Estimate the error of interpolation of function sinhx. 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 $[-1 , 1]$:
$$
\frac{|p(x) - f(x)|}{|f(x)|} \leq \frac{2^n}{n!}
$$  
 A: 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,
$$
|\Delta(x)| \leq \frac{\|f^{(n)}\|}{n!} \int_0^x |P_{x1}(\xi)| |d\xi| \leq
\frac{2}{n!} \int_0^x 2^{n-1} |d\xi| \leq \frac{2^n}{n!} |x| \leq \frac{2^n}{n!} |\sinh x|.
$$
