BOLZANO'S THEOREM
Analysis
1. STATING THE THEOREM

Bolzano's Theorem: If a function f(x) is defined and is continuous over a closed bounded interval [a, b] and the values of the endpoints a and b are of different signs, then at least one point c exists in the open interval (a, b) where the value of the function is 0 and the graph cuts the X-axis.

 

2.  PLAYING WITH THE THEOREM. EXERCISES.

The graphs of different polynomial functions of the fourth degree or less can be drawn in the following window in red. The functions are of the type f(x)= (px4+qx3 +rx2 +sx+t)/40, where the parameters p, q, r, s and t have been chosen so that their values are whole numbers between -15 and 15. (You may need to alter the scale at times to see the graph more clearly).

The parameters a and b to the right of x can be used to change the endpoints of the interval.

Point P (in red), whose coordinates are given, moves over the curve of f(x) when we alter the value of x. Two vertical blue lines can also be seen. One line joins point (a, 0) to (a, f(a)) and the other (b, 0) to (b, f(b)). The two points are indicated on the curve of the function (Q and R respectively) along with their coordinates.

 

All the function graphs in this window are differentiable and therefore continuous on the real line, especially in any interval [a, b]. They therefore satisfy the first hypothesis in Bolzano's theorem. 

 
1.- Try different graphs of functions that satisfy the theorem's hypotheses for different intervals and notice how the thesis is also satisfied. In other words, the graph comes into contact with the X-axis at a point c in the open interval.

Naturally, this does not actually prove the theorem.

2.- Draw the graph of the function f(x)=(x4 +4x3 -4x2 +4x-5)/40, (make p=1, q=4, r=-4, s=4, t=5). Are the hypotheses of Bolzano's theorem satisfied in the interval [-3, 2]? If so, where is point c in the interval (-3, 2) where the value of the function is 0 and the graph cuts the X-axis? Check your results by solving the equation.

There can be more than one point c, as we can see in the following example:

3.- The function f(x)=(x4 +3x3 -4x2 -12x)/40 satisfies the theorem's hypotheses in the interval [-4,1]. At which point(s) is the value of the function 0 and cuts the X-axis? Check your results by solving the equations.

If the theorem's hypotheses are not satisfied then the thesis may or may not be satisfied. Let's look at an example of both cases:

4.- The function f(x)=(x4 -4x2)/40 does not satisfy one of the hypotheses of Bolzano's theorem in the interval [-3,3]. Which one is it? Is the thesis satisfied? At which point(s) in the interval (-3, 3) does the graph cut the X-axis? Check your results by solving the equation.

5.- The function f(x)=(x4 +3x2 +15)/40 does not satisfy one of the hypotheses of Bolzano's theorem in the interval [-2,2]. Which one is it? Is the thesis satisfied? Check your results by solving the equation.

6.- You have seen the roots of different polynomial equations of the fourth degree. How many real roots does each of these equations have? What about cubic equations (of the third degree)? You should have noticed that equations of the fourth degree have an even number of real roots and cubic equations an odd number. However, watch out for double roots!

The second window is similar to the first, the only difference being that the graphs of rational functions can be drawn in this one. I.e. those graphs of the form: f(x)=(px+q)/(rx2 +sx+t), where p, r, s and t are whole numbers between -1 and 1.

 

This window allows us to focus on two other cases: those where the hypothesis of continuity is not satisfied.

 

As we are working with elementary functions, they only cease to be continuous at the points where they are not defined.

 
1.- Draw the graph of the function f(x)=(x+4)/x. (Make p=1, q=4, r=0, s=1, t=0). Which of the hypotheses of Bolzano's theorem is not satisfied in the interval [-3,3]? Is the thesis satisfied?

2.- Draw the graph of the function f(x)=x/(x2 -1). (Make p=1, q=0, r=1, s=0, t=-1). Which of the hypotheses of Bolzano's theorem is not satisfied in the interval [-2,2]? Is the thesis satisfied? If so, where is point c in the interval (-2,2) where the graph cuts the X-axis?

 
3. APPLICATIONS. PROBLEMS AND SOLUTIONS.

As stated in the introduction, Bolzano's theorem is mainly used to find intervals where the root of the equation is located (bounding the root) so as to find approximate values of the roots. Here are a few examples:
1.- Find four intervals on the real line, each one containing a root of the polynomial equation g(x)=x4-x3-13x2 +x+12.

Solution: The function g(x)=x4-x3-13x2 +x+12  satisfies the following: g(-4)=120>0; g(-2)=-18<0; g(0)=12>0; g(2)=-30<0 and g(5)=192>0. Therefore each root is found in these intervals:(-4, -2) (-2, 0) (0, 2) and (2, 5), taking into account that neither -2 nor 2 are roots of the equation.

These results can be checked in the first window, since the functions g(x) and g(x)/40 have the same roots. In the window you can check that the roots are -3, -1, 1 and 4. Each one is found in one of the intervals obtained as the solution to the problem.

As you know, a positive real number has two square roots but if you tried to show this in detail you would come up against a few problems. The following exercise deals with this.

2.- Show that any positive real number has at least one square root.

Solution: Let n>0. We have to show that a real number x exists so that x2=n. Therefore, x is the square root of n.

The equation x2=n is equal to x2-n=0, so we have to show that if g(x)=x2-n then g(x) has at least one real root. If n=0 has a root then we can expect n>0. Therefore g(0)=-n<0. Each value of n needs to give a positive value for g(x). If n>1 then this value is n, for example, as g(n)= n2-n>0 and if 0<n<1, it is 1, for example, as g(1)=1-n>0 (obviously 1 has a root and can be excluded).

We can forget about the values 0 and 1.

If n>1, g(0)=0-n<0 and g(n)= n2-n>0, then n has a root in [0,n].

If n<1, g(0)=0-n<0 and g(1)=1-n>0, then n has a root in [0,1]. 

In the third window the graph of the function g(x)= x2-n is drawn in red (the value of n can be altered). The X-coordinates of the two points where the graph cuts the X-axis are the two roots of n.

Use this window to check the results obtained from solving the problem.

 

3.- Find the value of the square root of 3 to within a hundredth.

Solution: Let g(x)=x2-3. The number we are looking for is where g(x) has a value of 0 and cuts the axis. We have to find an interval on the real line where one of the endpoints of g(x) has a positive value and the other a negative value.

Since g(1.7)=2.89-3<0 and g(1.8)=3.24-3>0, then the root is in the interval (1.7, 1.8).

Now we focus on the midpoint of the interval [1.7, 1.8], i.e. 1.75, and we divide the interval [1.7, 1.8] into two equal subintervals: [1.7, 1.75] and [1.75, 1.8]. Since g(1.75)=3.0625-3>0, we keep the subinterval [1.7, 1.75] as the values of g(x) at its endpoints are of opposite signs. As the root cannot be 1.75, it is found in the interval (1.7, 1.75).

We now divide the new interval into two more equal intervals: [1.7, 1.725] and [1.725, 1.75]. Since g(1.725)=2.975625-3<0, we keep the interval [1.725, 1.75] as the values of the function at its endpoints are of opposite signs. As the root we are looking for cannot be 1.725, it is therefore found in the interval (1.725, 1.75). 

This new interval is divided once again into two more equal intervals: [1.725, 1.7375] and [1.7375,1.75]. Since g(1.7375)=3.0229...-3>0, we keep the subinterval [1.725,1.7375] as the value of the function at its endpoints are of opposite signs. As the root cannot be 1.7375 then it must be found in the interval (1.725, 1.7375).

If we take the root of 3 to be 1.73125 (the midpoint of the last interval), the error of the calculation is less than 1.73125-1.725=0.00625. In other words, less than a hundredth. YOU SHOULD NOTICE THAT:

When n=3, a=1.7 and b=1.8 in the third window, we can see that the root is found in this interval. By increasing the scale (to more than 128) we can see that the root is nearer to 1.7 than to 1.8. However, this is the nearest approximation we can get.

The interesting aspect of this problem is that the approximate value of a square root can be calculated by finding values of a quadratic function (which is easier to work with).

However, in this present computer age, interest in this problem is purely theoretical, such as in the application of Bolzano's theorem.

 

THE BISECTION METHOD

The technique that we have used to find the square root of 3 is called the bisection method as each interval in which the root is located is divided into two equal parts, then the subinterval, where the values of the function at the endpoints are of opposite signs, is kept and bisected again. In this way, we are left with smaller and smaller intervals in which the root is located.

This method can also be used to prove the theorem as if we continue to bisect the intervals indefinitely, the point that all the intervals have in common (which exists) is the root we are looking for. 
       
           
  Valerio Chumillas Checa
 
Spanish Ministry of Education. Year 2001