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The scheme is always numerically stable and convergent but usually more numerically intensive than the explicit method as it requires solving a system of numerical equations on each time step. The errors are linear over the time step and quadratic over the space step: Δ u = O ( k ) + O ( h 2 ) . {\displaystyle \Delta u=O(k)+O(h^{2}).}
This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2: (+) In the case above, this gives the equation:
If an equation can be put into the form f(x) = x, and a solution x is an attractive fixed point of the function f, then one may begin with a point x 1 in the basin of attraction of x, and let x n+1 = f(x n) for n ≥ 1, and the sequence {x n} n ≥ 1 will converge to the solution x.
The step size is =. The same illustration for = The midpoint method converges faster than the Euler method, as .. Numerical methods for ordinary differential equations are methods used to find numerical approximations to the solutions of ordinary differential equations (ODEs).
An important application of divide and conquer is in optimization, [example needed] where if the search space is reduced ("pruned") by a constant factor at each step, the overall algorithm has the same asymptotic complexity as the pruning step, with the constant depending on the pruning factor (by summing the geometric series); this is known as ...
Karatsuba's basic step works for any base B and any m, but the recursive algorithm is most efficient when m is equal to n/2, rounded up. In particular, if n is 2 k, for some integer k, and the recursion stops only when n is 1, then the number of single-digit multiplications is 3 k, which is n c where c = log 2 3.
An important application is Newton–Raphson division, which can be used to quickly find the reciprocal of a number a, using only multiplication and subtraction, that is to say the number x such that 1 / x = a. We can rephrase that as finding the zero of f(x) = 1 / x − a. We have f ′ (x) = − 1 / x 2 . Newton's ...
For single digit numbers simply duplicate the number into the tens digit, for example: 1 × 11 = 11, 2 × 11 = 22, up to 9 × 11 = 99. The product for any larger non-zero integer can be found by a series of additions to each of its digits from right to left, two at a time. First take the ones digit and copy that to the temporary result.