I have found this question in Kincaid book, and it's asking about the interpolation error. I don't know how to find the requested polynomial. I would appreciate it if anyone could help me.
Let $p$ be a polynomial of degree $\leq n-1$ that interpolates the function $f(x) = \sinh x$ at any set of $n$ nodes in the interval $[-1 , 1]$, subject only to the condition that one of the nodes is $0$. Prove that the relative error obeys this inequality on $[-1 , 1]$: $$ \frac{|p(x) - f(x)|}{|f(x)|} \leq \frac{2^n}{n!} $$
Since $x = 0$ is one of the nodes, $$ p(0) = f(0) = 0. $$ Now, let's denote $\Delta(x) = p(x) - f(x)$. $$ \Delta(x) = \Delta(0) + \int_0^x \Delta'(\xi) d\xi = \int_0^x \Delta'(\xi) d\xi $$ Thus $$ |\Delta(x)| \leq \left|\int_0^x \Delta'(\xi) d\xi\right| \leq \int_0^x |\Delta'(\xi)| |d\xi| $$ Using formula from this paper Derivative error bounds for Lagrange interpolation $$ |\Delta'(x)| \leq |P_{x1}(x)| \frac{\|f^{(n)}\|}{n!} $$ where $\|f\| = \max_{x \in [-1, 1]} |f(x)|$ and $P_{x1}(x)$ is $$ P_{x1}(x) = (x - \xi_1) (x-\xi_2) \cdots (x - \xi_{n-1}), \quad \min_i x_i \leq \xi_i \leq \max_i x_i $$ This gives us the following bound: $$ |\Delta(x)| \leq \frac{\|f^{(n)}\|}{n!} \int_0^x |P_{x1}(\xi)| |d\xi|. $$ The $\|f^{(n)}\|$ term can be bounded by $\max(\sinh 1, \cosh 1) = \cosh 1 \approx 1.543 < 2$. Now $$ |P_{x1}(x)| = |x - \xi_1| |x-\xi_2| \cdots |x - \xi_{n-1}| \leq 2^{n-1} $$ since each of the terms is not greater than $2$.
Finally, $$ |\Delta(x)| \leq \frac{\|f^{(n)}\|}{n!} \int_0^x |P_{x1}(\xi)| |d\xi| \leq \frac{2}{n!} \int_0^x 2^{n-1} |d\xi| \leq \frac{2^n}{n!} |x| \leq \frac{2^n}{n!} |\sinh x|. $$