Search results
Results from the WOW.Com Content Network
If only one root, say r 1, is real, then r 2 and r 3 are complex conjugates, which implies that r 2 – r 3 is a purely imaginary number, and thus that (r 2 – r 3) 2 is real and negative. On the other hand, r 1 – r 2 and r 1 – r 3 are complex conjugates, and their product is real and positive. [23]
In the simple case of a function of one variable, say, h(x), we can solve an equation of the form h(x) = c for some constant c by considering what is known as the inverse function of h. Given a function h : A → B, the inverse function, denoted h −1 and defined as h −1 : B → A, is a function such that
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.
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.
The effect has been to fold up the u 4 term into a perfect square: (u 2 + a) 2. The second term, au 2 did not disappear, but its sign has changed and it has been moved to the right side. The next step is to insert a variable y into the perfect square on the left side of equation ( 2 ), and a corresponding 2 y into the coefficient of u 2 in the ...
where a = 5(4ν + 3) / ν 2 + 1 . Using the negative case of the square root yields, after scaling variables, the first parametrization while the positive case gives the second. The substitution c = −m / ℓ 5 , e = 1 / ℓ in the Spearman–Williams parameterization allows one to not exclude the special case a = 0 ...
For each solution (c 0, s 0) of this system, there is a unique solution x of the equation such that 0 ≤ x < 2 π. In the case of this simple example, it may be unclear whether the system is, or not, easier to solve than the equation.
Thomas' algorithm is not stable in general, but is so in several special cases, such as when the matrix is diagonally dominant (either by rows or columns) or symmetric positive definite; [1] [2] for a more precise characterization of stability of Thomas' algorithm, see Higham Theorem 9.12. [3]