How does one prove the Taylor's Theorem by the Cauchy's Mean Value Theorem? I am trying to prove there is a $c \in (a,b)$ such that
$$
f(b)=f(a)+f'(a)(b-a)+ \frac{1}{2} f''(c)(b-a)^2
$$
which I do believe is a specialized form of the Taylor's Theorem when you plug in $b$ and $a$ into the appropriate spots.  I think you can approach this by proving the Lagrangian remainder is the same, but it doesn't seem all that intuitive.  
 A: Define the function
$$
g(x)=f(x)-\left[f(a)+(x-a)f'(a)\right]-\frac{f(b)-\left[f(a)+(b-a)f'(a)\right]}{(b-a)^2}(x-a)^2
$$
Since $g(a)=0$ and $g(b)=0$, the Mean Value Theorem says there is a $c_1\in(a,b)$ so that $g'(c_1)=0$.
$$
g'(x)=f'(x)-f'(a)-\frac{f(b)-\left[f(a)+(b-a)f'(a)\right]}{(b-a)^2}(x-a)
$$
Since $g'(a)=0$ and $g'(c_1)=0$, the Mean Value Theorem says there is a $c_2\in(a,c_1)$ so that $g''(c_2)=0$. That is,
$$
0=f''(c_2)-2\frac{f(b)-\left[f(a)+(b-a)f'(a)\right]}{(b-a)^2}
$$
this, becomes
$$
f(b)=f(a)+(b-a)f'(a)+\frac12(b-a)^2f''(c_2)
$$
A: I answered the same question here using the function
$$g(x)=f(b)-f(x)-(b-x)f'(x)-\left(\frac{b-x}{b-a}\right)^2[f(b)-f(a)-(b-a)f'(a)]$$
In general, if
$$F_n(x)=f(b)-\sum_{k=0}^{n-1} \frac{f^{(k)}(x)(b-x)^k}{k!}$$
Then you can let
$$g_n(x)=F_n(x)-\left(\frac{b-x}{b-a}\right)^nF_n(a)$$
Differentiate, factor out $\frac{n(b-x)^{n-1}}{(b-a)^n}$, and apply the mean value theorem. This will prove the $(n-1)$th order analogue of the mean value theorem, i.e.
$$f(b)=\sum_{k=0}^{n-1} \frac{f^{(k)}(a)(b-a)^k}{k!}+\frac{f^{(n)}(c)(b-a)^n}{n!}$$
for some $a<c<b$. 
A: Let 
$$g(y)=f(b)-(f(y)+f'(y)(b-y))-z(b-y)^2$$
where z is chosen such that $g(a)=0$ i.e:
$$z(b-a)^2=f(b)-(f(a)+f'(a)(b-a))   \ \ \ \ (1) $$
Now we have:
$$g(a)=g(b)=0$$
It follows from Rolle's theorem that
$$ \exists c \in ]a,b[, g'(c)=0$$
This is the desired result noticing that the last equation set the value $\frac {f"(c)} {2}$ for $z$
