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Equations
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If the equation to be solved has a root of multiplicity 2 or greater, Newton s method may converge slowly or not at all. In this situation, convergence can sometimes be improved by a judicious choice of DampingFactor.
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DampingFactor factor is an option that controls the behavior of convergence in Newton s method. The size of each step taken in Newton s method is multiplied by the value of factor. The default is DampingFactor 1.
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FindRoot[(Exp[x] 1)2, {x, 2}, EvaluationMonitor n++] Print[n," iterations"]
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{x 6.95942
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10-9 }
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32 iterations
FindRoot[(Exp[x] 1)2, {x, 2}, DampingFactor 2, EvaluationMonitor n++] Print[n," iterations"]
{x 6.6703
10-17 }
8 iterations
FindRoot can also be used to determine the solution of simultaneous equations.
FindRoot[equations, { var1, a1} , { var2, a2} ,...] attempts to solve equations using initial values a1, a2, . . . for var1, var2, . . . , respectively. The equations are enclosed in a list: {equation1, equation2,...}. Alternatively, the equations may be separated by && (logical and).
Convergence of Newton s method for functions of several variables is much more sensitive to choice of starting values than its counterpart for single variables. Therefore, a good graph of the functions involved is quite helpful.
EXAMPLE 23 Solve the system of equations sin x + cos y = 1
e x + ln y = 2
First we graph the equations. ContourPlot[{Exp[x] + Log[y] 2, Sin[x] + Cos[y] 1}, {x, 0, 2}, {y, 0, 3}, Frame False, Axes True]
Equations
It appears that there is only one solution. We use x = 1, y = 1 for our initial guess. FindRoot[{Exp[x] + Log[y] 2, Sin[x] + Cos[y] 1}, {x, 1}, {y, 1}]
0.624295, y 1.14233}
If the function in an equation is such that its evaluation is costly, particularly if high precision is desired, there is another procedure that may be beneficial.
InterpolateRoot[lhs rhs, {x, a, b}] solves the equation lhs = rhs using initial values a and b.
Whereas FindRoot uses linear functions (straight lines) to approximate the root of the equation, InterpolateRoot uses polynomials of degree 3 or less. The result is that higher precision can be achieved with fewer function evaluations. InterpolateRoot is contained within the package FunctionApproximations` and must be loaded prior to use. As with FindRoot, the equation may be replaced by a function, in which case its zero is computed.
This example computes the zero (between 2 and 3) of the Bessel function2 J0(x), using a working precision of 1000 significant digits. For comparison purposes, the Mathematica function Timing is used. The actual numerical approximation is suppressed to save space. As a result, the value Null is returned. Delete the semicolon and run the command to see the actual result of the calculation.
EXAMPLE 24
FindRoot[BesselJ[0, x], {x, 2}, WorkingPrecision 1000]; //Timing
{0.219,
Null}
FunctionApproximations` InterpolateRoot[BesselJ[0, x], {x, 2, 3}, WorkingPrecision 1000]; //Timing
{0.046,
Null}
SOLVED PROBLEMS
6.12 Solve the equation 5 cos x = 4 x 3 . Make sure you find all solutions.
SOLUTION
Since 5 cos x = 4 x 3 if and only if 5 cos x 4 + x 3 = 0 , we introduce the function f ( x ) = 5 cos x 4 + x 3 and look for x-intercepts. (Although we could look for the intersection of two curves, it is easier to approximate where points intercept an axis.) f[x_] = 5 Cos [x] 4 + x3; Plot[f[x], {x, 1, 2}]
1.0
0.5 1
J0(x) is a solution of the differential equation x2 y" + x y ' + x2 y = 0.
Equations
It appears that there are three solutions, near 0.5, 0.8, and 1.6. FindRoot[f[x], {x, 0.5}]
0.576574} 0.797323} 1.61805}
FindRoot[f[x], {x, 0.8}]
FindRoot[f[x], {x, 1.6}]
6.13 Find a solution of the equation sin x = 2. (This problem may be omitted by those unfamiliar with functions of a complex variable.)
SOLUTION
Since 1 sin x 1 for all real x, this problem has no real solutions. We can force FindRoot to search for a complex solution by using a complex initial guess. FindRoot[Sin[x] 2,{x, I}]
1.5708 + 1.31696 }
6.14 Find a 20 significant digit approximation to the equation x + | sin (x 1) | = 5.
SOLUTION
First we plot the function f(x) = x + | sin (x 1) | 5. f[x_] = x + Abs[Sin[x 1]] 5; Plot[f[x], {x, 10, 10}]
5 5
It appears that the only solution lies between 4 and 5. FindRoot[f[x],{x,5}, AccuracyGoal 20, WorkingPrecision 25]
4.577640011987577295259374}
To 20 significant digits, the solution is 4.5776400119875772953 (last digit rounded up).
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