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Numerical Solution of Polynomial EquationsNumerical Solution of Differential Equations

3.9.6 Numerical Root Finding

NSolve gives you a general way to find numerical approximations to the solutions of polynomial equations. Finding numerical solutions to more general equations, however, can be much more difficult, as discussed in Section 3.4.2. FindRoot gives you a way to search for a numerical solution to an arbitrary equation, or set of equations.

Numerical root finding.

The curves for and intersect at one point.

In[1]:= Plot[{Cos[x], x}, {x, -1, 1}]

Out[1]=

This finds a numerical approximation to the value of at which the intersection occurs. The 0 tells FindRoot what value of to try first.

In[2]:= FindRoot[Cos[x] == x, {x, 0}]

Out[2]=

In trying to find a solution to your equation, FindRoot starts at the point you specify, and then progressively tries to get closer and closer to a solution. Even if your equations have several solutions, FindRoot always returns the first solution it finds. Which solution this is will depend on what starting point you chose. So long as you start sufficiently close to a particular solution, FindRoot will always return that solution.

The equation has an infinite number of solutions of the form . If you start sufficiently close to a particular solution, FindRoot will give you that solution.

In[3]:= FindRoot[Sin[x] == 0, {x, 3}]

Out[3]=

If you start with , you get a numerical approximation to the solution .

In[4]:= FindRoot[Sin[x] == 0, {x, 6}]

Out[4]=

This is what happens if FindRoot cannot find a solution to your equation.

In[5]:= FindRoot[Sin[x] == 2, {x, 1}]

Out[5]=

If you want FindRoot to search for complex solutions to this equation, then you have to give a complex starting value.

In[6]:= FindRoot[Sin[x] == 2, {x, I}]

Out[6]=

You can give FindRoot bounds on the region in which you want it to look for solutions.

This tells FindRoot to try values of x starting at 1, but never going outside the region 0.5 to 1.5. In this case, FindRoot finds no solutions in the region you specified.

In[7]:= FindRoot[Sin[x] == 0, {x, 1, 0.5, 1.5}]

Out[7]=

Picking good starting points is crucial in getting useful answers from FindRoot. To know how to pick good starting points, you need to understand a little about how FindRoot actually works.

In the simplest case, FindRoot uses Newton's method. To find a solution to an equation of the form , the method starts at , then uses knowledge of the derivative to take a sequence of steps toward a solution. Each new point that it tries is found from the previous point by the formula .

One important limitation of Newton's method is that it "gets stuck" if it ever gets to a point where the derivative of the function vanishes. You can usually avoid this problem by choosing starting points that have no special properties with respect to the equations you are trying to solve.

The derivative of is zero at the starting point . As a result FindRoot cannot decide whether to take its first step in the positive or the negative direction.

In[8]:= FindRoot[x^2 - 1 == 0, {x, 0}]

Out[8]=

If you start at a random point, however, FindRoot will usually succeed in finding a solution.

In[9]:= FindRoot[x^2 - 1 == 0, {x, Random[ ]}]

Out[9]=

FindRoot uses versions of Newton's method in many cases. Especially when there are several variables, the precise set of starting points which lead to a particular solution can become extremely complicated. The best policy is to try and start as close to the solution as possible, and to avoid any "special points".

This finds a solution to a set of simultaneous equations. It is a good idea to avoid taking the starting values for x and y to be equal, or in any other way "special".

In[10]:= FindRoot[{Sin[x] == Cos[y], x + y == 1},

{x, .1}, {y, .2}]

Out[10]=

If the functions that appear in your equations are sufficiently simple, then Mathematica will be able to find their derivatives symbolically. In all the examples of FindRoot that we have used so far, this is possible. As a result, FindRoot can use the formula for Newton's method directly.

If, on the other hand, FindRoot has to estimate the derivative of your functions numerically, then it must take another approach. In simple cases, the approach it uses is based on the "secant method". One feature of this method is that to get it started, you have to specify not just the first value to try, but rather the first two values.

This specifies the first two values of x to try.

In[11]:= FindRoot[Cos[x] == x, {x, {0, 1}}]

Out[11]=

If Mathematica cannot get an explicit formula for the derivative of what appears in your equation, you must specify the first two values to try. Here FindRoot finds a zero of the Riemann zeta function.

In[12]:= FindRoot[Zeta[1/2 + I t] == 0, {t, {12, 13}}]

Out[12]=

If you are finding a root of a function of one variable, and the first two points you tell FindRoot to try give values of the function with opposite signs, then FindRoot is guaranteed to find a root. (This is true so long as your function is real and satisfies some basic continuity conditions.)

Options for FindRoot.

There are several options you can use to control the operation of FindRoot. First, you can set MaxIterations to specify the maximum number of steps that FindRoot should use in attempting to find a solution. Even if FindRoot does not successfully find a solution in the number of steps you specify, it returns the most recent values it has tried. If you want to continue the search, you can then give these values as starting points.

To work out whether it has found an acceptable root, FindRoot evaluates your function and sees whether the result is zero to within the accuracy specified by the option AccuracyGoal. FindRoot will always print a warning message if it does not find a solution to within the specified accuracy. In doing internal computations, FindRoot uses the precision specified by the option WorkingPrecision. The default AccuracyGoal is 10 digits less than WorkingPrecision.

This specifies that the zeta function needs to be zero to 10-digit accuracy at the solution. The Riemann hypothesis asserts that the imaginary part of t should be exactly zero.

In[13]:= FindRoot[Zeta[1/2 + I t] == 0, {t, {12, 13}},

AccuracyGoal -> 10]

Out[13]=

Numerical Solution of Polynomial EquationsNumerical Solution of Differential Equations