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.  

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 $a<c<b$. 
