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In mathematics, divided differences is an algorithm, historically used for computing tables of logarithms and trigonometric functions. [citation needed] Charles Babbage's difference engine, an early mechanical calculator, was designed to use this algorithm in its operation. [1] Divided differences is a recursive division process.
The simplest method is to use finite difference approximations. A simple two-point estimation is to compute the slope of a nearby secant line through the points (x, f(x)) and (x + h, f(x + h)). [1] Choosing a small number h, h represents a small change in x, and it can be either positive or negative.
Given n + 1 points, there is a unique polynomial of degree ≤ n which goes through the given points. Neville's algorithm evaluates this polynomial. Neville's algorithm evaluates this polynomial. Neville's algorithm is based on the Newton form of the interpolating polynomial and the recursion relation for the divided differences .
The difference quotient as a derivative needs no explanation, other than to point out that, since P 0 essentially equals P 1 = P 2 = ... = P ń (as the differences are infinitesimal), the Leibniz notation and derivative expressions do not distinguish P to P 0 or P ń:
A finite difference is a mathematical expression of the form f (x + b) − f (x + a).If a finite difference is divided by b − a, one gets a difference quotient.The approximation of derivatives by finite differences plays a central role in finite difference methods for the numerical solution of differential equations, especially boundary value problems.
For example, given a = f(x) = a 0 x 0 + a 1 x 1 + ··· and b = g(x) = b 0 x 0 + b 1 x 1 + ···, the product ab is a specific value of W(x) = f(x)g(x). One may easily find points along W(x) at small values of x, and interpolation based on those points will yield the terms of W(x) and the specific product ab. As fomulated in Karatsuba ...
The divided difference methods have the advantage that more data points can be added, for improved accuracy. The terms based on the previous data points can continue to be used. With the ordinary Lagrange formula, to do the problem with more data points would require re-doing the whole problem.
Take each digit of the number (371) in reverse order (173), multiplying them successively by the digits 1, 3, 2, 6, 4, 5, repeating with this sequence of multipliers as long as necessary (1, 3, 2, 6, 4, 5, 1, 3, 2, 6, 4, 5, ...), and adding the products (1×1 + 7×3 + 3×2 = 1 + 21 + 6 = 28). The original number is divisible by 7 if and only if ...