Exercise 1.4.6 (Leibniz formula)

Question

Let $f(x)$ and $g(x)$ be $n$-times differentiable function. Show that the $n$th derivative of $f(x)g(x)$ is 
$$\sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x).$$

Richard Ganaye · Accepted Answer

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Note: It is preferable to say ``the function $f$'' than '``the function $f(x)$" ($f(x)$ is a real).
\begin{proof} 
Let $f$ and $g$ be $n$-times differentiable function on some interval $I$. We show by finite induction up to $n$ that, for all $x \in I$, and $p\leq n$,
$$(fg)^{(p)}(x) = \sum_{k=0}^p \binom{p}{k} f^{(k)}(x) g^{(p-k)}(x).$$

Let $\mathscr{P}(p)$ the property defined for $n \in \N$ and $0 \leq p \leq n$ by
$$\mathscr{P}(p) \iff \forall x \in I,\ (fg)^{(p)}(x) = \sum_{k=0}^p \binom{p}{k} f^{(k)}(x) g^{(p-k)}(x),$$
that is
$$\mathscr{P}(p) \iff \ (fg)^{(p)}= \sum_{k=0}^p \binom{p}{k} f^{(k)} g^{(p-k)}.$$

If $p = 0$, then $(fg)^{(p)}(x) = (fg)^{(0)}(x) = (fg)(x)$, and 
$$\sum_{k=0}^p \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x) =  \binom{0}{0} f^{(0)}(x) g^{(0)}(x) = f(x) g(x)\qquad (x \in I),$$
thus $\mathscr{P}(0)$ is true.

Assume  that $\mathscr{P}(p)$ is true for some $p \in \N, p<n$. So $(fg)^{(p)}= \sum_{k=0}^p \binom{p}{k} f^{(k)} g^{(p-k)}$. Since $f,g$ are $n$-times differentiable, and $p<n$, then $f^{(k)}$ and $g^{(p-k)}$ are differentiable on $I$ for all $k \in [\![0,p]\!]$, so $(fg)^{p}$ is differentiable on $I$, and
\begin{align*}
(fg)^{(p+1)}&= ((fg)^{(p)})'\
&=  \sum_{k=0}^p \binom{p}{k} (f^{(k)} g^{(p-k)})'\
&= \sum_{k=0}^p \binom{p}{k}  (f^{(k+1)} g^{(p-k)} + f^{(k)} g^{(p-k + 1)})\
&= \sum_{k=0}^p \binom{p}{k}  f^{(k+1)} g^{(p-k)} + \sum_{k=0}^p \binom{p}{k} f^{(k)} g^{(p-k + 1)}\
&= f^{(p+1)} + \sum_{k=0}^{p-1} \binom{p}{k}  f^{(k+1)} g^{(p-k)} + \sum_{k=1}^p \binom{p}{k} f^{(k)} g^{(p-k + 1)} + g^{(p+1)}\
&=f^{(p+1)} + \sum_{j=1}^p \binom{p}{j-1} f^{(j)} g^{(p-j+1)} + \sum_{k=1}^p \binom{p}{k} f^{(k)} g^{(p-k + 1)}  +  g^{(p+1)} \qquad (j = k+1)\
&=f^{(p+1)} + \sum_{k=1}^p \binom{p}{k-1} f^{(k)} g^{(p-k+1)} + \sum_{k=1}^p \binom{p}{k} f^{(k)} g^{(p-k + 1)}  +  g^{(p+1)}\
&= f^{(p+1)} + \sum_{k=1}^p \left[\binom{p}{k-1} + \binom{p}{k}ight] f^{(k)} g^{(p-k + 1)}  +  g^{(p+1)}\
&=f^{(p+1)} + \sum_{k=1}^p \binom{p+1}{k} f^{(k)} g^{(p-k + 1)}  +  g^{(p+1)}\
&=  \sum_{k=0}^{p+1} \binom{p+1}{k} f^{(k)} g^{(p+ 1 - k)} 
\end{align*}
This gives $\mathscr{P}(p+1)$. The induction is done, so $\mathscr{P}(p)$ is true up to $n$. In particular $\mathscr{P}(n)$ is true: $fg$ is $n$ times differentiable on $I$, and
$$\forall x \in I,\ (fg)^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x).$$
\end{proof}

Exercise 1.4.6 (Leibniz formula)

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Exercise 1.4.6 (Leibniz formula)

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