Suppose $x_0,x_1,\ldots,x_n$ are $n+1$ distinct numbers in the interval $[a,b]$ and $f\in C^{n+1}[a,b]$. Then for each $x$ in $[a,b]$, there is a number $\xi$ in $(a,b)$ such that
$$f(x) = P(x) + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)(x-x_1)\cdots(x-x_n)$$
where $P(x)$ is the Lagrange interpolating polynomial of degree at most $n$ with $f(x_k)=P(x_k)$.
I am trying to understand geometrically why the remainder term $f(x)-P(x)$ should have the form given above. I am looking for a conceptual argument similar to the following for the lagrange form of the taylor remainder. Let
$$R(x) = g(x) – g(0) – g^{(1)}(0)x- \frac{g^{(2)}(0)}{2} x^2 – \cdots – \frac{g^{(k)}(0)}{k!}x^k$$
be the taylor remainder of $g(x)$. For a fixed $h>0$ let
$$p(x)= \frac{R(h)}{h^{k+1}}x^{k+1}$$
Then we have
$$R(0)=p(0), R^{(1)}(0)=p^{(1)}(0), \ldots, R^{(k)}(0)=p^{(k)}(0)$$ and $p(h) = R(h)$. In addition $p^{(k+1)}(x)$ is constant. If $R^{(k+1)}(x)$ were always strictly greater than the constant $p^{(k+1)}$ then since $R$ and $p$ agree on initial conditions at $x=0$ we would expect geometrically that $R$ to be greater than $p$ after $x=0$ contradicting $p(h) = R(h)$. Likewise if $R^{(k+1)}(x)$ were always strictly less than the $p^{(k+1)}$ we would expect $R$ to be less than $p$ after $x=0$ contradicting $p(h) = R(h)$. Thus we expect that $R^{(k+1)}(x)$ takes on values above and below $p^{(k+1)}$ and as well as $p^{(k+1)}$ itself. And this gives the lagrange form of the taylor remainder. Of course the standard formal argument would use the generalized form of Rolle's theorem, but I didn't need Rolle's theorem to see why the lagrange form of the taylor remainder should be right. There should be a similar "geometric" argument to motivate the error term for the lagrange interpolation polynomial.
Best Answer
You are trying to understand why the remainder of Lagrange interpolation polynomial has the form $$\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)(x-x_1)\cdots(x-x_n)$$
Me too!And I find it.Indeed,Lagrange's work is based on Newtons'work.Newton has his polynomial interpolation method called Newton's interpolation,and the remainder of Newton's interpolation is in the form of $$R_n(x)=f[x,x_0,x_1,\cdots,x_n]\prod_{i=0}^n(x-x_i)$$ You'd better google Newton's method and learn it.
Ok,It seems that you are not satisfied with my answer,because you do not respond.So I give an update to my answer to provide you all the details.Let's first see the differential mean value theorem,it is stated as follows:
Let $f(x)$ be a real valued function defined on $\mathbf{R}$,and $f(x)$ is second differentiable on $\mathbf{R}$,and its second order derivative on $\mathbf{R}$ is continuous.Then there exists $\xi\in (x_0,x_1)$ such that $$f'(\xi)=\frac{f(x_1)-f(x_0)}{x_1-x_0}$$ We prove the differential mean value theorem by constructing a function $$g(x)=f(x)-[\frac{f(x_1)-f(x_0)}{x_1-x_0}(x-b)+f(x_1)]$$ $g(a)=g(b)=0$,so we can apply Rolle's theorem once to get the differential mean value theorem.Its picture is shown below:
Now we see a deeper example.
Let $f(x)$ be a real valued function defined on $\mathbf{R}$,and $f(x)$ is second differentiable on $\mathbf{R}$,and its second derivative on $\mathbf{R}$ is continuous.There is a polynomial of degree 2 passing through the points $(x_0,f(x_0)),(x_1,f(x_1)),(x_2,f(x_2))$.This unique polynomial is in the form of $$a_2x^2+a_1x+a_0$$ Now the picture becomes
(I have to admit that my drawing skill is bad,my picture is not very accurate)
Once again,we want to estimate the interpolation error $$f(x)-(a_2x^2+a_1x+a_0)$$ $a_2x^2+a_1x+a_0$ is not a good form.Lagrange has his form,but that's also not good for our purpose.Newton has his form,called Newton's interpolation,that's exactly what we need.According to Newton's interpolation,the interpolation polynomial is $$Q(x)=f(x_0)+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1)$$where $$f[x_0,x_1]=\frac{f(x_0)-f(x_1)}{x_0-x_1}$$,$$f[x_0,x_1,x_2]=\frac{f[x_0,x_1]-f[x_1,x_2]}{x_0-x_2}$$Now we investigate the function $$g(x)=f(x)-(f(x_0)+f[x_0,x_1](x-x_0)+f[x_0,x_1,x_2](x-x_0)(x-x_1))$$ Why we investigate this function?Because this function is in the form of $$f(x)-Q(x)$$ further more,it is easy to verify that $$g(x_0)=g(x_1)=g(x_2)=0$$(This is because these points are exactly the intersection points of the function $f(x)$ and the polynomial $Q(x)$)So apply Rolle's theorem twice ,we can get that $$f''(\xi)=2!f[x_0,x_1,x_2]$$
Similary,it is easy to verify that $$f^{(n)}(\xi)=n!f[x_0,x_1,\cdots,x_n]$$ Now go back to Lagrange's interpolstion polynomial,we just need to prove that $$f(x)=P(x)+f[x_0,\cdots,x_n,x](x-x_0)(x-x_1)\cdots (x-x_n)$$ This is just Newton's formula!Done.