Since
\(f\) is continuous and has
\(m\) continuous derivatives, we can use Taylor’s Theorem (
Theorem 1.6.4) to write:
\begin{equation*}
f(x) = \sum_{j=0}^{m-1}\frac{f^{(j)}(p)}{j!}(x-p)^{j} + \frac{f^{(m)}(p)}{m!}(x-p)^{m} + \frac{f^{(m+1)}(\xi(x))}{(m+1)!}(x-p)^{m+1}
\end{equation*}
for some \(\xi \in [x,p]\text{.}\) Since \(f^{(j)}(p)=0\) for \(j=0,1,\ldots,m-1\text{,}\) all terms in the summation formula above are zero and thus we can write
\begin{equation*}
f(x) = \frac{f^{(m)}(p)}{m!}(x-p)^{m} + \frac{f^{(m+1)}(\xi(x))}{(m+1)!}(x-p)^{m+1}
\end{equation*}
factoring out a \((x-p)^{m}\) term
\begin{equation*}
f(x) = (x-p)^{m} \biggl( \frac{f^{(m)}(p)}{m!}+ (x-p) \frac{f^{(m+1)}(\xi(x))}{(m+1)!}\biggr),
\end{equation*}
which has a nonzero limit as \(x \rightarrow p\text{,}\) therefore \(f(x)\) has a root at \(x=p\) of multiplicity \(m\text{.}\)
In converse, assume that \(f(x)\) has a root of multiplicity \(p\text{.}\) This means that we can write
\begin{equation*}
f(x) = (x-p)^{m} g(x).
\end{equation*}
with \(g(p)\neq 0\text{.}\) Obviously \(f(p)=0\text{.}\) Also if \(f(x)\) has \(m\) continuous derivatives, then \(g(x)\) does as well.
\begin{equation*}
\begin{aligned}f'(x)\amp= m (x-p)^{m-1}g(x) + (x-p)^{m} g'(x) \\ f''(x)\amp= m(m-1) (x-p)^{m-2}g(x) + 2 m (x-p)^{m-1}g'(x) + (x-p)^{m} g''(x)\\ f'''(x)\amp= m(m-1)(m-2) (x-p)^{m-3}g(x) + 3 m (m-1) (x-p)^{m-2}g'(x) \\\amp\qquad + 3 m(x-p)^{m-1}g''(x) + (x-p)^{m} g'''(x) \end{aligned}
\end{equation*}
In general, we can write
\begin{equation*}
f^{(j)}(x) = \sum_{i=0}^{j}k_{i} (x-p)^{m-j+i}g^{(i)}(x),
\end{equation*}
where \(k_{i}\) is a function of \(m\) (however the exact form it not needed).
Evaluating \(f^{(m-1)}\) at \(x=p\) results in
\begin{equation*}
f^{(m-1)}(p) = \sum_{i=0}^{m-1}k_{i} \cdot (p-p)^{i+1}g^{(i)}(p) = 0
\end{equation*}
since \((p-p)^{i}=0\) for \(i=0,1,2,\ldots,m-1\) and evaluating \(f^{(m)}(x)\) at \(x=p\) results in
\begin{equation*}
f^{(m)}(p) = \sum_{i=0}^{m} k_{i} (p-p)^{i} g^{(i)}(p) = k_{0} g^{(0)}(p) \neq 0
\end{equation*}
since \(g^{(0)}(p)=g(p) \neq 0\text{.}\) Therefore the theorem holds.
