I am reading about the Lagrange linear interpolation for approximating a function.
In the book, truncation error in linear interpolation is derived as follows :
Suppose $f(x)$ is a function which is atleast twice differentiable in $[a,b]$ and is approximated by 1-degree polynomial
$P(x)=f(x_o)+\frac{f(x_1)-f(x_o)}{x_1-x_o}(x-x_o)$ where $x_o,x_1\in[a,b]$
Fix a point $x\in [a,b]$ and construct a function $g(t)=f(t)-P(t)-[f(x)-P(x)]\frac{(t-x_o)(t-x_1)}{(x-x_0)(x-x_1)}$
We can see that $g(t)=0$ for $x_o,x,x_1$
So by Rolle's theorem, $g'(t)=0$ for some $x'\in [x_o,x]$ and $x''\in [x,x_1]$.
Again by Rolle's theorem, $g''(t)=0$ for some $\xi\in[x',x'']$
So, $g''(t)=f''(t)-2\frac{f(x)-P(x)}{(x-x_o)(x-x_1)}$
$\implies f''(\xi)-2\frac{f(x)-P(x)}{(x-x_o)(x-x_1)}=0$
So, $\boxed{f(x)-P(x)=\frac{1}{2}f''(\xi)(x-x_o)(x-x_1)}$
As $f,f'\;and\;f''$ are continuous on a compact interval so they are also compact thus having an upper and lower bound and thus the truncation error has an upper bound.
My approach
So, basically we are trying to define a function $g$ which is equal to $0$ for atleast $3$ points, so that by applying Rolle's theorem two times, we get a single point at which $g''=0$ and using this point we can find for truncation error.
By Mean value theorem, we know that $\exists\alpha\in[x_o,x_1]$ such that $f'(\alpha)=\frac{f(x_1)-f(x_o)}{x_1-x_o}=P'(x)$
Fix some $x\in[x_o,x_1]$.
Define $g(t)=(f(x)-P(x))((t-x_o)(t-x_1)(t-\alpha))+(f'(t)-P'(t))((t-x_o)(t-x_1))$
We can see that $g(t)=0$ for $x_0,\alpha,x_1$.
$\boxed{g''(t)=(f(x)-P(x))(6t-(x_o+x_1+\alpha))+f'''(t)((t-x_o)(t-x_1))+2f''(t)(2t-(x_o+x_1))+2(f'(t)-P'(t))}$
($P(t)$ is a polynomial of degree $1$ so $P''(t)=0$)
Again by applying Rolle's theorem twice, $\exists\beta\in[x_o,x_1]$ such that $g''(\beta)=0$.
$\implies\boxed{f(x)-P(x)=\frac{f'''(\beta)((\beta-x_o)(\beta-x_1))+2f''(\beta)(2\beta-(x_o+x_1))+2(f'(\beta)-P'(\beta))}{x_o+x_1+\alpha-6\beta}}$
Again we can find upper and lower bound to get an inequality for the truncation error.
We can construct $g(t)$ in many ways, so as to get the inequality. I have just started studying numerical analysis. I feel that the proofs for finding the bounds for truncation errors are somewhat artificial.
Is my approach correct? Can we have many inequalities for the bounds? Please help me in clarifying the doubt.