5
$\begingroup$

I have the vector space $C^0([0,1],ℝ)$ of continious functions from 0 to 1, with the inner product

$$ \langle f , g \rangle = \int_{0}^{1} f(x)g(x) \,dx $$

I have to find all the first degree polynomials $g(x)=ax+b$ that are orthogonal on $f(x)=6x$, which means that $\langle f , g \rangle = 0$.

I have solved it like this

$$\langle f,g\rangle = \int_{0}^{1} f(x)g(x) \,dx = \int_{0}^{1} (6x)(ax+b) \,dx = 6a \int_{0}^{1} x^2 \,dx + 6b\int_{0}^{1} x \,dx $$ $$ = 6a \left[ \frac{x^3}{3} \right]^1_0 + 6b\left[ \frac{1}{2}\right]^1_0 = 6a \cdot \frac{1}{3} + 3b = 2a+3b $$ $$ 2a+3b = 0$$ $$ a =- \frac{3b}{2}$$

Therefore $g(x)=ax+b$ $$g(x) =- \frac{3b}{2}x+b = b(-\frac{3}{2}x+1)$$

Is this solution correct?

$\endgroup$
2
  • 2
    $\begingroup$ Yes, everything is okay! $\endgroup$ Jan 8 at 13:56
  • 2
    $\begingroup$ Your solution looks good. $\endgroup$ Jan 8 at 14:06

1 Answer 1

2
$\begingroup$

To provide an external reference, the orthogonal polynomials for the interval $[0,1]$ with weight $w(x)=1$ are the shifted Legendre polynomials, i.e.,

$$\tilde{P}_n(x)=P_n(2x-1)$$ where $P_n(x)$ is the $n$th Legendre polynomial. (Both families are normalized by $P_n(1)=\tilde{P}_n(1)=1$.) In particular, the first two are $$\tilde{P}_0(x)=1,\quad \tilde{P}_1(x)=2x-1.$$ Hence the given linear function is $$f(x)=6x=3\tilde{P}_1(x)+3\tilde{P}_0(x).$$ The OP claims that this is orthogonal to $$g(x)=6x-4=3\tilde{P}_1(x)-\tilde{P}_0(x)$$ where I've rescaled $g(x)$ for convenience. To test this, the orthogonality condition for shifted Legendre polynomials is $$\langle \tilde{P}_n|\tilde{P}_m\rangle = \int_0^1 \tilde{P}_n (x)\tilde{P}_m(x)\,dx=\frac{2}{2n+1}\delta_{mn}$$ So the inner product between $f$ and $g$ yields

$$\langle f|g\rangle = \frac{(3)(3)}{3}+\frac{(3)(-1)}{1}=3-3=0$$ which verifies the orthogonality.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .