Let f satisfy the conditions of Cauchy's theorem on [0,x]

Let $f$ satisfy the conditions of Cauchy's theorem on $[0,x]$ and suppose $f(0) = f'(0) = \cdots = f^{(n-1)}(0) = 0$. Prove that $\dfrac{f(x)}{x^{n}} = \dfrac{f^{(n)}\bigl(\theta x\bigr)}{n!}$ with $0\lt \theta\lt 1$.

Step 1: The idea

Compare $f$ with $g(t) = t^{n}$. Both vanish at $0$, together with their first $n-1$ derivatives — for $g$ because $\bigl(t^{n}\bigr)^{(k)} = \dfrac{n!}{(n-k)!}t^{\,n-k}$ vanishes at $0$ for $k\lt n$, and for $f$ by hypothesis. That is exactly the situation in which Cauchy's theorem can be applied over and over.

Step 2: First application

On $[0,x]$, with $g(t) = t^{n}$ and $g'(t) = nt^{\,n-1}\neq0$ inside, there is an $x_{1}\in(0,x)$ with

\[ \frac{f(x)}{x^{n}} = \frac{f(x)-f(0)}{x^{n}-0} = \frac{f'\bigl(x_{1}\bigr)}{n\,x_{1}^{\,n-1}}. \]

Step 3: Second application

Now compare $f'$ with $nt^{\,n-1}$ on $\bigl[0,x_{1}\bigr]$; both vanish at $0$, so there is an $x_{2}\in\bigl(0,x_{1}\bigr)$ with

\[ \frac{f'\bigl(x_{1}\bigr)}{n\,x_{1}^{\,n-1}} = \frac{f''\bigl(x_{2}\bigr)}{n\bigl(n-1\bigr)x_{2}^{\,n-2}}, \qquad\text{hence}\qquad \frac{f(x)}{x^{n}} = \frac{f''\bigl(x_{2}\bigr)}{n\bigl(n-1\bigr)x_{2}^{\,n-2}}. \]

Step 4: The general step

After $k$ applications, with $0\lt x_{k}\lt x_{k-1}\lt \cdots\lt x_{1}\lt x$, \[\frac{f(x)}{x^{n}} = \frac{f^{(k)}\bigl(x_{k}\bigr)}{n\bigl(n-1\bigr)\cdots\bigl(n-k+1\bigr)\,x_{k}^{\,n-k}}. \tag{$k$}\]

Each step is legitimate because $f^{(k-1)}(0) = 0$ for $k\le n$, and the comparison function's derivative does not vanish inside.

Step 5: The last step

At $k = n$ the comparison function is $n!\,t$, its derivative is the constant $n!$, and the power $x_{n}^{\,n-n} = 1$: $\frac{f(x)}{x^{n}} = \frac{f^{(n)}\bigl(x_{n}\bigr)}{n!}, \qquad x_{n}\in(0,x)$.

Writing $x_{n} = \theta x$ with $\theta\in(0,1)$ gives the statement.

Step 6: Verification

\[f(t) = t^{n}:\ \text{all lower derivatives vanish at } 0,\ f^{(n)} \equiv n!, \text{ and } \frac{x^{n}}{x^{n}} = \frac{n!}{n!} = 1 \;\;\checkmark\]

\[f(t) = t^{3},\ n = 2:\ \frac{x^{3}}{x^{2}} = x \text{ and } \frac{f''(\theta x)}{2!} = \frac{6\theta x}{2} = 3\theta x \;\Longrightarrow\; \theta = \frac13 \;\;\checkmark\]

\[f(t) = \sin t-t,\ n = 3:\ f,f',f'' \text{ all vanish at } 0 \text{ and } \frac{\sin x-x}{x^{3}} = \frac{-\cos\theta x}{6} \xrightarrow[x\to0]{} -\frac16 \;\;\checkmark\]