Computational Lab in Physics:
Finding roots of nonlinear functions.
Steven Kornreich
www.beachlook.com
Picture above shows the function exp(x)*log(x)-cos(x) in the interval x=0-4. The
function crosses the ordinate axis at some value, called the root. We can find roots of
nonlinear equations numerically when an analytic solution is not possible.
Finding roots of “simple” polynomials

Quadratic equation



ax2+bx+c=0
x=(-b±(b2-4ac))/2a
Cubic equation

x3+ax2+bx+c=0
Scipione del Ferro
p
a
x   u
3u
3
a2
p b
3
2a 3  9ab
q c
27
2
3
q
q
p
u3 

2
4 27
Niccolo Fontana
Tartaglia
2
Many nonlinear functions cannot be given
in closed form: Need numerical approach.

Bisection Method


Simplest and most robust.
How does it work:

We want to find the roots of f(x), i.e. f(x)=0.


Start with an interval [a,b] such that f(x) changes
sign in the interval.


f(x) is continuous in [a,b]
f(a)f(b)<0
There must be at least one real root on the
interval [a,b].
3
Example: f(x)=ln(x)-cos(5x)

Interval:



Function at the boundaries:



a=0.2
b=2
f(a)=-2.149
f(b)=1.53
Therefore:





f(a)f(b)<0
At least one root.
From graph, we see there
are actually 3 roots.
Bisection will find one of the
roots.
Multiple bisections will find
the rest.
4
Bisection procedure

Divide the interval [a,b] into two equal
intervals. (Hence the name)


Middle point x1=(a+b)/2.
Three possibilities
 0, root in [a, x1 ]

f (a) f ( x1 )    0, root in [ x1 , b]
  0, root is x
1



If x1 is not the root, we have a new interval.
Bisect new interval and test.
5
Example Procedure: f(x)=ln(x)-cos(5x)


Interval [0.2,2]
1st Iteration:





[
]
2nd Iteration



f(a)=-2.149, f(b)=1.53
x1=(a+b)/2=1.1,
f(x1)=-0.61
New interval [1.1,2]
x2=1.55, f(x2) = 0.33
New interval [1.1,1.55]
3rd Iteration


x3= 1.325, f(x3) = -0.66
New Interval [1.325,1.55]
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Continue iterating…

When do you stop?

Convergence criteria:
 Want f(xi)=0 for the true root.





Usually set |f(xi)|<d for a small number.
Smaller d, closer to the true root.
Small d might require more iterations.
Interval size:
 alternately, can use |xi-xi-1|<d.
For our example

d=10-2, needs 9 iterations:


d=10-3, needs 12 iterations:


xr=1.49, f(xr)=6.9 x 10-3.
xr=1.489, f(xr)=-1.2 x 10-5.
d=10-5, needs 18 iterations:

xr=1.48892, f(xr)=6.9 x 10-6.
7
Bisection Method in ROOT
while (fabs(func->Eval(x_root))>delta) {
++iteration;
TF1* func = new TF1("func","log(x)cos(5*x)",0,3);
func->SetNpx(1000);
TCanvas* funcCnv = new
TCanvas("funcCnv","Function",500,500);
func->Draw();
TLine* xaxis = new TLine(0,0,3,0);
xaxis->Draw();
double lowLim=0.2;
double uppLim=2;
double x_root = uppLim;
size_t iteration = 0;
double delta = 1e-5; //
// Here is the calculation using the bisection method
x_root=(lowLim+uppLim)/2.0;
cout << "x_" <<iteration <<" = " << x_root;
cout << ", f(x)= " << func->Eval(x_root) << endl;
if (func->Eval(x_root)==0.0) break;
if (func->Eval(lowLim)*func->Eval(x_root)<0) {
uppLim=x_root;
}
else {
lowLim=x_root;
}
}
TMarker* rootMarker =
new TMarker(x_root,func->Eval(x_root),20);
rootMarker->SetMarkerColor(4);
rootMarker->SetMarkerSize(1.5);
rootMarker->Draw("same");
8
Newton’s Method
Use a Taylor expansion of the function:
f '( x0 )
2 f ''( x0 )
3 f '''( x0 )
f ( x)  f ( x0 )  ( x  x0 )
 ( x  x0 )
 ( x  x0 )
1
2!
3!
 Keep only first two terms

f ( x)  f ( x0 )  ( x  x0 ) f '( x0 )



Linear approximation, the root of this
approximation is then
x1 = x0 - f (x0 ) / f '(x0 )
If x1 is not a root, repeat the procedure
around x1.
For many classes of function, speeds up the
method.

Note: It also has weaknesses:


Very slow convergence if f’(x)~0 near the root.
Local minima cause wild jumps from one point
to the next.
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Example: f(x)=ln(x)-cos(5x)
For precision 10-5.
x_1 = 2, f(x)= 1.53221871
x_2 = 2.690155793, f(x)= 0.3558518309
x_3 = 2.606217121, f(x)= 0.06395045778
x_4 = 2.581850754, f(x)= 0.006717235596
x_5 = 2.578603106, f(x)= 0.0001236111542

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Combination of Newton+Bisection

Newton Method drawbacks:

When close to root:


If x1 is far, f’(x1) could have opposite sign
than near f’(x_root):


numerical derivative error ~ root value.
Subsequent iterations diverge!
Solution: combine Newton+Bisection
Select interval [a,b] containing root.
 If x2 in [a,b] for Newton method, ok.
 Otherwise use bisection to determine x2.

11
Assignments

Chapter 7, Section 7.5

1. and 2. Bisection Method in ROOT

Write a program for the bisection method in ROOT.





Use the function: f(x)=exp(x)ln(x)-cos(x)=0
Find a root between x=[0,4], precision 10-8.
Plot the function and put a blue, square marker at location of root.
use TF1’s for the function, use TMarker’s for the marker.
3. and 4. Write a program for Newton’s method in ROOT.




Compare the effectiveness of the bisection method and Newton’s
method for
3
x - 2x - 2 = 0
This has a single root in [-4,2]
Plot the function, and put a red circle at location of root.
To compare effectiveness, have your program print out each
iteration. Show how many iterations are needed to reach a
precision of 10-8.
12