Krdytkyu

I asked this question before, but I quickly deleted it so that I can think about it more. the question was how can I show that:
$exists c in [a,b]$ such that $ f(b)=f(a)+(b-a)f'(a)+frac{(b-a)^2}{2}f''(c)$ knowing that $f$ is a function that can be differentiated two times on $[a,b]$

Here is what I did:

I used the mean value theorem on $f$ in the interval $]a;b[$ then I got:
$$exists c_0 in ]a;b[:f(b) - f(a) = f'(c_0)(b-a)$$
rearranging gives: (1)
$$exists c_0 in ]a;b[:f'(c_0) = frac{f(b)-f(a)}{b-a}$$
Then I used the theorem on $f'$ but in the interval $]a;c_0[$ which gives me
$$exists c in ]a;c_0[:f'(c_0) - f'(a) = f''(c)(c_0-a)$$
by replacing $f'(c_0)$ in the last equation with (1) and multiplying by $(b-a)$ I got
$$exists c in ]a;c_0[:f(b)-f(a)-(b-a)f'(a)=f''(c)(c_0-a)(b-a)$$ and rearranging this gets me really close to what I am supposed to get:
$$exists c in ]a;c_0[:f(b)=f(a)+(b-a)f'(a)+f''(c)(c_0-a)(b-a)$$
and because $]a;c_0[ subset]a;b[$ I can write : $$exists c in ]a;b[:f(b)=f(a)+(b-a)f'(a)+f''(c)(c_0-a)(b-a)$$

If only that $(c_0-a)$ was $frac{(b-a)}{2}$ or something like that.
Am I on the right track? I'd appreciate any help on this, I am open to any suggestion.

note: I asked my teacher about this problem and he told me to write I new function in terms of $f(x)$, maybe that will work, but I don't know about that, it seemed to me that he doesn't know the answer either, he could be wrong.

Given $fin C([a,b])$ twice-differentiable and $xin(a,b]$, consider
$$
F(t) = f(t)+f'(t)(x-t), qquad G(t) = (t-x)^2.
$$
By Cauchy's mean value theorem (a simple consequence of Rolle's theorem), we have
$$tag{1}
frac{F(x)-F(a)}{G(x)-G(a)} = frac{F'(c)}{G'(c)}
$$
for some $cin(a,x)$. But
$$
F'(t)=f'(t)+f''(t)(x-t)-f'(t)=f''(t)(x-t)
$$
and
$$
G'(t) = 2(t-x),
$$
so $(1)$ translates to
$$
frac{f(x)-f(a)-f'(a)(x-a)}{-(a-x)^2} = frac{f''(c)(x-c)}{2(c-x)} = -frac{f''(c)}2.
$$
Rearranging the terms leads to
$$
f(x) = f(a)+f'(a)(x-a)+frac{f''(c)}2(x-a)^2.
$$

You can start with the Taylor-formula
$$f(x) = f(a) + int_a^x f'(t) , d x = f(a) + f'(a)(x-a) +int_a^x (f'(t) -f'(a)) , dt.$$
Next, we have
begin{align}
int_a^x (f'(t) -f'(a)) , dt &ge min_{t in [a,x]} Big(frac{f'(t)-f'(a)}{t-a}Big) int_a^x (t-a) , dt \
& = min_{t in [a,x]} Big(frac{f'(t)-f'(a)}{t-a}Big) frac{(x-a)^2}{2}
end{align}
and similar
$$ int_a^x (f'(t) -f'(a)) , dt le max_{t in [a,x]} Big(frac{f'(t)-f'(a)}{t-a}Big) frac{(x-a)^2}{2}.$$
Note that the function
$$g(t) = begin{cases} frac{f'(t)-f'(a)}{t-a} & text{if} t>a \ f''(a) & text{if} t=a end{cases}$$
is continuous. Thus, by the intermediate value theorem there exists a $t in [a,b]$ with $$g(t) frac{(x-a)^2}{2} = int_a^x (f'(t) -f'(a)) , dt.$$
If $t= a$ we are done. Otherwise we can apply the mean-value theorem in order to get a $c in (a,t)$ with $$f''(c) = g(t).$$

I found this solution that my professor showed me:
let $$varphi(x)=f(b)-f(x)-(b-x)f'(x)+frac{(b-a)^2}{2}lambda$$
notice that $varphi(b)=0$,
we will choose $lambda$ so that $varphi(a)=0=varphi(b)$
$$varphi(a)=0Leftrightarrowlambda=frac{-2}{(b-a)^2}big(f(b)-f(a)-(b-a)f'(a)big)$$
and because $varphi$ is continuous on $[a,b]$ and differentiable on $]a,b[$ and $varphi(a)=varphi(b)$ we can use Rolle's theorem, we will get:
$$exists c in ]a,b[:varphi'(c)=0$$
we can diffrentiate $varphi$ so we get
$varphi'(x)=-f'(x)+f'(x)-(b-x)f''(x)-frac{2(b-x)}{2}lambda$
simplifying that and substituting $lambda$ we get:
$$varphi'(x)=-(b-x)f''(x)+frac{2}{(b-a)}big(f(b)-f(a)-(b-a)f'(a)big)$$
and then we have:
$$exists c in ]a,b[:varphi'(c)=0 Leftrightarrow exists c in ]a,b[: -(b-c)f''(c)+frac{2}{(b-a)}big(f(b)-f(a)-(b-a)f'(a)big) = 0$$
and by rearranging the terms we can achieve what we want:
$$exists c in ]a,b[:f(b)=f(a)+(b-a)f'(a)+frac{(b-a)^2}{2}f''(c)$$