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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.

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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:

enter image description here

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

enter image description here

(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.

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Thanks. Clear and clean – Bogdan M. May 21 '14 at 20:53

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