Proof. We shall prove the theorem the way we proved the mean value theorem. We shall construct a suitable function \(F\) and then apply Rolle’s Theorem which we shall use repeatedly. Define a function \(F\) by \[F(x) = f(x) - f(a) - f'(a)\, (x - a) - \dfrac {f''(a)}{2!}\, (x - a)^2 + \cdots \cdots \cdots -\frac {f^{(n - 1)} (a)}{(n - 1)!}\, (x - a)^{n - 1} - K\, (x -a)^n\hspace {0.3cm} \cdots \cdots \hspace {0.3cm} (I)\] Where \(K\) is chosen so that \(F(b) = F(a)\) since \(F(a) = 0\), this gives \[K = \dfrac {f(b) - f(a) -\sum \limits _{r = 1}^{n - 1}\, \dfrac {f^{(r)} (a)}{r!}\, (b - a)^r}{(b - a)^n}\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (II)\] Now the function \(F\) is continuous on \([a,b]\) and is differentiable on the open interval \((a,b)\) and \(F(a) = 0 = F(b)\). By Rolle’s Theorem there is some \(c_1\) with \(a < c_1 < b\) such that \(F'(c_1) = 0\hspace {0.3cm} \cdots \cdots \hspace {0.3cm} (III)\). Then differentiating \((I)\) gives \[F'(x) = f'(x) - f'(a) - f''(a) \, (x-a)-\cdots \cdots \cdots - \dfrac {f^{(n-1)} (a)}{(n - 2)!}\, (x - a)^{n - 2}- K\, n\, (x - a)^{n - 1}\] giving \(F'(a) = 0\) If we assume that \(n\geq 2\), then we see that \(F'\) is continuous on \([a,c_1]\) and differentiable on \((a,c_1)\) and \(F'(a) = F'(c_1) = 0\). Again by Rolle’s Theorem there is
\(c_2\), \(\, a< c_2 < c_1 < b\) and \(F'(c_2) = 0\hspace {0.4cm}\cdots \cdots \cdots \hspace {0.3cm} (IV)\).
Note that \(F^k(a) = 0\) for \(1 \leq k \leq n - 1\). In particular if \(n\geq 3\), then \(F''(a) = 0 = F''(c_2)\). Applying Rolle’s theorem to \(F''\) on \([a,c_2]\) we obtain a number \(c_3\) such that \(a < c_3 < c_2 < c_1 <b\) and \(F'''(c_3) = 0\). Continuing this way and repeating the process a further times, we obtain a number \(c_n\) such that \(a < c_n < c_{n-1} < \cdots \cdots \cdots < c_3 < c_2 < c_1 < b\) and
\(F^n(c_n) = 0\hspace {0.3cm}\cdots \cdots \cdots \hspace {0.3cm} (V)\). From \((I)\) we have \(F^n(x) = f^n(x) - K\, n!\hspace {0.2cm} a < x <b\). Therefore, equation \((V)\) gives \(F^n(c) = n!\, K\,\) where \(\, c = c_n\,\). That is \(\, K = \dfrac {f^n (c)}{n!}\). But from \((II)\) \[ K = \dfrac {f(b) - f(a) - \sum \limits _{r = 1}^{n -1} \dfrac {f^r(a)}{r!}\, (b - a)^r}{(b - a)^n}\] So that \[\dfrac {f^n(c)\, (b - a)^n}{n!} = f(b) - f(a) - \sum ^{n - 1}_{r=1}\dfrac {f^r(a)}{r!}\, (b - a)^r\] Solve for \(f(b)\) to get \[f(b) = f(a) + \sum ^{n -1}_{r =1} \dfrac {f^r(a)}{r!}(b - a)^r + \dfrac {f^(c)}{n!}\,(b - a)^n\hspace {0.4cm} \text {as required}\]

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Solution. Expanding \(e^x\) into power series, we have \[e^x = 1 + x + \dfrac {x^2}{2!} + \dfrac {x^3}{3!} + \cdots \cdots \cdot + \dfrac {x^{n -1}}{(n - 1)!} + R_n(x)\] where \(R_n(x) = \dfrac {f^n(c)\, x^n}{n!}\) for some \(c\) such that \(0 < c < x\). But \(f^n(c) = e^c\). Thus
\(\begin {vmatrix} R_n(x)\\ \end {vmatrix} \leq \dfrac {e^{|x|}\, x^n}{n!}\longrightarrow 0\) as \(n\longrightarrow \infty \). That is the series converges to \(e^x\).