Derivatives

The ladder relations have become: $$ \left[ (1-x^2) \frac{d}{dx} - n\,x \right] T_n(x) = -n\; T_{n+1}(x) \\ \left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = +n\; T_{n-1}(x) $$ If we add these equations togther again, then: $$ 2 (1-x^2) \frac{dT_n}{dx} = - n \left[ T_{n+1}(x) - T_{n-1}(x) \right] $$ Giving for the derivative of a Chebyshev solution function: $$ \frac{dT_n}{dx} = \frac{n}{2(x^2-1)} \left[T_{n+1}(x) - T_{n-1}(x)\right] $$ Remember the recursion relationship: $$ T_{n+1}(x) + T_{n-1}(x) = 2 x\,T_n(x) $$ With help of this, we can do the following: $$ T_{n+1}(x) - T_{n-1}(x) = 2 x\,T_n(x) - 2 T_{n-1}(x) =\\ 2 x \left[ 2\,x T_{n-1}(x) - T_{n-2}(x) \right] - 2 T_{n-1}(x) =\\ 4 (x^2-1) T_{n-1}(x) + 2 T_{n-1}(x) - 2 x\, T_{n-2}(x) =\\ 4 (x^2-1) T_{n-1}(x) + 2 T_{n-1}(x) - \left[ T_{n-1}(x) + T_{n-3}(x) \right] =\\ 4 (x^2-1) T_{n-1}(x) + \left[ T_{n-1}(x) - T_{n-3}(x) \right] $$ Herewith: $$ \frac{dT_n}{dx} = \frac{n}{2(x^2-1)} \left[T_{n+1}(x) - T_{n-1}(x)\right] =\\ 2 n T_{n-1}(x) + \frac{n}{2(x^2-1)} \left[T_{n-1}(x) - T_{n-3}(x)\right] $$ So we have a recursion formula expressing the derivative of Chebyshev function into a sequence of lower order Chebyshev functions: $$ \frac{dT_n}{dx} = 2 n T_{n-1}(x) + 2 n T_{n-3}(x) + 2 n T_{n-5}(x) + \; .\,.\,. \; + \; ( \mbox{last term} ) $$ Where the last term can be one of these two: $$ \frac{n}{2(x^2-1)} \left[T_3(x) - T_1(x)\right] \quad \vee \quad \frac{n}{2(x^2-1)} \left[T_2(x) - T_0(x)\right] $$ If solutions of the Chebyshev differential equation are restricted to Chebyshev polynomials of the first kind, then we have $T_3(x) = 4x^3-3x\;$, $T_2(x) = 2x^2-1\;$, $T_1(x) = x\;$, $T_0(x) = 1\;$. Herewith: $$ \frac{n}{2(x^2-1)} \left[T_3(x) - T_1(x)\right] = 2 n . x \quad \vee \quad \frac{n}{2(x^2-1)} \left[T_2(x) - T_0(x)\right] = n $$ Let's concentrate on the value of the derivative at $x=1$. From the definition of the Chebyshev polynomials $T_n(x) = \mbox{cos[h]}(n\,\mbox{arccos[h]}(x))$ it is clear that, for all $n$ : $T_n(1) = 1$. And $x=1$ is the only place where all polynomials assume that maximum value, at the interval $\left[ -1,+1 \right]$. Thus the derivatives $dT_n(x)/dx$ assume their maximum value at that point as well. We can even calculate what the maximum is. For even $n$ we have $n/2$ terms $2n$, giving a total of $n/2.2n = n^2$. For odd $n$ we have $(n-1)/2$ terms equal to $2n$ and one term equal to $n$, giving a total of $(n-1)/2.2n+n = n^2$. Both sequences add up to $n^2$. Thus: $$ \left. \frac{dT_n(x)}{dx} \right|_{x=1} = n^2 $$