Question

∇^p(ao+a1*t+...an*t^p)=p!*an prove left hand side.. and note that a0,a1...an 0,1..n is in the subscript of a and ^p means power and * means multiply.

Answer 1

QUESTION:

$\nabla^{\wedge}p(a_0 + a_1*t + \ldots an*t^{\wedge}p) = p!^{*}an$ prove left hand side.. and note that a0,a1...an 0,1..n is in the

subscript of a and ^p means power and * means multiply.

SOLUTION:

$$\begin{array}{l} \nabla^ {p} \left(a _ {0} + a _ {1} t + a _ {2} t ^ {2} + \dots + a _ {n} t ^ {p}\right) = \nabla^ {p} \left(a _ {0}\right) + \nabla^ {p} \left(a _ {1} t\right) + \dots + \nabla^ {p} \left(a _ {n} t ^ {p}\right) = a _ {n} \nabla^ {p} \left(t ^ {p}\right) \\ \text{(since } \nabla^ {p} (a _ {0}) = 0, \quad \nabla^ {p} (a _ {1} t) = 0, \dots , \nabla^ {p} (a _ {n - 1} t ^ {p - 1}) = 0. \text{ Hence} \\ a _ {n} \nabla^ {p} (t ^ {p}) = a _ {n} \cdot p \cdot \nabla^ {p - 1} (t ^ {p - 1}) = a _ {n} \cdot p \cdot (p - 1) \nabla^ {p - 2} (t ^ {p - 2}) = \dots = a _ {n} \cdot p (p - 1) (p - 2) (p - 3) \dots 1 = a _ {n} p! \end{array}$$