Let $P(z)$ be a polynomial of degree $n$ with $|P(z)|\leq 1$ on $|z|=1$ and $P\_m(z)$ be a partial sum of $P(z).$ How large $P\_m(z)$ can be on $|z|=1?$ 



A problem on polynomials
The trivial upper bound $\max\_{|w|=1}|P\_{m}(w)|\leq \|1+z+\cdots+z^{m}\|\_{L^{1}(\mathbb{T})} \asymp C \log(m)$ that I wrote in the comment is actually sharp in the regime $m=n/2$. Here is the proof.


Notice that $P\_{m}(z)$ is convolution of $P(z)$ with $D\_{m}(z) = 1+z+\cdots+z^{m}$ on the unit circle, therefore, by the triangle inequality $\max\_{|z|=1}|P\_{m}(z)| \leq \frac{1}{2\pi}\int\_{-\pi}^{\pi}|1+e^{i\theta}+\cdots+e^{i m \theta}| d\theta \asymp C \log(m)$


Next, let us show that the upper bound is sharp in the regime $m=\frac{n}{2}$ where $n$ is large.


Indeed, consider the polynomial
$$
P(z) = z^{n} \overline{\left( 1+\frac{z}{1}+\cdots+\frac{z^{n}}{n}\right)} - z^{n}\left(1+\frac{z}{1}+\cdots+\frac{z^{n}}{n} \right) = z^{n}+\frac{z^{n-1}}{1}+\cdots+\frac{1}{n}-z^{n}-\frac{z^{n+1}}{1}-\cdots-\frac{z^{2n}}{n}
$$
it is of degree $2n$, and $\max\_{|z|=1}|P\_{n-1}(z)|\geq |P\_{n-1}(1)| \asymp \log(n)$. On the other hand let us show that $\max\_{|z|=1}|P(z)|\asymp 1$. Indeed, for $z=e^{ix}$ we have  

$$
|P(z)| = 2\left| \sum\_{k=1}^{n} \frac{\sin(kx)}{k}\right|\leq 2 \int\_{0}^{\pi} \frac{\sin(s)}{s}ds \asymp 1 \quad \text{for all} \quad n \geq 1, \; x \in [0, 2\pi)
$$
The last inequality I guess is known "since Democritus".


