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Assume $f(x)$ is continuous in $[a,b]$, and $f''$ in $(a,b)$, prove that for every $c\in (a,b)$, $\exists$ $\xi \in(a,b)$ such that

$$\dfrac{1}{2}f''(\xi)=\dfrac{f(a)}{(a-b)(a-c)}+\dfrac{f(b)}{(b-a)(b-c)}+\dfrac{f(c)}{(c-a)(c-b)}.$$

I don't know how to use Cauchy Mean Theorem to prove it.

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    Presumably, did you mean to say that $f''$ is _continuous_ on $(a,b)$? Also, the size of the post's title was too large for the main site to fit on a line, so I took the liberty of editing it.2012-02-03
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    @JavaMan It doesn't say it is continuous for $f^{''}$2012-02-03
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    What does $f''$ in $(a,b)$ mean then? Do you just mean that $f''$ is defined there?2012-02-03

2 Answers 2

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I assume that $f$ is continuous on $[a,b]$ and twice differentiable on $(a,b)$. Let $$ P(x):=f(x)-\left(f(a)\dfrac{(x-b)(x-c)}{(a-b)(a-c)}+f(b)\dfrac{(x-a)(x-c)}{(b-a)(b-c)}+f(c)\dfrac{(x-a)(x-b)}{(c-a)(c-b)}\right). $$ Then, apply Rolle's Theorem successively to $P$ and $P'$.

[In more detail: $P$ is continuous on $[a,b]$, twice differentiable on $(a,b)$, and $P(a)=P(c)=P(b)=0$. Therefore, by Rolle's Theorem, there exist $d$ in $(a,c)$ and $e$ in $(c,b)$ such that $P'(d)=P'(e)=0$. Since $P'$ is differentiable on $(a,b)$, it is continuous on $[d,e]$, and it is differentiable on $(d,e)$. So, we may apply Rolle's Theorem to $P'$ to find $\xi$ in $(d,e)$ with $P''(\xi)=0$. This $\xi$ is then as desired.]

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    If I want to use Cauchy mean Theorem, and prove $\dfrac{1}{2} f^{''}(\xi)=\dfrac{f(a)(c-b)+f(b)(a-c)+f(c)(b-a)}{(a-b)(b-c)(c-a)}$2012-02-03
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@David Moews

I have just come out the solution using Cauchy mean Theorem. $\dfrac{1}{2} f^{''}(\xi)=\dfrac{f(a)(c-b)+f(b)(a-c)+f(c)(b-a)}{(a-b)(b-c)(c-a)}$

so $\dfrac{1}{2} f^{''}(\xi)=\dfrac{f(a)(c-b)+f(b)(a-c)+f(c)(b-a)}{(a-b)(b-c)(c-a)}=\dfrac{f^{'}(\zeta_1)(a-b)+f(b)-f(a)}{(a-b)(2\zeta_1-a-b)}=\dfrac{f^{'}(\zeta_2)(a-b)+f(b)-f(a)}{(a-b)(2\zeta_2-a-b)}=\dfrac{(f^{'}(\zeta_1)-f^{'}(\zeta_2))(a-b)}{(a-b)(2\zeta_1-2\zeta_2)}$

where $a<\zeta_1

Then use Lagrange Theorem and that is