![]() By iteratively solving successive sets of linear equations, a solution to the nonlinear equations can be found. A set of linear equations is formed from the Jacobian matrix that can be solved to approximate a solution to the nonlinear equations. The zeros of the functions are found by the Newton-Raphson method, by which each function is differentiated with respect to each master unknown to form the Jacobian matrix. When equilibrium is satisfied, all of the functions relevant to a specific equilibrium calculation are equal to zero. These equations are derived primarily by substituting the equations for the moles of species (derived from mass-action equations in the previous section) into mole- and charge-balance equations. So, if we try $x_1 = 1., are used to describe heterogeneous equilibrium. The computations (to 3 decimal places) from these initial values are shown in the table below. With some more computations, we can narrow down further and find that the switch happens somewhere between $x_1 = 1.447$ and $x_1 = 1.448$. The outcome seems to switch somewhere between the initial estimates $x_1 = 1.44$ and $x_1 = 1.45$. On the other hand, taking $x_1 = 1.44$ behaves similarly to $1.4$: $x_n$ approaches the solution $x=1$. We find that the the initial estimate $x_1 = 1.45$ behaves similarly to $1.5$: the sequence $x_n$ approaches the solution $x = 0$. We might then ask: where, between the initial estimates $x_1 = 1.4$ and $x_1 = 1.5$, does the sequence switch from approaching $x=1$, to approaching $x=0$? And starting from $x_1 = 1.5$ we compute $x_2 = 0$, an exact solution, so all $x_3 = x_4 = \cdots = 0$. X = x_1 - \frac$ we compute, to 6 decimal places,Īnd the $x_n$ converge to the solution $x=1$. Setting $y=0$, we find the $x$-intercept as In our situation, the line has gradient $f'(x_1)$, and passes through $(x_1, y_1)$, so has equation Recall that, given the gradient $m$ of a line, and a point $(x_0, y_0)$ on it, the line has equation ![]() The tangent line we are looking for passes through the point $(x_1, y_1)$ and has gradient $f'(x_1)$. The key calculationĪs our introduction above just showed, the key calculation in each step of Newton's method is to find where the tangent line to $y=f(x)$ at the point $(x_1, y_1)$ intersects the $x$-axis. This method for finding a solution is Newton's method.Īs we'll see, Newton's method can be a very efficient method to approximate a solution to an equation - when it works. In the figure shown, $x_1, x_2, x_3$ rapidly approach the red solution point!Ĭontinuing in this way, we find points $x_1, x_2, x_3, x_4, \ldots$ approximating a solution. If not, then we draw the tangent line to $y=f(x)$ at $(x_2, y_2)$, and our next guess $x_3$ is the point where this tangent line intersects the $x$-axis. We calculate $y_2 = f(x_2)$ if it is zero, we're finished. This tangent line is a good linear approximation to $f(x)$ near $x_1$, so our next guess $x_2$ is the point where the tangent line intersects the $x$-axis, as shown above. How to find that better guess? The trick of Newton's method is to draw a tangent line to the graph $y=f(x)$ at the point $(x_1, y_1)$. If $f(x_1) = 0$, we are very lucky, and have a solution. That is, we want to find the red dotted point in the picture below. Roughly, the idea of Newton's method is as follows. As such, it requires calculus, in particular differentiation. It is based on the geometry of a curve, using the tangent lines to a curve. ![]() Newton's method for solving equations is another numerical method for solving an equation $f(x) = 0$. John Von Neumann Finding a solution with geometry Truth is much too complicated to allow anything but approximations.
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