Here is the question from my worksheet:

Consider the space of continuous functions on the interval $[-1,1]: $$\mathcal C^0([-1,1])$ The inner product on this space is defined as follows:

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

Given the polynomials $p_0(x)=1, p_1(x)=x, p_2(x)=x^2$, use the Gram-Schmidt process find to the polynomals $P_0,P_1,P_2$ such that $\langle P_i,P_j\rangle=0$ for $i \not=j$ and $P_i(1)=1$ for $i=0,1,2$. (You just have to find an orthogonal (not orthonormal) basis).

Here is what I have got so far:


$P_1=p_1(x)-\frac{\langle p_1(x),P_1\rangle}{\langle P_0,P_0\rangle}P_0=x-(\int_{-1}^1\tau d \tau)(\frac{1}{2})=x-0=x$

$P_2=p_2(x)-\frac{\langle p_2(x),P_0\rangle}{\langle P_0,P_0\rangle}P_0-\frac{\langle p_2(x),P_1\rangle}{\langle P_1,P_1\rangle}P_1=x^2-(\int_{-1}^1 \tau^2 d\tau) \frac{1}{2}-(\int_{-1}^1 \tau^3 d\tau )\frac{P_1}{\langle P_1,P_1\rangle}=x^2-\frac{1}{3}$

$$\boxed{P_0=1, P_1=x, P_2=x^2-\frac{1}{3}}$$

Now I can check the first property: $\langle P_i,P_j\rangle=0$ for $i\not=j$

$\langle P_0,P_1\rangle=\int_{-1}^1x dx=\frac{(1)^2}{2}-\frac{(-1)^2}{2}=0$

$\langle P_0,P_2\rangle=\int_{-1}^1 (x^2-\frac{1}{3})dx=(\frac{x^3}{3}-\frac{1}{3}x)_{x=-1}^{x=1}=(\frac{1}{3}-\frac{1}{3})-(-\frac{1}{3}--\frac{1}{3})=0$

$\langle P_1,P_2\rangle=\int_{-1}^1 (x^3)dx=(\frac{x^4}{4})_{x=-1}^{x=1}=\frac{1}{4}-\frac{1}{4}=0$

However the property $P_i(1)=1$ is not satisfied for every $i$ since:


Is this a mistake in the question or am I doing something wrong here?


1 Answer 1


It is clearly mentioned that $P_i$'s need not have norm $1$. Multiply the polynomials you have found by constants $c_0,c_1,c_2$ so that the new polynomials have the value $1$ when $x=1$. The new ones are still orthogonal.

  • $\begingroup$ That makes sense. I first thought about just adding a constant but that didn' work out so well. Multiplying makes much more sense (also in the context of linear algebra and what scalar multiplication does to dot products etc.). I can just write $P_2=\frac{3}{2}x-\frac{1}{2}$. Thank you for your help! $\endgroup$
    – qmd
    Jun 7, 2019 at 8:54

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