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?


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 at 8:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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