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Muller's method is a root-finding algorithm, a numerical method for solving equations of the form f(x) = 0.It was first presented by David E. Muller in 1956.. Muller's method proceeds according to a third-order recurrence relation similar to the second-order recurrence relation of the secant method.
The roots of the quadratic function y = 1 / 2 x 2 − 3x + 5 / 2 are the places where the graph intersects the x-axis, the values x = 1 and x = 5. They can be found via the quadratic formula. In elementary algebra, the quadratic formula is a closed-form expression describing the solutions of a quadratic equation.
For solving the cubic equation x 3 + m 2 x = n where n > 0, Omar Khayyám constructed the parabola y = x 2 /m, the circle that has as a diameter the line segment [0, n/m 2] on the positive x-axis, and a vertical line through the point where the circle and the parabola intersect above the x-axis.
The points P 1, P 2, and P 3 (in blue) are collinear and belong to the graph of x 3 + 3 / 2 x 2 − 5 / 2 x + 5 / 4 . The points T 1, T 2, and T 3 (in red) are the intersections of the (dotted) tangent lines to the graph at these points with the graph itself. They are collinear too.
Because (a + 1) 2 = a, a + 1 is the unique solution of the quadratic equation x 2 + a = 0. On the other hand, the polynomial x 2 + ax + 1 is irreducible over F 4, but it splits over F 16, where it has the two roots ab and ab + a, where b is a root of x 2 + x + a in F 16. This is a special case of Artin–Schreier theory.
Two other solutions are x = 3, y = 6, z = 1, and x = 8, y = 9, z = 2. There is a unique plane in three-dimensional space which passes through the three points with these coordinates, and this plane is the set of all points whose coordinates are solutions of the equation.
This is the case, for example, if f(x) = x 3 − 2x + 2. For this function, it is even the case that Newton's iteration as initialized sufficiently close to 0 or 1 will asymptotically oscillate between these values. For example, Newton's method as initialized at 0.99 yields iterates 0.99, −0.06317, 1.00628, 0.03651, 1.00196, 0.01162, 1.00020 ...
In either case the full quartic can then be divided by the factor (x − 1) or (x + 1) respectively yielding a new cubic polynomial, which can be solved to find the quartic's other roots. If a 1 = a 0 k , {\displaystyle \ a_{1}=a_{0}k\ ,} a 2 = 0 {\displaystyle \ a_{2}=0\ } and a 4 = a 3 k , {\displaystyle \ a_{4}=a_{3}k\ ,} then x = − k ...