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Given an arbitrary polynomial f(x) = aₙxⁿ + ··· +a₂x²+a₁x+a₀, find a formula that relates the coefficient a to the derivative f⁽ᵏ⁾ (0). Explain why your formula is correct.

User Myxaxa
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Final answer:

The formula that relates the coefficient a_k to the k-th derivative of a polynomial f(x) at x = 0 is f^(k)(0) = k!a_k. As we differentiate the polynomial k times, only the term containing x^k will contribute to the derivative at x = 0, yielding k! times the coefficient of that term.

Step-by-step explanation:

The formula that relates the coefficient ak to the derivative f(k)(0) of an arbitrary polynomial f(x) = anxn + … + a2x2 + a1x + a0 can be derived by differentiating the polynomial k times. When we take the k-th derivative of f(x), each term aixi where i < k will ultimately have a derivative of zero because the exponent will decrease by one with each differentiation until it is multiplied by x to the power of zero, rendering it a constant and therefore having a derivative of zero. For the term with xk, after k differentiations we will be left with k! times the coefficient ak as the remaining exponential factor will be x0 = 1. Thus, the formula is f(k)(0) = k!ak. The value of the derivative at x = 0 directly gives us the coefficient ak, after dividing by k! because all other terms become zero.

User BuzzBubba
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