[proofplan]
We derive the recurrence for the Chebyshev coefficients of $f'$ by expanding $f'(x) = \sum_{m=0}^{\infty} \breve{f}_m T_m'(x)$ using the derivative formulas from the [Derivatives of Chebyshev Polynomials](/theorems/1384), then reading off the coefficient of $T_n(x)$ in the resulting expression. The key observation is that $T_m'$ contributes to $T_n$ only when $n < m$ and $n + m$ is odd (from the parity structure of the derivative formulas), with a coefficient that is $m$ times a normalisation factor $c_n$ that equals $1$ for $n = 0$ and $2$ for $n \geq 1$.
[/proofplan]
[step:Expand $f'(x)$ using the derivative formulas for individual Chebyshev polynomials]
Differentiating the Chebyshev expansion $f(x) = \sum_{m=0}^{\infty} \breve{f}_m T_m(x)$ term by term:
\begin{align*}
f'(x) = \sum_{m=1}^{\infty} \breve{f}_m T_m'(x),
\end{align*}
where the $m = 0$ term vanishes since $T_0' = 0$. (Termwise differentiation is valid under appropriate convergence assumptions on the Chebyshev coefficients.)
By the [Derivatives of Chebyshev Polynomials](/theorems/1384), $T_m'$ is a finite linear combination of Chebyshev polynomials of lower degree:
**Even $m = 2p$:** $T_{2p}'(x) = (2p) \cdot 2\sum_{j=1}^{p} T_{2j-1}(x)$.
**Odd $m = 2p+1$:** $T_{2p+1}'(x) = (2p+1)\left[T_0(x) + 2\sum_{j=1}^{p} T_{2j}(x)\right]$.
In both cases, $T_m'$ involves only Chebyshev polynomials $T_n$ with $n < m$ and $n + m$ odd. Moreover, the coefficient of $T_n$ in $T_m'$ is:
\begin{align*}
[T_n] T_m'(x) = \begin{cases} \frac{m}{c_n} \cdot 2 & \text{if } n < m \text{ and } n + m \text{ is odd}, \\ 0 & \text{otherwise}, \end{cases}
\end{align*}
where the factor accounts for the normalisation: when $n = 0$ (which only arises in the odd case $m = 2p+1$), the $T_0$ term appears with coefficient $(2p+1) \cdot 1 = m$, while for $n \geq 1$, the $T_n$ term appears with coefficient $m \cdot 2$.
[/step]
[step:Read off the coefficient $\breve{f'}_n$ by collecting all contributions to $T_n(x)$]
The Chebyshev coefficient of $f'$ at index $n$ is obtained by collecting the coefficient of $T_n(x)$ from all terms $\breve{f}_m T_m'(x)$:
\begin{align*}
\breve{f'}_n = \sum_{m=1}^{\infty} \breve{f}_m \cdot [T_n] T_m'(x).
\end{align*}
From the previous step, $[T_n] T_m'(x)$ is nonzero only when $m > n$ and $m + n$ is odd. We compute $[T_n] T_m'(x)$ in two cases:
**Case $n = 0$:** Only odd $m = 2p+1$ contribute (since $0 + m$ odd requires $m$ odd). The coefficient of $T_0$ in $T_{2p+1}'$ is $(2p+1) \cdot 1 = m$. Therefore:
\begin{align*}
\breve{f'}_0 = \sum_{\substack{m = 1 \\ m \text{ odd}}}^{\infty} m \, \breve{f}_m = 1 \cdot \sum_{\substack{m = 1 \\ 0 + m \text{ odd}}}^{\infty} m \, \breve{f}_m = c_0 \sum_{\substack{m = 1 \\ 0 + m \text{ odd}}}^{\infty} m \, \breve{f}_m,
\end{align*}
with $c_0 = 1$.
**Case $n \geq 1$:** Only $m > n$ with $m + n$ odd contribute. The coefficient of $T_n$ in $T_m'$ is $2m$ (from the factor of $2$ in both the even and odd derivative formulas). Therefore:
\begin{align*}
\breve{f'}_n = \sum_{\substack{m = n+1 \\ n + m \text{ odd}}}^{\infty} 2m \, \breve{f}_m = 2 \sum_{\substack{m = n+1 \\ n + m \text{ odd}}}^{\infty} m \, \breve{f}_m = c_n \sum_{\substack{m = n+1 \\ n + m \text{ odd}}}^{\infty} m \, \breve{f}_m,
\end{align*}
with $c_n = 2$ for $n \geq 1$.
Combining both cases into a single formula, and noting that the $k$th derivative is obtained by applying the first-derivative recurrence $k$ times:
\begin{align*}
\breve{f^{(k)}}_n = c_n \sum_{\substack{m = n+1 \\ n + m \text{ odd}}}^{\infty} m \, \breve{f^{(k-1)}}_m, \qquad \forall k \geq 1,
\end{align*}
where $c_0 = 1$ and $c_n = 2$ for $n \geq 1$.
[guided]
Let us verify the formula on a concrete example. Consider $f(x) = T_3(x)$, so $\breve{f}_3 = 1$ and $\breve{f}_m = 0$ for $m \neq 3$. By the derivative formula, $T_3'(x) = 3[T_0(x) + 2T_2(x)]$, so $\breve{f'}_0 = 3$ and $\breve{f'}_2 = 6$.
Using the recurrence: For $n = 0$, the sum over $m > 0$ with $m$ odd gives $c_0 \cdot 3 \cdot 1 = 1 \cdot 3 = 3$. For $n = 2$, the sum over $m > 2$ with $m + 2$ odd gives $c_2 \cdot 3 \cdot 1 = 2 \cdot 3 = 6$. Both match.
The parity constraint $n + m$ odd is crucial: it means that even-indexed coefficients of $f$ contribute only to odd-indexed coefficients of $f'$, and vice versa. This is a direct consequence of the parity structure in the [Derivatives of Chebyshev Polynomials](/theorems/1384): even Chebyshev polynomials have odd derivatives and odd Chebyshev polynomials have even derivatives.
The factor $c_n$ (which equals $1$ for $n = 0$ and $2$ for $n \geq 1$) arises because the $T_0$ coefficient in the Chebyshev expansion has a different normalisation: in the standard Chebyshev expansion, the $n = 0$ mode does not carry the factor of $2$ that all other modes carry in the derivative formulas. This is the same $\frac{1}{2}$-factor that appears in discrete cosine transforms and Chebyshev interpolation.
The higher-derivative recurrence $\breve{f^{(k)}}_n$ is obtained by induction: applying the first-derivative formula to $f^{(k-1)}$ (whose Chebyshev expansion exists by the same argument) gives $\breve{f^{(k)}}_n$ in terms of $\breve{f^{(k-1)}}_m$.
[/guided]
[/step]
