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This is a problem in an upcoming lecture:

Show that the first two Chebyshev polynomials, $T_0(x) = 1$ and $T_1(x) = x$ are orthogonal with respect to the weighting function $r(x) = (1 − x^2)^{-\frac{1}{2}}$ for $x \in [−1,1]$. Determine constants $\alpha$ and $\beta$ such that the function $h(x) = 1 + \alpha x + \beta x^2$ is orthogonal to both $T_0$ and $T_1$ with respect to the weighting function $r(x)$ for $x \in [−1,1]$. How does $h(x)$ compare with $T_2(x)$?

I haven't yet done any of these types of problems (we just finished doing some Sturm-Liouville work), and there are no solutions available yet. I would greatly appreciate it if someone could please take the time to demonstrate how these problems are done with accompanying explanation, so that I may learn.

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  • $\begingroup$ Start with showing $\int_{-1}^1T_0(x)T_1(x)r(x)=0$. Next compute $\int_{-1}^1T_k(x)h(x)r(x)dx$ for $k=1,2$ as functions of $\alpha$ and $beta$. Finally compute $\alpha$ and $\beta$ to make those last two integrals zero. I presume you know $T_2$ so you can compare. $\endgroup$ – herb steinberg Aug 13 '18 at 0:44
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The inner product on the space of continuous functions from $[-1,1]$ to $\mathbb{R}$ is, as indicated in the statement, the weighted integral $$ \langle f,g\rangle = \int_{[-1,1]}f(x)g(x) r(x)dx \tag{1} $$ with $r(x) = (1-x^2)^{-1/2}$. (I leave out the task to prove that $r$ is indeed integrable on $[-1,1]$. It is, and everything is well-defined.)

Thus, you are asking to prove that $\langle T_0,T_1\rangle =0$, and to do so must compute $$ \langle T_0,T_1\rangle = \int_{[-1,1]}T_0(x)T_1(x) r(x)dx = \int_{[-1,1]}1\cdot x \cdot (1-x^2)^{-1/2}dx = \int_{[-1,1]} \frac{x}{\sqrt{1-x^2}}dx $$ and show it is zero. Can you see why? (Hint: look at the parity of the integrand; you integrate an odd function on a domain symmetric around $0$.)

For the second part, you need to find $\alpha,\beta$ such that $\langle h,T_0\rangle=0$ and $\langle h,T_1\rangle=0$. To do so, compute the two integrals (linearity of the integral will help); you will get two results as a function of $\alpha,\beta$. For these two values to both be zero, you need $\alpha,\beta$ to be the solution to a system of two equations: \begin{align} 0 &= \int_{[-1,1]} 1\cdot (1+\alpha x+\beta x) \frac{dx}{\sqrt{1-x^2}} = [...]\\ 0 &= \int_{[-1,1]} x\cdot (1+\alpha x+\beta x) \frac{dx}{\sqrt{1-x^2}} = [...]\\ \end{align} where the "[...]" denote what you get when computing these two integrals. Solve this system to find $\alpha,\beta$.

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Eigenfunctions belonging to different eigenvalues are orthogonal with respect to a given weight function $\sigma(x)$ in other words

$$\int_{a}^{b}\phi_{n}(x)\phi_{m}(x)\sigma(x) dx = 0 $$

Now, then it said $ x \in [-1,1]$ and $r(x) = (1-x^{2})^{\frac{-1}{2}}$

$$ \int_{-1}^{1} T_{0} h(x) r(x) dx = 0 \implies \int_{-1}^{1} (1-x^{2})^{\frac{-1}{2}} (1+\alpha x+\beta x^{2}) dx = \frac{\pi}{2}(b+2) \implies b=-2$$

$$ \int_{-1}^{1} T_{1} h(x) r(x) dx = 0 \implies \int_{-1}^{1} x(1-x^{2})^{\frac{-1}{2}} (1+\alpha x+\beta x^{2}) dx = \frac{\pi a}{2} \implies a=0$$

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