomial $1$ is obviously just $p_1$.

The polynomial $x$ is almost $p_2$, except $p_2$ has that constant $-\frac{1}{2}$. We want to get rid of that constant. Well, $p_1$ is also constant, so we can use that: scale $p_1$ so that it exactly matches the constant term we're trying to get rid of (but opposite sign), then add it. This gives $x=\frac{1}{2}p_1+p_2$.

But now what about $x^2$? Well, $x^2$ is almost $p_3+p_2$, except again, the constant term isn't quite right. So... using the above as an example, how do we correct the constant term to get $x^2$ in terms of this basis?

EDIT: I overlooked the bit about using orthogonality. Apologies. Here's a solution using orthogonality that's much more general than this specific circumstance:

There is a nice result (that you presumably have seen if this problem has been assigned) that says given an orthogonal basis $\{v_1,\ldots,v_n\}$ for a vector space with inner product $\langle\cdot, \cdot\rangle$, any vector $u$ in the space can be writen $u=\frac{\langle u, v_1\rangle}{|v_1|^2}\,v_1+\cdots+\frac{\langle u, v_n\rangle}{|v_n|^2}\,v_n\,.$ So for this problem, the $p_i$'s are the $v_i$'s, and the polynomials $1$, $x$, and $x^2$ each play the role of $u$.

  • 0
    Oh, wow, I feel dumb. I see my mistake now. Thanks a lot !2012-10-23
1

Here are nice notes. Write $x$ as

$ x=ap_1+bp_2+cp_3 \,,$

and use the orthogonality condition $=0$ to find the constants.

  • 0
    Thank you, everyone gave me the same answer so I did what you guys said and I got that $c_1$ = $\frac{1}{2}$ , $c_2$ = 1, but $c_3$ = 19 which is wrong because it should equal 0? Can you elaborate a bit more?2012-10-21