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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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What's that $f^n\times\left(\frac x{n+1}\right)$? BTW, try to format the equation so that it doesn't overflow the page. – kennytm Aug 9 '10 at 18:44
@ Kenny TM: Hey thanks a lot. I hope its fine now. I really dont know as to how i can make the last term come in the next line. Please help me with TeX command. – anonymous Aug 9 '10 at 18:59
Also, typically higher-order derivatives appear as $f^{(n-1)}$ or whatever order. Sorry to be a pain, but the title should be more specific about what this question relates to, i.e. it should mention Taylor's Expansion or whatever it is you are looking for here. – Tom Stephens Aug 9 '10 at 19:59
Is the last term really $x^{n+2} \times (n+2)!$ instead of $\frac{x^{n+2}}{(n+2)!}$? – kennytm Aug 11 '10 at 18:37

2 Answers 2

up vote 2 down vote accepted

@ 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 don't think so. The question is absolutely correct. Please wait until somebody gets a solution. – anonymous Aug 10 '10 at 7:48
Could you say where (which book if it is in English) did you get it? – Júlio César Aug 10 '10 at 12:13
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? – Júlio César Aug 10 '10 at 20:54

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