Does anyone know how this is proven?
The function $f(x)=0,2x^5$ is given. Prove that for an interpolation polynomial $p_4(x)$ that interpolates a function $f(x)$ at nodes $x_0=0$, $x_1=1$, $x_2=2$, $x_3=3$, $y_4=4$, the following estimate is valid:
The following hint is also given, which can be used without proof:
I'm assuming this is an interpolation error. I don't understand at all what needs to be applied for this proof.
In Newton divided differences, you have $$ f(x)=\begin{aligned}[t]f(x_0)&+f[x_0,x_1](x-x_0)\\ &+f[x_0,x_1,x_2](x-x_0)(x-x_1)\\ &+f[x_0,x_1,x_2,x_3](x-x_0)(x-x_1)(x-x_2)\\ &+f[x_0,x_1,x_2,x_3,x_4](x-x_0)(x-x_1)(x-x_2)(x-x_3)\\ &+f[x_0,x_1,x_2,x_3,x_4,x](x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4)\\ \end{aligned} \\ =P_4(x)+\frac1{5!}f^{(5)}(\xi)(x-x_0)(x-x_1)(x-x_2)(x-x_3)(x-x_4) $$ You should be able to combine the given hints into the claimed inequality.