After 30+ years out of college, I just cracked open a book on a subject that has always intrigued me: Introduction to Orthogonal Polynomials, by Theodore S. Chihara. I have my quibbles with his approach but let's take it on as he has presented it.
I am in the midst of proving that $\cos n\theta$ can be expressed as a polynomial in $\cos(\theta)$ alone. That's not my question, however; let's just assume it works for now. If I now let:
(1) $x = \cos\theta$
then $\cos n\theta$, being a polynomial in $\cos\theta$, can be expressed as
(2) $T[n](x)$
since I am substituting $x$ for $\cos\theta$. So far, so good.
Note also, that $\theta = \cos^{-1}x$ so that
(3) $$\frac{d\theta}{dx} = 1/\sqrt{1-x^2}$$
Back to the integral.
It was easy enough to prove that for integers $m$ and $n$ ($m \neq n$) then
(4) $$\int_0^\pi (\cos n\theta)(\cos m\theta)d\theta = 0$$
Now, since $x = \cos\theta$ then when
(5a) $\theta = 0$, $x = 1$
(5b) $\theta = \pi$, $x = -1$
Thus, if I plug in $x = \cos\theta$ into the polynomials $\cos m\theta$ and $\cos n\theta$ I would get $T[m](x)$ and $T[n](x)$ in the integral, and applying the chain rule with (3), the integral becomes:
(6) $$\int_{-1}^1 \frac{T[m](x) \cdot T[n](x)}{\sqrt{1-x^2}}dx $$
Not the limits on the integral; these are consistent with (5a-b).
(Whew! Finally getting to the point!)
Now my problem with the book, right on the first page of chapter 1: He has the limits of the integral on the polynomial form as $[-1,1]$.
For $m\neq n$ this is OK, since the integral will be 0. But for $m = n$, we have the perfect inner product for the vector space of this class of polynomials (Tchebichef) and, with the limits as Chihara has them, this would always be a negative number, something that should never happen in an inner product space.
Can someone point me to where I've missed a point?
Best Answer
This particular version of $\arccos$ is a decreasing function on $[0, \pi]$. Its derivative is negative; you took the wrong square root.