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Here's what I tried. Let $f(x)$ be the function we are approximating which is at least $m$ times differentiable on the interval $[x_0,x]$ with continuous derivatives and let $g(x)$ be its approximation and define $p(x)=f(x)-g(x)$ where $g(x)=f\left(x_0\right)+f'\left(x_0\right)\left(x-x_0\right)....+\frac{f^{\left(m-1\right)}\left(x_0\right)}{\left(m-1\right)!}\left(x-x_0\right)^{\left(m-1\right)}$

We apply the mean value theorem to $p(x)$ on the interval $[x_0,x]$ to get $$p'\left(c_1\right)=\frac{p\left(x\right)-p\left(x_0\right)}{x-x_0}=\frac{p\left(x\right)}{x-x_0}$$ so $p(x)=p'\left(c_1\right)(x-x_0)$ for some $c_1$ in $[x_0,x]$. Applying the theorem again to $p'\left(x\right)=f'\left(x\right)-g'\left(x\right)$ on the interval $[x_0,c_1]$ we get $p''\left(c_2\right)=\frac{p'\left(c_1\right)-p'\left(x_0\right)}{c_1-x_0}=\frac{p'\left(c_1\right)}{c_1-x_0}\ $ with $c_2$ in $[x_0,c_1]$, which gives in general, $$p^i\left(c_i\right)=p^{\left(i+1\right)}\left(c_{i+1}\right)\left(c_i-x_0\right) $$ with $c_{i+1}$ in $[x_0,c_i]$. Applying this $m-1$ times and noting that $p^{\left(m\right)}\left(c_m\right)=f^{\left(m\right)}\left(c_m\right)$, we get $$p\left(x\right)=\left(x-x_0\right)\left(c_1-x_0\right)....\left(c_{m-1}-x_0\right)f^{\left(m\right)}\left(c_m\right)$$ setting $a=\frac{\left(c_1-x_0\right)....\left(c_{m-1}-x_0\right)}{\left(x-x_0\right)^\left(m-1\right)}$ we see that $$p(x)=a\left(x-x_0\right)^mf^{\left(m\right)}\left(c_m\right)$$ with $c_m$ in $[x_0,x]$ which is precisely the Lagrange form, but I'm having a hard time dealing with $a$. Obviously, for fixed $x$, $a$ is a constant whose absolute value is less than $1$ which makes this even closer to the Lagrange form. Any help would be appreciated.

I am not asking for other proofs of this as I can easily find them, so please don't just throw proofs at me.

Km356
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  • You are applying mean value theorem on the wrong function. Taylor theorem can not be obtained by multiple applications of mean value theorem but rather via a single application on a suitable non-obvious function. – Paramanand Singh Aug 28 '19 at 13:57
  • See this answer https://math.stackexchange.com/a/1915029/72031 – Paramanand Singh Aug 28 '19 at 14:01
  • Excuse me, I just don't like how you so confidently assert that it CANNOT be obtained this way. There are many ways to prove the same statement and we can never be sure a certain statement can't be obtained in a certain manner. That said, the answer you linked is nice and thanks for referring me to it. – Km356 Aug 28 '19 at 14:43
  • Ok lets rephrase. I have not seen Taylor's theorem being proved using mean value theorem multiple times. Also the point was not to discourage you, rather it was to point out that this approach does not give the Lagrange form. – Paramanand Singh Aug 28 '19 at 14:47
  • This is a good post, which would be so much better if $g$ were called $P$ instead, and $p$ were $R$. ($P$ stands for "Polynomial" and $R$ for "Remainder"). – Giuseppe Negro Aug 28 '19 at 15:57
  • I often find myself throwing letters around when I'm trying to make an idea work :) – Km356 Aug 28 '19 at 16:00

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