# Thread: Proof of the Lagrange formula

1. ## Proof of the Lagrange formula

I'm not too sure how to do this question?

Suppose that $f$ is $(n+1)$-times differentiable on some open interval $I$ containing $0$ and that $f^{(n+1)}$ is continuous on $I$. Suppose that $x\in I$ and recall that the remainder term in Taylor's theorem is given by:

$R_{n+1}(x)=\frac{1}{n!}\int_0^x f^{n+1}(t)(x-t)^n dt$

Use the mean value theorem for integrals to deduce that:

$R_{n+1}(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}$

where c is some real number between $0$ and $x$

2. Originally Posted by acevipa
I'm not too sure how to do this question?

Suppose that $f$ is $(n+1)$-times differentiable on some open interval $I$ containing $0$ and that $f^{(n+1)}$ is continuous on $I$. Suppose that $x\in I$ and recall that the remainder term in Taylor's theorem is given by:

$R_{n+1}(x)=\frac{1}{n!}\int_0^x f^{n+1}(t)(x-t)^n dt$

Use the mean value theorem for integrals to deduce that:

$R_{n+1}(x)=\frac{f^{n+1}(c)}{(n+1)!}x^{n+1}$

where c is some real number between $0$ and $x$
There are many different versions of the MVT for integrals, one of them states that If
$f(x),g(x)$ are continuous on $[a,b]$ then

there exists a $c \in (a,b)$ such that

$\displaystyle \int_{a}^{b}f(x)g(x)dx=f(c)\int_{a}^{b}g(x)dx$

So in you case we have the interval $[0,x]$ and the functions

$f^{n+1}(t),\text{ and }\frac{(x-t)^n}{n!}$ this gives

$\displaystyle \int_{0}^{x}f^{n+1}(t)\frac{(x-t)^n}{n!}dt=f^{n+1}(c)\int_{0}^{x}\frac{(x-t)^n}{n!}dt=f^{n+1}(c)\frac{-(x-t)^{n+1}}{(n+1)!}\bigg|_{0}^{x}=...$ for some $c \in (0,x)$