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1.10:1.10-Бесселівська інтерполяція
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$B_3 = (\theta - \frac{1}{2}) \theta ( \theta - 1) / 3! = \theta (0.5 + \theta (-1.5 + \theta ))/6 \label{1.10.8} \tag{1.10.8}$ \[B_4 = \frac{1}{2} ( \theta + 1) \theta ( \theta - 1) ( \theta - 2) /... $B_3 = (\theta - \frac{1}{2}) \theta ( \theta - 1) / 3! = \theta (0.5 + \theta (-1.5 + \theta ))/6 \label{1.10.8} \tag{1.10.8}$ $B_4 = \frac{1}{2} ( \theta + 1) \theta ( \theta - 1) ( \theta - 2) / 4 ! = \theta ( 2 + \theta ( -1 + \theta ( -2 + \theta ))) / 48 \label{1.10.0} \tag{1.10.9}$ $B_5 = (\theta - \frac{1}{2} ) ( \theta + 1) \theta ( \theta - 1) (\theta - 2) / 5 ! = \theta ( -1 + \theta ( 2.5 + \theta^2 (-2.5 + \theta )))/120 \label{1.10.10} \tag{1.10.10}$