Cauchy form of remainder theorem

Question

I am currently studying Understanding analysis by Stephen Abbott. I am doing exercise of deriving Cauchy remainder theorem.

When considering $E_{N}(x,a)$ as a function of a , $E_{N}(x,x)=f(x)-f(a)$ .

I am able to show formula for $(E_{N})^{’}(x,a)$ asked in part (b) The only thing left is to apply Mean value theorem on interval [0,x]. So when we apply Mean value theorem on interval [0,x] , there exists $c \in (0,x)$ such that $\frac{E_{N}(x,x)-E_{N}(x,0)}{x-0}= (E_{N})^{’}(x,c)$ . That is $\frac{f(x)-f(a)-E_{N}(x,0)}{x-0}= \frac{f^{N+1}(c)(x-c)^{N}}{N!}$ . But in order to get required formula for remainder, $E_{N}(x,x)=f(x)-f(a)$ should be zero. My question is how it can be zero? It should be zero when x=a and we haven’t assumed x=a.

Answer 1

Actually, $E_{N}(x,x)=f(x)-f(x)=0$

$E_{N}(x,x)=f(x)-S_{N}(x,x)=f(x)-c_0=f(x)-f(x)=0$

Answer 2

It's just a matter of being careful with the definition. We have $$E_N(x, x) = f(x) - S_N(x,x),$$ where $$S_N(x,x) = c_0 + \sum_{n=1}^{N}c_n(x - x)^n = c_0, \text{ where } c_0 = \frac{f^{(0)}(x)}{0!} = f(x),$$ and so $$E_N(x, x) = f(x) - f(x) = 0.$$