Let \(f\) be a function whose first \(n+1\) derivatives are continuous on an interval \(I\) containing the number \(c\). Then for every \(x\) in the interval, there is a number \(u\) between \(x\) and \(c\) for which \begin{equation*} f( x) =f( c) +f^{\prime} ( c) ( x-c) +\dfrac{f^{\prime \prime} ( c) }{2!}( x-c) ^{2}+\cdots +\,\dfrac{f^{( n) }( c) }{n!}( x-c) ^{n}+R_{n}( x) \end{equation*}
where \begin{equation*} R_{n}( x) =\dfrac{f^{( n+1) }(u) }{( n+1) !}( x-c) ^{n+1} \end{equation*}
For a fixed number \(x\neq c\) in the interval \(I,\) there is a number \(L\) (depending on~\(x\)) for which \begin{eqnarray} f(x) &=& f(c) + \dfrac{f^{\prime} (c)}{1!}(x-c)+\dfrac{f^{\prime \prime} (c)}{2!} (x-c)^{2} + \cdots + \dfrac{f^{(n)}(c)}{n!}(x-c)^{n} + \dfrac{L}{(n+1)!}(x-c)^{n+1}\quad\tag{1} \end{eqnarray}
Define the function \(F\) to be \begin{eqnarray} F(t) &=& f(x)-f(t)-\dfrac{f^{\prime} (t)}{1!}(x-t)-\dfrac{f^{\prime \prime} (t)}{2!} (x-t)^{2}-\cdots -\;\dfrac{f^{(n)}(t)}{n!}(x-t)^{n} -\dfrac{L}{(n+1)!} (x-t)^{n+1}\quad\tag{2} \end{eqnarray}
The domain of \(F\) is \(c\leq t\leq x\) if \(x>c\) and \(x\leq t\leq c\) if \(x < c\). Since \(f(t),\) \(f^{\prime} ( t) \), \(f^{\prime \prime} ( t) ,\) \(\ldots\) , \(f^{(n)}(t)\) are each continuous, then \(F\) is continuous on its domain. Furthermore, \(F\) is differentiable and \begin{eqnarray*} \dfrac{dF}{dt} = F^{\prime} (t)&=&-f^{\prime} (t) + \left[ f^{\prime} (t)-\dfrac{f^{\prime \prime} (t)}{1!}(x-t)\right] +\left[ \dfrac{f^{\prime \prime} (t)}{1!}(x-t)-\dfrac{ f^{\prime \prime \prime} (t)}{2!}(x-t)^{2}\right]\\[4pt] && +\cdots +\left[ \dfrac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}-\dfrac{f^{(n+1)}(t)}{n!} (x-t)^{n}\right] +\dfrac{L}{n!}(x-t)^{n} \\[4pt] &=& -\dfrac{f^{(n+1)}(t)}{n!}(x-t)^{n}+\dfrac{L}{n!}(x-t)^{n}\qquad \end{eqnarray*}
for all \(t\) between \(x\) and \(c.\) From (1) and (2), we have \[ F(c) = f(x)-f(c)-\dfrac{f^{\prime }(c)}{1!}(x-c)-\dfrac{f^{\prime \prime }(c)}{ 2!}(x-c)^{2}-\cdots -\dfrac{f^{(n)}(c)}{n!}(x-c)^{n} -\dfrac{L}{(n+1)!} (x-c)^{n+1}=0 \]
Then \[ F(x)=\;f(x)-f(x)-\dfrac{f^{\prime} (x)}{1!}(x-x)-\cdots -\dfrac{f^{(n)}(x)}{n!} (x-x)^{n}-\dfrac{L}{(n+1)!}(x-x)^{n+1}=0 \]
B-13
Now apply Rolle's Theorem to \(F\). Then there is a number \(u\) between \(c\) and \(x\) for which \begin{equation*} F^{\prime} (u)=-\dfrac{f^{(n+1)}(u)}{n!}(x-u)^{n}+\dfrac{L}{n!}(x-u)^{n}=0 \end{equation*}
Solving for \(L\), we find \(L=f^{(n+1)}(u)\). Now let \(t=c\) and \( L=f^{(n+1)}(u) \) in (2) and solve for \(f(x).\) Then \begin{equation*} f( x) =f( c) + f^{\prime} ( c) ( x-c) +\dfrac{f^{\prime \prime} ( c) }{2!}( x-c) ^{2}+\cdots +\,\dfrac{f^{( n) }( c) }{n!}( x-c) ^{n}+R_{n}( x) \end{equation*}
where \[ R_{n}( x) =\dfrac{f^{( n+1) }(u) }{( n+1) !}( x-c) ^{n+1} \]