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Was reading "Taylor's Expansion" when i came across this interesting problem, interesting for me.

Suppose that $f^{n+2}$ is continuous on $[0,x]$. Prove that there is a $\theta \in (0,1)$ such that

\begin{align} f(x) &= f(0) + \frac{f'(0)}{1!}x + \cdots + \frac{f^{n-1}(0)}{(n-1)!}x^{n-1} \\ & \quad + \frac{f^{n}\left(\frac{x}{n+1}\right)}{n!}x^{n} + \frac{n}{2(n+1)}f^{n+2}(\theta x)x^{n+2}{(n+2)!} \end{align}

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    Is the last term really $x^{n+2} \times (n+2)!$ instead of $\frac{x^{n+2}}{(n+2)!}$?2010-08-11

2 Answers 2

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@ Julio Cesar: I don't know about the problem, but this is why i think that the result could be correct.

Theorem: Let $f$ and $g$ be continuous on $[a,b]$, and let $g$ have constant sign on $[a,b]$. Then there is a $c \in (a,b)$ such that $\int\limits_{a}^{b} f(x)g(x) \ dx = f(c) \int\limits_{a}^{b} g(x) \ dx$

Please refer "E.I. Poffald, Amer.Math.Monthly 97 (1990), 205-213" for a proof of this result.

Using this result, by Taylor's formula with integral remainder we have $f^{n}\Bigl(\frac{x}{n+1}\Bigr)=f^{n}(0) + f^{n+1}(0)\frac{x}{n+1} + \int\limits_{0}^{\frac{x}{n+1}} f^{n+2}(t) \Bigl(\frac{x}{n+1}-t\Bigr) \ dt$

Hence $f(0)+ \frac{f'(0)}{1!}x + \cdots + \frac{f^{n-1}(0)}{(n-1)!}x^{n-1} + \frac{f^{n}(\frac{x}{n+1})}{n!}x^{n} = \text{ Our sum with the Integral Remainder}$

Then one can define $g$ in this manner.

$g(t)= \frac{(x-t)^{n+1}}{n+1} - x^{n}\Bigl(\frac{x}{n+1}-t\Bigr) \quad ; t \in [0,x]$

We have g'(t)>0 for $t \in (0,x)$ and $g(0)=0$. Thus $g$ is positive on the open interval $(0,x)$. Then using our theroem we have $\int\limits_{0}^{\frac{x}{n+1}} f^{n+2}(t)(g(t))=f^{n+2}(c) \int\limits_{0}^{\frac{x}{n+1}} g(t) \ dt$

This is where i am stuck.

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Something is wrong with this formula. It leads to

e = 1 + 1 + \frac{e^{\frac{1}{3}}}{2} + e^\theta 8 > e.

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    I really don´t see your point bellow. I´m sorry. Really, there is somethig wrong. Probably some misstype in the formula from where you got it. Why do you believe it is true even with a counter-example?2010-08-10