صيغة ليبنيز - Formule de Leibniz
لتكن $\displaystyle{\displaylines{f}}$ و $\displaystyle{\displaylines{g}}$ دالتين قابلتان للإشتقاق $\displaystyle{\displaylines{p}}$ مرة على مجال $\displaystyle{\displaylines{I}}$.
نرمز بـ$\displaystyle{\displaylines{f^{(k)}}}$ للمشتقة من الدرجة $\displaystyle{\displaylines{k}}$ للدالة $\displaystyle{\displaylines{f}}$ واصطلاحا $\displaystyle{\displaylines{f^{(0)} = f}}$
بين أن :
$\displaystyle{\displaylines{(\forall n \in \mathbb{N} , \, n \leq p) (\forall x \in I) \ : \ (f \times g)^{(n)}(x) = \sum_{k=0}^{n} C_{n}^{k} f^{(k)}(x) \ g^{(n-k)}(x)}}$
راجع
درس الإشتقاق وتطبيقاته
البرهان بالترجع Démonstration par récurrenceمن أجل
$\displaystyle{\displaylines{n = 0}}$ لدينا :
$\displaystyle{\displaylines{(f \times g)^{(0)}(x) = (f \times g)(x) = \sum_{k=0}^{0} C_{0}^{k} f^{(k)}(x) \ g^{(0-k)}(x)}}$إذن الخاصية صحيحة من أجل
$\displaystyle{\displaylines{n= 0}}$نفترض أن :
$\displaystyle{\displaylines{ (f \times g)^{(n)}(x) = \sum_{k=0}^{n} C_{n}^{k} f^{(k)}(x) \ g^{(n-k)}(x) }}$ ولنبين أن :
$\displaystyle{\displaylines{ (f \times g)^{(n+1)}(x) = \sum_{k=0}^{n+1} C_{n+1}^{k} f^{(k)}(x) \ g^{(n+1-k)}(x) }}$لدينا :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = ((f \times g)^{(n)})^{'}(x)}}$ . إذن :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \left(\sum_{k=0}^{n} C_{n}^{k} f^{(k)}(x) \ g^{(n-k)}(x)\right)^{'}}}$وبالتالي :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \sum_{k=0}^{n} C_{n}^{k} \left( f^{(k+1)}(x) \ g^{(n-k)}(x) + f^{(k)}(x) \ g^{(n+1-k)}(x) \right)}}$إذن :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \sum_{k=0}^{n} C_{n}^{k} f^{(k+1)}(x) \ g^{(n-k)}(x) + \sum_{k=0}^{n} C_{n}^{k} f^{(k)}(x) \ g^{(n+1-k)}(x)}}$نقوم بتغيير المتغير في المجموع الأول :
$\displaystyle{\displaylines{p = k +1}}$ $\displaystyle{\displaylines{\sum_{k=0}^{n} C_{n}^{k} f^{(k+1)}(x) \ g^{(n-k)}(x) = \sum_{p=1}^{n+1} C_{n}^{p-1} f^{(p)}(x) \ g^{(n+1-p)}(x)}}$وبالتالي :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \sum_{k=1}^{n+1} C_{n}^{k-1} f^{(k)}(x) \ g^{(n+1-k)}(x) + \sum_{k=0}^{n} C_{n}^{k} f^{(k)}(x) \ g^{(n+1-k)}(x)}}$إذن :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \left( \sum_{k=1}^{n} (C_{n}^{k-1} +C_{n}^{k} ) \times \left( f^{(k)}(x) \ g^{(n+1-k)}(x) \right) \right) + C_{n}^{n} f^{(n+1)}(x) \ g^{(0)}(x) + C_{n}^{0} f^{(0)}(x) \ g^{(n+1)}(x)}}$لاحظ أن :
$\displaystyle{\displaylines{ C_{n}^{k-1} +C_{n}^{k} = C_{n+1}^{k} }}$ و
$\displaystyle{\displaylines{ C_{n}^{n} = C_{n}^{0} = 1}}$إذن :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \left( \sum_{k=1}^{n} C_{n+1}^{k} f^{(k)}(x) \ g^{(n+1-k)}(x) \right) + f^{(n+1)}(x) \ g^{(0)}(x) + f^{(0)}(x) \ g^{(n+1)}(x)}}$لاحظ أنه يمكننا إدخال الحدين الأخيرين في المجموع :
$\displaystyle{\displaylines{ k = 0 }}$ :
$\displaystyle{\displaylines{C_{n+1}^{0} f^{(0)}(x) \ g^{(n+1-0)}(x) = f^{(0)}(x) \ g^{(n+1)}(x)}}$$\displaystyle{\displaylines{ k = n +1 }}$ :
$\displaystyle{\displaylines{ C_{n+1}^{n+1} f^{(n+1)} \ g^{(n+1-(n+1))}(x) = f^{(n+1)}(x) \ g^{(0)}(x)}}$إذن :
$\displaystyle{\displaylines{(f \times g)^{(n+1)}(x) = \sum_{k=0}^{n+1} C_{n+1}^{k} f^{(k)}(x) \ g^{(n+1-k)}(x)}}$إذن حسب البرهان بالترجع لدينا :
$\displaystyle{\displaylines{(\forall n \in \mathbb{N} , \, n \leq p) (\forall x \in I) \ : \ (f \times g)^{(n)}(x) = \sum_{k=0}^{n} C_{n}^{k} f^{(k)}(x) \ g^{(n-k)}(x)}}$